quantitative techniques study pack 2006 wip

Acknowledgment ii

Acknowledgement

This study kit’s success is attributed first and foremost to Saint Jose Maria Escriva the founder of opus dei and the inspiration to the formation of Strathmore University and its values.

Much appreciation is also extended to Mr. Randhir Ahluwalia of Strathmore University for his contribution, Mr. Paul Maloba for his contributon and compilation of the book.

Acknowledgement is given to the books that were used in research for compilation of this book; Applied Mathematics for Business Economics and Social Sciences, and Quantitative Techniques 6th edition.

We gratefully acknowledge permission to quote from the past examination papers of the following bodies: Kenya Accountants and Secretaries National Examination Board (KASNEB); Chartered Institute of Management Accountants (CIMA); Association of Chartered Certified Accountants (ACCA).

ii

iii Acknowledgement

Instructions for Students

Quantitative techniques paper in section 4 of the CPA examination is an eight question paper devided intotwo sections.Section I has 5 questions and only 3 questions are to be attempted, section II has 3 questions and only 2 are to be attempted. All questions have a total of 20marks.

Section I contains mainly questions on calculus, probability, matrices, test of hypothesis, regression analysis, sampling and estimation and descriptive statistics.

Section II contains question from operations research and decision theory.

from past papers statitics the topicss frequently feature in exam are network analysis, matrices, test of hypothesis and calculus. Other topics that are also favorites of the examiner are regression analysis, time series, decision making techniques, linear programming, probability and sets theory. It is very important to clearly understand these topics through and through, emphasize more on working out examples and assignments from this book.The course demands rigorous practice of of questions to internalize concepts, that’s the only way to go about it. Work out reinforcing questions at the end of each study session and compare them with answers given in study session 9. comprehensive assignments are to be handed in to DLC department for marking.

STRATHMORE UNIVERSITY ● STUDY PACK

Instructions for Students iv

Contents

Acknowledgement...........................................................................iiInstructions for Students...............................................................iiiContents..........................................................................................ivCourse Description..........................................................................vIndex...............................................................................................viLESSON ONE...................................................................................11. Linear Algebra and Matrices.....................................................................................1LESSON TWO................................................................................452. Sets Theory and Calculus........................................................................................45LESSON THREE.............................................................................813. Descriptive Statistics and Index Numbers..............................................................81LESSON FOUR.............................................................................1184. Measures of Relationships and Forecasting..........................................................118LESSON FIVE...............................................................................1575. Probability.............................................................................................................157LESSON SIX.................................................................................1856. Sampling and Estimation......................................................................................185LESSON SEVEN...........................................................................2287. Decision Theory....................................................................................................228LESSON EIGHT............................................................................2498. Operation Research...............................................................................................249LESSON NINE..............................................................................3059. Revision Aid..........................................................................................................305

QUANTITATIVE TECHNIQUES

v Contents

Course Description

Quantitative techniques is a mathematical paper and fundamental to many proffessionalcourses it forms the basis of finance and acconting. The course dwells more on practical applicationtype questions and sometimes even requires one to make inference and decisions based on accurate analysis of information .This paper also formas a basis for section 5 & 6 papers, management accounting and Financial management.The book has been tailored to make you comfortable in a mathematical environment so as to excel in the accounting profession.


Course Description vi

Index


1 Linear Algebra and Matrices

LESSON ONE

Linear Algebra and MatricesContents

- Functions and graphs- Linear equations, higher order equations, inequalities and

simultaneous equations- Matrix algebra- Application of matrix algebra to input-output analysis and

elementary Markovian process.


Lesson One 2

1.1 Functions and graphsA function is a mathematical relationship in which the value of a single dependent variable are determined from the values of one or more independent variables. The following is an example of a function in which y is said to be a function of x.

y = a + bxIn the above example, both x and y are variables this is because they may assume different values throughout the analysis of the function. On the other hand, a and b are referred to as constants because they assume fixed values.The variable y is a dependant variable in the sense that its values are generated from an independent variable x.The collection of all the values of the independent variable for which the function is defined is referred to as the domain of the function corresponding to this we have the range of the function, which is the collection of all the values of the dependent variable defined by the functionThe fact that it is a function of x can also be denoted by the following general form

y = f(x)Functions of a single independent variable may either be linear or non linear.Linear functions can be represented by:

y = a + bxWhereas non – linear functions can be represented by functions such as:

i. y = α0 + α x + α2x3

ii. y2 = 3x + 18iii. y = 2x2 + 5x + 7iv. ax2 + bx + cy + d = 0v. xy = k

vi. y = ax

Graph of a functionA graph is a visual method of illustrating the behaviour of a particular function. It is easy to see from a graph how as x changes, the value of the f(x) is changing.The graph is thus much easier to understand and interpret than a table of values. For example by looking at a graph we can tell whether f(x) is increasing or decreasing as x increases or decreases.We can also tell whether the rate of change is slow or fast. Maximum and minimum values of the function can be seen at a glance. For particular values of x, it is easy to read the values of f(x) and vice versa i.e. graphs can be used for estimation purposesDifferent functions create different shaped graphs and it is useful knowing the shapes of some of the most commonly encountered functions. Various types of equations such as linear, quadratic,


Where α, a, b, c, d, k =


trigonometric, exponential equations can be solved using graphical methods.

EquationsAn equation is an expression with an equal sign (=)Equations are classified into two main groups linear equations and non linear equations. Examples of linear equations are

x + 13 = 157x + 6 = 0

Non linear equations in the variable x are equations in which x appears in the second or higher degrees. They include quadratic and cubic equations amongst others. For example

5x2 + 3x + 7 = 0 (quadratic equation)2x3 + 4x2 + 3x + 8 = 0 (cubic equation)

The solution of equations or the values of the variables for which the equations hold is called the roots of the equation or the solution set.

Solution of Linear EquationSupposing M, N, and P are expressions that may or may not involve variables, then the following constitute some rules which will be useful in the solution of linear equationsRule 1: Additional rule

If M = N then M + P = N + PRule 2: Subtraction rule

If M = N, Then M – P = N – PRule 3: multiplication rule

If M = N and P ≠ O then M x P = N x PRule 4: Division rule

If P x M = N and P ≠ OAnd N/P = Q Q being a raterial number then M = N/P

Examplei. Solve 3x + 4 = - 8

ii. Solve = - 4

Solutionsi. 3x + 4 = –8

3x + 4 – 4 = – 8 – 4 (by subtraction rule)3x = – 12 (simplifying)


Lesson One 4

(by division rule)

x = – 4 (simplifying)

ii.

y = –12 (simplifying)Solution of quadratic equationsSuppose that we have an equation given as follows

ax2 + bx + c = 0Where a, b and c are constants, and a≠ 0. such an equation is referred to as the general quadratic equation in x. if b = 0, then we have

ax2 + c = 0Which is a pure quadratic equation There are 3 general methods for solving quadratic equations; solution by factorization, solution by completing the square and solution by the quadratic formula.

Solution by FactorizationThe following are the general steps commonly used in solving quadratic equations by factorization

(i) Set the given quadratic equation to zero(ii) Transform it into the product of two linear factors(iii) Set each of the two linear factors equal to

zero(iv) Find the roots of the resulting two linear

equationsExampleSolve the following equation by factorization

i. 6x2 = 18xii. 15x2 + 16x = 15

Solutionsi. 6x2 = 18x

6x2 – 18x = 0...................................(step 1)6x(x – 3) = 0....................................(step 2)6x = 0..............................................(step 3)and x – 3 = 0



∴ x = 0 or x = 3 ..............................(by step 4)ii. 15x2 + 16x = 15

15x2 + 16x – 15 = 0......................... (step 1)(5x – 3) (3x +5) = 0.........................(step 2)(5x – 3) = 0} Step 3{3x + 5 = 0}

∴ x = - or + .............................(step 4)

Solution by Completing the SquareThe process of completing the square involves the construction of a perfect square from the members of the equation which contains the variable of the equation. Consider the equation – 9x2 – bx = 0The method of completing the square will involve the following steps

i. Make the coefficient of x2 unityii. Add the square of ½ the coefficient of x to both sides of the

equal sign. Theleft hand side is now a perfect square

iii. Factorize the perfect square on the left hand side.iv. Find the square root of both sidesv. Solve for x

ExampleSolve by completing the square.

i. 3x2 = 9xii. 2x2 + 3x + 1 = 0

Solutionsi. 3x2 = 9x or

(3x2 - 9x = 0)x2 - 3x = 0............................................. (Step 1)

..........................(Step 2)


Lesson One 6

...........................................(Step 3)

...........................................(Step 4)

∴

(= 3 or 0)

ii. 2x2 + 3x + 1 = 0 or (2x2 + 3x = -1)

…………………...….. (Step 1)

……… (Step 2)

…………………….. (Step 3)

Solution by Quadratic FormulaConsider the general quadratic equation

The roots of the equation are obtained by the following formula:

ExampleSolve for x by formula

5x2 + 2x – 3 = 0



Solutiona = 5, b = 2, c = - 3

InequalitiesAn inequality or inequation is an expression involving an inequality sign (i.e. >, <, ≤, ≥, i.e. greater than, less than, less or equal to, greater or equal to) The following are some examples of inequations in variable x.

3x + 3 > 5x2 – 2x – 12 < 0

The first is an example of linear inequation and the second is an example of a quadratic in equation.

Solutions of inequationsThe solutions sets of inequations frequently contain many elements. In a number of cases they contain infinite elements.

ExampleSolve and graph the following inequation

x – 2 > 2 ; (where x is a subset of w)

Solution x – 2 > 2 so x – 2 + 2 > 2 + 2

Thus, x>4The solution set is infinite, being all the elements in w greater than 4

0 1 2 3 4 5 6 7 8 9 10 11

ExampleSolve and graph

3x – 7 < - 13;


Lesson One 8

Solution

Rules for solving linear inequationsSuppose M, M1, N, N1 and P are expressions that may or may not involve variables, then the corresponding rules for solving inequations will be:Rule 1: Addition rule

If M > N and M1> N1

Then M + P > N + P and M1 + P >N1+ P

Rule 2: Subtraction RuleIf M < N and M1 ≥N1

Then M – P < N – P and M1 – P ≥N1– P

Rule 3: Multiplication ruleIf M ≥N and M1 > N1 and P≠ 0Then MP ≥NP; M1P > N1P

M(-P) ≤ N(-P) and M1(-P) < N1(-P)

Rule 4: DivisionIf M > N and M1< N1 and P≠ 0Then M/P > N/P: M1/P < N1/P

M/(-P) < N/(-P) : and M1/(-P) > N1/(-P)Rule 5: Inversion Rule


-4 -3 -2 -1 0 2 3 4

….. R Line


If M/P ≤ N/Q where P, Q ≠ 0M1/P > N1/QThen P/M ≥ Q/N and P/M1 < Q/N1

Note: The rules for solving equations are the same as those for solving equations with one exception; when both sides of an equation is multiplied or divided by a negative number, the inequality symbol must be reversed (see rule 3 & Rule 4 above).

ExampleSolve and graph the following:

i. 7 – 2x > - 11 ;ii. –5x + 4 ≤ 2x – 10 ;iii. –3 ≤ 2x + 1 < 7 ;

Solutions


Q

-3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11

line

Q

-4 -3 -2 -1 0 1 2 3 4 5 line

Lesson One 10

Linear inequation in two variables: relationsAn expression of the form

y ≥ 2x – 1is technically called a relation. It corresponds to a function, but different from it in that, corresponding to each value of the independent variable x, there is more than one value of the dependent variable yRelations can be successfully presented graphically and are of major importance in linear programming.

1.2 Linear simultaneous equations:Two or more equations will form a system of linear simultaneous equations if such equations be linear in the same two or more variables.For instance, the following systems of the two equations is simultaneous in the two variables x and y.

2x + 6y = 234x + 7y = 10

The solution of a system of linear simultaneous equations is a set of values of the variables which simultaneously satisfy all the equations of the system.

Solution techniquesa) The graphical technique

The graphical technique of solving a system of linear equations consists of drawing the graphs of the equations of the system on the same rectangular coordinate system. The coordinates of the point of intersection of the equations of the system would then be the solution.


Q

-4 -3 -2 -1 0 1 2 3 4 5 line


Example

The above figure illustrates:Solution by graphical method of two equations

2x + y = 8x + 2y = 10

The system has a unique solution (2, 4) represented by the point of intersection of the two equations.

b) The elimination technique This method requires that each variable be eliminated in turn by making the absolute value of its coefficients equal in the equations of the system and then adding or subtracting the equations. Making the absolute values of the coefficients equal necessitates the multiplication of each equation by an appropriate numerical factor. Consider the system of two equations (i) and (ii) below

2x – 3y = 8 …….. .................................(i). 3x + 4y = -5 ……..................................(ii).

Step 1Multiply (i) by 36x – 9y = 24 …… ...........................................(iii).Multiply (ii) By 26x + 8y = - 10 …… ........................................(iv).Subtract iii from iv.


(2,4)

-1 1 2 3 4 5 6 7 8 9 10 11 12 13

10 9 .

8 . 7 . 6

. 5

. 4

. 3

. 2

. 1

x + 2y = 102x + y = 8

Lesson One 12

17y = -34 …….................................................(v).y = -2

Step 2Multiply (i) by 48x – 12y = 32 ……. ........................................(vi)Multiply (ii) by 3 9x + 12y = -15 ….. ........................................(vii)Add vi to vii 17x = 17 ……..................................................(viii)

x = 1Thus x = 1, y = -2 i.e. {1,-2}

c) The substitution technique To illustrate this technique, consider the system of two equations (i). and (ii) reproduced below.....2x – 3y = 8 …….. (i)......3x + 4y = -5 …… (ii).The solution of this system can be obtained by

a) Solving one of the equations for one variable in terms of the other variable;

b) Substituting this value into the other equation(s) thereby obtaining an equation with one unknown only

c) Solving this equation for its single variable finallyd) Substituting this value into any one of the two original equations

so as to obtain the value of the second variable

Step 1Solve equation (i) for variable x in terms of y

2x – 3y = 8x= 4 + 3/2 y (iii)

Step 2Substitute this value of x into equation (ii). And obtain an equation in y only

3x + 4y = -53 (4 + 3/2 y) + 4y = -5



8 ½ y = - 17 ……. (iv)Step 3Solve the equation (iv). For y

8½y = -17y = -2

Step 4Substitute this value of y into equation (i) or (iii) and obtain the value of x

2x – 3y = 82x – 3(-2) = 8x = 1

Example Solve the following by substitution method

2x + y = 83x – 2y = -2

SolutionSolve the first equation for y

y = 8 – 2x Substitute this value of y into the second equation and solve for x

3x – 2y = -23x – 2 (8-2x) = -2x = 2

Substitute this value of x into either the first or the second original equation and solve for y

2x + y = 8(2) (2) + y = 8y = 4

1.3 MATRICES A matrix is a rectangular array of items or numbers. These items or numbers are arranged in rows and columns to represent some information.The position of an element in one matrix is very important as well be seen later; therefore an element is located by the number of the row and column which it occupies.The size of a matrix is defined by the number of its rows (m) and column (n).


Lesson One 14

are (2 x 2) and (3 x 3) matrices since A has 2 rows and 2 columns and B has 3 rows and 3 columns.A matrix A with three rows and four columns is given by one of:

Properties of matricesEqual MatricesTwo matrices A and B are said to be equal, that is

If and only if they are identical if they both have the same number of rows and columns and the elements in the corresponding locations in the two matrices should be the same, that is, aij = bij for all i. And j.

Example

The following matrices are equal

Column Matrix or column vectorA column matrix, also referred to as column vector is a matrix consisting of a single column.



For example x =

Row matrix or row vectorIt is a matrix with a single row

For example

Transpose of a MatrixThe transpose of an mxn matrix A is the nxm matrix AT obtained by interchanging the rows and columns of A.

A = aij

The transpose of A i.e. AT is given byAT = aij = aji

mxn nxm

ExampleFind the transposes of the following matrices

Solution


Lesson One 16

Square MatrixA matrix A is said to be square when it has the same number of rows as columns

e.g.

2 5 A = 3 7 is a square matrix of order 2

B = n × n is a square matrix of the order n

Diagonal matricesIt is a square matrix with zeros everywhere in the matrix except on the principal diagonale.g.

An identity of unity matrixIt is a diagonal matrix in which each of the diagonal elements is a positive one (1)e.g.



A null or zero matrixA null or zero matrix is a matrix whose elements are all equal to zero.Sub matrixThe sub matrix of the matrix A is another matrix obtained from A by deleting selected row(s) and/or column(s) of the matrix A.

are both sub matrices of A

OPERATION ON MATRICESMatrix addition and subtractionWe can add any number of matrices (or subtract one matrix from another) if they have the same sizes. Addition is carried out by adding together corresponding elements in the matrices. Similarly subtraction is carried out by subtracting the corresponding elements of two matrices as shown in the following exampleExample: Given A and B, calculate A + B and A – B


Lesson One 18

If it is assumed that A, B, C are of the same order, the following properties are fulfilled:a) Commutative law: A + B = B + Ab) Associative law: (A + B) + C = A + (B + C) = A + B + C

Multiplying a matrix by a numberIn this case each element of the matrix is multiplied by that number

Example

Matrix Multiplicationa) Multiplication of two vectors



Let row vector A represent the selling price in shillings of one unit of commodity P, Q, R respectively and let column vector B represent the number of units of commodities P, Q, R sold respectively. Then the vector product A B will be equal to the total sales valuei. e. A B = Total sales value

Rules of multiplicationi. The row vector must have the same number of elements as the

column vectorii. The first vector is a row vector and the second is a column

vectoriii. The corresponding elements in each vector are multiplied

together and the results obtained are added. This addition is always a single numberGoing back to the example given before


Lesson One 20

b) Multiplication of two matricesRules

i. Multiplication is only possible if the first matrix has the same number of columns as the second matrix has rows. That is if A is the order a×b, then B has to be of the order b×c. If the A×B = D, then D must be of the order a×c.

ii. The general method of multiplication is that the elements in row m of the first matrix are multiplied by the corresponding elements in columns n of the second matrix and the products obtained are then added giving a single number.

We can express this rule as follows

Let A =

Then A B = D =

A = 2 x 2 matrix B = 2 x 3 matrix D = 2 x 3 matrixWhere

d11 = a11 b11 + a12 b21

d12 = a11 b12 + a12 b22

Example I

Example IIMatrix X gives the details of component parts used in the make up of two products P1 and P2 matrix Y gives details of products made on each day of the week as follows:



Use matrix multiplication to find the number of component parts used on each day of the week.Solution:After careful consideration, it will be easy to decide that the correct order of multiplication is YXX (Note the order of multiplication). This multiplication is compatible and also it gives the desired answer.

5 x 2 matrix 2 x 3 matrix = 5 x 3 matrix

InterpretationOn Monday, number of component parts A used is 7, B is 14 and C is 8. in the same way, the number of component parts used for other days can be interpreted.

The determinant of a square matrixThe determinant of a square matrix det (A) or A is a number associated to that matrix. If the determinant of a matrix is equal to zero, the matrix is called singular matrix otherwise it is called non-singular matrix. The determinant of a non square matrix is not defined.

Determination of a 2 x 2 matrix


Lesson One 22

ii. Determinant of a 3 x 3 matrix

iii. Determinant of a 4 x 4 matrix

Simplify 3 x 3 determinants as in ii and then evaluate the 4 x 4 determinants.

Inverse of a matrix

If for an n ( n square matrix A, there is another n ( n square matrix B such that there product is the identity of the order n X n, In, that is A X B = B X A = I, then B is said to be inverse of A. Inverse if generally written as A-1

Hence AA-1 = I

Note: Only non singular matrices have an inverse and therefore the inverse of a singular matrix is non defined.

General method for finding inverse of a matrix



In order to introduce the rule to calculate the determinant as well as the inverse of a matrix, we should introduce the concept of minor and cofactor.

The minor of an element

Given a matrix A = (aij), the minor of an element aij in row i and column j (call it mij), is the value of the determinant formed by deleting row i and column j in matrix A.

Example

Let A = EMBED Equation.DSMT4 µ §

The minors are,

Similarly

The cofactor of an elementThe cofactor of any element aij (known as cij) is the signed minor associated with that element.The sign is not changed if (i+j) is even and it is changed if (i+j)is odd. Thus the sign alternated whether vertically or horizontally, beginning with a plus in the upper left hand corner.


Lesson One 24

i.e. 3 x 3 signed matrix will have signs

Hence the cofactor of element a11 is m11 = -3, cofactor of a12 is –m12 = +2 the cofactor of element a13 is +m13 = 3 and so on.

Matrix of cofactors of A =

in general for a matrix M =

Cofactor of a is written as A, cofactor of b is written as B and so on.Hence matrix of cofactors of M is written as

=

The determinant of a n×n matrixThe determinant of a n×n matrix can be calculated by adding the products of the element in any row (or column) multiplied by their cofactors. If we use the symbol ∆ for determinant.

Then ∆ = aA + bB + cC or

= dD + eE + fF e.t.c

Note: Usually for calculation purposes we take ∆ = aA + bB + cC

Hence in the example under discussion∆ = (4 –3) + (2 2) + (3 3) = 1

The ad joint of a matrix

The ad joint of matrix is written as



i.e. change rows into columns and columns into rows

(transpose)

The inverse of the matrix

is written as

Where ∆ = aA + bB + cC

Hence inverse of

is found as follows

∆ = (4 –3) + (2 2) + (3 ( 3) = 1

A = -3 B = 2 C = 3

D = 9 E = -6 F = -8

G = -16 H = 11 I = 14

EMBED Equation.DSMT4 µ §

(note: Check if A ( A-1 = A-1 A = 1)

Solution of simultaneous equationsIn order to determine the solutions of simultaneous equations, we may use either of the following 2 methods

i. The cofactor methodii. Cramers rule

The cofactor methodThis method requires that we obtain


Lesson One 26

a) The minors and cofactorsb) The adjoint of the matrixc) The inverse of the matrixd) Multiply the original by the inverse on both sides of the matrix

equation

ExampleSolve the following

a) 4x1 + x2 – 5x3 = 8-2x1 + 3x2 + x3 = 123x1 – x2 + 4x 3 = 5

b) 4x1 + 3x3 + 5x3 = 27x1 + 6x2 + 2x3 = 193x1 + x2 + 3x3 = 15

c) 4x1 + 2x2 + 6x3 = 283x1 + x2 + 2x3 = 2010x1 + 5x2 + 15x3 = 70

d) 2x1 + 4x2 – 3x3 = 123x1 – 5x2 + 2x3 = 13-x1 + 3x2 + 2x3 = 17

Solution a) From a, we have

A X bWe need to determine the minors and the cofactors for the above matrix

DefinitionA minor is a determinant of a sub matrix obtained when other elements are detected as shown below.A cofactor is the product of (-1) i + j and a minor where



i = Ith row i = 1, 2, 3 …….j = Jth row j = 1, 2, 3 …….

Cofactor of 4 (a11) = (-1) 1+1

Cofactor of -2 (a21) = (-1) 2+1

Cofactor of 3 (a31) = (-1) 3+1

Cofactor of 1 (a12) = (-1) 1+2

Cofactor of 3 (a22) = (-1) 2+2

Cofactor of -1 (a23) = (-1) 2+3

Cofactor of -5 (a13) = (-1) 1+3

Cofactor of +1 (a23) = (-1) 2+3

Cofactor of 4 (a33) = (-1) 3+3

The matrix of C of cofactors is

CT = = Adjoint of the original matrix of coefficients

The original matrix of coefficients


Lesson One 28

=

Therefore

= (48 + 3 – 10) – (-45 – 4 – 8)= 41 + 57= 98

The inverse of the matrix of coefficients, see (*) will be

by multiplying the inverse on both sides of * we have,



X1 = 2, X2 = 5, X3 = 1

c) 4x1 + 2x2 + 6x3 = 283x1 + x2 + 2x3 = 2010x1 + 5x2 + 15x3 = 70

= (60 + 40 + 90) – (60 + 40 + 90)= 0

Hence the solutions of x1, x2, and x3 do no exist. The equations are independent Now work out part (b) on your own.Cramers Rule in Solving Simultaneous Equations Consider the following system of two linear simultaneous equations in two variables.

a11 x1 + a12 x2 = b1 ……………(i)a21 x1 + a22 x2 = b2 ……………(ii)

after solving the equations you obtain


Lesson One 30

Solutions of x1 and x2 obtained this way are said to have been derived using Cramers rule, practice this method over and over to internalize it. It is advisable for exam situation since it is shorter.ExampleSolve the following systems of linear simultaneous equations by Cramers’ rule:

i) 2x1 – 5x2 = 7x1 + 6x2 = 9

ii) x1 + 2x2 + 4x3 = 42x1 + x3 = 33x2 + x3 = 2

Solutions i. 2x1 – 5x2 = 7

x1 + 6x2 = 9can be expressed in matrix form as

and applying cramers’ rule



(ii) can be expressed in matrix form as

and by cramers’ rule

Solving simultaneous Equations using matrix algebra


Lesson One 32

i. Solve the equations2x + 3y = 133x + 2y = 12

in matrix format these equations can be written as

pre multiply both sides by the inverse of the matrix

and inverse of the matrix is

Pre multiplication by inverse gives

Therefore x = 2 y = 3

ii. Solve the equations 4x + 2y + 3z = 45x + 6y + 1z = 22x + 3y = -1

Solution:Writing these equations in matrix format, we get

Pre-multiply both sides by the inverse



the inverse of A as found before is A-1 =

hence x = 22 y = -15 z = -18(Note: under examination conditions it may be advisable to check the solution by substituting the value of x, y, z into any of the three original equations)

MARKONIKOV CHAIN

Probability Transition Matrices (Brand switching)These are matrices in which the individual elements are in the form of probabilities.The probabilities represent the probability of one event following another event i.e. the probability of transition from one event to the nextThe probabilities of the various changes applied to the initial state by matrix multiplication, give a forecast of the succeeding state.Normally a transition matrix is defined with its columns adding upto one and state vectors as column vectors.In this case the succeeding state is found by pre-multiplying the transition matrix by the proceeding state (column) vectorIf the transition matrices are given with their rows adding up to one, then the succeeding state is found by post multiplying the transition matrix by the preceding state (row) vector.

Example 1The probability transition matrix of the switching probabilities, consider that two brands G and X share the market in the ratio of 60% to 40% respectively of customers. If in every week 70% of G’s customers retain the brand but 30% switch to product x where as 80% of X’s customers retain brand but 20% percent switch to brand G. Analyse the exchange in share market per week.

G X

State the system next week

Column vector for initial market share


Lesson One 34

Share next week

Share week after

and so onThis process can continue till equilibrium is reached.

Let the market share be

∴

0.7G + 0.2 X = G or 0.3G + 0.8X = X0.2X = 0.3G or 0.3X = 0.2X

i.e. =

Hence G’s share is 60% and X’s share is 40%Example 2A marketing division toothpaste manufacturing company has worked out the following transition probability matrices concerning the behaviors of customers before and after an advertising campaign.

Transition probability matrix(before advertising campaign)

TO

FROMOur Brand(State I)

Another Brand(Sate II)

Our brand (State I) 0.8 0.2Another Brand (sate II) 0.4 0.6

Transition probability matrix(After advertisement)

TOOur Brand Another Brand



FROM (State I) (Sate II)Our brand (State I) 0.9 0.1Another Brand (sate II) 0.5 0.5

If the advertising campaign costs Shs 20,000 per year, would it be worthwhile for the company to undertake the campaign?You may suppose there are 60,000 buyers of toothpaste in the market and for each customer average annual profit of the company is Shs 2.50

SolutionLet P1 be the fraction share of our brand and P2 be the fraction share of another brand

Before Advertising

Thus:

After Advertising

Thus:

If there are 60,000 buyers

Before Advertising

this implies that,

customers will buy our brand

= Sh.100,000

After Advertising

this implies that,

customers will buy our brand

= Sh.105,000the difference between advertising and not advertising is

105,000 – 100,000 = Sh.5,000 in favour of advertising, Thus the advertising campaign is worthwhile.


Lesson One 36

INPUT – OUTPUT ANALYSISThe input output analysis is a topic which requires application of matricesThe technique analyses the flow of inputs from one sector of the economy to the other sectors thus the technique is quite useful in studying the interdependence of sectors within a single economy.The input – output analysis was first developed by Prof Leontief hence the Leontief matrix has been developed. See the following example

Example (B)INPUT OUTPUT TABLE

TO Final TotalFROM Agric Industr

yService

Demand

Demand (output)

Agric 300 360 320 1080 2060Industry 450 470 410 800 2130Service 610 500 520 270 1900Primary inputs 700 800 650 - -Total inputs 2060 2130 1900 - -

NB: In the above table, one should be able to interpret the table e.g. of the total demand of 2060 metric tones from the agriculturalsector; 300 is produced for the agricultural sector, 360 for industrial sector, 320 for the service sector and 1080 metric tones makes up the final demand.The final demand is the additional demand besides the sectoral demand which is normally made by other users e.g. government, foreign countries, other manufacturers not included in the other sectors.For production if items besides the inputs from other sectors namely labour capital e.t.c

Technical coefficients : (‘to’ sectors)

Agriculture 300 = = 0.14

450 = = 0.22

610 = = 0.30

Industry 360 = = 0.7

470 = = 0.22

500 = = 0.23



Service 320 = = 0.17

410 = = 0.22

520 = = 0.27

The matrix of technical coefficients is:TO Final Total

FROM Agric Industry

Service Demand

Demand (output)

Agric 0.14 0.7 0.17 1080 (y1)

2060

Industry 0.22 0.22 0.22 800 (y2) 2130Service 0.30 0.23 0.27 270(y3) 1900Primary inputs

x x x - -

2060(x1

)2130(x2

)1900(x3

)- -

From the above table, we may develop the following equations0.14x1 + 0.7x2 + 0.17x3 + y1 = x1

0.22x1 + 0.22x2 + 0.22x3 + y2 = x2

0.30x1 + 0.23x2 + 0.27x3 + y3 = x3

Let the coefficient matrix be represented by

Equation (*) may be written as AX + Y = X Y = X – AX Y = X (I-A)

⇒ (I – A)–1 Y = X


Lesson One 38

The matrix I – A is known as Leontief Matrix

Technical CoefficientsThese show the units required from each sector to make up one complete product in a given sector e.g. in the above matrix of coefficients it may be said that one complete product from the agricultural sector requires 0.14 units from the agricultural sector itself, 0.22 from the industrial sector and 0.30 from the service sectorNB: The primary inputs are sometimes known as “value added”

Example 1Determine the total demand (x) for the industry 1, 2, 3 given the matrix of technical coefficients (A), Capital and the final demand vector B.

From the input – output analysis

X = (I – A)–1 B, Where X = is the demand vector

The matrix of cofactors of is



=

The transpose (adjoint) of the above matrix is

=

= 0.495 – (0.008 + 0.126 + 0.18

= 0.809

Therefore X =

The total demand from the three industries 1, 2 and 3 is 30 from 1, 37.7 from 2 and 21.9 from 3.

Example 2Three clients of Disrup, Ltd P, Q and Rare direct competitors in the retail business. In the first week of the year P had 300 customers Q had 250 customers and R had 200 customers. During the second week, 60 of the original customers of P transferred to Q and 30 of the original customers of P transferred to R. similarly in the second week 50 of the original customers of Q transferred to P with no transfers to R and 20 of the original customers of R transferred to P with no transfers to Q.

Required


Lesson One 40

a) Display in a matrix the pattern of retention and transfers of customers from the first to the second week

(4 marks)b) Re-expres the matrix that you have obtained in part (a) showing the

elements as decimal fractions of the original numbers of customers of P, Q and R (2 marks) Refer to this re expressed matrix as B

c) Multiply matrix B by itself to determine the proportions of the original customers that have been retained or transferred to P, Q and R from the second to the third week. (4 marks)

d) Solve the matrix equation (xyz)B = (xyz) given that x + y + z = 1 (8 marks)

e) Interpret the result that you obtain in part (d) in relation to the movement of customers between P, Q and R

(2marks)(Total 20 marks)

Solutiona). Think of each row element as being the point from which the

customer originated and each column element as being the destination e.g. 210 customers move from P to P, 60 move from P to Q and 30 move from P to R. The sum of the elements of the first row totalling 300, that is the number of customers originally with P.

Hence required matrix is

b). The requirement of this part is to express each element as a decimal fraction of its corresponding row total. The second row, first element is therefore 50/250, that is 0.2 and the second element is therefore 200/250 that is 0.8.

Hence B =

c).

The result can be checked by the normal rules of matrix multiplication.



d).

This produces from the first row0.7x + 0.2y + 0.1z = x

Which reduces to 0.2y + 0.1z = 0.3xOr 2y + z = 3x ………………………...(i)

Or The second row produces, 0.2x + 0.8y = y

Reducing to 0.2x = 0.2 yX = y ………………………..(ii)

Or The third row produces 0.1x + 0.9z = z

Reducing to 0.1x = 0.1z X = z ………. (iii)

At this point you will notice that condition h (ii) and condition (iii) produce 2x + x = 3x when substituted into condition (i), we therefore need extra condition x + y + z = 1 to solve the problem.

Thus x + x + x = 1Or 3x = 1

That is x =

Leading to x = , y = , z =

e). In proportion terms this solution means that P, Q, and R will in the long term each have one third of the total customers

Example 4There are three types of breakfast meal available in supermarkets known as brand BM1, brand BM2 and Brand BM3. In order to assess the market, a survey was carried out by one of the manufacturers. After the first month, the survey revealed that 20% of the customers purchasing brand BM1 switched to BM2 and 10% of the customers purchasing brand BM1 switched to BM3. similarly, after the first month of the customers purchasing brand BM2, 25% switched to BM1 and 10% switched to BM3 and of the customers purchasing brand BM3 0.05% switched to BM1 and 15% switched to BM2


Lesson One 42

Requiredi. Display in a matrix S, the patterns of retention and transfers of

customers from the first to the second month, expressing percentage in decimal form. (2marks)

ii. Multiply matrix S by itself (that is form S2) (5 Marks)

iii. Interpret the results you obtain in part ii with regard to customer brand loyalty (3 marks)

SolutionThe objective of the first part of the question was to test the candidate’s ability to formulate and manipulate a matrix, then interpret the result of such manipulation.a. i. The matrix showing the pattern of retention and transfer from the

first to the second month is

(The second element in the first row shows the 20% movement from BM1 to BM2 and so on)

i. The product of matrix S with itself is demonstrated as follows

Where for example second element in the first row, that is 0.2850 is the result of multiplying the corresponding elements of the first row of S by the second column of S and summing the product.

0.2850 = 0.70 × 0.20 + 0.20 × 0.65 + 0.10 × 0.15= 0.14 + 0.13 + 0.015 e.t.c.

ii. The resulting matrix may be interpreted in the following wayOf the original customers who buy BM1, 54.5% will remain loyal to the brand in month three, 28.5% will have switched to BM2 and 17% will have switched to BM3.Of the original customers who buy BM2, 48.7% will remain loyal to the brand in month three 34.25% will have switched to BM1 and 17% will have switched to BM3



Of the original customers who buy BM3, 66% will remain loyal to the brand in month three 11.25% will have switched to BM1 and 22.75% will have switched to BM2Alternatively In month three 54.5% of the customers buying BM1 are original customers. 34.5% came from BM2 originally and remaining 11.25% have switched from BM3 and so on.

MARKOV CHAINS/PROCESSESThe Markov processes are defined as a set of trials which follow a certain sequence which depend on a given set of probabilities known as transition probabilities. These probabilities indicate how a particular activity or product moves from one state to another.

Applications of Markov Chains in BusinessThe Markov processes or chains are frequently applied as follows:-1. Brand SwitchingBy using the transitional probabilities we can be able to express the manner in which consumers switch their tastes from one product to another.

2. Insurance industryMarkov analysis may be used to study the claims made by the insured persons and also decide the level of premiums to be paid in future.

3. Movement of urban populationBy formulating a transition matrix for the current population in the urban areas, one can be able to determine what the population will be in say 5 years.

4. Movement of customers from one bank to another.It is a fact that customers tend to look for efficient banks. Therefore at a certain time when a given bank installs such machinery as computers it will tend to attract a number of customers who will move from certain banks to efficient ones.

PROPERTIES OF MARKOV CHAINS1. Each outcome in a markov process belongs to a state space or transition matrix. E.g.


Lesson One 44

Where S1, S2, S3 are states and P11 P12 e.t.c are

probabilities

2. The outcome of each trial depends on the immediate preceding activities but not on the previous activities

BASIC TERMS IN MARKOV CHAINSa) Probability VectorThis is a row matrix whose elements are non-negative and also they add up to 1 e.g. u = 0.2, 0.1, 0.2, 0.5)Example Consider u =( ¾ , 0, - ¼ , 1/2 ) Not because – ¼ is negative

v = ( ¾ , ½ , 0, ¼ ) Not because the sum of the element <1w= ( ¼ , ¼ , 0, ½ ) Adds up to 1, each element is non

negative. Therefore it’s a prob, vector

State the ones which are probability vectorsb) Stochastic matrixA matrix whose row elements are all non negative and also add up to 1.

Example (i) M =

Example ii) = Consider the following matrices

A =

A is not stochastic matrix because the element in the 2nd row and 3rd

column is negative.



B is not Stochastic matrix because the elements in the second row do not add up to 1C is stochastic matrix because each element is non negative and they add up to 1 in each row.

c) Regular stochastic matrixA matrix P is said to be regular stochastic matrix if all the elements in Pm are all positive, where m is a power, m = 1, 2, 3 e.t.c

Let A = Where A is a Stochastic Matrix

A2 =

A3 =

Since the elements in A2 and A3 are all positive then A is regular Stochastic matrix.

ABSORBING STATESA state Si (I = 1, 2, 3 …) of a markov chain is called absorbing if the system remains in the state, Si once it enters there. Thus a state, Si is absorbing if and only if the ith row of the transition matrix p has a 1 on the main diagonal and zeroes every where else. See the following example.The following matrix, P is a transition matrix of the markov chain.

The States S2 and S5 are absorbing states since the 2nd and 4th rows have 1 on the main diagonal.


REINFORCING QUESTIONSWork out these questions and compare with answers given in lesson 9

QUESTION ONEDetermine a) f(1) b) f(-2) for the following functions

1. f(x) = –5x + 2 2. f(x) = – x2 + 3x+10

3. f(t) = 10 – t + t3 4. f(u) =

QUESTION TWOSolve the following simultaneous equations.

a)

b)

c)

QUESTION THREEBecause of inreasing cost increasing cost energy, the population within Maueni district seem to be shifting from the north to the south the transition matrix S describes the migration behaviour observed between the regions.

determine whether the populations will attain an equillibrium condition and if so, the population of the two regions.QUESTION FOURA simple hypothetical economy of three industries A, B and C is represented in the following table (data in millions of shillings).

User

A B C Final Demand

Total Output


ProducerA 80 100 100 40 320B 80 200 60 60 400C 80 100 100 20 300

Determine the output vector for the economy if the final demand changes to 60 for A, 60 for B and 60 for C

QUESTION FIVEA tea blender uses two types of tea, T1, and T2, to produce two blends, B1

and B2 for sale. B1 uses 40% of available T1 and 60% of the available T2

whilst B2 uses 50% of the available T1 and 25% of the available T2.

Required:a) Given that t1 kilos of T1 and t2 of T2 are made available to produce b1

kilos of B1 and b2 kilos of B2. Express the blending operation in the matrix format.

b) If 400 kilos of T1 and 700 kilos of T2 were made available for blending, what quantities of B1 and B2 would be produced?

c) If 600 kilos of B1 and 450 kilos of B2 were produced, use a matrix method to determine what quantities of T1 and T2 would be used to produce the blends.

QUESTION SIX

Let A =

a) Find A2 and A3

b) If F(x) = x3 – 3x2 – 2x + 4IFind F (A)

c) Find the inverse of matrix A

QUESTION SEVENA child’s toy is marketed in three sizes standard size contains 10 squares (S), 15 triangles (T) and 6 hexagons (H). The deluxe set contains 15 S, 20 T, 6 H and 4 octagons (O). The super set contains 24 T, 8 H, 16 H, 16 S and 6 (O). Squares cost 12 pence to produce, triangles cost 8p, hexagons cost 18p and octagons 22p.

The standard set is sold at £6, the deluxe set for £10 and super for £15. The manufacturer produces 100 standard sets, 80 deluxe sets and 50 super sets per week.

Use matrix form and matrix multiplications to find:

STRATHMORE UNIVERSITY ● STUDY PACK 47

The cost of producing each set.The number of each shape required each weekTotal expenditure on shapes each week

QUESTION EIGHTMatrix N below shows the number of items of type A, B, and C in warehouses Y and W. Matrix p shows the cost in pence per day of storing (S) and maintaining (M) one item each of A, B and C

a) Evaluate the matrix (N×P) and say what it represents.

b) Stock movement occurs as follows:

At the start of the day 1:Withdrawal of 2 type B from warehouse Y, 20 of type A from warehouse W.

At the start of day 2:Delivery of 7 type B and 10 of type C to warehouse Y and 15 of type B to warehouse W.

Evaluate the total cost of storage and maintenance for days 1 and 2.

c) Write down without evaluating a matrix expression which could be used to evaluate the storage and maintenance cost of items A, B and C for the period from day 1 to 4. Allow for the stock movements on days 1 and 2, as described in part (b). There were no stock movements on days 3 and 4.


49 Sets Theory and Calculus

LESSON TWO

Sets Theory and Calculus- Sets and set theory- Calculus- Differentiation and integration of polynomial, exponential and

logarithmic functions- Application of calculus to economic models

2.1 Sets and set theoryA set is a collection of distinct objects. We may consider all the ocean in the world to be a set with the objects being whales, sea plants, sharks, octopus etc, similarly all the fresh water lakes in Africa can form a set. Supposing A to be a set

A = {4, 6, 8, 13}The objects in the set, that is, the integers 4, 6, 8 and 13 are referred to as the members or elements of the set. The elements of a set can be listed in any order. For example,

A = {4, 6, 8, 13} = {8, 4, 13, 6}Sets are always precisely defined. Each element occurs once and only once in a set.The notion is used to indicate membership of a set. ∉ represents non membership. However, in order to represent the fact that one set is a subject of another set, we use the notion . A set “S” is a subject of another set “T” if every element in “S” is a member of “T”ExampleIf A = {4, 6, 8, 13} then

i) 4 {4, 6, 8, 13} or 4 A; 16 ∉ Aii) {4, 8} A; {5, 7} A; A A

Methods of set representationCapital letters are normally used to represent sets. However, there are two different methods for representing members of a set:

i. The descriptive method andii. The enumerative method

The descriptive method involves the description of members of the set in such a way that one can determine the elements of the set without difficulty.The enumerative method requires that one writes out all the members of the set within the curly brackets.For example, the set of numbers 0, 1, 2, 3, 4, 5, 6 and 7 can be represented ass follows


Lesson Two 50

P = {0, 1, 2, 3, 4, 5, 6, 7} , enumerative methodP = {X/x = 0, 1, 2…7} descriptive methodOr P = {x/0 ≤ x ≤7} where x is an interger.



Finite and infinite setsA set can be classified as a finite or infinite set, depending upon the number of elements it has. A finite set has a finite number of elements whereas an infinite set has an infinite number of elements.For example, set P below has ten elements and is therefore a finite set. Set S, on the other hand, is an infinite set since it has an infinite number of elements.

P = {2, 4, 6…20}S = {1, 3, 5…}

Universal setThe term refers to the set that contains all the elements that an analyst wishes to study. The notation U or ξ is generally used to denote universal setsThe null set or empty setThis is a set which contains no elements. It is normally designated by a Greek letter Ø, or { }. The sets Ø and { Ø } are not the same thing since the former has no elements in it, while the later has one element in it, namely zeroEqual or equivalent setsTwo sets C and D are said to be equal if every member of set C belongs to D and every member of set D also belongs to CComplement of a setThe complement of set A is written as A΄. This set contains all those elements of universal set which are not in A Intersection and union

denotes the intersection of B and C. it is the set containing all those elements, which belong to both B and C

If B = {5, 8, 11, 20, 25} and C = {1, 3, 5, 7, 9, 11, 13}Then = {5, 11}

= {1, 3, 5, 7, 8, 9, 11, 13, 20 25}

SET OPERATIONS AND SOME LAWS OF SET THEORYTHE VENN DIAGRAMA simple way of representing sets and relations between sets is by means of the Venn diagram. Venn diagram consists of a rectangle that represents the universal set. Subjects of the universal set are represented by circles drawn within the rectangle, or the universe.Suppose that the universal set is designated by U and the sets A, B and C are subject of U.The Venn diagram below can be used to illustrate the sets as follows


Lesson Two 52

Venn diagram below representing the intersection of set A and B or = C is illustrated as follows

Intersection of sets

U

Example: You are given the universal set

T = {1, 2, 3, 4, 5, 6, 7, 8}And the following subjects of the universal set:

A = {3, 4, 5, 6,}B = {1, 3, 4, 7, 8}

Determine the intersection of A and BSolutionThe intersection of A and B is the subject of T, containing elements that belong to both A and B

= {3, 4, 5, 6,} {1, 3, 4, 7, 8}= {3, 4}

or

U


U

AB

C

CA B

T

5 3 1A 6 4 7 B

8


ExampleConsider the following universal set T and its subjects C, D and E

T = {0, 2, 4, 6, 8, 10, 12}C = {4, 8,}D = {10, 2, 0}E = {0}

Find ii)iii)

Solutionii) = {10, 2, 0} {0} = {0}

= Shaded area

ii) = {4, 8} {10, 2, 0} {0} = { } = Ø

Mutually exclusive or disjointed setsTwo sets are said to be disjointed or mutually exclusive if they have no elements in common. Sets P and R below are disjointed


T

D E

T

D C

4; 8 E

2; 10 0

53

Lesson Two 54

Disjointed sets are represented by a null set in this caseP R = Ø

The union of setsVenn diagram representing the union of sets A and B or A B = Shaded area is illustrated below;-

A B = Shaded areaExampleConsider the universal set T and its subsets A, B and C below:

T = {a, b, c, d e, f} A = {a, d}B = {b, c, f} C = {a, c, e, f}

Findii)iii)iv)v)

Solutioni) = {a, d} {b, c, f} = {a, b, c, d, f} ii) = {a, d} {a, c, e, f} = {a, c, d, e, f}iii) = {b, c, f} {a, c, e, f} = {a, b, c, e, f} iv) = {a, d} {b, c, f } {a, c, e, f} = {a, b, c, d, e, f}

= T


T

P R

U

A B


Complement of a setVenn diagram representing the complement of a set say A represented by A1 is illustrated below.

A1= shaded area

Example For the universal set T = {1, 2, 3, 4, 5} and its subset A ={2, 3} and B ={5, }

Find ii) A1

iii) (A1)1

iv) (B1)1

Solutioni) A1 ={2, 3}1 = {1, 4, 5}ii) (A1)1 =({2, 3}1)1 = {1, 4, 5}1={2, 3} = Aiii) (B1)1=({5}1)1 = {1, 2, 3, 4}1={5} = B

Some Laws of Set AlgebraFrom the following Venn diagram where T is the universal set and A its subset, we can deduce a number of laws.

i) A Ø = Aii) A T = Tiii) A A = A


U

A1

T

A

55

Lesson Two 56

iv) A A= Av) A T= Avi) A A1 = Tvii) A A1=Øviii) (A1)1 = A

Solving Problems Using Venn DiagramsExample 1Of the 20 girls in a form, 16 play hockey 12 play tennis and 4 play basketball. Every girl plays at least one game and two play all the three. How many play two and only two games.Solution

N(ξ) = 20

H = Hockey T = Tennis B = Basketball

Those who play two and only two games = x + y + z

Using the diagram(14 – x – y) + (10 – y – z) +(2 – x – z ) + 2 + x + y+ z = 2028 –x –y –z = 20

x + y + z = 8Example 2250 members of a certain society have voted to elect a new chairman. Each member may vote for either one or two candidates. The candidate elected is the one who polls most votesThree candidates x, y z stood for election and when the votes were counted, it was found that

- 59 voted for y only, 37 voted for z only- 12 voted for x and y, 14 voted for x and z


n(T) = 12n(H) = 16

n(B) = 4

16 – x – y – 2

y 12 – y –z – 2

2 x z

4 –x –z –2


- 147 voted for either x or y or both x and y but not for z- 102 voted for y or z or both but not for x

Requiredi) How may voters did not voteii) How many voters voted for x onlyiii) Who won the elections

Solution

P + 12 + 59 = 147 giving P = 76Q + 59 + 37 = 102 giving Q = 6

i) Those who did not vote= 250 – ( 76 + 12 + 14 + 59 + 6 + 37)= 250 – 204 = 46

ii) x = 76 + 12 + 14 = 102y = 12 + 59 + 6 = 77z = 37 + 14 + 6 = 57

iii) X won the election

2.2 CALCULUSCalculus is a branch of mathematics which explains how one variable changes in relationship to another variable. It enables us to find the rate of change of one variable with respect to another variable.


N(ξ) = 250

X Y

Z

P

Tail end

14 Q

37

57

Lesson Two 58

Examplei. The rate at which business revenue is increasing at a particular

stage when volume of sales is increasing.ii. The rate at which costs are changing at a particular stage when

volume of sales is giveniii. The evaluation of ‘rate of change’ can help us to identify when

the change in one variable reaches a maximum or minimum.iv. Calculus may be used in production management when the

production manager wants to know a) How much is to be manufactured in order to maximize the

profits, revenues e.t.cb) How much is to be produced in order to minimize the

production costs

Calculus is divided into two sections namely:

Differentiation and integrationDifferentiation deals with the determination of the rates of change of business activities or simply the process of finding the derivative of a function.Integration deals with the summation or totality of items produced over a given period of time or simply the reverse of differentiation

The derivative and differentiationThe process of obtaining the derivative of a function or slope or gradient is referred to as derivation or differentiation.The derivative is denoted by or f΄(x) and is given by dividing the change in y variable by the change in x variable.The derivative or slope or gradient of a line AB connecting points (x,y) and (x+dx, y + dy) is given by

Where dy is a small change in y and dx is a small change in x variables.

Illustration


(x,y) = A

B = (x + dx, y +dy) dy

y dx

(y + dy)

Line AB


x (x + dx)

Rules of Differentiation1. The constant function rule

If given a function y = k where k is a constant then

= 0

ExampleFind the derivative of (i) y = 5

Solutioni. y = 5 d y = 0

dx

Illustration

y

5 y = 5

derivative of a constant function x

2. Power function rule

Given a function

Example

Find d y for;


dy

5 0 00

dyslopedx

59

Lesson Two 60

dx(i). y = x7

(ii). y = x2ˠ

(iii). y = x-3

(iv). y = x

Solutioni. y = x7

d y = 7x 7-1 = 7x6

dxii. y = x2ˠ

d y = 2ˠ x(2ˠ - 1)

dxiii. y = x-3

d y = -3x –3-1 = -3x-4

dxiv. y = x

d y = 1x 1-1 = 1.x0= 1 (since x0=1)dx

3. Power function multiplied by a constantIf given y = Axr, then d y = rAxr-1

dx

4. The sum ruleThe derivative of the sum of two or more functions equals the sum of the derivatives of the functions.For instance If H(x) = h(x) + g(x)Then d y or H´(x) = h´(x) + g´(x)

dx

5. The difference ruleThe derivative of the difference of two or more functions equals the difference of the derivatives of the functions If H (£) = h(x) – g(x)

Then H´(£) = h´(x) – g´(x)

Examples.



Find the derivatives of i. y = 3x2 + 5x + 7ii. y = 4x2 – 2xb

Solutioni. y = 3x2 + 5x + 7

ii. y = 4x2 – 2xb

6. The product rule – both factors are functionsThe derivative of the product of two functions equals the derivative of the first function multiplied by the second function PLUS the derivative of the second function multiplied by the first function.given that

Then

Example Find dy for

dxi. y = x2(x)ii. y = (x2+ 3) (2x3+ x2- 3)

Solutioni. y = x2(x)


Lesson Two 62

Note that y = x2(x) = x3. Directly differentiating this we get 3x2.

ii. y = (x2+ 3) (2x3+ x2- 3)

7. Quotient RuleThe derivative of the quotient of two functions equals the derivative of the numerator times the denominator MINUS the derivative of the denominator times the numerator, all which are divided by the square of the denominator

If given H (x) =

then

For exampleFind dy for

dx

i.

ii.



Solutions

i.

=

=

ii.

ExampleA farmer of a large farm of poultry announced that egg production per month follows the equation;

w = 3m 3 – m 2 m2 + 10

Where w – Total no of eggs produced per monthm – amount in kilograms of layers mash feed.

Required Determine the rate of change of w with respect to m (i.e. the rate at which the number of eggs per month increase or decrease depending on the rate at which the kilos of layers marsh are increased).

SolutionLet u = 3m3 – m2

∴ du = 9m2 – 2mdm

Let v = m2 + 10∴ dv = 2m

dm

∴


Lesson Two 64

8. Chain RuleThis rule is generally applied in the determination of the derivatives of composite functions, which can be defined as a function in which another function can be considered to have taken the place of the independent variable. The composite function is also referred to as a function of a function.It is normally of the form y = (2x2 + 3)3. If we let u = (2x2 + 3), then y = u3.In order to differentiate such an equation we use the formula

Solutiony = (2x2 + 3)3

Let u = 2x2 + 3∴ du = 2x

dx

Let y = u3

∴ = 3u2

dy = dy . du = 3u2 x 4x = 12xu2

dx du dx= 12x(2x2 + 3)2

ExampleConsider the function

y = (x2 + 16x + 5)2

which can be decomposed into y = u2 and u = x2 + 16x + 5. in this case y is a function of (x2 + 16x + 5)Hence y = f(u) and u = g(x)

dy = dy . du dx du dx



= (2u) (2x +16)

= 2 (x2 + 16x + 5) (2x + 16)

9. The derivative of a function raised to power r; the composite function rule.

The derivative of a function raised to power r equals to the power r times the function which is raised by power (r-1), all of which is multiplied by the derivative of the function

If y = [g(x)]r

Then dy = r[g(x)]r-1 . g´(x) dx

For example

Find

Solution

Differentiation of an implicit functionAn Implicit function is one of the y = x2 y + 3x2 + 50. it is a function in which the dependent variable (y) appears also on the right hand side.To differentiate the above equation we use the differentiation method for a product, quotient or function of a function.

Solutiony = x2 y + 3x2 + 50


Lesson Two 66

Partial derivativesThese derivatives are used when we want to investigate the effect of one independent variable on the dependent variable.For example, the revenues of a farmer may depend on two variables namely; the amount of fertilizer applied and also the type of the natural soil.Let ㄫ = 30x2y + y2 + 50x + 60yWhere ㄫ = annual revenue in £ ‘000’

x = type of soily = amount of fertilizer applied

RequiredDetermine the rate of change of the ㄫ with respect to x and y

Solution

ㄫ = 30x2y + y2 + 50x + 60yDifferentiating ㄫ with respect to x keeping y constant we have

d ㄫ = 60xy + 50dx

Differentiating ㄫwith respect to y keeping x constant we have

d ㄫ = 30x2 + 2y + 60 dy

Maxima, minima and points of inflexiona) Test for relative maximum

Consider the following function of x whose graph is represented by the figure below

y = f(x)dy = f´(x)dx


y f xE

D

C

B

A

x3x2x1x

0dydx

0dydx


Relative maximum pointThe graph of the function slopes upwards to the right between points A and C and hence has a positive slope between these two points. The function has a negative slope between points C and E. At point C, the slope of the function is Zero.

Between points X1 and X2 Where X1 ≤ X < X2

and between X2 and X3 Where X2 < X ≤ X3.

Thus the first test of the maximum points require that the first derivative of a function equals zero or

The second text of a maximum point requires that the second derivative of a function is negative or

Example Determine the critical value for the following functions and find out the critical value that constitutes a maximum

y = x3 – 12x2 + 36x + 8

Solution y = x3 – 12x2 + 36x + 8then dy = 3x2 – 24x + 36 +0

dx


Lesson Two 68

iii. The critical values for the function are obtained by equating the first derivative of the function to zero, that is:

dy = 0 or 3x2 – 24x + 36 = 0dxHence (x-2) (x-6) = 0And x = 2 or 6

The critical values for x are x = 2 or 6 and critical values for the function are y = 40 or 8ii. To ascertain whether these critical values of x will give rise to a maximum, we apply the second text, that is

d 2 y < 0 d2xdy = 3x2 – 24x + 36 and dxd 2 y = 6x - 24 d2x

a) When x = 2 Then d 2 y = -12 <0

d2x

b) When x = 6Then d 2 y = +2 > 0

d2xHence a maximum occurs when x = 2, since this value of x satisfies the second condition. X = 6 does not give rise to a local maximum

b) Tests for relative minimumThere are two tests for a relative minimum point

i. The first derivative, that is dy = f´(x) = 0dx

ii. The second derivative, that is d 2 y = f´(x) > 0dx2

Example:For the function

h(x) = 1/3 x3 + x2 – 35x + 10 Determine the critical values and find out whether these critical values are maxima or minima. Determine the extreme values of the function

Solutioni. Critical values

h(x) = 1/3 x3 + x2 – 35x + 10 and h´(x) = x2 + 2x – 35

by first text,then h´(x) = x2 + 2x – 35 = 0or (x-5) (x+7) = 0



Hence x = 5 or x = -7

ii. The determinant of the maximum and the minimum points requires that we test the value x = 5 and –7 by the second textH´´(x) = 2x + 2a) When x = -7 h´´(x) = -12 <0b) When x = 5 h´´(x) = 12>0

There x = -7 gives a maximum point and x = 5 gives a minimum point.

iii. Extreme values of the functionh(x) = 1/3 x3 + x2 – 35x + 10 when x = -7, h(x) = 189 2/3 when x = 5, h(x) = -98 1/3 The extreme values of the function are h(x) = 189 2/3 which is

a relative maximum and h(x) = -98 1/3 , a relative minimum

c) Points of inflexion Given the following two graphs, points of inflexion can be determined at points P and Q as follows:

y y=g(x)

P

k1 x


Lesson Two 70

Diagram (i)

y

y =f(x)

Q

k2 xThe points of inflexion will occur at point P when

g´´(x) = 0 at x = k1

g´´(x) < 0 at x < k1

g´´ (x) > 0 at x > k1

and at point Q when f´´(x) = 0 at x = k1

f´´(x) > 0 at x < k1

f´´(x) < 0 at x > k1

ExampleFind the points of inflexion on the curve of the function

y = x3

SolutionThe only possible inflexion points will occur where

From the function given

Equating the second derivative to zero, we have 6x = 0 or x = 0

We test whether the point at which x = 0 is an inflexion point as follows

When x is slightly less than 0, which means a downward

concavity



When x is slightly larger than 0, which means an upward

concavityTherefore we have a point of inflexion at point x = 0 because the concavity of the curve changes as we pass from the left to the right of x = 0Illustration

y

y=x3

point of

inflexion

0 x

Example1. The weekly revenue Sh. R of a small company is given by

Where x is the number of units produced.

Requiredi. Determine the number of units that maximize the revenueii. Determine the maximum revenueiii. Determine the price per unit that will maximize revenue

Solutioni. To find maximum or minimum value we use differential calculus

as follows


Lesson Two 72

Therefore at x = 18, the value of R is a maximum. Similarly at x = -18, the value of R is a minimum. Therefore, the number of units that maximize the revenue = 18 unitsii. The maximum revenue is given by

R = 14 + 81 + 18 – (18) 3 12= Shs. 986

ii. The price per unit to maximize the revenue is 986 = 54.78 or Shs.54.7818

2.3 INTEGRATIONIt is the reversal of differentiationAn integral can either be indefinite (when it has no numerical value) or definite (have specific numerical values)It is represented by the sign ʃf(x)dx.

Rules of integrationi. The integral of a constant

ʃadx = ax +cwhere a = constantExample Find the following

a) ʃ23dx



b) ʃɤ2dx. (where ɤ is a variable independent of x, thus it is treated as a constant).

Solutioni. ʃ23dx = 23x + cii. ʃɤ2dx. = ɤ2 x + c

ii. The integral of x raised to the power n

Example Find the following integrals

a) ʃx2dx b) ʃx-5/2 dx

Solution

iii). Integral of a constant times a function

Example Determine the following integrals

i. ʃax3dx ii. ʃ20x5dx

Solution

iv). Integral of sum of two or more functionsʃ{f(x) + g(x)} dx = ʃf(x)dx + ʃg(x) dxʃ{f(x) + g(x) + h(x)}dx = ʃf(x)dx + ʃg(x)dx + ʃh(x)dx

ExampleFind the following

i. ʃ(4x2 + ½ x-3) dxii. ʃ(x3/4 + 3/7 x- ½ + x5)

Solution


Lesson Two 74

5. Integral of a differenceʃ{f(x) - g(x)} dx = ʃf(x)dx - ʃg(x) dx

Definite integrationDefinite integrals involve integration between specified limits, say a and b

The integral Is a definite integral in which the limits of

integration are a and bThe integrals is evaluated as follows

1. Compute the indefinite integral ʃf(x)dx. Supposing it is F(x) + c

2. Attach the limits of integration3. Substitute b(the upper limit) and then substitute a (the

lower limit) for x.4. Take the difference and the result is the numerical value for

the definite integral.

Applying these steps to the definite integral

Example

Evaluate i. (3x 2 + 3)dx

ii. (x + 15)dx

Solution

a. (3x 2 + 3)dx = [(x 3 + 3x + c)]



= (27 + 9 + c) – (1 + 3 + c)= 32

b. (x + 15)dx = [( ½ x2 + 15x + c)]

= (12 ½ + 75 + c) – (0 + 0 + c)= 87 ½

The numerical value of the definite integral f(x)dx can be interpreted as the area bounded by the function f(x), the horizontal axis, and x=a and x=b see figure below

y = f(x)

f(x)

0 a b xarea under curve

Therefore f(x)dx = A or area under the curve

Example1. You are given the following marginal revenue function

Find the corresponding total revenue functionSolutionTotal revenue


0

20

25

30

35

40

45

50

55

60

65

70A

75

Lesson Two 76

Example 2A firm has the following marginal cost function

Find its total cost function.

Solution The total cost C is given by

C = ʃMC.dq= ʃ(a + a1q + a2q2).dq

Note: Exams focus: Note the difference between marginal function and total function. You differentiate total function to attain marginal function, this is common in exams, total profit = total revenue – total cost.

Example 3. Your company manufacturers large scale units. It has been shown that the marginal (or variable) cost, which is the gradient of the total cost curve, is (92 – 2x) Shs. thousands, where x is the number of units of output per annum. The fixed costs are Shs. 800,000 per annum. It has also been shown that the marginal revenue which is the gradient of the total revenue is (112 – 2x) Shs. thousands.

Required i. Establish by integration the equation of the total cost curveii. Establish by integration the equation of the total revenue curveiii. Establish the break even situation for your companyiv. Determine the number of units of output that would

a) Maximize the total revenue and b) Maximize the total costs, together with the maximum total

revenue and total costs

Solutioni. First find the indefinite integral limit points of the marginal

cost as the first step to obtaining the total cost curveThus ʃ(92 – 2x) dx = 92x – x2 + cWhere c is constant

Since the total costs are the sum of variable costs and fixed costs, the constant term in the integral represents the fixed costs, thus if Tc are the total costs then,

Tc = 92x – x2 + 800or Tc = 800 + 92x - x2



ii. As in the above case, the first step in determining the total revenue is to form the indefinite integral of the marginal revenueThus ʃ(112 - 2x) dx = 112x – x2 + cWhere c is a constant

The total revenue is zero if no items are sold, thus the constant is zero and if Tr represents the total revenue, then

Tr = 112x – x2 iii. At break even the total revenue is equal to the total costs

Thus 112x – x2 = 800 + 92x - x2 20x = 800 x = 40 units per annum

iv. a) Tr = 112x – x2

at the maximum point

x = 56 units per annum

Since this confirms the maximum

The maximum total revenue is Shs. (112 x 56 – 56 x 56) x 1000= Shs. 3,136,000

ii. Tc = 800 + 92 x – x2

At this maximum point


Lesson Two 78

92 – 2x = 092 = 2xx = 46 units per annum

since

this confirms the maximum

the maximum costs are Shs. (800 + 92 x 46 - 46 x 46) x 1000= Shs. 2,916,000



REINFORCING QUESTIONS

QUESTION ONEFind the derivative of

a) y = 6x – x

b)

c)

d)

QUESTION TWOA cost function is

Ksh.(c) = Q2 – 30Q + 200Where Q = quantity of units produced

Find the point of minimum cost.

QUESTION THREE250 members of a certain society have voted to elect a new chairman. Each member may vote for either one or two candidates. The candidate elected is the one who polls most votes.

Three candidates x, y, z stood for election and when the votes were counted, it was found that,

59 voted for y only, 37 voted for z only12 voted for x and y, 14 voted for x and z147 voted for either x or y or both x and y but not for z102 voted for y or z or both but not for x.

Required:i) How many voters did not vote?

ii) How many voters voted for x only?iii) Who won the election?

QUESTION FOURThe weekly revenue Ksh.R of a small company is given by:R = 14 + 81x – x 3 where x is the number of units produced

12

Required:


Lesson Two 80

a) Determine the number of units that maximize the revenue.b) Determine the maximum revenue.c) Determine the price per unit that will maximize the revenue



QUESTION FIVEA furniture firm has two operating departments; Production and sales. The firms’s operating costs are split between these two departments with the resultant period of fixed costs of Shs.20,000 and Shs.6,000 respectively. The production department has a basic variable cost per unit of Shs.6 plus additional variable cost per unit of Shs.0.0002 which relates to all the manufactured items during the period. The sales department has a variable cost per unit of Shs.2. The sales department receives the finished goods from the production department and pay the basic variable cost per unit plus 80% of the same.

NB: Demand Q is given by the following function:

Q = 40,000 – 2,000P, where P is the selling price of the sales department.

Required:a) Calculate the quantity that maximizes the profits of the production

department.b) Calculate the selling price that maximizes the profits of the sales

department.c) Determine the firm’s profit as a result of adopting the quantity and

selling prices in i and ii.

d) Determine the quantity and selling price that maximize the ship’s profit. What is the amount of this profit?

QUESTION SIXa) Describe how quadratic equations can be used in decision making.

b) The demand for a commodity is given by p = 400 – q. The average total cost of producing the commodity is given by

where p is the price in shillings and q is the quantity in kilograms.

Requiredi) What does in the ATC equation represent economically?

(1 mark)ii) Determine the output that leads to maximum profit and the profit

at the level of output. (9

marks)

c) Alpha industries sells two products, X and Y, in related markets, with demand functions given by:


Lesson Two 82

Px – 13 + 2X + Y = 0Py– 13 + X + 2Y = 0

The total cost, in shillings, is given by:TC = X + Y

Required:Determine the price and the output for each good which will

maximize profits. (7 marks)(Total: 20

marks)



QUESTION SEVENa) The following table shows the Fixed Cost (F) and the variable cost (V)

of producing 1 unit of X and 1 unit of Y:

ProductX Y

Cost F 5 8 (Shs ‘000’)Cost V 4 12

When x units of X and y units of Y are produced, the total fixed cost is Shs.640,000 and total variable cost is Shs.820,000. Express this information as a matrix equation and hence find the quantities of x and y produced using matrix algebra. (10 marks)

The marginal productivity of an industrial operation (the production of electric furnaces) is given by:

Where x is capitalization in millions of shillings. Given that, when the capitalization is Shs. Million they can produce 62 of the furnaces per week.

Required:a) How many furnaces they will be able to produce if their capitalization

increased to Shs 10 million.b) What does the term marginal of productivity mean?

(10 marks)(Total: 20 marks)

Compare your solutions with those given in lesson 9


Lesson Two 84

COMPREHESIVE ASSIGNMENT ONEWork out these question for three hours (exam condition) then hand

them in to DLC for marking

Instructions:Answer any THREE questions from SECTION I and TWO questions from SECTION II. Marks allocated to each question are shown at the end of the question. Show all your workings

SECTION IQUESTION ONEa) Explain the importance of set theory in business.

(4 marks)b) By use of matrix algebra, develop the Leontief inverse matrix.

(8 marks)c) Digital Ltd. Manufactures and sells floppy disks at Nairobi Industrial

Area. The average total cost (ATC) and Average Revenue (AR) (in thousands

of shillings) of producing x floppy disks are given by the following functions:

ATC =

And

AR = 800 – 2x2

Where: x is the number of floppy disks produced

Required:i) The profit function (3

marks)ii) The number of floppy disks required to maximize profit

(3 marks)iii) The maximum profit (2

marks)(Total: 20

marks)

QUESTION TWODefine the following terms as used in Markov analysis:

Markov process (2 marks)Equilibrium or steady state (2 marks)Absorbing state (2 marks)



Closed state (2 marks)

The manufacturer of Tamu Soft drinks has been facing stiff competition on its main brand Tamu-cola soda. The management is considering an extensive advertising and rebranding campaign for Tamu-Cola soda. If the current branding remains, the transition matrix of consumer between Tamu-Cola and other brands will be as follows:

The advertising and rebranding campaign is expected to cost Sh.20 million each year.

There are 40 million consumers of soft drinks in the market and for each consumer the average profitability is Sh.5 annually.

Required:i) The equilibrium state proportion of consumers using Tamu-Cola

before the advertising campaign.(4 marks)

ii) The equilibrium state proportion of consumers using Tamu-Cola after the advertising campaign.

(4 marks)iii) The expected annual profit increase or decrease after the advertising

campaign. Would you recommend the advertising campaign?(4 marks)

(Total: 20 marks)

QUESTION THREEa) A market researcher investigating consumers’ preference for three

brands of beverages namely: coffee, tea and cocoa, in Ongata town gathered the following information:

From a sample of 800 consumers, 230 took coffee, 245 took tea and 325 took cocoa, 30 took all the three beverages, 70 took coffee and cocoa, 110 took coffee only, 185 took cocoa only.

Required:i) Present the above information in a Venn Diagram.

(4 marks)ii) The number of customers who took tea only.

(2 marks)iii) The number of customers who took coffee and tea only.

(2 marks)


ToTamu-Cola

Others

FromTamu-Cola

0.85 0.15

Others 0.25 0.75

85

Lesson Two 86

iv) The number of customers who took tea and cocoa only.(2 marks)

v) The number of customers who took none of the beverages.(2 marks)

b)i) Explain the importance of the Chi-square significance test

(2 marks)ii) The number of books borrowed from Millennium town library

during a particular week was recorded as shown below:

Days of the week

Monday

Tuesday

Wednesday

Thursday

Friday

Total

Number of books borrowed

132 110 128 105 150 625

Required:Test the hypothesis that the number of books borrowed does not depend on the day of the week at the 1% significance level. (6 marks)

(Total: 20 marks)

QUESTION FOURa) The general multiple linear regression equation is expressed as:

Where Yi is the response variable Xi are the explanatory variables βo is the constant β1 are the parameters, and is the error term

Required:Express the above multiple linear regression equation in a matrix form. Clearly indicate the size of each vector column and the matrix.

(10 marks)

b) Mambo Company Ltd. Manufactures five products V, W, X Y and Z. The company has divided its sales team into three regions; A, B and C. The Matrix Q below represent the expected sales quantities in thousands for each product in each sales region for the coming year.



Region

A B C50 20 35 V40 30 10 W

Q = 25 42.5 5 X products10 15 35 Y25 17.5 22.5 Z

Each product is manufactured using combinations of four standard components. The matrix T below indicates the number of units of each component used in producing each product.

Components

1 2 3 41 1 2 0 V0 1 1 2 W

T = 3 2 1 1 X products

0 0 3 1 Y1 2 3 1 Z

The manufacture of each component requires the use of certain resources. The matrix M below indicates the quantities of the three standard parts and the number of production labour hours and assembly labour hours used to produce one unit of each component.

Part 1

Part 2

Part 3

ProductionsLabour hours

Assembly hours

2 1 0 2 1 10 3 2 4 3 2

M = 1 2 1 1 6 3 components

2 5 4 1 2 4


Lesson Two 88

The costs of the resources in matrix M are Part 1 Sh.20, Part 2 Sh.10, Part 3 Sh.30 while each labour hour in the production and assembly departments cost Sh.15 and Sh.5 respectively.

Required:i) The total expected demand for each product. (2

marks)ii) The quantities of each component needed in the production

process. (3 marks)iii) The quantities of each resource required in the production.

(3 marks)iv) The total cost of producing the required units of each product.

(2 marks)(Total: 20

marks)

QUESTION FIVEThe Young Children’s Fund (YCF) is planning its annual fund-raising campaign for its December school holiday camp for disadvantaged children. Campaign expenditures will be incurred at a rate of Sh.10,000 per day. From past experience, it is known that contributions will be high during the early stages of the campaign and will tend to fall off as the campaign continues. The function describing the rate at which contributions are received is:

C (t) = 100t2 + 200,000Where t = days of the campaignC (t) = rate at which contributions are received in shillings per day

The fund wants to maximize the net precedes from the campaign.

Required:i) The number of days the campaign should be conducted to maximize

thenet proceeds. (3

marks)ii) The total campaign expenditure (2

marks)iii) The total contributions expected to be collected (5

marks)iv) Net proceeds from the campaign (1

mark)

The national office of a car rental company is planning its maintenance for the next year. The company’s management are interested in determining the company’s needs for certain repair parts. The company rents saloon cars, station wagons and double cab pick-ups. The matrix N shown below indicates the number of each type of vehicle available for renting in the four regions of the country.



Saloons Station wagons

Double cabs

160 400 500 Coast N =

150 300 200 Central

100 100 150 Western120 400 300 Highland

s

Four repair parts of particular interest, because of their cost and frequency of replacement, are fan belts, spark plugs, batteries and tyres. On the basis of studies of maintenance records in different parts of the country, the management have determined the average number of repair parts needed per car during a year.

These are summarized in matrix R below:

Saloons Station wagons

Double cabs

17 16 15 Fan belts N =

12 8 5 Plugs

9 7 5 Batteries4 7 6 Tyres

Required:i) The total demand for each type of car. (3

marks)ii) The total number of each repair part required for the fleet.

(3 marks)iii) If matrix C below contains the cost per unit in shillings for fan belts,

spark plugs, batteries and tyres, calculate the total cost s for all repair parts. C = (1250,800,6500,8000). (3 marks)

(Total: 20 marks)

SECTION IIQUESTION SIXa) Define the following terms as used in input-output analysis:

i) Transactions table. (2 marks)

ii) Primary inputs. (2 marks)

iii) Technical coefficients. (2 marks)

b) Briefly explain the importance of input-output analysis.(4 marks)


Lesson Two 90

c) A small economy has three main industries which are steel, motor vehicles and construction. The industries are interdependent. Each unit of steel output requires 0.2 units from steel, 0.3 units from motor vehicles and 0.4 units from construction. A unit of motor vehicles output requires 0.2 units from steel, 0.4 units from motor vehicles and 0.2 units from construction. A unit of construction output requires 0.3 units from steel, 0.4 units from motor vehicles and 0.1 units from construction. The final demand is 20 million units from steel. 50 million units from motor vehicles and 30 million units from construction.

Required:i) The technical coefficient matrix. (2

marks)ii) Total output of each industry, given that the Lentief’s inverse matrix

is

1__ 0.192

0.46 0.24 0.260.43 0.60 0.410.30 0.24 0.42

(3 marks)

iii) If the final demand from steel drops by 2 million units, and that from motor vehicles increases by 10 million units, but there is no change in the final demand from construction, what would be the change in the total output of constructions? (5 marks)

(Total: 20 marks)

QUESTION SEVENa) Explain the purpose of Venn diagram

(3 marks)

b) A market study taken at a local sporting goods store, Maua Wahome Stores showed that of the 200 people interviewed, 60 owned tents, 100 owned sleeping bags, 80 owned camping stoves, and 40 owned both tents and camping stoves and 40 owned both sleeping bags and camping stoves.

Required:If 20 people interviewed owned a tent, a sleeping bag and a camping stove, determent how many people owned only a camping stove. In this case, is it possible for 30 people to own both a tent and a sleeping bag, but not a campaign stoves? (6 marks)



“Under One Thousand Shillings” Corner Store is planning to open a new store on the corner of Main and Crescent Streets. It has asked the ‘Tomorrow’s Marketing company’ to do a market study of randomly selected families within a five kilometers radius of the store,.the questions it wishes ‘Tomorrow’s Marketing Company’ to ask each home-owner are:

i) Family incomeii) Family size

iii) Distance from home to the store siteiv) Whether or not the family owns a car or uses public transport

Required:For each of the four questions, develop a random variable of interest to “Under One Thousand Shillings” Corner Store. Denote which of these are discrete and which are continuous random variables.

(11 marks)(Total: 20

marks)

QUESTION EIGHTTwo CPA students were discussing the relationship between average cost and total cost. One student said that since average cost is obtained by dividing the cost function by the number of units Q, it follows that the derivative of the average cost is the same as marginal cost, since the derivative of Q is 1.

Required:Comment on this analysis. (4 marks)

Gatheru and Karibu Certified Public Accountants have recently started to give business advice to their clients. Acting as consultants, they have estimated the demand curve of a client’s firm to be;

AR = 200 – QWhere AR is average revenue in millions of shillings and Q is the

output in units.

Investigations of the clients firm’s cost profile shows that marginal cost (MC) is given by:

MC = Q2 – 28Q + 211 (in millions of shillings)

Further investigations have shown that the firm’s cost when not producing output is Sh.10 million.

Required:i) The equation of total cost. (5

marks)


Lesson Two 92

ii) The equation of total revenue (2 marks)

iii) An expression for profit (2 marks)

iv) The level of output that maximizes profit. (5 marks)

v) The equation of marginal revenue. (2 marks)

(Total: 20 marks)


93 Descriptive Statistics and Index Numbers

LESSON THREE

Descriptive Statistics and Index Numbers

Contents- Application of statistics- Measures of centra tendency

i. Arithmetic meanii. Medianiii. Modeiv. Geometric meanv. Harmonic mean

- Measures of dispersioni. Simple rangeii. Quartile deviationiii. Mean deviationiv. Standard deviationv. Coefficient of mean deviationvi. Coefficient of quartile deviation

- Skew ness and Kurtosis- Indices


Lesson Three 94

3.1 Descriptive Statisticsa) Statistics

Definition: Statistics viewed as a subject is a process of collecting, tabulating and analyzing numerical data upon which significant conclusions are drawn.Statistics may also be defined as numerical data, which has been, collected from a given source and for a particular purpose e.g. population statistics from the ministry of planning, Agricultural statistics from the ministry of AgricultureStatistics may also refer to the values, which have been obtained from statistical calculations e.g. the mean, mode, range e.t.c.

b) Application of statistics1. Quality ControlUsually there is a quality control departments in every industry which is charged with the responsibility of ensuring that the products made do meet the customers standards e.g. the Kenya bureau of standards (KeBS) is one of the national institutions which on behalf of the government inspects the various products to ensure that they do meet the customers specification.The KeBS together with other control department have developed quality control charts. They use these charts to check whether the products are up to standards or not.

2. Statistics may be used in making or ordering economic order quantities (EOQ). It is important for a business manager to realize that it is an economic cost if one orders a large quantity of items which have to be stored for too long before they are sold. This is because the large stock holds a lot of capital which could otherwise be used in buying other items for sale.It is also important to realize that the longer the items are stored in the stores the more will be the storage costsOn the other hand if one orders a few items for sale he will incur relatively low storage expenses but may not be able to satisfy all the clients. These may lose their customers if the goods are out of stock. Therefore it is advisable to work out the EOQ which will be sufficient for the clients in a certain period before delivery.The EOQ will also ensure that minimal costs are incurred in terms of storage3. ForecastingStatistics is very important for business managers when predicting the future of a business for example if a given business situation involves a dependent and independent variables one can develop an equation which can be used to predict the output under certain given conditions.4. Human resource management



Statistics may be used in efficient use of human resources for example we may give questionnaires to workers to find out where the management is weakBy compiling the statistics of those who were signing it may be found useful to analyze such data to establish the causes of resignation thus whether it is due to frustration or by choice.

3.2 Measures of Central TendencyThese are statistical values which tend to occur at the centre of any well ordered set of data. Whenever these measures occur they do not indicate the centre of that data. These measures are as follows:

i. The arithmetic meanii. The modeiii. The medianiv. The geometric meanv. The harmonic mean

1. The arithmetic meanThis is commonly known as average or mean it is obtained by first of all summing up the values given and by dividing the total value by the total no. of observations.

i.e. mean =

Where x = no. of values ∑ = summation n = no of observations

ExampleThe mean of 60, 80, 90, 120

The arithmetic mean is very useful because it represents the values of most observations in the population.The mean therefore describes the population quite well in terms of the magnitudes attained by most of the members of the population

Computation of the mean from grouped Data i.e. in classes.


Lesson Three 96

The following data was obtained from the manufacturers of electronic cells. A sample of electronic cells was taken and the life spans were recorded as shown in the following table.

Life span hrs No. of cells (f)

Class MP(x) X – A = d fd

1600 – 1799 25 1699.5 -600 -150001800 – 1999 32 1899.5 -400 -128002000-2199 46 2099.5 -200 -92002200 – 2399 58 2299.5(A) 0 0

2400 – 2599 40 2499.5 200 80002600 – 2799 30 2699.5 400 120002800 – 2999 7 2899.5 600 4200A = Assumed mean, this is an arbitrary number selected from the data, MP = mid point

= 2299.5 +-53.78

= 2245.72 hours

Example 2 – (use of the coded method)The following data was obtained from students who were registered in a certain college.The table shows the age distributionAge (yrs) No. of Students

(f)mid points (x)

x-a = d D/c = u fu

15 – 19 21 17 -15 -3 -6320 – 24 35 22 -10 -2 -7025 – 29 38 27 -5 -1 -3830 – 34 49 32(A) 0 0 035 – 39 31 37 +5 + 3140 – 44 19 42 +10 +2 38

193 -102

Required calculate the mean age of the students using the coded method



= 29.36 years

NB. The following statistical terms are commonly used in statistical calculations. They must therefore be clearly understood.

i) Class limitsThese are numerical values which limits uq extended of a given class i.e. all the observations in a given class are expected to fall within the interval which is bounded by the class limits e.g. 15 & 19 are class limits as in the table of the example above.

ii) Class boundariesThese are statistical boundaries, which separate one class from the other. They are usually determined by adding the lower class limit to the next upper class limit and dividing by 2 e.g. in the above table the class

boundary between 19 and 20 is 19.5 which is = .

iii) Class Mid pointsThis are very important values which mark the center of a given class. They are obtained by adding together the two limits of a given class and dividing the result by 2.

iv) Class interval/widthThis is the difference between an upper class boundary and lower class boundary. The value usually measures the length of a given class.

2. The mode

- This is one of the measures of central tendency. The mode is defined as a value within a frequency distribution which has the highest frequency. Sometimes a single value may not exist as such in which case we may refer to the class with the highest frequency. Such a class is known as a modal class

- The mode is a very important statistical value in business activities quite often business firms tend to stock specific items which are heavily on demand e.g. footwear, clothes, construction materials (beams, wires, iron sheets e.t.c.

- The mode can easily be determined form ungrouped data by arranging the figures given and determining the one with the highest frequency.

- When determining the values of the mode from the grouped data we may use the following methods;-

i. The graphical method which involves use of the histogram


Lesson Three 98

ii. The computation method which involves use of formula

ExampleIn a social survey in which the main purpose was to establish the intelligence quotient (IQ) of resident in a given area, the following results were obtained as tabulated below:

IQ No. of residents Upper class bound CF1 – 20 6 20 621 – 40 18 40 2441 – 60 32 fo 60 5661 – 80 48 f1 80 10481 – 100 27 f2 100 131101 – 120 13 120 144121 – 140 2 140 146

Required Calculate the modal value of the IQ’s tabulated above using

i. The graphical method andii. Formular



Graphical method

50

40

30

20

10

20 40 60 80 100 120 140Value of the mode

Computation method

Where L = Lower class boundary of the class containing the modef0 = Frequency of the class below the modal classf1 = Frequency of the class containing the modef2 = frequency of the class above the modal classc = Class interval

= 69.143. The median- This is a statistical value which is normally located at the center of a

given set of data which has been organized in the order of magnitude or size e.g. consider the set 14, 17, 9, 8, 20, 32, 18, 14.5, 13. When the data is ordered it will be 8, 9, 13, 14, 14.5, 17, 18, 20, 32The middle number/median is 14.5


Lesson Three 100

- The importance of the median lies in the fact that it divides the data into 2 equal halves. The no. of observations below and above the median are equal.

- In order to determine the value of the median from grouped data. When data is grouped the median may be determined by using the following methods

i. Graphical method using the cumulative frequency curve (ogive)

ii. The formula

Example Referring to the table in 105, determine the median using the methods aboveThe graphical method

IQ No of resid UCB Cumulative Frequency0 – 20 6 20 620 – 40 18 40 2440 – 60 32 60 5660 – 80 48 80 10480 – 100 27 100 131100 – 120 13 120 144120 – 140 2 140 146

146

xv

160

140

120

100

80

60

40

20

20 40 60 80 100 120 140 160

Value of the median



ii ComputationThe formula used is

Where L = Lower class boundary of the class containing the medianN = No of observationsCfo= cfbm = Cumulative frequency of the class before that

containing the median

F1 fmc = Frequency of the class containing the median

= 60 + 7.29 = 67.29

4. Geometric mean- This is a measure of central tendency normally used to measure

industrial growth rates. - It is defined as the nth root of the product of ‘n’ observations or

values -

Example In 1995 five firms registered the following economic growth rates; 26%. 32% 41% 18% and 36%.Required Calculate the GM for the above values


Lesson Three 102

No. Log26 1.415032 1.505241 1.612818 1.253336 1.5563

7.3446Therefore Log of GM = 1/5 x 7.3446 = 1.46892So GM = Antilog of 1.46892

= 29.43

5. Harmonic meanThis is a measure of central tendency which is used to determine the average growth rates for natural economies. It is defined as the reciprocal of the average of the reciprocals of all the values given by HM.

Example The economic growth rates of five countries were given as 20%, 15%, 25%, 18% and 5% Calculate the harmonic mean

6. Weighted mean- This is the mean which uses arbitrarily given weights- It is a useful measure especially where assessment is being done yet

the conditions prevailing are not the same. This is particularly true when assessment of students is being done given that the subjects being taken have different levels of difficulties.

Examples



The following table shows that marks scored by a student doing section 3 and 4 of CPA

Subject Scores (x) Weight (w) wxSTAD 65 50 3250BF 63 40 2520FA2 62 45 2340LAW 80 35 2800QT 69 55 3795FA3 55 60 3300

w = 285 wx = 18005

Weighted mean

Merits and demerits of the measures of central tendencyThe arithmetic mean (a.m)Merits

i. It utilizes all the observations givenii. It is a very useful statistic in terms of applications. It has

several applications in business management e.g. hypothesis testing, quality control e.t.c.

iii. It is the best representative of a given set of data if such data was obtained from a normal population

iv. The a.m. can be determined accurately using mathematical formulas

Demerits of the a.m.i. If the data is not drawn from a ‘normal’ population, then the

a.m. may give a wrong impression about the populationii. In some situations, the a.m. may give unrealistic values

especially when dealing with discrete variables e.g. when working out the average no. of children in a no. of families. It may be found that the average is 4.4 which is unrealistic in human beings

The modeMerits


Lesson Three 104

i. It can be determined from incomplete data provided the observations with the highest frequency are already known

ii. The mode has several applications in businessiii. The mode can be easily definediv. It can be determined easily from a graph

Demeritsi. If the data is quite large and ungrouped, determination of the

mode can be quite cumbersomeii. Use of the formula to calculate the mode is unfamiliar to most

business peopleiii. The mode may sometimes be non existent or there may be two

modes for a given set of data. In such a case therefore a single mode may not exist

The medianMerits

i. It shows the centre of a given set of dataii. Knowledge of the determination of the median may be

extended to determine the quartilesiii. The median can easily be definediv. It can be obtained easily from the cumulative frequency curvev. It can be used in determining the degrees of skew ness (see

later)

Demeritsi. In some situations where the no. of observations is even, the

value of the median obtained is usually imaginaryii. The computation of the median using the formulas is not well

understood by most businessmeniii. In business environment the median has got very few

applications

The geometric meanMerits

i. It makes use of all the values given (except when x = 0 or negative)

ii. It is the best measure for industrial growth rates

Demeritsi. The determination of the GM by using logarithms is not familiar

process to all those expected to use it e.g managersii. If the data contains zeros or –ve values, the GM ceases to exist

The harmonic mean and weighted mean



Merits – same as the arithmetic meanDemerits – same as the arithmetic mean

3.3 Measures of Dispersion- The measures of dispersion are very useful in statistical work

because they indicate whether the rest of the data are scattered around the mean or away from the mean.

- If the data is approximately dispersed around the mean then the measure of dispersion obtained will be small therefore indicating that the mean is a good representative of the sample data. But on the other hand, if the figures are not closely located to the mean then the measures of dispersion obtained will be relatively big indicating that the mean does not represent the data sufficiently

- The commonly used measures of dispersion area) The rangeb) The absolute mean deviationc) The standard deviationd) The semi – interquartile and quartile deviatione) The 10th and 90th percentile rangef) Variance

a) The range- The range is defined as the difference between the highest and the

smallest values in a frequency distribution. This measure is not very efficient because it utilizes only 2 values in a given frequency distribution. However the smaller the value of the range, the less dispersed the observations are from the arithmetic mean and vice versa

- The range is not commonly used in business management because 2 sets of data may yield the same range but end up having different interpretations regarding the degree of dispersion

b) The absolute mean deviation- This is a useful measure of dispersion because it makes use of all

the values given see the following examples

Example 1In a given exam the scores for 10 students were as follows

Student Mark (x)

A 60 1.8B 45 16.8C 75 13.2D 70 8.2E 65 3.2F 40 21.8G 69 7.2H 64 2.2


Lesson Three 106

I 50 11.8J 80 18.2

Total 618 104.4

RequiredDetermine the absolute mean deviation

Mean, = = 61.8

Therefore AMD =

Example 2The following data was obtained from a given financial institution. The data refers to the loans given out in 1996 to several firms

Firms (f)

Amount of loan per firm

(x)

fx

3 20000 60000 4157.9 12473.70

4 60000 240000 35842.1 143368.40

1 15000 15000 9157.9 9157.95 12000 60000 12157.9 60789.5

06 14000 84000 10157.9 60947.4

0Σf = 19 Σfx =

459000286736.

90

Required Calculate the mean deviation for the amount of items given



NB if the absolute mean deviation is relatively small it implies that the data is more compact and therefore the arithmetic mean is a fair sample representative.

c) The standard deviation- This is one of the most accurate measures of dispersion. It has the

following advantages;i. It utilizes all the values givenii. It makes use of both negative and positive values if they occuriii. The standard deviation reflects an accurate impression of how

much the sample data varies from the mean. This is because its suitability can also be tested using other statistical methods

ExampleA sample comprises of the following observations; 14, 18, 17, 16, 25, 31Determine the standard deviation of this sampleObservation.

x

14 -6.1 37.2118 -2.1 4.4117 -3.1 9.6116 -4.1 16.8125 4.9 24.0131 10.9 118.81

Total 121 210.56

standard deviation,

= 5.93

Alternative methodx X2

14 19618 32417 28916 25625 625


Lesson Three 108

31 961Total 121 2651

= 5.93

Example 2The following table shows the part-time rate per hour of a given no. of laborers in the month of June 1997.

Rate per hr (x) Shs

No. of labourers (f)

fx fx2

230 7 1610 370300400 6 2400 960000350 2 700 245000450 1 450 202500200 8 1600 320000150 11 1650 247500

Total 35 8410 2345300

Calculate the standard deviation from the above table showing how the hourly payment were varying from the respective mean

∴ standard deviation,

=

=

=

= 96.29

Example 3 – Grouped dataIn business statistical work we usually encounter a set of grouped data. In order to determine the standard deviation from such data, we use any of the three following methods

i. The long methodii. The shorter method



iii. The coded methodThe above methods are used in the following examples

Example 3.1 The quality controller in a given firm had an accurate record of all the iron bars produced in may 1997. The following data shows those records

i. Using long methodBar lengths (cm)

No. of bars(f)

Class mid point (x)

fx fx2

201 – 250 25 225.5 5637.5 1271256.25251 – 300 36 275.5 9918 2732409 301 – 350 49 325.5 15949.5 5191562.25351 – 400 80 375.5 30040 11280020401 – 450 51 425.5 21700.5 9233562.75451 – 500 42 475.5 19971 9496210.50501 - 550 30 525.5 15765 8284507.50

313 118981.50 47489526

Calculate the standard deviation of the lengths of the bars

∴ standard deviation, σ =

=

= 84.99 cm

ii. Using the shorter method

Bar lengths (cm)

No. of bars(f)

mid point (x)

x-A = d fd Fd2

201 – 250 25 225.5 -150 -3750 562500251 – 300 36 275.5 -100 -3600 360000301 – 350 49 325.5 -50 -2450 122500351 – 400 80 375.5 (A) 0 0 0401 – 450 51 425.5 50 2550 127500451 – 500 42 475.5 100 4200 420000501 - 550 30 525.5 150 4500 675000Total 313 1450 2267500

Calculate the standard deviation using the shorter method quagmire


Lesson Three 110

∴ Standard deviation, σ =

=

=

=

= 84.99 cm

iii. Using coded method

Bar lengths (cm)

(f) mid point (x) x-A = d d/c = u fu fu2

201 – 250 25 225.5 -150 -3 -75 225251 – 300 36 275.5 -100 -2 -72 144 301 – 350 49 325.5 -50 -1 -49 49351 – 400 80 375.5 (A) 0 0 0 0401 – 450 51 425.5 50 1 51 51451 – 500 42 475.5 100 2 84 168501 - 550 30 525.5 150 3 90 270

313 29 907

C = 50 where c is an arbitrary number, try picking a different figure say 45 the answer should be the same.

Standard deviation using the coded method. This is the most preferable method among the three methods

= 50 × 1.6997

= 84.99



Variance Square of the standard deviation is called variance.

d) The semi interquartile range- This is a measure of dispersion which involves the use of

quartile. A quartile is a mark or a value which lies at the boundary of a division when any given set of data is divided into four equal divisions

- Each of such divisions normally carries 25% of all the observations

- The semi interquartile range is a good measure of dispersion because it shows how the rest of the data are generally spread around the mean

- The quartiles normally used are three namely;i. The lower quartile (first quartile Q1) this usually binds the

lower 25% of the dataii. The median (second quartile Q2)iii. The upper quartile (third quartile Q3)

The semi-interquartile range,

Example 1The weights of 15 parcels recorded at the GPO were as follows:16.2, 17, 20, 25(Q1) 29, 32.2, 35.8, 36.8(Q2) 40, 41, 42, 44(Q3) 49, 52, 55 (in kgs)Required

Determine the semi interquartile range for the above data

Example 2 (Grouped Data)The following table shows the levels of retirement benefits given to a group of workers in a given establishment.

Retirement benefits £ ‘000

No of retirees (f)

UCB cf

20 – 29 50 29.5 5030 – 39 69 39.5 11940 – 49 70 49.5 18950 – 59 90 59.5 27960 – 69 52 69.5 331


Lesson Three 112

70 – 79 40 79.5 37180 – 89 11 89.5 382

Required i. Determine the semi interquartile range for the above dataii. Determine the minimum value for the top ten per cent.(10%)iii. Determine the maximum value for the lower 40% of the retirees

Solution The lower quartile (Q1) lies on position

= 29.5 + 6.63= £36.13

The upper quartile (Q3) lies on position

= 287.25

∴ the value of Q3 = 59.5 + × 10

= 61.08

The semi interquartile range =



= 12.475= £12,475

ii. The top 10% is equivalent to the lower 90% of the retireesThe position corresponding to the lower 90%

= 0.9 x 383

= 344.7

∴ the benefits (value) corresponding to the minimum value for top 10%

= 69.5 + x 10

= 72.925= £ 72925

iii. The lower 40% corresponds to position

= (382 + 1)

= 153.20

∴ retirement benefits corresponding to its position

= 39.5 + x 10

= 39.5 + 4.88 = 44.38= £ 44380

e. The 10th – 90th percentile rangeThis is a measure of dispersion which uses percentile. A percentile is a value which separates one division from the other when a given data is divided into 100 equal divisions.This measure of dispersion is very important when calculating the co-efficient of skewness (see later)

Example


Lesson Three 114

Using the above data for retirees calculate the 10th - 90th percentile. The tenth percentile 10th percentile lies on position

(382 + 1) = 0.1 x 383

= 38.3∴ the value corresponding to the tenth percentile

= 19.5 + 7.66 = 27.16

The 90th percentile lies on position

= 344.7∴ the value corresponding to the 90th percentile

= 69.5 + x 10

= 69.5 + 3.425= 72.925

∴ the required value of the 10th – 90th percentile = 72.925 – 27.16 = 45.765

Relative measures of dispersionDefinition:A relative measure of dispersion is a statistical value which may be used to compare variations in 2 or more samples.The measures of dispersion are usually expressed as decimals or percentages and usually they do not have any other units

ExampleThe average distance covered by vehicles in a motor rally may be given as 2000 km with a standard deviation of 5 km.In another competition set of vehicles covered 3000 km with a standard deviation of 10 kmsNB: The 2 standard deviations given above are referred to as absolute measures of dispersion. These are actual deviations of the measurements from their respective meanHowever, these are not very useful when comparing dispersions among samples.



Therefore the following measures of dispersion are usually employed in order to assess the degree of dispersion.

i. Coefficient of mean deviation

ii. Coefficient of quartile deviation

Where Q1 = first quartile Q3 = third quartile

iii. Coefficient of standard deviation

iv. Coefficient of variation

=

Example (see information above)First group of cars: mean = 2000 kms

Standard deviation = 5 kms∴ C.O.V = 5 x 100

2000

= 0.25%

Second group of cars: mean = 3000 kms Standard deviation = 10kms

∴ C.O.V = 1 0 x 100 3000

= 0.33%

ConclusionSince the coefficient of variation is greater in the 2nd group, than in the first group we may conclude that the distances covered in the 1st group are much closer to the mean that in the 2nd group.


Lesson Three 116

Example 2In a given farm located in the UK the average salary of the employees is £ 3500 with a standard deviation of £150The same firm has a local branch in Kenya in which the average salaries are Kshs 8500 with a standard deviation of Kshs.800Determine the coefficient of variation in the 2 firms and briefly comment on the degree of dispersion of the salaries in the 2 firms.First firm in the UK

C.O.V = 150 x 100 3500

= 4.29%

Second firm in Kenya C.O.V = 800 x 100

8500

= 9.4%Conclusively, since 4.29% < 9.4% then the salaries offered by the firm in UK are much closer to the mean given them in the case to the local branch in Kenya

COMBINED MEAN AND STANDARD DEVIATIONSometimes we may need to combine 2 or more samples say A and B. It is therefore essential to know the new mean and the new standard deviation of the combination of the samples.

Combined meanLet m be the combined meanLet x1 be the mean of first sampleLet x2 be the mean of the second sampleLet n1 be the size of the 1st sampleLet n2 be the size of the 2nd sampleLet s1 be the standard deviation of the 1st sampleLet s2 be the standard deviation of the 2nd sample

ExampleA sample of 40 electric batteries gives a mean life span of 600 hrs with a standard deviation of 20 hours.



Another sample of 50 electric batteries gives a mean lifespan of 520 hours with a standard deviation of 30 hours.If these two samples were combined and used in a given project simultaneously, determine the combined new mean for the larger sample and hence determine the combined or pulled standard deviation.

Size x s40(n1) 600 hrs(x1) 20hrs (s1)50 (n1) 520 hrs (x2) 30 hrs (s2)

Combined standard deviation

SKEWNESS- This is a concept which is commonly used in statistical decision

making. It refers to the degree in which a given frequency curve is deviating away from the normal distribution- There are 2 types of skew ness namely

i. Positive skew nessii. Negative skew ness

1. Positive Skewness- This is the tendency of a given frequency curve leaning towards

the left. In a positively skewed distribution, the long tail extended to the right.

In this distribution one should note the following i. The mean is usually bigger than the mode and medianii. The median always occurs between the mode and meaniii. There are more observations below the mean than above the

meanThis frequency distribution as represented in the skewed distribution curve is characteristic of the age distributions in the developing countries


Lesson Three 118

2. Negative Skewness This is an asymmetrical curve in which the long tail extends to the left

NB: This frequency curve for the age distribution is characteristic of the age distribution in developed countries

- The mode is usually bigger than the mean and median- The median usually occurs in between the mean and mode- The no. of observations above the mean are usually more than

those below the mean (see the shaded region)

MEASURES OF SKEWNESS- These are numerical values which assist in evaluating the

degree of deviation of a frequency distribution from the normal distribution.

- Following are the commonly used measures of skew ness.1. Coefficient Skewness

=

2. Coefficient of skewness

= NB: These 2 coefficients above are also known as Pearsonian measures of skewness.

3. Quartile Coefficient of skewness

=

Where Q1 = 1st quartile Q2 = 2nd quartile Q3 = 3rd quartile

NB: The Pearsonian coefficients of skewness usually range between –ve 3 and +ve 3. These are extreme value i.e. +ve 3 and –ve 3 which therefore indicate that a given frequency is negatively skewed and the amount of skewness is quite high.


frequency

Mod

e

Med

ian

Mea

n

Mea

n

Med

ian

Mod

e

Normal distribution

frequency Positively skewed frequency curve Negatively skewed

frequency curve

Long tail


Similarly if the coefficient of skewness is +ve it can be concluded that the amount of skew ness of deviation from the normal distribution is quite high and also the degree of frequency distribution is positively skewed.

ExampleThe following information was obtained from an NGO which was giving small loans to some small scale business enterprises in 1996. the loans are in the form of thousands of Kshs.

Loans Units (f)

Midpoints(x)

x-a=d d/c= u

fu Fu2 UCB cf

46 – 50

32 48 -15 -3 -96 288 50.5 32

51 – 55

62 53 -10 -2 -124 248 55.5 94

56 – 60

97 58 -5 -1 -97 97 60.5 191

61 –65 120 63 (A) 0 0 0 0 0 066 –70 92 68 5 +1 92 92 70.5 40371 –75 83 73 10 +2 166 332 75.5 48676 – 80

52 78 15 +3 156 468 80.5 538

81 – 85

40 83 20 +4 160 640 85.5 57.8

86 – 90

21 88 25 +5 105 525 90.5 599

91 – 95

11 93 30 +6 66 396 95.5 610

Total 610 428 3086

Required Using the Pearsonian measure of skew ness, calculate the coefficients of skew ness and hence comment briefly on the nature of the distribution of the loans.

Arithmetic mean = Assumed mean +

= 63 +

= 66.51


It is very important to note that the method of obtaining arithmetic mean (or any other statistic) by minusing assumed mean (A) from X and then deviding by c can be abit confusing, if this is the case then just use the straight forward method of:

119

Lesson Three 120

The standard deviation = c ×

=5 ×

= 10.68

The Position of the median lies m =

= = 305.5

= 60.5 + × 5

= 60.5 + × 5

Median = 65.27Therefore the Pearsonian coefficient

=

= 0.348CommentThe coefficient of skewness obtained suggests that the frequency distribution of the loans given was positively skewed This is because the coefficient itself is positive. But the skewness is not very high implying the degree of deviation of the frequency distribution from the normal distribution is small

Example 2Using the above data calculate the quartile coefficient of skewness

Quartile coefficient of skewness =

The position of Q1 lies on =

∴ actual value Q1 =55. 5 +



The position of Q3 lies on =

∴ actual value Q3 =70.55 + × 5

Q2 position: i.e. 2 = 305.5

Actual Q2 value

The required coefficient of skew ness

=

ConclusionSame as above when the Pearsonian coefficient was used

KURTOSIS- This is a concept, which refers to the degree of peaked ness of

a given frequency distribution. The degree is normally measured with reference to normal distribution.

- The concept of kurtosis is very useful in decision making processes i.e. if is a frequency distribution happens to have either a higher peak or a lower peak, then it should not be used to make statistical inferences.

- Generally there are 3 types of kurtosis namely;-i. Leptokurticii. Mesokurtic iii. Platykurtic

Leptokurtica) A frequency distribution which is lepkurtic has generally a

higher peak than that of the normal distribution. The coefficient of kurtosis when determined will be found to be more than 3. thus frequency distributions with a value of more than 3 are definitely leptokurtic

b) Some frequency distributions when plotted may produce a curve similar to that of the normal distribution. Such frequency distributions are referred to as mesokurtic. The degree of kurtosis is usually equal to 3

c) When the frequency curve contacted produces a peak which is lower that that of a normal distribution when such a curve is said to be platykurtic. The coefficient of such is usually less than 3

- It is necessary to calculate the numerical measure of kurtosis. The commonly used measure of kurtosis is the percentile coefficient of kurtosis. This coefficient is normally determined using the following equation


Lesson Three 122

Percentile measure of kurtosis, K (Kappa) =

ExampleRefer to the table above for loans to small business firms/unitsRequiredCalculate the percentile coefficient of Kurtosis

P90 =

= 0.9 (611) = 549.9

The actual loan for a firm in this position

(549.9) = 80.5 + x 5 = 81.99

P10 = (n + 1) = 0.1 (611) = 61.1

The actual loan value given to the firm on this position is

50.5 + x 5 = 52.85

= 0.9 (611) = 549.9

∴ percentile measure of kurtosis

K(Kappa) = ½

= ½

= 0.26Since 0.26 < 3, it can be concluded that the frequency distribution exhibited by the distribution of loans is platykurticKurtosis is also measured by moment statistics, which utilize the exact value of each observation.

i. M1 the first moment = M1 = = Mean M1 or M1

M2 =

M3 =



M4 =

3. M2 second moment about the mean M2 or f2 M2 = M2 – M12

4. M3 third moment about the mean M3 (a measure of the absolute skew ness)

M3 = M3 – 3M2M1 + 2M13

5. M4 fourth moment about the mean M4 (a measure of the absolute Kurtosis)

M4 = M4– 4M3M1 + 6M2M12 + 3M14An alternative formula

M4 = Where m is mean

Moment coefficient of Kurtosis

ExampleFind the moment coefficient of the following distribution

x f12 114 416 618 1020 722 2

X f xf (x-m) (x-m)2 (x-m)2f (x-m)4f12 1 12 -5.6 31.36 31.36 983.4514 4 56 -3.6 12.96 51.84 671.8516 6 96 -1.6 2.56 15.36 39.3218 10 180 .4 0.16 1.60 0.25620 7 140 2.4 5.76 40.32 232.2422 2 44 4.4 19.36 38.72 749.62

30 528 179.20 2,676.74

M = = 17.6

σ2 = = 5.973

σ4 = 35.677


Lesson Three 124

M4 = = = 89.22

Moment coefficient of Kurtosis = = 2.5

Note Coefficient of kurtosis can also be found using the method of assumed mean.

3.4 IndicesAn index number is an attempt to summarize a whole mass of data into one figure. The single figure shows how one year differs from another year.It is a statistical devise used to measure the change in the level of prices, wages output and other variables at given times, relative to their level at an earlier time which is taken as the base for comparison purposes

A simple price index = × 100 (an unweighted price index)

A simple quantity index = × 100 (an unweighted quantity index)

Where pn is the price of a commodity in the current year (the year for which the price index to be calculated)

Where po is the price of the same commodity in the base year (the year for comparison purposes)

Similarly Qn and Qo are defined in the same way

AGGREGATE PRICE INDEX NUMBERS AND QUANTITY INDEX NUMBERS

PRICE INDEX QUANTITY INDEXLASPEYRE’S INDEX

× 100 × 100

PAASCHE’S INDEX × 100 × 100

Value index = × 100

MODIFIED FORM OF THE LASPEYRE’S PRICE INDEX NUMBER



Laspeyre’s Price index

Where w0 are the proportions of the total expected in the basic period. This formula is frequently used to calculate retail price index.

CHANGING THE BASE OF THE INDEXFor comparison purposes if two series have different base years, it is difficult to compare them directly. In such cases, it is necessary to change the base year of one of the series (or both) so that both have the same base.It is also necessary to keep the index relevant to current conditions hence the need to change the base from time to time.

Example;Year 1985 1986 1987 1988 1989 1990 1991 1992Price index 100 104 108 109 112 120 125 140

Suppose we wish to change the base year to 1989We recalculate each index by expressing it as a percentage of 1989

Previous index Recalculated index1985 100

× 100 = 89.3

1986 104 × 100 = 92.9

1987 108 × 100 = 96.4

1988 109 × 100 = 97.3

1989 (new base year) 112 × 100 = 100

1990 120 × 100 = 107.1

1991 125 × 100 = 111.6

1992 140 × 100 = 125.0

When changing the base year, it is advisable to update the weights used in the base year.

CHAIN BASED INDEX NUMBERS


Lesson Three 126

A chain based index is one where the index is calculated every year using the previous year as the base year. This type of index measures rate of change from year to year.This method is suitable where weights are changing rapidly and items are constantly being brought into the index and unwanted items taken out. It can be a price or quantity index

Previous index Recalculated chainbased index

fixedbased index

1985 100 100 100(1985 base year1986 104

× 100 = 104 × 100 = 104

1987 108 × 100 = 103.8 × 100 = 108

1988 109 × 100 = 100.9 × 100 = 109

1989 112 × 100 = 102.8 × 100 = 112

1990 120 × 100 = 107.1 × 100 = 120

1991 125 × 100 = 104.2 × 100 = 125

1992 140× 100 = 112 × 100 = 140

The Fisher’s indexThe Fisher’s index acts as a compromise between Laspeyre’s index and Paasche’ index. It is calculated as a geometric mean of the two indexes.

Retail price indexIt is weighted average of price relatives based upon an average household in the base year. The items consumed are divided into groups such as food, housing, transport, alcoholic drinks, footwear, fuel, light, water, household goods, services e.t.c. each item included in the index is given a weighting and a price relative to the base is calculated. Modified form of laspeyre’s price index formula is used as a weighted arithmetic mean of price relatives.

I.e. Retail Price index

The index is used by the Government as a guide in determining the minimum wages, pension rates unemployed benefits (in UK e.t.c). Trade unions use it as a basis for their wages claims.

Deflation



Indexes may be used to deflate time series so that comparisons between periods may be made in real termsIt is a process of reducing a value measured in current period prices to its equivalent in the base period prices. The deflated value is what would have been necessary to purchase the same amount of goods as the present value can purchase in the current period

Deflation Factor = × 100

Deflation of a time seriesYear Average monthly

earnings (shs)Retail index

Real earnings

1 5,000 100 5000 = 50002 5,500 120

5,500 × =

4,583.33 6,000 140

6,000 × =

4,285.74 6,500 170

6,500 × =

3,823.55 7,200 200

7,200 × =

3,600.0

The technique of index number constructionWhen preparing index numbers it is important to define

a) The exact purpose of the indexb) How the items are to be selectedc) The choice of the weightsd) The choice of the basee) The type of average to be used

The base year should be as close to the normal trend as possible. The best methods should be used for collection of data. The items should be selected in such a way that they are a fair representation of all the relevant items.Due consideration should be given to the weighting of all items selected

The index of industrial productionIt is a quantity index compiled by the government. It measures changes in the volume of production in major industries. The index is a good indication of the state of national economy.It covers the following major industries in the UK

i. Mining and quarrying ii. Manufacturing such as food, drinks and tobacco, chemicals,

metal manufacture, engineering e.t.c


Lesson Three 128

iii. Textileiv. Constructionv. Gas electricity, water e.t.c

It excludes agriculture, fishing, trade, transport, finance and other such industries.Each industries order is given a weighting. The weighting is based on average monthly production in each industry in a fixed base year. It gives each item its relative importance amongst all other items and thus gives a better estimate of the index for comparison purposes.

The Geometric Index (Industrial Share index)This index is an index of 30 selected top industrial companies. It is calculated by taking an unweighted geometric mean of the price relatives of the selected shares.

ExampleThe share prices of ordinary shares of four companies on 1st January 1990 and 1st January 1991 were as follows.

Share Price on 1.1.1990

Price on 1.1.1991

Company A

Shs 10 Shs 12

Company B

Shs 12 Shs 15

Company C

Shs 20 Shs 25

Company D

Shs 5 Shs 6

Using an unweighted geometric index, calculate the index of share prices at 1.1.1991 if 1.1.1990 is the base date, index 100

Solution

percentage increase = 22.5% index = 122.5

InflationThe inflation rate for a given period can be calculated using the following formula;

Inflation = × 100

Marshal Hedge Worth Index



Marshal Hedge worth index = × 100

Tests For An Ideal Index Number1. Factor Reversal TestThis test indicates that when the price index is multiplied with a quantity index i.e. factors are reversed), it should result in the value index.2. The time reversal testIf we reverse the time subscripts of a price or quantity index, the result should be reciprocal of the original index.


Lesson Three 130

LESSON 3 REINFORCING QUESTIONS

QUESTION ONEa) Distinguish between discrete and continuous data.b) What is dispersion and what is the formula for the standard

deviation?c) What is the measure of relative dispersion?d) Draw diagrams showing positive and negative skewness

QUESTION TWOThe managers of an import agency are investigating the length of time that customers take to pay their invoices, the normal terms for which are 30 days net. They have checked the payment record of 100 customers chosen at random and have compiled the following table:

Payment in Number of customers

5 to 9 days 410 to 14 days 1015 to 19 days 1720 to 24 days 2025 to 29 days 2230 to 34 days 1635 to 39 days 840 to 44 days 3

Required:a) Calculate the arithmetic mean.b) Calculate the standard deviationc) Construct a histogram and insert the modal value.d) Estimate the probability that an unpaid invoice chosen at random will

be between 30 and 39 days old.

QUESTION THREEThe price of the ordinary 25p shares of Manco PLC quoted on the stock exchange, at the close of the business on successive Fridays is tabulated below

126 120 122 105 129 119 131 138125 127 113 112 130 122 134 136128 126 117 114 120 123 127 140124 127 114 111 116 131 128 137127 122 106 121 116 135 142 130

Requireda) Group the above date into eight classes. (4

marks)



b) Calculate cumulative frequency, the median value, quartile values and the semi-quartile range. (4

marks)c) Calculate the mean and standard deviation of your frequency

distribution. (7 marks)d) Compare and contrast the values that you have obtained for:

i) The median and meanii) The semi-interquartile range and the standard deviation

(5 marks)(Total: 20

marks)


Lesson Three 132

QUESTION FOURDefine the coefficient of variation.

The following table gives profits (in ten thousands of shillings) of two supermarkets over a duration of one year.

Month Supermarket A Supermarket BJanuary 65 28February 48 33March 15 20April 28 23May 41 69June 59 45July 41 53August 10 15September 24 35October 56 57November 92 99December 120 136

Required:i) Compute the coefficient of variation for each supermarket.

ii) Indicate for which supermarket the variability of profits is relatively greater.

QUESTION FIVEProdco PLC manufactures an item of domestic equipment which requires a number of components which have varied as various modifications of the model have been used. The following table shows the number of components required together with the price over the last three years of production.

COMPONENT

1981 1982 1983

Prices Quantity

Prices Quantity

Prices Quantity

A 3.63 3 4.00 2 4.49 2B 2.11 4 3.10 5 3.26 6C 10.03 1 10.36 1 12.05 1D 4.01 7 5.23 6 5.21 5

Required:a) Establish the base weighted price indices for 1982 and 1983 based on

1981 for the item of equipment. (8 marks)b) Establish the current weighted price indices for 1982 and 1983 based

on



1981 for the item of equipment. (8 marks)c) Using the results of (a) and (b) as illustrations, compare and contrast

Laspeyre’s and Paasche price index numbers.(4 marks)

(Total: 20 marks)


Lesson Three 134

QUESTION SIXa) A company manufacturing a product known as 257 uses five

components in its assembly.

The quantities and prices of the components used to produce a unit of K257 in 1982, 1983 and 1984 are tabulated as follows:

COMPONENT

1982 1983 1984

Quantity

Prices Quantity

Prices Quantity

Prices

A 10 3.12 12 3.17 14 3.20B 6 11.49 7 11.58 5 11.67C 5 1.40 8 1.35 9 1.31D 9 2.15 9 2.14 10 2.63E 50 0.32 53 0.32 57 0.32

Required:i) Calculate Laspyere’s type price index number for the cost of one unit

of K257 for 1983 and 1984 based on 1982. (6

marks)

ii) Calculate Paasche type price index numbers for the cost of one unit of K257for 1983 and 1984 based on 1982. (6

marks)

iii) Compare and contrast the Laspeyre and Paasche price-index numbers you have obtained in (i) and (ii) (3

marks)

A number of employers manufacturing plastic components used in plumbing have formed themselves into an association for the purpose of negotiating with the trade union for this industrial sector.

The negotiations cover pay and contributions in this sector.

Required:Explain the usefulness of an index of Industrial Production and an index of retail prices to both sides in a series of pay negotiations.

(5 marks)(Total: 20

marks)

QUESTION SEVENThe data given below indicates the prices and production of some horticulatural products in Central Territory:



Produce Production(1000 boxes)

Price per box (Shs)

1980 1990 1980 1990Cabbages 48,600 62,000 100 150Tomatoes 22,000 37,440 220 310Onions 47,040 61,430 180 200Spinach 43,110 55,720 130 170


Lesson Three 136

Required:Calculate the increase or decrease in prices from 1980 on the basis of the following indices:

a) Mean relativesb) Laspeyres indexc) Paasche indexd) Marshall – Hedgeworth indexe) Fishers index.



137 Measures of Relationships and Forecasting

LESSON FOUR

Measures of Relationships and Forecasting- Correlation- Regression analysis- Multiple Linear Regression- Time series analysis and forecasting


Lesson Four 138

4.1 Correlation and Regression

CorrelationThis is an important statistical concept which refers to interrelationship or association between variables.The purpose of studying correlation is for one to be able to establish a relationship, plan and control the inputs (independent variables) and the output (dependent variables)In business one may be interested to establish whether there exists a relationship between the

i. Amount of fertilizer applied on a given farm and the resulting harvest

ii. Amount of experience one has and the corresponding performance

iii. Amount of money spent on advertisement and the expected incomes after sale of the goods/service

There are two methods that measure the degree of correlation between two variables these are denoted by R and r.

(a) Coefficient of correlation denoted by r, this provides a measure of the strength of association between two variables one the dependent variable the other the independent variable r can range between +1 and – 1 for perfect positive correlation and perfect negative correlation respectively with zero indicating no relation i.e. for perfect positive correlation y increase linearly with x increament.

(b) Rank correlation coefficient denoted by R is used to measure association between two sets of ranked or ordered data. R can also vary from +1, perfect positive rank correlation and -1 perfect negative rank correlation where O or any number near zero representing no correlation.

SCATTER GRAPHS- A scatter graph is a graph which comprises of points which have

been plotted but are not joined by line segments- The pattern of the points will definitely reveal the types of

relationship existing between variables- The following sketch graphs will greatly assist in the

interpretation of scatter graphs.



Perfect positive correlation

y

Dependant variable x

x

x

x

x

x

x

x

Independent variableNB: For the above pattern, it is referred to as perfect because the points may easily be represented by a single line graph e.g. when measuring relationship between volumes of sales and profits in a company, the more the company sales the higher the profits.

Perfect negative correlation

y x

Quantity sold x

X

x

x

x

x

x

x

10 20 Price X


Lesson Four 140

This example considers volume of sale in relation to the price, the cheaper the goods the bigger the sale.

High positive correlation

y

Dependant variable xx xx

x x xx xx

xx xx

x xxx

xx

independent variable x

High negative correlation

y

quantity sold x x xx

x xx

x xx x

xx x

price



No correlation

y

600 x x x x x

x x x

400 x x x x x

x x x x

200 x x x x x

x x x x

0

10 20 30 40 50 xh) Spurious Correlations

- in some rare situations when plotting the data for x and y we may have a group showing either positive correlation or –ve correlation but when you analyze the data for x and y in normal life there may be no convincing evidence that there is such a relationship. This implies therefore that the relationship only exists in theory and hence it is referred to as spurious or non sense e.g. when high passrates of student show high relation with increased accidents.

Correlation coefficient- These are numerical measures of the correlations existing between

the dependent and the independent variables- These are better measures of correlation than scatter graphs

(diagrams)- The range for correlation coefficients lies between +ve 1 and –ve

1. A correlation coefficient of +1 implies that there is perfect positive correlation. A value of –ve shows that there is perfect negative correlation. A value of 0 implies no correlation at all

- The following chart will be found useful in interpreting correlation coefficients


Lesson Four 142

__ 1.0 } Perfect +ve correlation

} High positive correlation

__ 0.5}

} Low positive correlation

__0 }

} Low negative correlation

__-0.5}

} High negative correlation

__-1.0} Perfect –ve correlation

There are usually two types of correlation coefficients normally used namely;-

Product Moment Coefficient (r)It gives an indication of the strength of the linear relationship between two variables.

r =

note that this formula can be rearranged to have different outlooks but the resultant is always the same.

ExampleThe following data was observed and it is required to establish if there exists a relationship between the two.X 15 24 25 30 35 40 45 65 70 75Y 60 45 50 35 42 46 28 20 22 15



SolutionCompute the product moment coefficient of correlation (r)X Y X2 Y2 XY15 60 225 3,600 90024 45 576 2,025 1,08025 50 625 2,500 1,25030 35 900 1,225 1,05035 42 1,225 1,764 1,47040 46 1,600 2,116 1,84045 28 2,025 784 1,26065 20 4,225 400 1,30070 22 4,900 484 1,54075 15 5,625 225 1,125

r =

r =

=

The correlation coefficient thus indicates a strong negative linear association between the two variables.

Interpretation of r – Problems in interpreting r values

NOTE: A high value of r (+0.9 or – 0.9) only shows a strong association

between the two variables but doesn’t imply that there is a causal relationship i.e. change in one variable causes change in the other it is possible to find two variables which produce a high calculated r yet they don’t have a causal relationship. This is known as spurious or nonsense correlation e.g. high pass rates in QT in Kenya and increased inflation in Asian countries.

Also note that a low correlation coefficient doesn’t imply lack of relation between variables but lack of linear relationship between the variables i.e. there could exist a curvilinear relation.

A further problem in interpretation arises from the fact that the r value here measures the relationship between a single independent variable and dependent variable, where as a particular variable may be dependent on several independent variables (e.g. crop yield may be dependent on fertilizer used, soil exhaustion, soil acidity level,


Lesson Four 144

season of the year, type of seed etc.) in which case multiple correlation should be used instead.

The Rank Correlation Coefficient (R)Also known as the spearman rank correlation coefficient, its purpose is to establish whether there is any form of association between two variables where the variables arranged in a ranked form.

R = 1 -

Where d = difference between the pairs of ranked values.n = numbers of pairs of rankings

ExampleA group of 8 accountancy students are tested in Quantitative Techniques and Law II. Their rankings in the two tests were.Student Q. T. ranking Law II

rankingd d2

A 2 3 -1 1B 7 6 1 1C 6 4 2 4D 1 2 -1 1E 4 5 -1 1F 3 1 2 4G 5 8 -3 9H 8 7 1 1

d = Q. T. ranking – Law II ranking

R = 1 -

= 0.74Thus we conclude that there is a reasonable agreement between student’s performances in the two types of tests.

NOTE: in this example, if we are given the actual marks then we find r. R varies between +1 and -1.

Tied RankingsA slight adjustment to the formula is made if some students tie and have the same ranking the adjustment is

where t = number of tied rankings the adjusted formula

becomes



R = 1 -

ExampleAssume that in our previous example student E & F achieved equal marks in Q. T. and were given joint 3rd place.

SolutionStudent Q. T. ranking Law II

rankingd d2

A 2 3 -1 1B 7 6 1 1C 6 4 2 4D 1 2 -1 1E 3 ½ 5 -1 ½ 2 ¼F 3 ½ 1 2 ½ 6 ¼G 5 8 - 3 9H 8 7 1 1

R = 1- = 1 -

= 0.68NOTE: It is conventional to show the shared rankings as above, i.e.

E, & F take up the 3rd and 4th rank which are shared between the two as 3½ each.

ii. Coefficient of DeterminationThis refers to the ratio of the explained variation to the total variation and is used to measure the strength of the linear relationship. The stronger the linear relationship the closer the ratio will be to one.

Coefficient determination = Explained variationTotal variation

Example (Rank Correlation Coefficient)In a beauty competition 2 assessors were asked to rank the 10 contestants using the professional assessment skills. The results obtained were given as shown in the table below


Lesson Four 146

Contestants 1st assessor

2nd assessor

A 6 5B 1 3C 3 4D 7 6E 8 7F 2 1G 4 8H 5 2J 10 9K 9 10



REQUIREDCalculate the rank correlation coefficient and hence comment briefly on the value obtained

d d2A 6 5 1 1B 1 3 -2 4C 3 4 -1 1D 7 6 1 1E 8 7 1 1F 2 1 1 1G 4 8 -4 16H 5 2 3 9J 10 9 +1 1K 9 10 -1 1

Σd2 = 36∴ The rank correlation coefficient R

R = 1 -

= 1 -

= 1 - = 1 – 0.22

= 0.78Comment: since the correlation is 0.78 it implies that there is high positive correlation between the ranks awarded to the contestants. 0.78 > 0 and 0.78 > 0.5

ExampleContestant

1st

assessor

2nd

assessord d2

A 1 2 -1 1B 5 (5.5) 3 2.5 6.25C 3 4 -1 1D 2 1 1 1E 4 5 -1 1F 5 (5.5) 6.5 -1 1G 7 6.5 -0.5 0.25H 8 8 0 0

Σd2 = 11.25Required: Complete the rank correlation coefficient


Lesson Four 148

∴ R = 1 - = 1 -

= 1 –

= 1 – 0.13= 0.87

This implies high positive correlation

Example (Rank Correlation Coefficient)Sometimes numerical data which refers to the quantifiable variables may be given after which a rank correlation coefficient may be worked out.Is such a situation, the rank correlation coefficient will be determined after the given variables have been converted into ranks. See the following example;

Candidates

Math r Accounts r d d2

P 92 1 67 5 -4 16Q 82 3 88 1 2 4R 60 5(5.5

)58 7(7.5) -2 4

S 87 2 80 2 0 0T 72 4 69 4 0 0U 60 5(5.5

)77 3 -2.50 6.25

V 52 8 58 7(7.5) 0.5 0.25W 50 9 60 6 3 9X 47 10 32 10 0 0Y 59 7 54 9 -2 4

Σd2 = 43.5

∴ Rank correlation r = 1 -

= 1 - = 1 –

= 0.74 (High positive correlation between mathematics

marks and accounts)

Example(Product moment correlation)The following data was obtained during a social survey conducted in a given urban area regarding the annual income of given families and the corresponding expenditures.



Family (x)Annual income £

000

(y)Annual expenditure £

000

xy x2 Y2

A 420 360 151200 176400

129600

B 380 390 148200 144400

152100

C 520 510 265200 270400

260100

D 610 500 305000 372100

250000

E 400 360 144000 160000

129600

F 320 290 92800 102400

84100

G 280 250 70000 78400 62500H 410 380 155800 16810

014440

0J 380 240 91200 14440

057600

K 300 270 81000 90000 72900Total 4020 3550 1504400 17066

0013429

00RequiredCalculate the product moment correlation coefficient briefly comment on the value obtainedThe produce moment correlation

r =

Workings:

= = 402

r =

= 0.89

Comment: The value obtained 0.89 suggests that the correlation between annual income and annual expenditure is high and positive. This implies that the more one earns the more one spends.

4.2 REGRESSION- This is a concept, which refers to the changes which occur in the

dependent variable as a result of changes occurring on the independent variable.


Lesson Four 150

- Knowledge of regression is particularly very useful in business statistics where it is necessary to consider the corresponding changes on dependant variables whenever independent variables change

- It should be noted that most business activities involve a dependent variable and either one or more independent variable. Therefore knowledge of regression will enable a business statistician to predict or estimate the expenditure value of a dependant variable when given an independent variable e.g. consider the above example for annual incomes and annual expenditures. Using the regression techniques one can be able to determine the estimated expenditure of a given family if the annual income is known and vice versa

- The general equation used in simple regression analysis is as follows

y = a + bxWhere y = Dependant variable

a= Interception y axis (constant)b = Slope on the y axisx = Independent variablei. The determination of the regression equation such as given

above is normally done by using a technique known as “the method of least squares’.

Regression equation of y on x i.e. y = a + bx



y x x Line of best fit

x x

x x

x x

x x

x x

xThe following sets of equations normally known as normal equation are used to determine the equation of the above regression line when given a set of data.

Σy = an + bΣxΣxy = aΣx + bΣx2

Where Σy = Sum of y valuesΣxy = sum of the product of x and y Σx = sum of x valuesΣx2= sum of the squares of the x valuesa = The intercept on the y axisb = Slope gradient line of y on x

NB: The above regression line is normally used in one way only i.e. it is used to estimate the y values when the x values are given.Regression line of x on y i.e. x = a + by

- The fact that regression lines can only be used in one way leads to what is known as a regression paradox

- This means that the regression lines are not ordinary mathematical line graphs which may be used to estimate the x and y simultaneously

- Therefore one has to be careful when using regression lines as it becomes necessary to develop an equation for x and y before doing the estimation.

The following example will illustrate how regression lines are used

ExampleAn investment company advertised the sale of pieces of land at different prices. The following table shows the pieces of land their acreage and costs


Lesson Four 152

Piece of land

(x)Acreage Hectares

(y) Cost £ 000

xy x2

A 2.3 230 529 5.29B 1.7 150 255 2.89C 4.2 450 1890 17.64D 3.3 310 1023 10.89E 5.2 550 2860 27.04F 6.0 590 3540 36G 7.3 740 5402 53.29H 8.4 850 7140 70.56J 5.6 530 2969 31.36

Σx =44.0 Σy = 4400 Σxy= 25607 Σx2 = 254.96

RequiredDetermine the regression equations of

i. y on x and hence estimate the cost of a piece of land with 4.5 hectares

ii. Estimate the expected average if the piece of land costs £ 900,000

Σy = an + bΣxyΣxy = a∑x + bΣx2

By substituting of the appropriate values in the above equations we have 4400 = 9a + 44b …….. (i)25607 = 44a + 254.96b ……..(ii)

By multiplying equation …. (i) by 44 and equation …… (ii) by 9 we have 193600 = 396a + 1936b …….. (iii)230463 = 396a + 2294.64b ……..(iv)

By subtraction of equation …. (iii) from equation …… (iv) we have 36863 = 358.64b102.78 = b

by substituting for b in …….. (i)4400 = 9a + 44( 102.78)4400 – 4522.32 = 9a –122.32 = 9a-13.59 = a

Therefore the equation of the regression line of y on x isY = 13.59 + 102.78x

When the acreage (hectares) is 4.5 then the cost (y) = -13.59 + (102.78 x 4.5)= 448.92= £ 448, 920

Note that Where the regression equation is given by

y= a + bxWhere a is the intercept on the y axis and

b is the slope of the line or regression coefficient n is the sample size



then,

intercept a =

Slope b =

ExampleThe calculations for our sample size n = 10 are given below. The linear regression model is y = a + bxTable

Distance x miles

Time y mins xy x2 y2

3.5 16 56.0 12.25 2562.4 13 31.0 5.76 1694.9 19 93.1 24.01 3614.2 18 75.6 17.64 3243.0 12 36.0 9.0 1441.3 11 14.3 1.69 1211.0 8 8.0 1.0 643.0 14 42.0 9.0 1961.5 9 13.5 2.25 814.1 16 65.6 16.81 256Σx = 28.9 Σy = 136 Σxy =

435.3Σx2 =

99.41Σy2= 1972

The Slope b =

= 2.66

and the intercept a =

= 5.91We now insert these values in the linear model giving

y = 5.91 + 2.66xor Delivery time (mins) = 5.91 + 2.66 (delivery distance in miles)The slope of the regression line is the estimated number of minutes per mile needed for a delivery. The intercept is the estimated time to prepare for the journey and to deliver the goods, that is the time needed for each journey other than the actual traveling time.


Lesson Four 154

PREDICTION WITHIN THE RANGE OF SAMPLE DATAWe can use the linear regression model to predict the mean of dependant variable for any given value of independent variableFor example if the sample model is given by

Time (min) = 5.91 + 2.66 (distance in miles)Then if the distance is 4.0 miles then our estimated mean time is

Ý = 5.91 + 2.66 x 4.0 = 16.6 minutes

4.3 Multiple Linear Regression ModelsThere are situations in which there is more than one factor which influence the dependent variableExampleCost of production per week in a large department depends on several factors;

i. Total numbers of hours workedii. Raw material used during the weekiii. Total number of items produced during the weekiv. Number of hours spent on repair and maintenance

It is sensible to use all the identified factors to predict department costsScatter diagram will not give the relationship between the various factors and total costsThe linear model for multiple linear regression if of the type; (which is the line of best fit).

y = α + b1x1 +b2x2 +………… + bnxn We assume that errors or residuals are negligible.In order to choose between the models we examine the values of the multiple correlation coefficient r and the standard deviation of the residuals α.A model which describes well the relationship between y and x’s has multiple correlation coefficient r close to ±1 and the value of α which is small.

ExampleOdino chemicals limited are aware that its power costs are semi variable cost and over the last six months these costs have shown the following relationship with a standard measure of output.

Month Output (standard units)

Total power costs £ 000

1 12 6.22 18 8.03 19 8.64 20 10.45 24 10.26 30 12.4

Requiredi. Using the method of least squares, determine an appropriate

linear relationship between total power costs and output



ii. If total power costs are related to both output and time (as measured by the number of the month) the following least squares regression equation is obtainedPower costs = 4.42 + (0.82) output + (0.10) monthWhere the regression coefficients (i.e. 0.82 and 0.10) have t values 2.64 and 0.60 respectively and coefficient of multiple correlation amounts to 0.976Compare the relative merits of this fitted relationship with one you determine in (a). Explain (without doing any further analysis) how you might use the data to forecast total power costs in seven months.

Solutiona) Output (x) Power costs

(y)x2 y2 xy

12 6.2 144 38.44 74.4018 8.0 324 64.00 144.0019 8.6 361 73.96 163.4020 10.4 400 108.16 208.0024 10.2 576 104.04 244.8030 12.4 900 153.76 372.00Σx = 123 Σy = 55.8 Σx2 = 2705 Σy2 =

542.36Σxy=

1,206.60

b =

=

= = 0.342

a = (Σy – bΣx)

= (55.8 – 0.342) 123

= 2.29 (Power costs) = 2.29 + 0.342 (output)

b. For linear regression calculated above, the coefficient of correlation r is


Lesson Four 156

r =

=

= 0.96

This show a strong correlation between power cost and output. The multiple correlation when both output and time are considered at the same time is 0.976.We observe that there has been very little increase in r which means that inclusion of time variable does not improve the correlation significantlyThe value for time variable is only 0.60 which is insignificant as compared with a t value of 2.64 for the output variableIn fact, if we work out correlation between output and time, there will be a high correlation. Hence there is no necessity of taking both the variables. Inclusion of time does improve the correlation coefficient but by a very small amount.If we use the linear regression analysis and attempt to find the linear relationship between output and time i.e.

Month Output 1 122 183 194 205 246 30

The value of b and a will turn out to be 3.11 and 9.6 i.e. relationship will be of the form

Output = 9.6 + 3.11 × monthFor this equation forecast for 7th month will be

Output = 9.6 + 3.11 × 7= 9.6 + 21.77 = 31.37 units

Using the equation , Power costs = 2.29 + 0.34 × output= 2.29 + 0.34 × 31.37= 2.29 + 10.67 = 12.96 i.e. £ 12,960

Non Linear RelationshipsIf the scatter diagram and the correlation coefficient do not indicate linear relationship, then the relationship may be non – linearTwo such relationships are of peculiar interest



Both of these can be reduced to linear model. Simple or multiple linear regression methods are then used to determine the values of the coefficients

i. Exponential model

Take log of both sideslog y = log a + log bx

log y = log a + xlog bLet log y = Y and log a = A and log b = B

Thus we get Y = A + Bx. This is a linear regression model

ii. Geometric model

using the same technique as abovelog y = log a + blog xY = A + bX

Where Y = log yA = log a X = log x

Using linear regression technique (the method of least squares), it is possible to calculate the value of a and b

TIME SERIES AND ANALYSISThis is the mathematical or statistical analysis on past data arranged in a periodic sequence.Decision making and planning in an organization involves forecasting which is one of the time series analysis.

Impediments in time series analysisAccuracy of data in reflectinga) Drastic changes e.g. in the advent of a major competitor, period of

war or sudden change of taste.b) For long term forecasting internal and external pressures makes

historical data less effective.


Lesson Four 158

1. Moving AveragePeriodical data e.g. monthly sales may have random fluctuation every month despite a general trend being evident. Moving average helps in smoothing away these random changes.

A moving average is the forecast for a period that takes the average of the previous periods.

Example:The table below represents company sales, calculate 3 and 6 monthly moving averages, for the data

Months SalesJanuary 1200February 1280March 1310April 1270May 1190June 1290July 1410August 1360September 1430October 1280November 1410December 1390

Solution.These are calculated as follows

April’s forecast = =

May’s forecast = =

And so on…

Similarly for 6 monthly moving average

July forecast = =

And so on…3 months moving average

6 months moving average

April 1263May 1287June 1257July 1250 1257August 1297 1292



September

1353 1305

October 1400 1325November 1357 1327December 1373 1363

Note:When plotting moving average on graphs the points are plotted as the midpoint of the period of the average, e.g. in our example the forecast for April (1263) is plotted on mid Feb.Characteristics of moving average

1) The more the number of periods in the moving average, the greater the smoothing effect.

2) Different moving averages produce different forecasts.3) The more the randomness of data with underlying trend being

constant then the more the periods should be involved in the moving averages.

Limitations of moving averages.1) Equal weighing with disregard to how more recent data is more

relevant.2) Moving average ignores data outside the period of the average

thus it doesn’t fully utilise available data.3) Where there is an underlying seasonal variation, forecasting

with unadjusted moving average can be misleading.

2. Exponential smoothingThis is a weighted moving average technique, it is given by:

New forecast = Old forecast + (Latest Observation – Old forecast)Where = Smoothing constant

This method involves automatic weighing of past data with weights that decrease exponentially with time.

ExampleUsing the previous example and smoothing constant 0.3 generate monthly forecasts

Months Sales Forecasts: = 0.3January 1200February 1280 1200March 1310 1224April 1270 1250May 1190 1256June 1290 1233July 1410 1250August 1360 1283September 1430 1327October 1280 1358November 1410 1335December 1390 1357


Lesson Four 160

SolutionSince there were no forecasts before January we take Jan to be the forecast for February.

Feb – 1200For March;

March forecast = Feb forecast + 0.3 ( Feb sales – Feb forecast)= 1200 + 0.3 (1280 – 1200)=1224

Note: The value lies between 0 and 1. The higher the value, the more the forecast is sensitive to the

current status.

Characteristics of exponential smoothing More weight is given to the most recent data. All past data are incorporated unlike in moving averages. Less data is needed to be stored unlike in periodic moving averages.

Decomposition of time seriesTime series has the following characteristics.a) A long term trend (T) –tendency of the whole series to rise and fall.b) Seasonal variation (S) – short term periodic fluctuations in values.

e.g. in Kenya maize yield is high in November and low in March or matatus have better business on Friday and very low on Sundays.

c) Cyclical variation (C) – These are medium term changes caused by factors which apply for a while then disappear, and come back again in a repetitive cycle. e.g. drought hits Kenya every 7 years.Note that cyclic variation has a longer term than seasonal variation e.g. seasonal variation may occur once every year while cyclic variation occurs once every several years.

d) Random residual variation (R) – These are non-recurring random variations e.g. war, fire, coup e.t.c.

For accurate forecasts these aspects are qualified separately (i.e. T,C,S and R) from data. This is known as time decomposition or time series analysisThe separate elements are then combined to produce a forecast.

Time series models:

Additive ModelTime series value = T +S +C +R

Where S, C and R are expressed in absolute value.This model is best suited where the component factors are independent e.g. where the seasonal variation is unaffected by trend.

Multiplicative Model:



Time series value = T × S× C × RWhere S, C and are expressed as percentage or proportions.

This model is best applied where characteristics interact e.g. where high trends increase seasonal variations. Multiplicative model is more commonly used in practice.

Of the four elements of time series the most important are trend and seasonal variation. The following illustration shows how the trend (T) and seasonal variation (S) are separated out from a time series and how the calculated T and S values are used to prepare forecast. The process of separating out the trend and seasonal variation is known as deseasonalising the data.

There are two approaches to this process: one is based on regression through the actual data points and the other calculates the regression line through moving average trend points. The method using the actual data is demonstrated first followed by the moving average method.


Lesson Four 162

1. Time series analysis: trend and seasonal variation using regression on the data

The following data will be used to illustrate how the trend and seasonal variation are calculated.Example 1

Sales of widgets in ‘000sQuarter 1

Quarter 2

Quarter 3

Quarter 4

Year 1 20 32 62 292 21 42 75 313 23 39 77 484 27 39 92 53

It will be apparent that there is a strong seasonal element in the above data (low in Quarter 1 and high in Quarter 3) and there is a generally upward trend.

The steps in analyzing the data and preparing a forecast are:

Step 1: Calculate the trend in the data using the least squares method.

Step 2: Estimate the sales for each quarter using the regression formula established in step 1.

Step 3: Calculate the percentage variation of each quarter’s actual sales from the estimates, obtained in step 2.

Step 4: Average the percentage variations from step 3. This establishes the average seasonal variations.

Step 5: Prepare forecast based on trend percentage seasonal variations.

Solution

Step 1Calculate the trend in the data by calculating the linear regression line y = a + bx.

x (quarters)

x (sales) xy x2

1 20 20 12 32 64 4

Year 1 3 62 186 94 29 116 16



5 21 105 256 42 252 36

Year 2 7 75 525 498 31 248 64

9 23 207 8110 39 390 100

Year 3 11 77 847 12112 38 576 144

13 27 351 16914 39 546 196

Year 4 15 92 1380 22516 53 848 256

x=136 y= 710 xy= 6661 x2=1496

Least square equations

y = an + bx

xy = ax + bx2

710 = 16a + 136b6661 = 136a + 1496b626 = 340bb = 1.84 and substituting we obtaina = 28.74Trend line = 28.74 + 1.84x

Steps 2 and 3Use the trend line to calculate the estimated sales for each quarter.

For example, the estimate for the first quarter in year 1 isestimate = 28.74 + 1.84 (1) = 30.58

The actual value of sales is then expressed as a percentage of this estimate. For example, actual sales in the first quarter were 20 so the seasonal variation is

x (quarters)

y (sales) Trend

1 20 30.58 652 32 32.42 99

Year 1

3 62 34.26 181


Lesson Four 164

4 29 36.10 80

5 21 37.94 556 42 39.78 106

Year 2

7 75 41.62 180

8 31 43.46 71

9 23 45.30 5110 39 47.14 83

Year 3

11 77 48.98 157

12 48 50.82 94

13 27 52.66 5114 39 54.50 72

Year 4

15 92 56.34 163

16 53 58.18 91

Trend estimates and percentage variations table.

Step 4

Average the percentage variations to find the average seasonal variations.Q1 Q2 Q3 Q4% % % %65 99 181 8055 106 180 7151 83 157 9451 72 163 91

222 360 681 336 4 = 56% 90% 170% 84%

These then are the average variations expected from the trend for each of the quarters; for example, on average the first quarter of each year will be 56% of the value of the trend. Because the variations have been averaged, the amounts over 100% (Q3 in this example). This can be checked by adding the average and verifying that they total 400% thus:

56% + 90% + 170% + 84% = 400%.

On occasions, roundings in the calculations will make slight adjustments necessary to the average variations.

Step 5

Prepare final forecasts based on the trend line estimates from “trend estimates and percentages variation table” (i.e. 30.58, 32.42, etc) and



the averaged seasonal variations from the table above. (i.e. 56%, 90%, 170% and 84%)

The seasonally adjusted forecast is calculated thus:

Seasonally adjusted forecast = Trend estimate × Seasonal variation%

X (quarters)

Y (sales) Seasonally adjusted

forecast

1 20 17.12Year 1 2 32 29.18

3 62 58.244 29 30.32

5 21 21.24Year 2 6 42 35.80

7 75 70.758 31 36.51

9 23 25.37Year 3 10 39 42.43

11 77 83.2712 48 42.69

13 27 29.49Year 4 14 39 49.05

15 92 95.7816 53 48.87

Seasonally adjusted forecastsThe forecasts are compared with the actual data to get some idea of how good extrapolated forecasts might be. With further analysis they enable us to quantify the residual variations.

Extrapolation using the trend and seasonal factors Once the formulae above have been calculated, they can be used to forecast (extrapolate) future sales. If it is required to estimate the sales for the next year (i.e. Quarters 17, 18, 19 and 20 in our series) this is done as follows:

Quarter 17 Basic trend = 28.74 + 1.84 (17)= 60.02

Seasonal adjustment for a first quarter = 56%Adjusted forecast = 60.02 × 56%

= 33.61


Lesson Four 166

A similar process produces the following figures:

Adjusted forecasts Quarter 18 = 55.6719 = 108.2920 = 55.05

Notes:a) Time series decomposition is not an adaptive forecasting system

like moving averages and exponential smoothing.b) Forecasts produced by such an analysis should always be treated

with caution. Changing conditions and changing seasonal factors make long term forecasting a difficult task.

c) The above illustration has been an example of a multiplicative model. This is the seasonal variations were expressed in percentage or proportionate terms. Similar steps would have been necessary if the additive model had been used except that the variations from the trend would have been the absolute values. For example, the first two variations would have been

Q1: 20 – 30.58 = absolute variation = -10.58Q2: 32 – 32.42 = absolute variation = - 0.42

And so on.

The absolute variations would have been averaged in the normal way to find the average absolute variation, whether + or -, and these values would have been used to make the final seasonally adjusted forecasts.

2. Trend and seasonal variation using moving averagesWhen the correlation coefficient is low the method of calculating the regression line through the actual data points should not be used. This is because the regression line is too sensitive to changes in the data values.

In such circumstances, calculating a regression line through the moving average trend points is more robust and stable.

Example 1 is reworked below using this method and, because there are many similarities to the earlier method, only the key stages are shown.

x y 3 point moving

average (1)

Trend line (2)

1 20 34.38 582 32 38 35.70 903 62 41 37.02 1674 29 37.3 38.34 765 21 30.7 39.66 536 42 46 40.98 102



7 75 49.3 42.30 1778 31 43 43.62 719 23 31 44.94 5110 39 46.3 46.26 8411 77 54.7 47.58 16212 48 50.7 48.90 9813 27 38 50.22 5414 39 52.7 51.54 7615 92 61.3 52.86 17416 53 54.18 98

Trend estimates and percentage variations utilizing moving averages

The first three moving average is calculated as follows

The next calculated:

The regression line y = a + bx of the moving average values is calculated in the normal manner and results in the following:

y = 33.06 + 1.32x

This is used to calculate the trend line:

e.g. For Period 1:y = 33.06 + 1.32(1) = 34.38 For Period 2:y = 33.06 + 1.32 (2) = 35.70

The percentage variations are averaged as previously shown, resulting in the following values:

Q1 Q2 Q3 Q4Average seasonal variation %

54 89 170 86

The trend line and the average seasonal variations are then used in a similar manner to that previously described.

For example, to extrapolate future sales for the next year (i.e. quarters 17, 18, 19 and 20) is as follows:

Quarter 17 Forecast sales = (33.06 + 1.32(17)) × 0.54 =

29.97

A similar process produces the following figures:


Lesson Four 168

Quarter 18 = 50.5719 = 98.8420 = 51.13

Forecast errorsDifferences between actual results and predictions may arise from many reasons. They may arise from random influences, normal sampling errors, choice of the wrong forecasting system or alpha value or simply that the future conditions turn out to be radically different from the past. Whatever the cause(s) management wish to know the extent of the forecast errors and various methods exist to calculate these errors.

A commonly used technique, appropriate to time series, is to calculate the mean squared error of the deviations between forecast and actual values then choose the forecasting system and/or parameters which gives the lowest value of mean squared errors, i.e. akin to the ‘least squares’ method of establishing a regression line.

Longer- term forecastingMoving averages, exponential smoothing and decomposition methods tend to be used for short to medium term forecasting. Longer term forecasting is usually less detailed and is normally concerned with forecasting the main trends on a year to year basis. Any of the techniques of regression analysis described in the preceding chapters could be used depending on the assumptions about linearity or non- linearity, the number of independent variables and so on. The least squares regression approach is often used for trend forecasting.

Forecasting using least squares

Example 2Data have been kept of sales over the last seven years

Year 1 2 3 4 5 6 7Sales (in ‘000 units

14 17 15 23 18 22 27

It is required to forecast the sales for the 8th year



SolutionYears (x) Sales (y) xy x2

1 14 14 12 17 34 43 15 45 94 23 92 165 18 90 256 22 132 367 27 189 49

x=28 y = 136 xy=596 x2= 140

136 = 7a + 28b

596 = 28a + 140b b = 1.86

And substituting in one of the equations we obtaina = 12

Regression line = y = 12 + 1.86x

Or, Sales in (‘000s of units) = 12.00 + 1.86 (no of years)

We use this expression for forecasting, for 8th year sales = 12 + 1.86 (8)=26.88 i.e. 26,888 units


Lesson Four 170


QUESTION ONEa) What is meant by correlation?b) Why is the co-efficient of determination calculated?c) Define R. (coefficient of rank correlation)

QUESTION TWOExplain the difference between Linear model, exponential model and geometric model, and write down their formulas

QUESTION THREEAn analysis of representatives’ car expenses shows that the expenses are dependent on the miles travelled (x) and the type of journey (x). the general form is:

y = a + b1x1 + b2x2

Calculations have produced the following values (where y is expenses per month)

y = £86 + 0.37x1 + 0.08x2

r2x1 = 0.78

r2x2 = 0.16

R = 0.88

Interpret these values

QUESTION FOUR

Month Actual Sales (units)January 450February 440March 460April 410May 380June 400July 370August 360September 410October 450November 470December 490January 460



Required.Provide 3 month, 6 month and 12 month moving average.

QUESTION FIVEThe manager of a company is preparing revenue plans for the last

quarter of 1993/94 and for the first three quarters of 1994/5. The data below refer to one of the main products:

Revenue

April-June July-Sept Oct-Dec Jan-March

Quarter 1 Quarter 2 Quarter 3 Quarter 4£‘000 £‘000 £‘000 £‘000 £‘0001990/91 49 37 58 671991/92 50 38 59 681992/93 51 40 60 701993/94 50 42 61 -

Required:a) Calculate the four-quarterly moving average trend for this set of data.b) Calculate the seasonal factors using either the additive model or the

multiplicative model, but not both.c) Explain, but do not calculate how you would use the results in parts

(a) and (b) of this question to forecast the revenue for the last quarter of 1993/4 and for the first three quarters of 1994/95.


Lesson Four 172

QUESTION SIXA company has a fleet of vehicles and is trying to predict the annual maintenance cost per vehicle. The following data have been supplied for a sample of vehicles:

Vehicle number

Age in years Maintenance cost

Per annum£ X 10

(x) (y)1 2 602 8 1323 6 1004 8 1205 10 1506 4 847 4 908 2 689 6 104

10 10 140

Required:a) Using the least squares technique calculate the values of a and b in

the equation y = a + bx, to allow managers to predict the likely maintenance cost, knowing the age of the vehicle.

b) Prepare a table of maintenance costs covering vehicles from 1 to 10 years of age, based on your calculations in (a).

c) Estimate the maintenance costs of a 12-year-old vehicle and comment on the validity of making such an estimate.

QUESTION SEVENA company is building a model in order to forecast total costs based on the level of output. The following data are available for last year:


Month Output Costs‘000 units

£000

(x) (y)January 16 170February 20 240March 23 260April 25 300May 25 280June 19 230July 16 200August 12 160September

19 240

October 25 290November

28 350

December 12 200


Required:a) State two possible reasons for the large variation in output per

month.b) Plot a graph of output and costs, and comment on the relationship

observed.c) Using the least square technique, calculate the values of a an b in the

equation y = a + bx in order to predict costs given the output, and explain the meaning of the calculated values.

QUESTION EIGHTYour company has been selling data base and spreadsheets for the last four years and has found the business to vary with season. The quarterly sales figures for the last four years are shown in table 6b1 and table 6b2 shows the deviation from the trend at the appropriate periods

Table 6b1Quarterly sales in £ 000s

Year Q1 Q2 Q3 Q41983 360 530 3541984 304 430 750 3951985 340 500 660 5091986 374 590 710 5211987 440

Table 6b2Seasonal deviation from trend in £ 000s

Year Q1 Q2 Q3 Q41983 -421984 -128 -37 276 -931985 -145 12 153 -151986 -165 43 153

Requiredi. establish the trend figures from the data in the two tablesii. establish the seasonal variations for the four year periodiii. using your results from parts (i) and (ii) forecast sales for 1987

quarter 2

QUESTION NINE1. The directors of your company wish to make a serious study of the

heating costs of the **** block. The data for the last sixteen quarterly periods are tabulated as follows.Heating costs in £


Lesson Four 174

QuarterYear Q1 Q2 Q3 Q41980 15601981 1730 1554 1504 16301982 1950 1595 1540 17001983 1860 1709 1574 17901984 1910 1721 1640



Requireda) Assuming the additive model calculate the trend of heat costs

using the method of moving averagesb) Estimate the seasonal deviations from trendc) Estimate the heating costs for quarter IV of 1984 and comment

on any factors affecting the reliability of your forecast



Lesson Four 176

COMPREHESIVE ASSIGNMENT TWOWork out these question for three hours (exam condition) then hand



SECTION IQUESTION ONEa) In the just concluded higher education seminar at Shoppers Paradise,

Nairobi, the College of Business Administration of Highland University states in some of its promotional material that the average graduate of the college earns over Sh.3 million a year. Assume, for simplicity, that only four people have graduated to date; Sam, Tom, Jackie and Mary who earn Sh.1.6 million, Sh.1.8 million, Sh.1.8 million and Sh.2 million respectively in a year.

Required:Compute the mean, median and the mode. Is the college’s claim correct? (3 marks)

b) Let us change our assumption about the number of graduates in (a) above and instead assume that five people have graduated. They consist of the four listed above and Suki who earns Sh.5.3 million per year.

Required:Compute the mean, median and the mode for the five graduates. Is the college’s claim correct?

(3 marks)

c) Changing our assumption one more time about the number of graduates, let us assume that six people have graduated. They consist of the four original ones, Suki, who earns Sh.5.3 million a year; and Bob who earns Sh.6.7 million a year.

Required:i) Compute the mean, median and mode for the six graduates. Is the

college’s claim correct?(2 marks)

ii) Comment on what happened to the mean, median and mode as you moved from part (a) to (b) to (c) of this problem.

(2 marks)iii) What do the results in (ii) above suggest about the relative stability

of the mean, median and the mode?(2 marks)



iv) How do you feel about the ethics of this college in claiming that their average graduates earn over Sh.3 million a year?

(2 marks)

d) Genuine athletic Company Ltd., manufactures weight-lifting equipment. The company’s top-of-the-line equipment are used in events such as the Olympics and other prestigious professional weight-lifting competitions. Consequently, it is very important that if a barbell plate is stamped say, “50 kilogrammes”, it weighs very close to 50 kilogrammes. In addition, a barbell plate must have a hole just slightly larger than 1 centimeter in diameter so that it will slip onto the 1 centimeter diameter bar easily but fit smoothly when it is in place.

A recent sampling of barbell plates of 10 and 50 kilogrammes revealed the following information:

i) Weights of 10 kilogram plates had an arithmetic mean of 10.013 kilogrammes and a standard deviation of 0.124 kilogrammes.

ii) Weights of 50 kilogram plates had an arithmetic mean of 50.032 kilogrammes and a standard deviation of 0.465 kilogrammes.

iii) Diameters of holes in the 10 kilogram plates had an arithmetic mean of 1.22 centimeters and a standard deviation of 0.187 centimeters.

iv) Diameter of holes in the 50 kilogram plates had an arithmetic mean of 1.20 centimeters and a standard deviation of 0.183 centimeters.

Required:Determine whether the production process associated with one size of barbell plant produced more variable results than the production process associated with the other size.

(6 marks)(Total: 20

marks)

QUESTION TWOa) The index of industrial production in the Utopian country by July 2001

is given below:

July 2001 IndexSector Weight (1994 = 100)Mining and quarrying

41 361

Manufacturing:Food, drink and tobacco

77 106

Chemicals 66 109Metal 47 72


Lesson Four 178

Engineering 298 86Textiles 67 70Other manufacturing 142 91Construction 182 84- Gas, electricity and water

80 115

Required:i) Calculate the index of industrial production for all industries and

manufacturing industries.(6 marks)

ii) Comment on your results. (4 marks)

b) Explain some of the uses of index numbers.(5 marks)

c) What are some of the limitations of index numbers?(5 marks)

(Total: 20 marks)QUESTION THREELeisure Publishers Ltd. recently published 20 romantic novels by 20 different authors. Sales ranged from just over 5,000 copies for one novel about 24,000 copies for another novel. Before publishing, each novel had been assessed by a reader who had given it a rating between 1 and 10. The managing director suspects that the main influence on sales is the cover of the book. The illustrations on the front covers were drawn either by artist A or artist B. the short description on the back cover of the novel was written by either editor C or editor D.

A multiple regression analysis was done using the following variables:Y Sales (million of shillings)X1 1 if front cover is by artist A

2 if front cover is by artist BX2 readers’ ratingX1 1 if the short description of the novel is by editor C

2 if the short description of the novel is by editor D

The computer analysis produced the following results:

Correlation coefficient r = 0.921265Standard error of estimate = 2.04485

Analysis of varianceDegrees of freedom

Sum of squares

Mean square F ratio

Regression

3 375.37 125.12 29.923

Residue 16 66.903 1.1814



Individual analysis of variablesVariable Coefficient Standard

errorF Value

Constant 15.7588 2.54389 38.3751 -6.25485 0.961897 42.2842 0.0851136 0.298272 0.0814283 5.86599 0.922233 40.457741

Correlation coefficients1 -0.307729 0 -0.674104

1 0.123094 0.3108381 0.627329

1

Required:a) The regression equation. (3

marks)b) Does the regression analysis provide useful information? Explain.

(3 marks)c) Explain whether the covers were more important for sales than

known quality of the novels.(4 marks)

d) State with 95% confidence the difference in sales of a novel if its cover illustrations were done by artist B instead of artist A.

(5 marks)e) State with 95% confidence the difference in sales of a novel if its

short description was by editor D and not editor C.(5 marks)

(Total: 20 marks)QUESTION FOURa) Explain the difference between regression and correlation analysis .

(4 marks)

b) Explain why the existence of a significant correlation does not imply causation. (2 marks)

c) A bakery bakes cakes under the brand name ‘super cakes’. Irene Juma, the manageress does not know the cost of each cake. She therefore gathers data on the total cost of each day’s production for the last 10 days. The results are shown in the table below;


Lesson Four 180

Day Number of cakes

(‘00’ units)

Total cost (Sh. ‘000’)

1 22.5 23.02 21.0 21.63 27.5 23.34 21.5 24.05 30.0 28.26 20.0 22.47 24.0 23.18 26.5 25.39 18.3 20.1

10 17.0 16.5

Required:i) Estimate the total cost function using the ordinary least squares

method. State the fixed cost and unit cost.(11 marks)

ii) If each cake is sold at Sh.10, determine the break even number of cakes. (3 marks)

(Total: 20 marks)QUESTION FIVEDifferentiate between additive model and the multiplicative model as used in time series analysis.

(4 marks)

The sales data of XYZ Ltd. (in millions of shillings) for the years 2001 and 2004 inclusive are as given below:

QuarterYear 1 2 3 42001 40 64 124 582002 42 84 150 622003 46 78 154 962004 54 78 184 106

Required:i) The trend in the data using the least squares method.

(8 marks)ii) The estimated sales for each quarter of year 2004.

(4 marks)iii) The percentage variation of each quarter’s actual sales for year 2004.

(4 marks)(Total: 20

marks)SECTION II

QUESTION SIX



a) Explain the following terms as used in index numbers:i) Price index (2

marks)ii) Quantity index (2

marks)iii) Composite index (2

marks)iv) Value index (2

marks)

b) The following prices and quantities reflect the average weekly consumption pattern of a certain family for the years 2001 and 2002.


Lesson Four 182

Year 2001 Year 2002Price (p0) Quantity

(q0)Price (p1) Quantity

(q1)Item Sh. Sh.Oranges (kg) 15 2 25 1Milk (Litres) 30 2 35 2Bread (Loafs) 30 3 40 3Eggs (Dozens) 50 1 65 1

Required:i) Price relatives for each item (4

marks)ii) Laspeyres price index (4

marks)iii) Paasche price index (4

marks)(Total: 20

marks)QUESTION SEVENExplain three methods of fitting a trend in time series analysis.

(6 marks)

The quarterly sales data for Chuce hardware are given below:

QuarterYear 1 2 3 4

(Sh. Million)

(Sh. Million)

(Sh. Million)

(Sh. Million)

2000 8.5 10.4 7.5 11.82001 9.5 12.2 8.8 13.62002 10.4 13.5 9.7 13.12003 9.5 11.7 8.4 12.92004 10.9 13.7 10.1 15.0

Required:(a) The centred four-quarter moving averages. (6

marks)(b) The specific seasonal variation for each quarter

(3 marks)(c) The typical seasonal indices (3

marks)(d) Explain the third quarter typical seasonal index

(2 marks)(Total: 20

marks)QUESTION EIGHT



a) A machine produces circular bolts and as a quality control test, 250 bolts were selected randomly and the diameter of their heads measured as follows:

Diameter of head (cm)

Number of components

0.9747 - 0.9749 20.9750 - 0.9752 60.9758150.9753

- 0.9755 8

0.9759-0.9756

- 0.9761 42

0.9762 - 0.9764 680.9765 - 0.9767 490.9768 - 0.9770 250.9771 - 0.9773 180.9774 - 0.9776 120.9777 - 0.9779 40.9780 - 0.9782 1

Required:b)

i) Determine whether the customer is getting reasonable value if the label on the circular bolt advertises that the average diameter of the head is 0.97642 cm. (8 marks)

ii) In what situation would weighted mean be used?(3 marks)

iii) Describe briefly how to estimate the median on a grouped frequency distribution graphically?

(3 marks)iv) Why is the mode not used extensively in statistical analysis?

(3 marks)“The standard deviation is the natural partner to the mean”.

Explain (3 marks)(Total: 20

marks)


Probability 184

LESSON FIVE

Probability

Contents - Probability theory- Bayes Theorem and conditional probability- Permutations and combinations- Discrete probability distributions- Continuous probability distribution


185 Lesson Five

5.1 PROBABILITY- Probability is a very popular concept in business management.

This is because it covers the risks which may be involved in certain business situations. It is a fact that when a business investment is being arranged, the outcome is usually uncertain. Therefore the concept of probability may be used to describe the degree of uncertainty of a particular business outcomeProbability may therefore be defied as the chances of a given event occurring. Numerically, probability values range between 0 and 1. a probability of 0 implies that the event cannot occur at all. A probability of 1 implies that the event will certainly occur.Therefore other events have their probabilities with values lying between 0 and 1

- The formular used to determine probability is as followProbability (x) =

Application of Probability in Business1. Business games of chance e.g. Raffles Lotteries e.t.c.2. Insurance firms: this is usually done when a new client or property is

being insured. The company has to be certain about the chances of the insured risks occurring.

3. Business decision making regarding viability of projects thus the projects with a greater probability has greater chances.

ExampleA bag contains 80 balls of which 20 are red, 25 are blue and 35 are white. A ball is picked at random what is the probability that the ball picked is:(i) Red ball(ii) Black ball(iii) Red or Blue ball.

Solution(i) Probability of a red ball =

=

(ii) Probability of black ball =

=

(iii) P(R or B) =

= Note: in probability or is replaced by a plus (+) sign. See addition

rule.


Probability 186

Common termsEvents: an event is a possible outcome of an experiment or a result of a trial or an observation.

Mutually exclusive eventsA set of events is said to be mutually exclusive if the occurance of any one of the events precludes the occurrence of any of the other events e.g. when tossing a coin, the events are a head or a tail these are said to be mutually exclusive since the occurrence of heads for instance implies that tails cannot and has not occurred.It can be represented in venn diagram as.

E1 E2 = Ø

E1 E2 ≠ Ø

Consider a survey in which a random sample of registered voters is selected. For each voter selected their sex and political party affiliation are noted. The events “KANU” and “woman” are not mutually exclusive because the selection of KANU does not preclude the possibly that the voter is also a woman.

Independent EventsEvents are said to be independent when the occurance of any of the events does not affect the occurrence of the other(s).

e.g. the outcome of tossing a coin is independent of the outcome of the preceeding or succeeding toss.


E1 E2

E1 E2

Non-mutually exclusive events (independent events)

187 Lesson Five

ExampleFrom a pack of playing cards what is the probability of;(i) Picking either a ‘Diamond’ or a ‘Heart’ → mutually exclusive(ii) Picking eigher a ‘Flower’ or an ‘Ace’ → indepent events

Solutions.(i) P(Diamond or Heart)

= P(Diamond) + P(Heart)= = 0.5

(ii) P(Flower or Ace)= P(Flower) + P(Ace) – P(Flower and Ace)=

= = 0.31

Note: that the formula used incase of independent events is different to the one of mutually exclusive.

Rules of Probability(a) Additional Rule – This rule is used to calculate the probability of

two or more mutually exclusive events. In such circumstances the probability of the separate events must be added.

ExampleWhat is the probability of throwing a 3 or a 6 with a throw of a

die?

SolutionP(throwing a 3 or a 6) =

(b) Multiplicative ruleThis is used when there is a string of independent events for which individual probability is known and it is required to know the overall probability.

ExampleWhat is the probability of a 3 and a 6 with two throws of a die?

SolutionP(throwing a 3) and P(6)

= P(3) and P(6) =

Note: 1) In probability ‘and’ is replaced by ‘x’ – multiplication.


Probability 188

2) P(x) and P(y) ≠ P(x and y) note that these two are different. The first implies P(x) happening and P(y), but if the order of which happened first is unimportant then we have p(x and y).In the example above:

P(3) and P(6) = butP(3 and 6) = P(3 followed by 6) or P(6 followed by 3)

= [P(3) P(6)] or [P(6) P(3)]=


189 Lesson Five

(c) Conditional probabilityThis is the probability associated with combinations of events but given that some prior result has already been achieved with one of them.Its expressed in the form of

P(x|y) = Probability of x given that y has already occurred.

P(x|y) = → conditional probability formula.

Example:In a competitive examination. 30 candidates are to be selected. In all 600 candidates appear in a written test, and 100 will be called for the interview.(i) What is the probability that a person will be called for the interview?(ii) Determine the probability of a person getting selected if he has

been called for the interview?(iii) Probability that person is called for the interview and is selected?

Solution:Let event A be that the person is called for the interview and event B that he is selected.

(i) P(A) = =

(ii) P(B|A) =

(iii) P(AB) = P(A) × P(B|A)

=

Example:From past experience a machine is known to be set up correctly on 90% of occasions. If the machine is set up correctly then 95% of good parts are expected but if the machine is not set up correctly then the probability of a good part is only 30%.On a particular day the machine is set up and the first component produced and found to be good. What is the probability that the machine is set up correctly.

Solution:This is displayed in the form of a probability tree or diagram as follows:


CS = 0.9

IS = 0.1

GP = 0.95

BP = 0.05

GP = 0.3

BP = 0.7

CS GP

CS BP

IS GP

IS BP

CS – Correct Setting

IS – Incorrect Setting

GP – Good Product

BP – Bad Product189

Probability 190

P(CSGP) = 0.9 × 0.95 = 0.855P(CSBP) = 0.9 × 0.05 = 0.045P(ISGP) = 0.1 × 0.3 = 0.03P(ISBP) = 0.1 × 0.7 = 0.07

1.00

- Probability of getting a good part (GP) = CSGP or ISGP= CSGP + ISGP= 0.855 + 0.03 = 0.885

Note: Good parts may be produced when the machine is correctly set up and also when its incorrectly setup. In 1000 trials, 855 occasions when its correctly setup and good parts produced (CSGP) and 30 occasions when its incorrectly setup and good parts produced (ISGP).

- Probability that the machine is correctly set up after getting a good part.

=

Or

= P(CS|GP) =

ExampleIn a class of 100 students, 36 are male and studying accounting, 9 are male but not studying accounting, 42 are female and studying accounting, 13 are female and are not studying accounting.Use these data to deduce probabilities concerning a student drawn at random.

Solution:Accounting

ANot accounting Total

Male M 36 9 45Female F 42 13 55Total 78 22 100

P(M) =


191 Lesson Five

P(F) =

P(A) =

P =

P(M and A) = P(A and M) = = 0.36P(M and ) = 0.09P(F and ) = 0.13These probabilities can be express differently as;P(M) = P(M and A) or P(M and )

= 0.36 + 0.09 = 0.45

P(F) = P(F and A) or P(F and )= 0.42 + 0.13 = 0.55

P(A) = P(A and M) + P(A and F) = 0.36 + 0.42 = 0.78

P = P( and M) + P( and F) = 0.09 + 0.13 = 0.22

Now calculate the probability that a student is studying accounting given that he is male.

This is a conditional probability given as P(A|M)

P(A|M) =

From the formula above we get that,P(A and M) = P(M) P(A|M) ……………….. (i)Note that P(A|M) ≠ P(M|A)

Since P(M|A) = this is known as the Bayes’ rule.

Bayes’ rule/Theorem

This rule or theorem is given by

P(A|B) =

It’s used frequently in decision making where information is given the in form of conditional probabilities and the reverse of these probabilities must be found.

ExampleAnalysis of questionnaire completed by holiday makers showed that 0.75 classified their holiday as good at Malindi. The probability of hot weather in the resort is 0.6. If the probability of regarding holiday as good given hot weather is 0.9, what is the probability that there was hot weather if a holiday maker considers his holiday good?


Probability 192

SolutionP(A|B) =

Let H = hot weatherG = GoodP(G) = 0.75 P(H) = 0.6 and P(G|H) = 0.9 (Probability of regard holiday as good given hot weather)

Now the question requires us to getP(H|G) = Probability of (there was) hot weather given that the holiday has been rated as good).

=

= 0.72.

Worked examples on probability1. A machine comprises of 3 transformers A, B and C. The machine may operate if at least 2 transformers are working. The probability of each transformer working are given as shown below;

P(A) = 0.6, P(B) = 0.5, P(C) = 0.7A mechanical engineer went to inspect the working conditions of those transformers. Find the probabilities of having the following outcomes

i. Only one transformer operatingii. Two transformers are operatingiii. All three transformers are operatingiv. None is operatingv. At least 2 are operatingvi. At most 2 are operating

SolutionP(A) =0.6 P( ) = 0.4 P(B) = 0.5 P(~B) = 0.5P(C) = 0.7 P( ) = 0.3

i. P(only one transformer is operating) is given by the following possibilities

1st 2nd 3rd

P (A ) = 0.6 x 0.5 x 0.3 = 0.09P ( B ) = 0.4 x 0.5 x 0.3 = 0.06P ( C) = 0.4 x 0.5 x 0.7 = 0.14

∴ P(Only one transformer working)= 0.09 + 0.06 + 0.14 = 0.29

ii. P(only two transformers are operating) is given by the following possibilities.

1st 2nd 3rd

P (A B ) = 0.6 x 0.5 x 0.3 = 0.09P (A C) = 0.6 x 0.5 x 0.7 = 0.21


193 Lesson Five

P ( B C) = 0.4 x 0.5 x 0.7 = 0.14∴ P(Only two transformers are operating)

= 0.09 + 0.21 + 0.14 = 0.44iii. P(all the three transformers are operating).

= P(A) x P(B) x P(C)

= 0.6 x 0.5 x 0.7

= 0.21

iv. P(none of the transformers is operating). = P( ) x P( ) x P( )

= 0.4 x 0.5 x 0.3

= 0.06

v. P(at least 2 working). = P(exactly 2 working) + P(all three working)

= 0.44 + 0.21

= 0.65

vi. P(at most 2 working). = P(Zero working) + P(one working) + P(two working)= 0.06 + 0.29 + 0.44= 0.79

5.3 Permutations and Combinations

DefinitionPermutation

- This is an order arrangement of items in which the order must be strictly observed

Example

Let x, y and z be any three items. Arrange these in all possible permutations

1st 2nd 3rd

X Y ZX Z YY X Z Six different permutationsY Z XZ Y XZ X Y


Probability 194

NB: The above 6 permutations are the maximum one can ever obtain in a situation where there are only 3 items but if the number of items exceeds 3 then determining the no. of permutations by outlining as done above may be cumbersome. Therefore we use a special formula to determine such permutations. The formula is given below

The number of permutations of ‘r’ items taken from a sample of ‘n’ items

may be provided as nPr = where; ! = factorial

e.g.

i. 3P3 =

= note; 0! = 1

= = 6

ii. 5P3 =

== 60

iii. 7P5 =

=

=

= 2520

ExampleThere are 6 contestants for the post of chairman secretary and treasurer. These positions can be filled by any of the 6. Find the possible no. of ways in which the 3 positions may be filled.SolutionChairman Secretary Treasurer 6 5 4Therefore the no of ways of filing the three positions is 6 x 5 x 4 = 120

6P3 =

=

=

= 120


195 Lesson Five

CombinationsDefinition A combination is a group of times in which order is not important. For a combination to hold at any given time it must comprise of the same items but if a new item is added to the group or removed from the group then we have a new combination

Example3 items x, y and z will have 6 different permutations but only one combination. The following formular is usually used to determine the no. of combinations in a given situation.

Example

i.

= 8

ii.

= 15

iii.

= 56

ExampleThere is a committee to be selected comprising of 5 people from a group of 5 men and 6 women. If the selection is randomly done. Find the possibility of having the following possibilities (combinations)

i. Three men and two womenii. At least one man and at least one woman must be in the

committeeiii. One particular man and one particular woman must not be in

the committee (one man four women)

Solutioni. The committee size = 5 people

The group size = 5m + 6w


Probability 196

∴ assuming no restrictions the committee can be selected in 11C5

the committee has to consist of 3m & 2w∴ these may be selected as follows.5C3 × 6C2

P(committee 3m and 2w)

note that this formula can be fed directly to your scientific

calculator and attain a solution.

=

ii. P(at least one man and at least one woman must be in the committee)The no. of possible combinations of selecting the committee without any woman = 5C5

The probability of having a committee of five men only

the probability of having a committee of five women only

= ∴ P(at least one man and at least one woman)

= 1 – {P(no man) + P(no woman)}

= 1 –

= 1 –

= 1 –


197 Lesson Five

= iii. P(one particular man and one particular woman must not be in

the committee would be determined as followsThe group size = 5m + 6wCommittee size = 5 people

Actual groups size from which to Select the committee = 4m + 5wCommittee = 1m + 4w

The committee may be selected in 9C5

The one man may be selected in 4C1 waysThe four women may be selected in 5C4 ways

∴ P(committee of 4w1man).

=

5.4 DISCRETE PROBABILITY DISTRIBUTIONS

BINOMIAL PROBABILITY DISTRIBUTIONBinomial probability distribution is a set of probabilities for discrete events. Discrete events are those whose results or outcomes can be counted. Binomial probabilities are commonly encountered in business situations e.g. in quality control activities the binomial probabilities are frequently used especially when determining the probability of having a certain no. of defective items in a given consignment.

- The binomial probability distribution is usually characterized by the fact that the binomial events have to fulfill the following properties

i. Each event has 2 possible outcomes only known as success or failure

ii. The probability of each outcome is independent of the previous outcomes

iii. The sample size is generally fixediv. The probabilities of success and failure tend to approach 0.5 if

the sample size increases (in the event when an unbiased coin is thrown a number of times)


Probability 198

v. The probabilities are given by the following equation

Where p = Probability of successr = no. of successesn = sample sizeq = 1 – P = Probability of failure

Example 1A medical survey was conducted in order to establish the proportion of the population which was infected with cancer. The results indicated that 40% of the population were suffering from the disease.A sample of 6 people was later taken and examined for the disease. Find the probability that the following outcomes were observed

a) Only one person had the diseaseb) Exactly two people had the diseasec) At most two people had the diseased) At least two people had the diseasee) Three or four people had the disease

SolutionP(a persona having cancer) = 40% = 0.4 = PP(a person not having cancer) = 60% = 0.6 = 1 – p = qa) P(only one person having cancer)

= 6C1 (0.4)(0.6)5

= (0.4)1(0.6)5

= 0.1866Note that from the formula

nCrprqn-r: n = sample size = 6p = 0.4r = 1 = only one person having cancer

b) P(2 people had the disease) = 6C2 (0.4)2 (0.6)4

= (0.4) 2 (0.6)5

= (0.4) 2 (0.6)5

= 15 × (0.4) 2 (0.6)5

= 0.311c) P(at most 2) = P(0) + P(1) + P(2) = P(0) or P(1) or P(2)

So we calculate the probability of each and add them up.P(0) = P(nobody having cancer)


199 Lesson Five

= 6C0 (0.4) 0(0.6)6

= (0.4) 0(0.6)6

= (0.6)6

= 0.0467The probabilities of P(1) and P(2) have been worked out in part (a) and (b)Therefore P(at most 2) = 0.0467 + 0.1866 + 0.311 = 0.5443d) P(at least 2)

= P(2) + P(3) + P(4) + P(5) + P(6)= 1 – [P(0) + P(1)] This is a shorter way of working out the solution since

[P(0) + P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1]= 1 – (0.0467 + 0.1866)= 0.7667

e) P(3 or 4 people had the disease)= P(3) +P(4)= 6C3(0.4)3(0.6)3 + 6C4(0.4)4(0.6)2

= (0.4) 3(0.6)3 + (0.4) 4(0.6)2

= 6 × 5 × 4 × 3! (0.4) 3(0.6)3 + 6 × 5 × 4! (0.4) 4(0.6)2

3 × 2 × 1 × 3! 2 × 1 × 4!= 20(0.4)3(0.6)3 + 15(0.4)4(0.6)2 = (20 × 0.013824) + (15 × 0.009216)= 0.27648 + 0.13824

= 0.41472


Probability 200

Example 2An insurance company takes a keen interest in the age at which a person is insured. Consequently a survey conducted on prospective clients indicated that for clients having the same age the probability that they

will be alive in 30 years time is . This probability was established using

the actuarial tables. If a sample of 5 people was insured now, find the probability of having the following possible outcomes in 30 years

a) All are aliveb) At least 3 are alivec) At most one is alived) None is alivee) At least 1 is alive

Sample size = 5

a)

b)


201 Lesson Five

c)

d)

e)

POISSON PROBABILITY DISTRIBUTION- This is a set of probabilities which is obtained for discrete events

which are described as being rare. Occasions similar to binominal distribution but have very low probabilities and large sample size.

Examples of such events in business are as follows:i. Telephone congestion at midnightii. Traffic jams at certain roads at 9 o’clock at nightiii. Sales boomiv. Attaining an age of 100 years (Centureon)- Poisson probabilities are frequently applied in business situations

in order to determine the numerical probabilities of such events occurring.

- The formula used to determine such probabilities is as follows

Where x = No. of successes⋋ = mean no. of the successes in the sample (⋋ =

np)e = 2.718

Example 1A manufacturer assures his customers that the probability of having defective item is 0.005. A sample of 1000 items was inspected. Find the probabilities of having the following possible outcomes

i. Only one is defectiveii. At most 2 defective iii. More than 3 defective

P(x) =

(⋋ = np = 1000 × 0.005) = 5i. P(only one is defective) = P(1) = P(x = 1)


Probability 202

= Note that 2.718-5=

=

=

= 0.0337ii. P(at most 2 defective) = P(x ≤ 2)

= P(0) + P(1) + P(2)P(x = 0) =

= 2.718-5

=

= = 0.00674

P(1) = 0.0337

P(2) =

=

= 0.08427

P(x≤2) = 0.00674 + 0.0337 + 0.08427

= 0.012471

iii. P(more than 3 defective) = P(x > 3)

= 1 –

BINOMIAL MATHEMATICAL PROPERTIES1. The mean or expected value = n × p = np

Where; n = Sample Size p = Probability of success

2. The variance = npqWhere; q = probability of failure = 1 - p

3. The standard deviation =

ExampleA firm is manufacturing 45,000 units of nuts. The probability of having a defective nut is 0.15Calculate the following

i. The expected no. of defective nutsii. The variance and standard deviation of the defective nuts in a

daily consignment of 45,000

SolutionSample size n = 45,000P(defective) = 0.15 = p


203 Lesson Five

P(non defective) = 0.85 = qi. ∴ the expected no of defective nuts

= 45,000 × 0.15 = 6,750ii. The variance = npq

= 45000 × 0.85 × 0.15= 5737.50The standard deviation = = = 75.74

POISSON MATHEMATICAL PROPERTIES1. The mean or expected value = np = λ

Where; n = Sample Sizep = Probability of success

2. The variance = np = ⋋3. Standard deviation = =

ExampleThe probability of a rare disease striking a given population is 0.003. A sample of 10000 was examined. Find the expected no. suffering from the disease and hence determine the variance and the standard deviation for the above problem

SolutionSample size n = 10000P(a person suffering from the disease) = 0.003 = p

∴ expected number of people suffering from the disease Mean = λ = 10000 × 0.003

= 30= np = ⋋

variance = np = 30Standard deviation = = ⋋

= = 5.477

5.5 PROBABILITY DISTRIBUTION FOR CONTINUOUS RANDOM VARIABLES.In a continuous distribution, the variable can take any value within a specified range, e.g. 2.21 or 1.64 compared to the specific values taken by a discrete variable e.g 1 or 3. The probability is represented by the area under the probability density curve between the given values.The uniform distribution, the normal probability distribution and the exponential distribution are examples of a continuous distribution

- The normal distribution is a probability distribution which is used to determine probabilities of continuous variables

Examples of continuous variables are o Distanceso Timeso Weights


Probability 204

o Heightso Capacity e.t.c

- Usually continuous variables are those, which can be measured by using the appropriate units of measurement.

- Following are the properties of the normal distribution1. The total area under the curve is = 1 which is equivalent to

the maximum value of probability

Normal probability

Distribution curve

Line of symmetry

2. The line of symmetry divides the curve into two equal halves3. The two ends of the normal distribution curve continuously

approach the horizontal axis but they never cross it4. The values of the mean, mode and median are all equal

NB: The above distribution curve is referred to as normal probability distribution curve because if a frequency distribution curve is plotted from measurements of a given sample drawn from a normal population then a graph similar to the normal curve must be obtained.

- It should be noted that 68% of any population lies within one standard deviation, ±1σ

- 95% lies within two standard deviations ±2σ- 99% lies within three standard deviations ±3σ

Where σ = standard deviation


Age (Yrs)

Tail endTail end

205 Lesson Five

STANDARDIZATION OF VARIABLES- Before we use the normal distribution curve to determine

probabilities of the continuous variables, we need to standardize the original units of measurement, by using the following formular.

Z =Where χ = Value to be standardized

Z = Standardization of xµ = population meanσ = Standard deviation

Example A sample of students had a mean age of 35 years with a standard deviation of 5 years. A student was randomly picked from a group of 200 students. Find the probability that the age of the student turned out to be as follows

i. Lying between 35 and 40ii. Lying between 30 and 40iii. Lying between 25 and 30iv. Lying beyond 45 yrsv. Lying beyond 30 yrsvi. Lying below 25 years

Solution(i). The standardized value for 35 years

Z = = = 0

The standardized value for 40 years


0 Z

205

Probability 206

Z = = = 1

∴ the area between Z = 0 and Z = 1 is 0.3413 (These values are

checked from the normal tables see appendix)The value from standard normal curve tables.When z = 0, p = 0And when z = 1, p = 0.3413Now the area under this curve is the area between z = 1 and z = 0

= 0.3413 – 0 = 0.3413∴ the probability age lying between 35 and 40 yrs is 0.3413

(ii). 30 and 40 years

Z = = = = -1

Z = = = 1

∴ the area between Z = -1 and Z = 1 is

= 0.3413 (lying on the positive side of zero) + 0.3413 (lying on the negative side of zero)P = 0.6826

∴ the probability age lying between 30 and 40 yrs is 0.6826(iii). 25 and 30 years

Z = = = = -2

Z = = = -1

∴ the area between Z = -2 and Z = -1Probability area corresponding to Z = -2

= 0.4772 (the z value to check from the tables is 2)Probability area corresponding to Z = -1

= 0.3413 (the z value for this case is 1)∴ the probability that the age lies between 25 and 30 yrs

= 0.4772 – 0.3413 (The area under this curve)P = 0.1359

iv). P(beyond 45 years) is determined as follow = P(x > 45)

Z = = = = + 2

Probability corresponding to Z = 2 = 0.4772 = probability of

between 35 and 45∴ P(Age > 45yrs) = 0.5000 – 0.4772

= 0.0228


207 Lesson Five

The exponential distributionThe exponential distribution is of particular importance because of the wide ranging nature of the practical situations in which it is used.

Examples1. The length of time until an electronic device fails2. The time required to wait for the first emission of a particle

from a radio active source3. The length of time between successive accidents in a large

factory Assume that a probability density function f(x) is valid between the values a and b, then

(i)..

(ii).The mean of the distribution

(iii) The variance of the distribution = E(x2) – [E(x)]2

Where

Example of continuous probability distribution functionThe distribution of a random variable x has a probability density function f(x) given by

f(x) = kx for 0 ≤ x ≤1f(x) = 0 elsewhereWhere k is constant

Required.i. Show the value of k is 2ii. Find the mean of f(x)iii. Find the variance of f(x)

Solutioni) ii

)


Probability 208

iii)

Exponential distribution

ExampleThe mean life of an electrical component is 100 hours and its life has an exponential distribution.Find

a. The probability that it will last less than 60 hoursb. The probability that it will last more than 90 hours

SolutionA continuous random variable X has an exponential distribution, if for some constant k >0 it has the probability density function

The function f(x) is positive for all values of x and the area under the curve

The mean of an exponential distribution with parameter k is and its variance is ExampleThe mean of an exponential distribution is 100, find;a) P(x<60)b) P(x>90)

solution.

a)


209 Lesson Five

b)

The students t distributionThe students t distribution was presented by W. S. Gosset in 1908 under the pen name of ‘student’. The t distribution is of great importance in the so called small sample tests and is profoundly used in statistical inferenceThe t distribution has a single parameter, known as the number of degrees of freedom. It is denoted by the Greek symbol ℧ (read as nu). It can be interpreted as the number of useful items of information generated by a sample of given size. The degrees of freedom are sample size less one (v = n-1)

Properties of t distribution1. The t distribution ranges from – ∞ to ∞ first as does the

normal distribution2. The t distribution like the standard normal distribution is

bell shaped and symmetrical around mean zero3. The shapes of the t distribution changes as the number of

degrees of freedom changes4. The t distribution is more platykurtic that the normal

distribution5. The t distribution has a greater dispersion than the standard

normal distribution. As n gets larger the t distribution approaches the normal distribution when n = 30 the difference is very small


Probability 210

Relation between the t distribution and standard normal distribution is shown in the following diagram

Standard normal distribution

T distribution n = 15

T distribution n = 5

- 4 - 3 - 2 -1 0 1 2 3 4

Note that the t distribution has different shapes depending on the size of the sample. When the sample is quite small the height of the t distribution is shorter than the normal distribution and the tails are wider.

Assumptions of t distribution1. The sample observations are random2. Samples are drawn from normal distribution3. The size of sample is thirty or less n ≤ 30

Application of t distribution- Estimation of population mean from small samples- Test of hypothesis about the population mean- Test of hypothesis about the difference between two means

Chi Square distributionChi square was first used by Karl Pearson in 1900. It is denoted by the Greek letter χ2. it contains only one parameter, called the number of degrees of freedom (d-f), where the term degree of freedom represents the number of independent random variables that express the chi square

Properties 1. Its critical values vary with the degree of freedom. For every

increase in the number of degrees of freedom there is a new χ 2

distribution.2. This possesses additional property so that when χ 2

1 and χ 22 are

independent and have a chi square distribution with n1 and n2 degrees of from χ 2

1 + χ 22 will also be distributed as a chi

square distribution with n1 + n2 degrees of freedom


211 Lesson Five

3. Where the degrees of freedom is 3.0 and less the distribution of χ 2 is skewed. But, for degrees of freedom greater than 30 in a distribution, the values of χ 2 are normally distributed4. The χ 2 function has only one parameter, the number of

degrees of freedom.

P(x) ℧ = 1

℧ = 2

℧ = 3

℧ = 4

℧ = 5

0 1 2 3 4 5 6 7 8 9 10

. χ 2

5. χ 2 distribution is a continuous probability distribution which has the value zero at its lower limit and extends to infinity in the positive direction. Negative value of χ2 is not possible because the differences between the observed and expected frequencies are always squared

F distribution or Variance ratio distributionIt was developed by R. A Fisher in 1924 and is usually defined in terms of the ratio of the variances of two normally distributed populationsIt is used to test the hypothesis that the two normally distributed populations have two equal variancesF distribution ratio of the variances between two normally distributed

population may be expressed as

With ℧1 = n1–1 and ℧2 = n2–1 degrees of freedom


Probability 212

Where normal population means are unknownn1 – sample size of independent random 1n2 – sample size of independent random 2

- Sample variance of 1– sample variance of 2 - Population variance of 1 Population variance of 2

and are given by

as the unbiased estimator of d12

as the unbiased estimator of d22

if = then the statistic F =

F – Distribution with n1–1 and n2–1 degrees of freedom. F distribution depends on the degrees of freedom ℧1 for the numerator and ℧2 for the denominator. It has parameters ℧1 and ℧2 such that for different values of ℧1 and ℧2 will have different distributions.

Properties1. The shape of the f distribution depends upon the number of

degrees of freedom2. The mean and variance of the f distribution are

Mean = ℧ 1 for ℧2 >2-v2 - 2

for ℧2 > 4

3. The f distribution is positively skewed and its skewness decreases with increases in ℧1 and ℧2

4. The value of f must be positive or zero since variances are squares and can never assume negative values

Assumptionsa) All sample observations are randomly selected and independent b) The total variance of the various sources of variance should be

additive. c) The ratio of S1

2 to S22 should be equal to or greater than 1


213 Lesson Five

d) The population for each sample must be normally distributed with identical mean of variance

e) F value can never be negative


Probability 214


QUESTION ONEThe quality controller, Mr. Brooks, at Queensville Engineers has become aware of the need for an acceptance sampling programme to check the quality of bought-in components. This is of particular importance for a problem the company is currently having with batches of pump shafts bought in from a local supplier. Mr. Brooks proposes the following criteria to assess whether or not to accept a large batch of pump shafts from this supplier.

From each batch received take a random sample of 50 shafts, and accept that batch if no more than two defectives are found in the sample.

Mr. Brooks needed to calculate the probability of accepting a batch Pa , when the proportion of defectives in the batch, p, is small (under 10%, say)

Required:a) Explain why the Poisson distribution is appropriate to

invstigate this situation.b) Using the Poisson distribution, determine the probability of

accepting a batch Pa, containing p=2% defectives if the method is used.Determine Pa, for p = 0%, 2%, 5%, 10%, 15%

QUESTION TWOA woven cloth is liable to contain faults and is subjected to an inspection procedure. Any fault has a probability of 0.7 that it will be detected by the procedure, independent of whether any other fault is detected or not.

Required:a) If a piece of cloth contains three faults, A, B and C,

i) Calculate the probability that A and C are detected, but that B is undetected;

ii) Calculate the probability that any two of A, B and C be detected, the other fault being undetected;

iii) State the relationship between your answers to parts (i) and (ii) and give reasons for this.

b) Suppose now that, in addition to the inspection procedure given above, there is a secondary check which has a probability of 0.6 of detecting each fault missed by the first inspection procedure. This probability of 0.6 applies independently to each and every fault undetected by the first procedure.

i) Calculate the probability that a piece of cloth with one fault has this fault undetected by both the inspection procedure and the secondary check;


215 Lesson Five

ii) Calculate the probability that a piece of cloth with two faults has one of these faults detected by either the inspection procedure or the secondary check, and one fault undetected by both;

iii) Of the faults detected, what proportion are detected by the inspection procedure and what proportion by the secondary check?


Probability 216

QUESTION THREEA company has three production sections S1,S2 and S3 which contribute 40%,35% and 25%, respectively, to total output. The following percentages of faulty units have been observed:

S1 2% (0.02)S2 3% (0.03)S3 4% (0.04)

There is a final check before output is dispatched. Calculate the probability that a unit found faulty at this check has come from section 1, S1

QUESTION FOURAssuming a Binomial Distribution what is the probability of a salesman making 0,1,2,3,4,5 or 6 sales in 6 visits if the probability of making a sale on a visit is 0.3?

Do not use tables for this question.

QUESTION FIVERecords show that 60% of students pass their examinations at first attempt. Using the normal approximation to the binomial, calculate the probability that at least 65% of a group of 200 students will pass at the first attempt.

QUESTION SIXA batch of 5000 electric lamps has a mean life of 1000 hours and a standard deviation of 75 hours. Assume a normal distribution.

a) How many lamps will fail before 900 hours?b) How many lamps will fail between 950 and 1000 hours?c) What proportion of lamps will fail before 925 hours?d) Given the same mean life, what would the standard

deviation have to be to ensure that no more than 20% of lamps fail before 916 hours?


217 Lesson Six

LESSON SIX

Sampling and Estimation- Sampling techniques- Central limit theorem- Sampling distribution of statistical parameters- Test of hypothesis

6.1 Methods of Samplinga . Random or probability sampling methods

they includei. Simple random samplingii. Stratified samplingiii. Systematic samplingiv. Multi stage sampling

b. Non random probability sampling methods these consist of

i. Judgment samplingii. Quota samplingiii. Cluster sampling

Simple Random SamplingThis refers to the sampling technique in which each and every item of the population is given an equal chance of being included in the sample. Since selection of items in the sample depends entirely on chance, this method is also called chance selection or representative sampling.It is assumed that if the sample is chosen at random and if the size of the sample is sufficiently large, it will represent all groups in the populationRandom sampling is of 2 types; sampling with replacement and sampling without replacementSampling is said to be with replacement when from a finite population a sampling unit is drawn observed and then returned to the population before another unit is drawn. The population in this case remains the same and a sampling unit might be selected more than onceIf on the other hand a sampling unit is chosen and not retuned to the population after it has been observed the sampling is said to be without replacement.Random samples may be selected by the help of lottery method or table of random numbers (such as tippet’s table of random numbers, fischer and Yates numbers or Kendall and Babington Smith numbers.)

Stratified samplingIn this case the population is divided into groups in such a way that units within each group are as similar as possible in a process called stratification. The groups are called strata. Simple random samples from


Sampling and Estimation 218

each of the strata are collected and combined into a simple. This technique of collecting a sample from a population is called stratified sampling. Stratification may be by age, occupation income group e.t.c.


219 Lesson Six

Systematic SamplingThis sampling is a part of simple random sampling in ascending or descending orders. In systematic sampling a sample is drawn according to some predetermined object. Suppose a population consists of 1000 units, then every tenth, 20th or 50th item is selected. This method is very easy and economical. It also saves a lot of time

Multistage samplingThis is similar to stratified sampling except division is done on geographical/location basis, e.g. a country can be divided into provinces and then survey is done in 4 towns in each province. This helps to cut traveling costs for a surveyor.

Cluster SamplingThis is where a few geographical regions e.g. a location, town or village are selected at random and say every single household or shop in that area is interviewed. This again cuts on costs.

Judgment SamplingHere the interviewer selects whom to interview believing that their view is more fundamental since they might be directly affected e.g. to find out effects of public transport one may chose to interview only people who don’t own cars and travel frequently to work.

6.2 THE CENTRAL LIMIT THEOREMThe theory was introduced by De Moivre and according to it; if we select a large number of simple random samples, say from any population and determine the mean of each sample, the distribution of these sample means will tend to be described by the normal probability distribution with a mean µ and variance σ2/n. This is true even if the population itself is not normal distribution. Or the sampling distribution of sample means approaches to a normal distribution irrespective of the distribution of population from where the sample is taken and approximation to the normal distribution becomes increasingly close with increase in sample sizes

Types of distribution

Population distributionIt refers to the distribution of the individual values of population. Its mean is denoted by ‘µ’

Sample distributionIt is the distribution of the individual values of a single sample. Its mean is generally written as “ ”. it is not usually the same as µ

Distribution of Sample Means or sampling distributionA sample of size n is taken from the parent population and mean of the sample is calculated. This is repeated for a number of samples so that we



have a distribution of sample means, which approaches a normal distribution.

Standard errors of the meanThe series of sample means , , …….. is normally distributed or nearly so (according to the central limit theorem). It can be described by its mean and its standard deviation. This standard deviation is known as the standard error.

Standard error of the mean =

Note: this formula is satisfactory for larger samples and a large population i.e. n > 30 and n > 5% of N.- The word ‘error’ is in place of ‘deviation’ to emphasize that variation

among sample means is due to sampling errors.- The smaller the standard error the greator the precision of the

sample value.

6.3 Statistical inferenceIt is the process of drawing conclusions about attributes of a population based upon information contained in a sample (taken from the population).It is divided into estimation of parameters and testing of hypothesis. Symbols for statistic of population parameters are as follows.

Sample Statistic

Population Parameter

Arithmetic mean µStandard deviation s σNumber of items n N

Statistical estimationIt is the procedure of using statistic to estimate a population parameterIt is divided into point estimation (where an estimate of a population parameter is given by a single number) and interval estimation (where an estimate of a population is given by a range in which the parameter may be considered to lie) e.g. a bus meant to take a class of 100 students (population N) for trip has a limit to the maximum weight of 600kg of which it can carry, the teacher realizes he has to find out the weight of the class but without enough time to weigh everyone he picks 25 students selected at random (sample n = 25). These students are weighed and their average weight recorded as 64kg ( - mean of a sample) with a standard deviation (s), now using this the teacher intends to estimate the average weight of the whole class (µ – population mean) by using the statistical parameters standard deviation (s), and mean of the sample ( ).


221 Lesson Six

Characteristic of a good estimator (i) Unbiased: where the expected value of the statistic is equal

to the population parameter e.g. if the expected mean of a sample is equal to the population mean

(ii) Consistency: where an estimator yields values more closely approaching the population parameter as the sample increases

(iii) Efficiency: where the estimator has smaller variance on repeated sampling.

(iv) Sufficiency: where an estimator uses all the information available in the data concerning a parameter

Confidence IntervalThe interval estimate or a ‘confidence interval’ consists of a range (an upper confidence limit and lower confidence limit) within which we are confident that a population parameter lies and we assign a probability that this interval contains the true population valueThe confidence limits are the outer limits to a confidence interval. Confidence interval is the interval between the confidence limits. The higher the confidence level the greater the confidence interval. For exampleA normal distribution has the following characteristic

i. Sample mean ± 1.960 σ includes 95% of the populationii. Sample mean ± 2.575 σ includes 99% of the population

1. LARGE SAMPLESThese are samples that contain a sample size greater than 30(i.e. n>30)

(a) Estimation of population meanHere we assume that if we take a large sample from a population then the mean of the population is very close to the mean of the sampleSteps to follow to estimate the population mean includes

i. Take a random sample of n items where (n>30)ii. Compute sample mean ( ) and standard deviation (S)iii. Compute the standard error of the mean by using the following

formularS =

where S = Standard error of meanS = standard deviation of the samplen = sample size

iv. Choose a confidence level e.g. 95% or 99%v. Estimate the population mean as under

Population mean µ = ± (appropriate number) ×S‘Appropriate number’ means confidence level e.g. at 95% confidence level is 1.96 this number is usually denoted by Z and is obtained from the normal tables.



ExampleThe quality department of a wire manufacturing company periodically selects a sample of wire specimens in order to test for breaking strength. Past experience has shown that the breaking strengths of a certain type of wire are normally distributed with standard deviation of 200 kg. A random sample of 64 specimens gave a mean of 6200 kgs. Find out the population mean at 95% level of confidence

SolutionPopulation mean = ± 1.96 SNote that sample size is alredy n > 30 whereas s and are given thus step i), ii) and iv) are provided.Here: = 6200 kgs

S = = = 25

Population mean = 6200 ± 1.96(25)= 6200 ± 49= 6151 to 6249

At 95% level of confidence, population mean will be in between 6151 and 6249

FINITE POPULATION CORRECTION FACTOR (FPCF)If a given population is relatively of small size and sample size is more than 5% of the population then the standard error should be adjusted by multiplying it by the finite population correction factor

FPCF is given by =

where N = population sizen = sample size

ExampleA manager wants an estimate of sales of salesmen in his company. A random sample 100 out of 500 salesmen is selected and average sales are found to be Shs. 75,000. if a sample standard deviation is Shs. 15000 then find out the population mean at 99% level of confidence

SolutionHere N = 500, n = 100, = 75000 and S = 15000Now Standard error of mean

= S = x

= x


223 Lesson Six

= x

= (0.895)

S = 1342.50 at 99% level of confidence

Population mean = ± 2.58 S=shs 75000 ± 2.58(1342.50)=shs 75000 ± 3464= Shs 71536 to 78464

b) Estimation of difference between two meansWe know that the standard error of a sample is given by the value of the standard deviation (σ)divided by the square root of the number of items in the sample ( ).But, when given two samples, the standard errors is given by

=

Also note that we do estimate the interval not from the mean but from the difference between the two sample means i.e. .The appropriate number of confidence level does not changeThus the confidence interval is given by;

± Confidence level = ± Z

ExampleGiven two samples A and B of 100 and 400 items respectively, they have the means = 7 ad = 10 and standard deviations of 2 and 3 respectively. Construct confidence interval at 70% confidence level?

SolutionSample A B

= 7 = 10n1 = 100 n2 = 400S1 = 2 S2 = 3

The standard error of the samples A and B is given by

=

= =

= ¼ = 0.25



At 70% confidence level, then appropriate number is equal to 1.04 (as read from the normal tables)

= 7 – 10 = - 3 = 3We take the absolute value of the difference between the means e.g. the value of = absolute value of X i.e. a positive value of X.Confidence interval is therefore given by

= 3± 1.04 (0.25 ) From the normal tables a z value of 1.04 gives a value of 0.7.

= 3± 0.26

= 3.26 and 2.974

Thus 2.974 ≤ X ≤ 3.26

Example 2A comparison of the wearing out quality of two types of tyres was obtained by road testing. Samples of 100 tyres were collected. The miles traveled until wear out were recorded and the results given were as followsTyres T1 T2Mean = 26400 miles = 25000 miles Variance S2

1= 1440000 miles S22= 1960000 miles

Find a confidence interval at the confidence level of 70%

Solution = 26400 = 25000

Difference between the two means = (26400 – 25000)

= 1,400Again we take the absolute value of the difference between the two meansWe calculate the standard error as follows

=

=

= 184.4Confidence level at 70% is read from the normal tables as 1.04 (Z = 1.04).Thus the confidence interval is calculated as follows

= 1400 ± (1.04) (184.4)


225 Lesson Six

= 1400 ± 191.77

or (1400 – 191.77) to (1400 + 191.77)

1,208.23 ≤ X ≤ 1591.77

c) Estimation of population proportionsThis type of estimation applies at the times when information cannot be given as a mean or as a measure but only as a fraction or percentageThe sampling theory stipulates that if repeated large random samples are taken from a population, the sample proportion “p’ will be normally distributed with mean equal to the population proportion and standard error equal to

Sp = = Standard error for sampling of

population proportionsWhere n is the sample size and q = 1 – p.The procedure for estimating a proportion is similar to that for estimating a mean, we only have a different formula for calculating standard.

Example 1In a sample of 800 candidates, 560 were male. Estimate the population proportion at 95% confidence level.

SolutionHere

Sample proportion (P) = = 0.70

q = 1 – p = 1 – 0.70 = 0.30n = 800

=

Sp = 0.016

population proportion = P ± 1.96 Sp where 1.96 = Z.= 0.70 ± 1.96 (0.016)

= 0.70 ± 0.03

= 0.67 to 0.73

= between 67% to 73%

Example 2



A sample of 600 accounts was taken to test the accuracy of posting and balancing of accounts where in 45 mistakes were found. Find out the population proportion. Use 99% level of confidence

Solution Here

n = 600; p = = 0.075

q = 1 – 0.075 = 0.925

Sp = =

= 0.011

Population proportion = P ± 2.58 (Sp)= 0.075 ± 2.58 (0.011)

= 0.075 ± 0.028

= 0.047 to 0.10

= between 4.7% to 10%

d) Estimation of difference between population proportions Let the two proportions be given by P1 and P2, respectivelyThen the difference (absolute) between the two proportions is given by (P1 – P2)The standard error is given by

= where p = and q = 1 - p

Then given the confidence level, the confidence interval between the two population proportions is given by

(P1 – P2) ± Confidence level

= (P1 – P2) ± Z

Where P = always remember to convert P1 & P2 to P.

2. SMALL SAMPLES(a) Estimation of population mean If the sample size is small (n<30) the arithmetic mean of small samples are not normally distributed. In such circumstances, students t distribution must be used to estimate the population mean.


227 Lesson Six

In this case Population mean µ = ±

= Sample mean

=

S = standard deviation of samples = for small samples.

n = sample sizev = n – 1 degrees of freedom.The value of t is obtained from students t distribution tables for the required confidence level

ExampleA random sample of 12 items is taken and is found to have a mean weight of 50 grams and a standard deviation of 9 gramsWhat is the mean weight of population

a) with 95% confidenceb) with 99% confidence

Solution

S = 9; v = n – 1 = 12 – 1 = 11;

µ = x’ ±

At 95% confidence level

µ = 50 ± 2.262

= 50 ± 5.72 grams

Therefore we can state with 95% confidence that the population mean is between 44.28 and 55.72 gramsAt 99% confidence level

µ = 50 ± 3.25

= 50 ± 8.07 grams

Therefore we can state with 99% confidence that the population mean is between 41.93 and 58.07 grams

Note: To use the t distribution tables it is important to find the degrees of freedom (v = n – 1). In the example above v = 12 – 1 = 11



From the tables we find that at 95% confidence level against 11 and under 0.05, the value of t = 2.201

6.4 Hypothesis Testing

Definition- A hypothesis is a claim or an opinion about an item or issue. Therefore it has to be tested statistically in order to establish whether it is correct or not correct- Whenever testing an hypothesis, one must fully understand the 2 basic hypothesis to be tested namely

i. The null hypothesis (H0)ii. The alternative hypothesis(H1)

The null hypothesisThis is the hypothesis being tested, the belief of a certain characteristic e.g. Kenya Bureau of Standards (KBS) may walk to a sugar making company with an intention of confirming that the 2kgs bags of sugar produced are actually 2kgs and not less, they conduct hypothesis testing with the null hypothesis being: H0 = each bag weighs 2kgs. The testing will set out to confirm this or to refute it.

The alternative hypothesisWhile formulating a null hypothesis we also consider the fact that the belief might be found to be untrue hence we will reject it. We therefore formulate an alternative hypothesis which is a contradiction to the null hypothesis, thus when we reject the null hypothesis we accept the alternative hypothesis.In our example the alternative hypothesis would be

H1 = each bag does not weigh 2kg

Acceptance and rejection regionsAll possible values which a test statistic may either assume consistency with the null hypothesis (acceptance region) or lead to the rejection of the null hypothesis (rejection region or critical region)The values which separate the rejection region from the acceptance region are called critical values

Type I and type II errorsWhile testing hypothesis (H0) and deciding to either accept or reject a null hypothesis, there are four possible occurrences.a) Acceptance of a true hypothesis (correct decision) – accepting the null

hypothesis and it happens to be the correct decision. Note that statistics does not give absolute information, thus its conclusion could be wrong only that the probability of it being right are high.

b) Rejection of a false hypothesis (correct decision).c) Rejection of a true hypothesis – (incorrect decision) – this is called

type I error, with probability = α.d) Acceptance of a false hypothesis – (incorrect decision) – this is called

type II error, with probability = β.


229 Lesson Six

Levels of significanceA level of significance is a probability value which is used when conducting tests of hypothesis. A level of significance is basically the probability of one making an incorrect decision after the statistical testing has been done. Usually such probability used are very small e.g. 1% or 5%

0.5000 0.4900

1% provision for errors

0 Critical value

NB: If the standardized value of the mean is less than –1.65 we reject the null hypothesis (H0) and accept the alternative Hypothesis (H1) but if the standardized value of the mean is more than –1.65 we accept the null hypothesis and reject the alternative hypothesis

The above sketch graph and level of significance are applicable when the sample mean is < (i.e. less than the population mean)


0.45

5% = 0.05

Crititical value = -1.65

0

Critical region

229


The following is used when sample mean > population mean

Acceptance region

Critical region (rejection region)

5% = 0.05

0 Z = 1.65 (critical value)

NB: If the sample mean standardized value < 1.65, we accept the null hypothesis but reject the alternative. If the sample mean value > 1.65 we reject the null hypothesis and accept the alternative hypothesis The above sketch is normally used when the sample mean given is greater than the population mean

Accept null hyp( reject Alternative hyp)

Reject null hyp (accept alt hyp) Reject null hyp (accept alt hyp)

0.05% = 0.05 0.495 0.495 0.5% = 0.05

-2.58 +2.58

NB: if the standardized value of the sample mean is between –2.58 and +2.58 accept the null hypothesis but otherwise reject it and therefore accept the alternative hypothesis


231 Lesson Six

TWO TAILED TESTSA two tailed test is normally used in statistical work(tests of significance) e.g. if a complaint lodged by the client is about a product not meeting certain specifications i.e. the item will generate a complaint if its measurements are below the lower tolerance limit or above the upper tolerance limit

Region of acceptance for

H0

Critical region Critical region

15cm 17 ½ cm

NB: Alternative hypothesis is usually rejected if the standardized value of the sample mean lies beyond the tolerance limits (15cm and 17 ½ cm).

ONE TAILED TESTThis is a test where the alternative hypothesis (H1:) is only concerned with one of the tails of the distribution e.g. to test a business complaint if the complaint is above the measurements of item being shorter than is required.E.g. a manufacturer of a given brand of bread may state that the average weight of the bread is 500 gms but if a consumer takes a sample and weighs each of the pieces of bread and happens to have a mean of 450 gms he will definitely complain about the bread which is underweight. The statistical analysis to be done will concentrate on the left tail of the normal distribution in which one will have to establish whether 450 gms being less than 500g is statistically significant. Such a test therefore is referred to as one tailed test.



left

On the other hand the test may compuliate on the right hand tail of the normal distribution when this happens the major complaint is likely to do with oversize items bought. Therefore the test is known as one tailed as the focus is on one end of the normal distribution.

Number of standard errors

Two tailed test

One tailed test

5% level of significance

1.96 1.65

1% level of significance

2.58 2.33

HYPOTHESIS TESTING PROCEDUREWhenever a business complaint comes up there is a recommended procedure for conducting a statistical test. The purpose of such a test is to establish whether the null hypothesis or alternative hypothesis is to be accepted.The following are steps normally adopted

1. Statement of the null and alternative hypothesis2. Statement of the level of significance to be used.3. Statement about the test statistic i.e. what is to be tested e.g. the

sample mean, sample proportion, difference between sample means or sample proportions

4. Type of test whether two tailed or one tailed.5. Statement on critical values using the appropriate level of

significance6. Standardizing the test statistic7. Conclusion showing whether to accept or reject the null

hypothesis


233 Lesson Six

STANDARD HYPOTHESIS TESTSIn principal, we can test the significance of any statistic related to any probability distribution. However we will be interested in a few standard cases. The sample statistics mean, proportion and variance, are related to the normal, t, F, and chi squared distributionsThus



1. Normal testTest a sample mean ( ) against a population mean (µ) (where samples size n > 30 and population variance σ2 is known) and sample proportion, P(where sample size np >5 and nq >5 since in this case the normal distribution can be used to approximate the binomial distribution

2. t testTests a sample mean ( ) against a population mean and especially where the population variance is unknown and n < 30.

3. Variance ratio test or f testIt is used to compare population variances and it is used with samples of any size drawn from normal populations.

4. Chi squared testIt can be used to test the association between attributes or the goodness of fit of an observed frequency distribution to a standard distribution

Example 1A certain NGO carried out a survey in a certain community in order to establish the average at which the girls are married. The results of the survey indicated that the marriage age for the girls is 19 yearsIn order to establish the validity of the mean marital age, a sample of 50 women was interviewed and the average age indicated that they got married at the age of 16 years. However the different ages at which they were married differed with the standard deviation of 2.1yearsThe sample data indicates that the marital age is less 19 years. Is this conclusion true or not ?

RequiredConduct a statistical test to either support the above conclusion drawn from the sample statistics i.e. the marriage age is less than 19 years, use a level of significance of 5%

Solution1. Null hypothesis

H0: μ (mean marital age) = 19 yearsAlternative hypothesis H1: μ (mean marital age) < 19 years

2. The level of significance is 5%3. The test statistics is the sample mean age, = 16 years4. The critical value of the one tailed test (one tailed because the

alternative hypothesis is an inequality) at 5% level of significance is –1.65


235 Lesson Six

Acceptance region

Rejection region

- 1.65 0

5. The standardizes value of the sample mean is

Z = where =

Where, = Sample meanµ = Population meanS = sample standard deviationn = sample sizez = standard value (as per computation)

The standard value Z must fall within the acceptance region for us to accept the null hypothesis. Thus it must be > - 1.65 otherwise we accept the alternative hypothesis.

Z = = - 10.1

6. Since –10.1 < -1.65, we reject the null hypothesis but accept the alternative hypothesis at 5% level of significance i.e. the marriage age in this community is significantly lower than 19 years

Example 2A foreign company which manufactures electric bulbs has assured its customers that the lifespan of the bulbs is 28 month with a standard deviation of 4monthsRecently the company embarked on a quality improvement research for their product. After the research using new technology, a sample of 70 bulbs was tested and they gave a mean lifespan of 30.2 months



Does this justify the research undertaken? Use 1% level of significance to conduct a statistical test in order to establish the truth about the above question.Testing procedure

1. Null hypothesis H0: µ = 28Alternative hypothesis H1: µ > 28

2. The level of significance is 1% (one tailed test)3. The test statistics is the sample mean age, x’ = 30.2 4. The critical value of the one tailed test at 5% level of significance

is + 2.33

0.4900

1% = 0.01

2.335. The standardized value of the sample mean is

Z = = = 4.6

6. Since 4.6 > 2.33, we reject the null hypothesis but accept the alternative hypothesis at 1% level of significance i.e. the new sample mean life span is statistically significant higher than the population meanTherefore the research undertaken was worth while or justified

Example 3A construction firm has placed an order that they require a consignment of wires which have a mean length of 10.5 meters with a standard deviation of 1.7 mThe company which produces the wires delivered 90 wires, which had a mean length of 9.2 m., The construction company rejected the consignment on the grounds that they were different from the order placed.

Required Conduct a statistical test to indicate whether you support or not support the action taken by the construction company at 5% level of significance.


237 Lesson Six

SolutionNull hypothesis µ = 10.5 mAlternative hypothesis µ ≠ 10.5 m Level of significance be 5%The test statistics is the sample mean = 9.2m The critical value of the two tailed test at 5% level of significance is ± 1.96 (two tailed test).



- 1.96 +1.96The standardized value of the test Z =

Z = = = - 7.25

Since 7.25 < 1.96, reject the null hypothesis but accept the alternative hypothesis at 5% level of significance i.e. the sample mean is statistically different from the consignment ordered by the construction company. Therefore support the action taken by the construction company

TESTING THE DIFFERENCE BETWEEN TWO SAMPLE MEANS (LARGE SAMPLES)A large sample is defined as one which contains 30 or more items (n≥30) Where n is the sample sizeIn a business those involved are constantly observant about the standards or specifications of the item which they sell e.g. a trader may receive a batch of items at one time and another batch at a later time at the end he may have concluded that the two samples are different in certain specifications e.g. mean weight mean lifespan, mean length e.t.c. further it may become necessary to establish whether the observed differences are statistically significant or not. If the differences are statistically significant then it means that such differences must be explained i.e. there are known causes but if they are not statistically significant then it means that the difference observed have no known causes and are mainly due to chanceIf the differences are established to be statistically significant then it implies that the complaints, which necessitated that kind of test, are justifiedLet X1 and X2 be any two samples whose sizes are n1 and n2 and mean 1

and 2. Standard deviation S1 and S2 respectively. In order to test the


239 Lesson Six

difference between the two sample means, we apply the following formulas

Z = where =

Example 1An agronomist was interested in the particular fertilizer yield output. He planted maize on 50 equal pieces of land and the mean harvest obtained later was 60 bags per plot with a standard deviation of 1.5 bags. The crops grew under natural circumstances and conditions without the soil being treated with any fertilizer. The same agronomist carried out an alternative experiment where he picked 60 plots in the same area and planted the same plant of maize but a fertilizer was applied on these plots. After the harvest it was established that the mean harvest was 63 bags per plot with a standard deviation of 1.3 bags

RequiredConduct a statistical test in order to establish whether there was a significant difference between the mean harvest under the two types of field conditions. Use 5% level of significance.

Solution

H0 : µ1 = µ2

H1 : µ1 ≠ µ2

Critical values of the two tailed test at 5% level of significance are 1.96The standardized value of the difference between sample means is given by Z where

Z = where S =

Z =

= 11.11



- 1.96 0 +1.96Since 11.11 < -1.96, we reject the null hypothesis but accept the alternative hypothesis at 5% level of significance i.e. the difference between the sample mean harvest is statistically significant. This implies that the fertilizer had a positive effect on the harvest of maizeNote: You don’t have to illustrate your solution with a diagram.


241 Lesson Six

Example 2An observation was made about reading abilities of males and females. The observation lead to a conclusion that females are faster readers than males. The observation was based on the times taken by both females and males when reading out a list of names during graduation ceremonies.In order to investigate into the observation and the consequent conclusion a sample of 200 men were given lists to read. On average each man took 63 seconds with a standard deviation of 4 secondsA sample of 250 women were also taken and asked to read the same list of names. It was found that they on average took 62 seconds with a standard deviation of 1 second.

RequiredBy conducting a statistical hypothesis testing at 1% level of significance establish whether the sample data obtained does support earlier observation or not

Solution H0: µ1 = µ2 H1: µ1 ≠ µ2 Critical values of the two tailed test is at 1% level of significance is 2.58.

Z =

Z = = 3.45

Acceptance region

Rejection region

- 2.58 0 +2.58 +3.45Since 3.45 > 2.33 reject the null hypothesis but accept the alternative hypothesis at 1% level of significance i.e. there is a significant



difference between the reading speed of Males and females, thus females are actually faster readers.

TEST OF HYPOTHESIS ON PROPORTIONSThis follows a similar method to the one for means exept that the standard error used in this case:

Sp =

Z score is calculated as, Z = Where P = Proportion found

in the sample.Π – the hypothetical

proportion.

ExampleA member of parliament (MP) claims that in his constituency only 50% of the total youth population lacks university education. A local media company wanted to acertain that claim thus they conducted a survey taking a sample of 400 youths, of these 54% lacked university education.

Required:At 5% level of significance confirm if the MP’s claim is wrong.

Solution.Note: This is a two tailed tests since we wish to test the hypothesis that

the hypothesis is different (≠) and not against a specific alternative hypothesis e.g. < less than or > more than.

H0 : π = 50% of all youth in the constituency lack university education.

H1 : π ≠ 50% of all youth in the constituency lack university education.

Sp = = = 0.025

Z = = 1.6

at 5% level of significance for a two-tailored test the critical value is 1.96 since calculated Z value < tabulated value (1.96).

i.e. 1.6 < 1.96 we accept the null hypothesis.Thus the MP’s claim is accurate.

HYPOTHESIS TESTING OF THE DIFFERENCE BETWEEN PROPORTIONS

Example


243 Lesson Six

Ken industrial manufacturers have produced a perfume known as “fianchetto.” In order to test its popularity in the market, the manufacturer carried a random survey in Back rank city where 10,000 consumers were interviewed after which 7,200 showed preference. The manufacturer also moved to area Rook town where he interviewed 12,000 consumers out of which 1,0000 showed preference for the product.

RequiredDesign a statistical test and hence use it to advise the manufacturer regarding the differences in the proportion, at 5% level of significance.

Solution H0 : π1 = π2

H1 : π1 ≠ π2

The critical value for this two tailed test at 5% level of significance = 1.96.

Now Z =

But since the null hypothesis is π1 = π2, the second part of the numerator disappear i.e. π1 - π2 = 0 which will always be the case at this level.

Then Z =

Where;

Sample 1 Sample 2Sample size n1 =

10,000n2 =

12,000Sample proportion of success

P1 =0.72 P2 = 0.83

Population proportion of success.

Π1 Π2

Now =

Where P =

And q = 1 – pin our case



P =

=

= 0.78 q = 0.22

= 0.00894

Z = = 12.3

Since 12.3 > 1.96, we reject the null hypothesis but accept the alternative. the differences between the proportions are statistically significant. This implies that the perfume is much more popular in Rook town than in Back rank city.

HYPOTHESIS TESTING ABOUT THE DIFFERENCE BETWEEN TWO PROPORTIONSIs used to test the difference between the proportions of a given attribute found in two random samples.The null hypothesis is that there is no difference between the population proportions. It means two samples are from the same population.Hence H0 : π1 = π2

The best estimate of the standard error of the difference of P1 and P2 is given by pooling the samples and finding the pooled sample proportions (P) thus

P =

Standard error of difference between proportions

And Z =

Example In a random sample of 100 persons taken from village A, 60 are found to be consuming tea. In another sample of 200 persons taken from a village B, 100 persons are found to be consuming tea. Do the data reveal significant difference between the two villages so far as the habit of taking tea is concerned?


245 Lesson Six

SolutionLet us take the hypothesis that there is no significant difference between the two villages as far as the habit of taking tea is concerned i.e. π1 = π2

We are given P1 = 0.6; n1 = 100P2 = 0.5; n2 = 200

Appropriate statistic to be used here is given by

P =

=

= 0.53q = 1 – 0.53

= 0.47

=

=

= 0.0608

Z =

= 1.64

Since the computed value of Z is less than the critical value of Z = 1.96 at 5% level of significance therefore we accept the hypothesis and conclude that there is no significant difference in the habit of taking tea in the two villages A and B

t distribution (student’s t distribution) tests of hypothesis (test for small samples n < 30)For small samples n < 30, the method used in hypothesis testing is exactly similar to the one for large samples exept that t values are used from t distribution at a given degree of freedom v, instead of z score, the standard error Se statistic used is also different.Note that v = n – 1 for a single sample and n1 + n2 – 2 where two sample are involved.

a) Test of hypothesis about the population meanWhen the population standard deviation (S) is known then the t statistic is defined as

t = where

Follows the students t distribution with (n-1) d.f. where



= Sample meanμ = Hypothesis population meann = sample size

and S is the standard deviation of the sample calculated by the formula

S = for n < 30

If the calculated value of t exceeds the table value of t at a specified level of significance, the null hypothesis is rejected.

ExampleTen oil tins are taken at random from an automatic filling machine. The mean weight of the tins is 15.8 kg and the standard deviation is 0.5kg. Does the sample mean differ significantly from the intended weight of 16kgs. Use 5% level of significance.

Solution Given that n = 10; = 15.8; S = 0.50; μ = 16; v = 9

H0 : μ = 16H1 : μ ≠ 16

=

t =

=

= -1.25The table value for t for 9 d.f. at 5% level of significance is 2.26. the computed value of t is smaller than the table value of t. therefore, difference is insignificant and the null hypothesis is accepted.

b) Test of hypothesis about the difference between two meansThe t test can be used under two assumptions when testing hypothesis concerning the difference between the two means; that the two are normally distributed (or near normally distributed) populations and that the standard deviation of the two is the same or at any rate not significantly different.

Appropriate test statistic to be used is

t = at (n1 + n2 – 2) d.f.

The standard deviation is obtained by pooling the two sample standard deviation as shown below.


247 Lesson Six

Sp =

Where S1 and S2 are standard deviation for sample 1 & 2 respectively.

Now = and =

=

Alternatively = Sp

ExampleTwo different types of drugs A and B were tried on certain patients for increasing weights, 5 persons were given drug A and 7 persons were given drug B. the increase in weight (in pounds) is given below

Drug A 8 12 16 9 3Drug B 10 8 12 15 6 8 11

Do the two drugs differ significantly with regard to their effect in increasing weight? (Given that v= 10; t0.05 = 2.23)

SolutionH0 : μ1 = μ2

H1 : μ1 ≠ μ2

t =

Calculate for , and S

X1 X1 – (X1 – )2 X2 (X2 – ) (X2 – )2

8 -1 1 10 0 012 +3 9 8 -2 413 +4 16 12 +2 49 0 0 15 +5 253 -6 36 6 -4 16

8 -2 411 +1 1

ΣX1 = 45

Σ(X1– ) = 0

Σ (X1 – )2= 62

ΣX2= 70

Σ (X2 – ) = 0

Σ (X2– )2= 54



X1 = = = 9 X2 =

S1 = = 3.94 S2 =

Sp =

= 3.406

or 3.406

= 1.99

t = =

= 0.50

Now t0.05 (at v = 10) = 2.23 > 0.5

Thus we accept the null hypothesis.

Hence there is no significant difference in the efficacy of the two drugs in the matter of increasing weight

ExampleTwo salesmen A and B are working in a certain district. From a survey conducted by the head office, the following results were obtained. State whether there is any significant difference in the average sales between the two salesmen at 5% level of significance.


249 Lesson Six

A BNo. of sales 20 18Average sales in shs 170 205Standard deviation in shs 20 25

Solution H0 : μ1 = μ2

H1 : μ1 ≠ μ2

Where

Sp =

= Sp

Where: =170, = 205, n1 = 20, n2 = 18, S1 = 20, S2 = 25, V = 36

Sp =

= 22.5

= 7.31

t =

= 4.79t0.05(36) = 1.9 (Since d.f > 30 we use the normal tables)

The table value of t at 5% level of significance for 36 d.f. when d.f. >30, that t distribution is the same as normal distribution is 1.9. since the value computed value of t is more than the table value, we reject the null hypothesis. Thus, we conclude that there is significant difference in the average sales between the two salesmen

Testing the hypothesis equality of two variancesThe test for equality of two population variances is based on the variances in two independently selected random samples drawn from two normal populationsUnder the null hypothesis



F = Now under the H0 : it follows that

F = which is the test statistic.

Which follows F – distribution with V1 and V2 degrees of freedom. The larger sample variance is placed in the numerator and the smaller one in the denominatorIf the computed value of F exceeds the table value of F, we reject the null hypothesis i.e. the alternate hypothesis is accepted

ExampleIn one sample of observations the sum of the squares of the deviations of the sample values from sample mean was 120 and in the other sample of 12 observations it was 314. test whether the difference is significant at 5% level of significance

SolutionGiven that n1 = 10, n2 = 12, Σ(x1 – )2 = 120

Σ(x2 – )2 = 314

Let us take the null hypothesis that the two samples are drawn from the same normal population of equal variance

H0 : H1:

Applying F test i.e.

F =

=

=

=

since the numerator should be greater than denominator

F =


251 Lesson Six

The table value of F at 5% level of significance for V1 = 9 and V2 = 11. Since the calculated value of F is less than the table value, we accept the hypothesis. The samples may have been drawn from the two population having the same variances.

Chi square hypothesis tests (Non-parametric test)(X2)They include amongst others

i. Test for goodness of fitii. Test for independence of attributesiii. Test of homogeneityiv. Test for population variance

The Chi square test (χ2) is used when comparing an actual (observed) distribution with a hypothesized, or explained distribution.

It is given by; χ2 = Where O = Observed frequency

E = Expected frequencyThe computed value of χ2 is compared with that of tabulated χ2 for a given significance level and degrees of freedom.

i. Test for goodness of fitThis tests are used when we want to determine whether an actual sample distribution matches a known theoretical distributionThe null hypothesis usually states that the sample is drawn from the theoretical population distribution and the alternate hypothesis usually states that it is not.

Example Mr. Nguku carried out a survey of 320 families in Ateka district, each family had 5 children and they revealed the following distributionNo. of boys 5 4 3 2 1 0No. of girls 0 1 2 3 4 5No. of families 14 56 110 88 40 12

Is the result consistent with the hypothesis that male and female births are equally probable at 5% level of significance?

Solution If the distribution of gender is equally probable then the distribution conforms to a binomial distribution with probability P(X) = ½.Therefore

H0 = the observed number of boys conforms to a binomial distribution with P = ½

H1 = The observations do not conform to a binomial distribution.On the assumption that male and female births are equally probable the probability of a male birth is P = ½ . The expected number of families can be calculated by the use of binomial distribution. The probability of male births in a family of 5 is given by P(x) = 5cX Px q5-x (for x = 0, 1, 2, 3, 4, 5,)



= 5cX ( ½ )5 (Since P = q = ½ )To get the expected frequencies, multiply P(x) by the total number N = 320. The calculations are shown below in the tables

x P(x) Expected frequency = NP(x)

0 5c0 ( ½ )5 = 320 × = 101 5c1 ( ½ )5 = 320 × = 502 5c2 ( ½ )5 = 320 × = 1003 5c3 ( ½ )5 = 320 × = 1004 5c4 ( ½ )5 = 320 × = 505 5c5 ( ½ )5 = 320 × = 10

Arranging observed and expected frequencies in the following table and calculating x2

O E (O – E) 2 (O – E) 2 /E14 10 16 1.6056 50 16 0.72110 100 100 1.0088 100 144 1.4440 50 100 2.0012 10 4 0.40

Σ(0 – E) 2 /E = 7.16

χ2 =

= 7.16

The table of χ2 for V = 6 – 1 = 5 at 5% level of significance is 11.07. The computed value of χ2 = 7.16 is less than the table value. Therefore the hypothesis is accepted. Thus it can be concluded that male and female births are equally probable.

ii) Test of independence of attributesThis test disclosed whether there is any association or relationship between two or more attributes or not. The following steps are required to perform the test of hypothesis.

1. The null and alternative hypothesis are set as follows H0: No association exists between the attributesH1: an association exists between the attributes


253 Lesson Six

2. Under H0 an expected frequency E corresponding to each cell in the contingency table is found by using the formula

E =

Where R = a row total, C = a column total and n = sample size

3. Based upon the observed values and corresponding expected frequencies the χ2 statistic is obtained using the formular

χ2 =

4. The characteristic of this distribution are defined by the number of degrees of freedom (d.f.) which is given by

d.f. = (r-1) (c-1), Where r is the number of rows and c is number of columns corresponding to a chosen level of significance, the critical value is found from the chi squared table

5. The calculated value of χ2 is compared with the tabulated value χ2 for (r-1) (c-1) degrees of freedom at a certain level of significance. If the computed value of χ2 is greater than the tabulated value, the null hypothesis of independence is rejected. Otherwise we accept it.

ExampleIn a sample of 200 people where a particular devise was selected, 100 were given a drug and the others were not given any drug. The results are as follows

Drug No drug TotalCured 65 55 120Not cured 35 45 80Total 100 100 200Test whether the drug will be effective or not, at 5% level of significance.

SolutionLet us take the null hypothesis that the drug is not effective in curing the disease.Applying the χ2 testThe expected cell frequencies are computed as follows

E11 = = = 60

E12 = = = 60

E21 = = = 40



E22 = = = 40

The table of expected frequencies is as follows60 60 12040 40 80100 100 200

O E (O – E) 2 (O – E) 2 /E65 60 25 0.41755 60 25 0.62535 40 25 0.41745 40 25 0.625

Σ(O – E) 2 /E = 2.084

Arranging the observed frequencies with their corresponding frequencies in the following table we get

χ2 =

= 2.084

V= (r –1) (c-1) = (2 – 1) (2 –1) = 1; = 3.841

The calculated value of χ2 is less than the table value. The hypothesis is accepted. Hence the drug is not effective in curing the disease.

Test of homogeneity It is concerned with the proposition that several populations are homogenous with respect to some characteristic of interest e.g. one may be interested in knowing if raw material available from several retailers are homogenous. A random sample is drawn from each of the population and the number in each of sample falling into each category is determined. The sample data is displayed in a contingency tableThe analytical procedure is the same as that discussed for the test of independence

ExampleA random sample of 400 persons was selected from each of three age groups and each person was asked to specify which types of TV programs be preferred. The results are shown in the following table

Type of programAge group A B C Total Under 30 120 30 50 20030 – 44 10 75 15 100


255 Lesson Six

45 and above 10 30 60 100Total 140 135 125 400Test the hypothesis that the populations are homogenous with respect to the types of television program they prefer, at 5% level of significance.

SolutionLet us take hypothesis that the populations are homogenous with respect to different types of television programs they preferApplying χ2 testO E (O – E) 2 (O – E) 2 /E120 70.00 2500.00 35.714310 35.00 625.00 17.857110 35.00 625.00 17.857130 67.50 1406.25 20.833375 33.75 1701.56 50.416630 33.75 14.06 0.416650 62.50 156.25 2.50015 31.25 264.06 8.449960 31.25 826.56 26.449

Σ(O – E) 2 /E = 180.4948

χ2 =

The table value of χ2 for 4d.f. at 5% level of significance is 9.488The calculated value of χ2 is greater than the table value. We reject the hypothesis and conclude that the populations are not homogenous with respect to the type of TV programs preferred, thus the different age groups vary in choice of TV programs.



SUMMARY OF FORMULAE IN HYPOTHESIS

Testing(a) Hypothesis testing of mean

For n>30

Z = Where at level of significance.

For n < 30

t = where

at n – 1 d.f level of significance

(b) Difference between means (Independent samples)For n > 30

Z =

Where

At = level of significanceFor n < 30

t = at n1 + n2 – 2 d.f

where

and

(c) Hypothesis testing of proportions

Z =

Where: Sp =

p = Proportion found in sampleq = 1 – p

= hypothetical proportion

(d) Difference between proportions


257 Lesson Six

Z =

Where:

p =

q = 1 – P(e) Chi-square test

X2 =

Where O = observed frequency

E = = expected frequency

(f) F – test (variance test)

F =

here the bigger value between the standard deviations make the numerator.




QUESTION ONEA firm purchases a very large quantity of metal off-cuts and wishes to know the average weight of an off-cut. A random sample of 625 off-cuts is weighed and it is found that the mean sample weight is 150 grams with a sample standard deviation of 30 grams. What is the estimate of the population mean and what is the standard error of the mean? What would be the standard error if the sample size was 1225?

QUESTION TWOA sample of 80 is drawn at random from a population of 800. The sample standard deviation was found to be 6 grams.- What is the finite population correction factor?- What is the approximation of the correction factor?- What is the standard error of the mean?

QUESTION THREEState the Central Limit Theorem

QUESTION FOURa) What is statistical inference? b) What is the purpose of estimation?c) What are the properties of good estimators? d) What is the standard error of the mean? e) What are confident limits? f) When is the Finite Population Correction Factor used? What is the

formula? g) How are population proportions estimated? h) What are the characteristics of the t distribution?

QUESTION FIVEA market research agency takes a sample of 1000 people and finds that 200 of them know of Brand X. After an advertising campaign a further sample of 1091 people is taken and it’s found that 240 know of Brand X.It is required to know if there has been an increase in the number of people having an awareness of Brand X at the 5% level.

QUESTION SIXThe monthly bonuses of two groups of salesmen are being investigated to see if there is a difference in the average bonus received. Random samples of 12 and 9 are taken from the two groups and it can be assumed that the bonuses in both groups are approximately normally distributed and that the standard deviations are about the same. The same level of significance is to be used.


259 Lesson Six

The sample results were

QUESTION SEVENTorch bulbs are packed in boxes of 5 and 100 boxes are selected randomly to test for the number of defectives

Number ofDefectives

Numberof boxes

Totaldefectives

0 40 01 37 372 17 343 5 154 1 45 0 0

100 90The number of any individual bulb being a reject is

and it is required to test at the 5% level whether the frequency of rejects conforms to a binomial distribution.

QUESTION EIGHTa) Define type I and type II errors.b) What is a two-tail test?c) What is the best estimate of the population standard deviation when

the two samples are taken

QUESTION NINEExpress Packets guarantee 95%of their deliveries are on time. In a recent week 80 deliveries were made and 6 were late and the management says that, at the 95%level there has been a significant improvement in deliveries.

Can the MD’s statement be supported?If not, at what level of confidence can it be supported?A batch of weighing machines has been purchased and one machine is selected at random for testing. Ten weighing tests have been conducted and the errors found are noted as follows:

Test Errors (gms)1 4.62 8.23 2.14 6.35 5.0


n1=12 n2=9

x1=£1060 x2=£970

s1=£63 s2=£76

259


6 3.67 1.48 4.19 7.010 4.5

The purchasing manager has previously accepted machines with a mean error of 3.8 gms and asserts that these tests are below standard.Test the assertions at 5% level.

Compare your answers with those given in lesson 9 of the study pack


261 Lesson Six

COMPREHESIVE ASSIGNMENT THREEWork out these question for three hours (exam condition) then hand



Time allowed: Three hoursSECTION I

QUESTION ONEa) Explain what is meant by the following terms as used in statistical

inference:

i) Statistical hypothesis; (2 marks)

ii) Test of a hypothesis; (2 marks)

iii) Type I error; (2 marks)

iv) Type II error; (2 marks)

v) Level of significance (2 marks)

b) Cross Lines Group (CLG) has two factories in different parts of the country. Their Resources, including the labour force skills are regarded as identical and both factorieswere built at the same time.

A random sample of output data during a given period has been taken from each factory and converted to standard hours of output per employee. The data are given below:

Factory 1

42 50 43 39 41 49 52 41 46 48

Factory 2

39 45 36 42 52 37 43 41 40 39

You are given that for factory 1 mean = 45.1 and variance = 20.10 and that for factory 2 mean = 41.4 and variance = 21.16.

Required:i) Test the hypothesis that the mean of standard hours for employees in

the two factories is the same. (7

marks)



ii) Comment briefly on the conditions of the test and interpret the outcome. (3 marks)

(Total: 20 marks)

QUESTION TWOa) State clearly what is meant by two events being statistically

independent. (2 marks)

b) In a certain factory which employs 500 men, 2% of all employees have a minor accident in a given year. Of these, 30% had safety instructions whereas 80% of all employees had no safety instructions.

RequiredFind the probability of an employee being accident-free given that he had:i) No safety instructions (5

marks)ii) Safety instructions (5

marks)

c) An electric utility company has found that the weekly number of occurrences of lightning striking the transformers is a Poisson distribution with mean 0.4.

Required:i) The probability that no transformer will be struck in a week.

(3 marks)ii) The probability that at most two transformers will be struck in a week

(5 marks)(Total: 20

marks)

QUESTION THREEExplain the difference between the paired t-test and the two-sample t-test (4 marks)

Trendy Tyres Ltd. Has introduced a new brand of tyres which in their advertisements claim to be superior to their only competitor brand. The Roadmaster Tyres. The brand manager of Roadmaster Tyres disputes this claim which he says is an advertisement gimmick. The brand managers of the two companies agree to run a road test for the brands. Ten (10) saloon cars of uniform weight and identical specifications are to be used for the test. Each car is fitted with both brands of tyres: One brand at the front the other brand at the rear. The cars cover a distance of 5,000 kilometers and the trend wear is recorded as follows:

Trend tyres Roadmaster tyrescentimeters Centimeters


263 Lesson Six

1 1.08 1.122 1.06 1.093 1.24 1.164 1.20 1.245 1.17 1.236 1.21 1.257 1.18 1.208 1.10 1.159 1.22 1.1910 1.60 1.13

Required:i) Determine whether Trendy Tyres Ltd.’s claim is true using α = 0.01

(15 marks)ii) What are the assumptions you have made in (i) above?

(1 mark)(Total: 20

marks)

QUESTION FOURKenear Commercial bank Ltd. commissioned a research whose results indicated that automatic teller machines (ATM) reduces the cost of routine banking transactions.

Following this information, the bank installed an ATM facility at the premises of Joy Processing Company Ltd., which for the last several months has exclusively been, used by Joy’S 605 employees. Survey on the usage of the ATM facility by 100 of the employees in a month indicated the following:



Number of times ATM used

Frequency

0 201 322 203 134 105 5

Required:a) An estimate of the proportion of Joy’s employees who do not use the

ATM facility in a month(2 marks)

b)i) Determine the 95% confidence interval for the estimate in (a) above

(5 marks)ii) Can the bank be certain that at least 40% of Joy’s employees will

use the ATM facility?(1 mark)

c) The number of ATM transactions on average an employee of Joy makes per month

(3 marks)

d) Determine the 95% confidence interval of the mean number of transactions made by an employee in a month.

(6 marks)e) Is it possible that the population mean number of transactions is four?

Explain. (3 marks)(Total: 20

marks)

QUESTION FIVEState any five problems encountered in the construction of the consumer price index. (5 marks)

An investment analyst gathered the following data on the 91-day Treasury bill rates for the years 2003 and 2004.

Month Treasury bill rates (%)2003 2004

January 3.2 5.5February 3.0 5.2March 2.8 4.3April 2.5 3.6May 2.9 3.3June 3.4 2.7July 3.7 2.4August 4.0 2.0


265 Lesson Six

September 3.8 2.3October 4.2 2.8November 4.5 3.1December 5.1 3.7

The analyst would like to determine if on average there was a significant change in the Treasury bill rates over the two years.

Required:i) The mean and variance of the Treasury bill rates for each year.

(10 marks)ii) Determine if there is a significant difference in the average Treasury

bill rates (use a significance level of 1%)(5 marks)

Note:

(Total: 20 marks)

SECTION IIQUESTION SIXa) Describe the characteristics of the following distributions:

i) Binomial distribution (3 marks)

ii) Poisson distribution. (3 marks)

b) High Grade Meat Ltd. Produces beef sausages and sells them to various supermarket. In order to satisfy the industry’s requirement, the firm may only produces 0.2 per cent of sausages below a weight of 80 grammes. The sausage producing machine operates with a standard deviation of 0.5 grammes. The weights of the sausages are normally distributed.

The firm’s weekly output is 300,000 sausages and the sausage ingredients cost Sh.5.00 per 100 grammes. Sausages with weights in excess of 82 grammes require additional ingredients costing Sh.2.50 per sausage.

Required:i) The mean weight at which the machine should be set.

(4 marks)ii) The firm’s weekly cost of production

(10 marks)(Total: 20

marks)



QUESTION SEVENa) The past records of Salama Industries indicate that about 4 out of 10

of the company’s orders are for export. Further, their records indicate that 48 per cent of all orders are for export in one particular financial quarter. They expect to satisfy about 80 orders in the next financial quarter.

Required:Determine the probability that they will break their previous export record. (7 marks)Explain why you have used the approach you have chosen to solve part (i) above. (2 marks)

b) Gear Tyre Company has just developed a new steel-belted radial tyre that will be sold through a national chain of discount stores. Because the tyre is a new product, the company’s management believes that the mileage guarantee offered with the tyre will be an important factor in the consumer acceptance of the product. Before finalizing the tyre mileage guarantee policy, the actual road test with the tyres shows that the mean tyre mileage is μ = 36,500 kilometers and the standard deviation is σ = 5,000 kilometers. In addition, the data collected indicate that a normal distribution is a reasonable assumption.

Required:i) Gear Tyre Company will distribute the tyres if 20 per cent of the tyres

manufactured can be expected to last more than 40,000 kilometres. Should the company distributed the tyres?

(4 marks)

ii) The company will provide a discount on a new set of tyres if the mileage on the original tyres does not exceed the mileage stated on the guarantee.

What should the guarantee mileage be if the company wants no more

than 10% of the tyres to be eligible for the discount?(4 marks)

c) Explain briefly some of the advantages of the standard normal distribution. (3 marks)

(Total: 20 marks)

QUESTION EIGHTa) Explain the following terms used in statistical inference:

i) Null hypothesis (2 marks)


267 Lesson Six

ii) Parametric test (2 marks)

iii) Coefficient of correlation (2 marksiv) Rank correlation coefficient (2

marks)

b) State four areas that the chi-square distribution is used(4 marks)

c) In an analysis of the results of telecommunication students, the examining board classified the results as either credit, pass or discontinued. Further, the board analyzed the students’ method of study which was either full-time, part-time or private. An employee of the board cross-classified the examination results and the method of study of 300 students. He then computed a test statistic of 42.28

Required:i) State the null and alternative hypotheses that should be tested.

(4 marks)ii) What conclusion can be drawn from the results of the data? (use α

= 0.05) (4 marks)(Total: 20

marks)


Lesson Eight 268

LESSON SEVEN

Decision Theory- Decision theory- Decision trees and sequential decisions- Game theory

7.1 Decision Theory

Types of decisionsThere are many types of decision making

1. Decision making under uncertaintyThese refer to situations where more than one outcome can result from any single decision

2. Decision making under certainty Whenever there exists only one outcome for a decision we are dealing with this category e.g. linear programming, transportation assignment and sequencing e.t.c.

3. Decision making using prior dataIt occurs whenever it is possible to use past experience (prior data) to develop probabilities for the occurrence of each data

4. Decision making without prior dataNo past experience exists that can be used to derive outcome probabilities in this case the decision maker uses his/her subjective estimates of probabilities for various outcomes

Decision making under uncertaintySeveral methods are used to make decision in circumstances where only the pay offs are known and the likelihood of each state of nature are known

a) Maximin MethodThis criteria is based on the “conservative approach’ to assume that the worst possible is going to happen. The decision maker considers each strategy and locates the minimum pay off for each and then selects that alternative which maximizes the minimum payoff

Illustration Rank the products A B and C applying the Maximin rule using the following payoff table showing potential profits and losses which are expected to arise from launching these three products in three market conditions


269 Decesion Theory

(see table 1 below)


Lesson Eight 270

Pay off table in £ 000’sBoom condition

Steady state Recession Mini profits row minima

Product A +8 1 -10 -10Product B -2 +6 +12 -2Product C +16 0 -26 -26

Table 1Ranking the MAXIMIN rule = BAC

b) MAXIMAX methodThis method is based on ‘extreme optimism’ the decision maker selects that particular strategy which corresponds to the maximum of the maximum pay off for each strategy

IllustrationUsing the above example

Max. profits row maxima Product A +8Product B +12Product C +16

Ranking using the MAXIMAX method = CBA

c) MINIMAX regret methodThis method assumes that the decision maker will experience ‘regret’ after he has made the decision and the events have occurred. The decision maker selects the alternative which minimizes the maximum possible regret.

IllustrationRegret table in £ 000’s

Boom condition

Steady state Recession

Mini regret row maxima

Product A 8 5 22 22Product B 18 0 0 18Product C 0 6 38 38

A regret table (table 2) is constructed based on the pay off table. The regret is the ‘opportunity loss’ from taking one decision given that a certain contingency occurs in our example whether there is boom steady state or recessionThe ranking using MINIMAX regret method = BAC

d) The expected monetary value methodThe expected pay off (profit) associated with a given combination of act and event is obtained by multiplying the pay off for that act and event combination by the probability of occurrence of the given event. The


271 Decesion Theory

expected monetary value (EMV) of an act is the sum of all expected conditional profits associated with that act

ExampleA manager has a choice between i. A risky contract promising shs 7 million with probability 0.6 and

shs 4 million with probability 0.4 and ii. A diversified portfolio consisting of two contracts with

independent outcomes each promising Shs 3.5 million with probability 0.6 and shs 2 million with probability 0.4

Can you arrive at the decision using EMV method?

SolutionThe conditional payoff table for the problem may be constructed as below.

(Shillings in millions)Event Ei

Probability (Ei)

Conditional pay offs decision

Expected pay off decision

(i) Contract (ii)

Portfolio(iii)

Contract (i) x (ii)

Portfolio (i) x (iii)

Ei 0.6 7 3.5 4.2 2.1E2 0.4 4 2 1.6 0.8

EMV 5.8 2.9

Using the EMV method the manager must go in for the risky contract which will yield him a higher expected monetary value of shs 5.8 million

e) Expected opportunity loss (EOL) methodThis method is aimed at minimizing the expected opportunity loss (OEL). The decision maker chooses the strategy with the minimum expected opportunity loss

f) The Hurwiz methodThis method was the concept of coefficient of optimism (or pessimism) introduced by L. Hurwicz. The decision maker takes into account both the maximum and minimum pay off for each alternative and assigns them weights according to his degree of optimism (or pessimism). The alternative which maximizes the sum of these weighted payoffs is then selected

g) The Laplace methodThis method uses all the information by assigning equal probabilities to the possible payoffs for each action and then selecting that alternative which corresponds to the maximum expected pay off

ExampleA company is considering investing in one of three investment opportunities A, B and C under certain economic conditions. The payoff matrix for this situation is economic condition


Lesson Eight 272

Investment opportunities

1 £ 2 £ 3 £

A 5000 7000 3000B -2000 10000 6000C 4000 4000 4000

Determine the best investment opportunity using the following criteriai. Maximin ii. Maximaxiii. Minimaxiv. Hurwicz (Alpha = 0.3

SolutionEconomic condition

Investment opportunities

1 £ 2 £ 3 £ Minimum £

Maximum £

A 5000 7000 3000 3000 7000B -2000 10000 6000 -2000 10000C 4000 4000 4000 4000 4000

i. Using the Maximin rule Highest minimum = £ 4000Choose investment C

ii. Using the Maximax rule Highest maximum = £ 10000Choose investment B

iii. Minimax Regret rule

1 2 3 Maximum regret

A 0 3000 3000 3000B 7000 0 0 7000C 1000 6000 2000 6000

Choose the minimum of the maximum regret i.e. £3000Choose investment Aiv. Hurwicz rule: expected values

For A (7000 x 0.3) + (3000 x 0.7) = 2100 + 2100 = £4200For B (10000 x 0.3) + (-2000 x 0.7) = 3000 + 1400 = £ 1600For C (4000 x 0.3) + (4000 x 0.7) = 1200 + 2800 = £ 4000Best outcome is £ 4200 choose investment A

Value of perfect informationIt relates to the amount that we would pay for an item of information that would enable us to forecast the exact conditions of the market and act accordingly.


273 Decesion Theory

The expected value of perfect information EVPI is the expected outcome with perfect information minus the expected outcome without perfect information namely the maximum EMV

ExampleFrom table 1 above and given that the probabilities are Boom 0.6, steady state 0.3 and recession 0.1 then When conditions of the market are; boom launch product C: profit = 16When conditions of the market are; steady state launch product B: profit = 6When conditions of the market are; recession launch product B: profit = 12The expected profit with perfect information will be

(16 x 0.6) + (6 x 0.3) + (12 x 0.1) = 12.6our expected profit choosing product C is 7the maximum price that we would pay for perfect information is 12.6 – 7 = 5.6

7.2 DECISION TREES AND SUB SEQUENTIAL DECISIONSA decision tree is a graphic display of various decision alternatives and the sequence of events as if they were branches of a tree.

- The symbol and indicates the decision point and the situation of uncertainty or event respectively. The node depicted by a square is a decision node while outcome nodes are depicted by a circle.

- Decision nodes: points where choices exist between alternatives and managerial decisions is made based on estimates and calculations of the returns expected.

- Outome nodes are points where the events depend on probabilities

Illustration of a tree diagram Event

111

event B1

ACT E1 D2 B2 112

A1 E2 C1

D1 D3

A2 C2 121


131

122

273

Lesson Eight 274

For example 111 represents the payoff of the act event combination A1 – E1 – B1When probabilities of various events are known they are written along the corresponding branches. Joint probabilities are obtained by multiplying the probabilities along the branches

ExampleKauzi Agro mills ltd (KAM) is considering whether to enter a very competitive market. In case KAM decided to enter this market it must either install a new forging process or pay overtime wages to the entire workers. In either case, the market entry could result in

i. high salesii. medium salesiii. low salesiv. no sales

a) Construct an appropriate tree diagram b) Suppose the management of KAM has estimated that if they

enter the market there is a 60% chance of their stakeholders approving the installation of the new forge. (this means that there is a 40% chance of using overtime) a random sample of the current market structure reveals that KAM has a 40% chance of achieving high sales, a 30% chance of achieving medium sales, a 20% chance of achieving low sales and a 10% chance of achieving no sales. Construct the appropriate probability tree diagram and determine the joint probabilities for various branches

c) Market analysts of KAM have indicated that a high level of sales will yield shs 1,000,000 profit; a medium level of sales will result in a shs 600000 profit a low level of sales will result in a shs 200000 profit and a no sales level will cause KAM a loss of shs 500000 apart from the cost of any equipment. Entering the market will require a cash outlay of either shs 300000 to purchase and install a forge or shs 10000 for overtime expenses should the second option be selected.Draw the appropriate decision tree diagram

Solutiona) The tree diagram for this problem is illustrated as follows

The 1st stage of drawing a tree diagram is to show all decision points and outcome points done from left to right, concentrate first on the logic of the problem and on probabilities or values involved. This is called forward pass.The resultant is the figure below:


Outcome/

0

1

2

3

4

5

6

7

8

9

10

11

12

Act Act/event

Install forge

Use overtime

High sales

Medium sales

Low sales

No sales

High sales

Medium sales

Low sales

No sales

stop

Do not enter market

275 Decesion Theory

The entire sample space of act event choices is available to KAM are summarized in the table shown belowPath Summary of alternative Act event sequence0 – 1 – 3 – 5 Enter market, install forge, high sales0 – 1 – 3 – 6 Enter market, install forge, medium sales0 – 1 – 3 – 7 Enter market, install forge, low sales0 – 1 – 3 – 8 Enter market, install forge, no sales0 – 1 – 4 – 9 Enter market, use overtime, high sales0 – 1 – 4 –10 Enter market, use overtime, medium sales0 – 1 – 4 – 11 Enter market, use overtime, low sales 0 – 1 – 4 – 12 Enter market, use overtime, no sales0 – 2 Do not enter the market

b) The appropriate probability tree is shown in the figure below. The alternatives available to the management of KAM are identified. The joint probabilities are the result of the path sequence that is followed. For example, the sequence ‘enter market install forge, low sales’ yields (0.6) (0.2) = 0.12 = probability to install forge and get low sales.


Tree diagram

0

1

2

3

4

Use overtime

Don’t enter market

Enter Market

0.6

0.4

Install forge

(10,000)

(300,000)

0.4

0.3

0.2

0.1

0.4

0.30.2

0.1

HS = 0.24 = 1,000,000

MS = 0.18 = 600,000

LS = 0.12 = 200,000

NS = 0.06 = - 500,000

HS = 0.16 = 1,000,000MS = 0.12 = 600,000

LS = 0.08 = 200,000NS = 0.04 = - 500,000

Pay offs

275

Lesson Eight 276

(c) The overall decision is determined after analysis of the expected values at various points so the correct decision (with the highest expected value is made. The stage is worked from right to left and is known as the backward pass.- The expected value for a decision is the highest pay off

value where as the E.V for an outcome is the summation of probability x pay off value of each branch. In both cases any expenditure incurred due to the selection of the said option is deducted.

- In our caseNode 3 =

- 300,000E.V. = 615,000 – 300,000 = 315,000

Node 4 =

- 10,000E.V. = 615,000 – 10,000 = 605,000

Node 1 = (0.6 × 315,000) + (0.4 × 605,000)E.V. = 431,000

Node 0 = The highest of (0;431,000)Since not entering the market has a 0 expected

value= 431,000 = thus the decision should be to

enter the market.

This is represented as below in a tree diagram.


277 Decesion Theory

BAYES THEORY AND DECISION TREESIt makes an application of Bayes’ Theorem to solve typical decision problems. This is examined a lot so it is important to clearly understand it.

Example:Magana Creations is a company producing Ruy Lopez brand of cars. It is contemplating launching a new model, the Guioco. There are several possibilities that could be opted for.- Continue producing Ruy Lopez which has profits declining at 10% per

annum on a compounding basis. Last year its profit was Shs. 60,000.- Launch Guioco without any prior market research. If sales are high

annual profit is put at Shs. 90,000 with a probability which from past data is put at 0.7. Low sales have 0.3 probability and estimated profit of Shs. 30,000.

- Launch Guioco with prior market research costing Shs. 30,000 the market research will indicate whether future sales are likely to be ‘good’ or ‘bad.’ If the research indicates ‘good’ then the management will spend Shs. 35,000 more on capital equipment and this will increase annual profits to Shs. 100,000 if sales are actually high. If however sales are actually low, annual profits will drop to Shs. 25,000. Should market research indicate ‘good’ and management not


0

1

3

4

Use overtime

Don’t enter market

Enter Market

EV = 431,000

Install forge

EV = 605,000

EV = 315,000

0.4

0.3

0.2

0.1

0.4

0.30.2

0.1

1,000,000

600,000

200,000

- 500,000

1,000,000600,000

200,000

- 500,000

0

0.6

0.4

277

Lesson Eight 278

spend more on promotion the profit levels will be as for 2nd scenario above.

- If the research indicate ‘bad’ then the management will scale down their expectations to give annual profit of Shs. 50,000 when sales are actually low, but because of capacity constrints if sales are high profit will be Shs. 70,000.Past history of the market research company indicated the following results.

Actual salesHigh Low

Predicted sales level

Good 0.8* 0.1Bad 0.2 0.9

*When actual sales were high the market research company had predicted good sales level 80% of the time.

Required:Use a time horizon of 6 years to indicate to the management of the company which option theory should adopt (Ignore the time value of money).

Solution(a) First draw the decision tree diagram


2 A

B

C

1

D

E

Ruy Lopez(option 1)

GUIOCO(option 2)

Market Research (option 3) Good

Bad

Extra 35,000

No extra

High 0.7

Low 0.3

P(H|G)

P(L|G)0.95

0.05P(H|G)

0.95P(L|G)0.05P(H|B)0.34P(L|

B)0.66

60,000 (declining)

90,000

30,000

100,000

25,000

90,000

30,000

70,000

50,000

279 Decesion Theory

Computations; note how probability figures are arrived at.- The decision tree dictates that the following probabilities need to be

calculated.

P(G)P(B)


For market research

279

Lesson Eight 280

P(H|G)P(L|G)P(H|B)P(L|B)

P(G|H) = 0.8P(B|H) = 0.2P(G|L) = 0.1P(B|L) = 0.9P(H) = 0.7P(L) = 0.3

Good P(G&H) = P(H) × P(G|H)

0.7 × 0.8 = 0.56

P(G&L) = P(L) × P(G|L)

0.3 × 0.1 = 0.03Bad B&H = P(H) × P(B|H)

0.7 × 0.2 = 0.14P(B&L) = P(L) ×

P(B|L)0.3 × 0.9 = 0.27

High 0.7 Low 0.3

P(G) = P(G and H) + P(G and L)= 0.56 + 0.03 = 0.59

P(B) = P(B and H) + P(B and L)= 0.14 + 0.27 = 0.41Note that P(G) + P(B) = 0.59 + 0.41 = 1.00

From Bayes’ rule;

Evaluating financial outcome:Option 1:Last year Shs. 60,000 profits

Year Shs.1 = 60,000 × 0.9

=54,000.

02 = 60,000 × 0.92 48,000.


For sales

Give

281 Decesion Theory

= 03 = 60,000 × 0.93

=43,740.

04 = 60,000 × 0.94

=39,366.

05 = 60,000 × 0.95

=35,429.

56 = 60,000 × 0.96

= 31,88

6.5253,02

2.0Option 2

Expected value of GiuocoNode (A): 0.7(90,000 × 6) + 0.3(30,000 × 6)= 378,000 + 54,000 = Shs. 432,000

Note that the figures a multiplied by 6 to account for the 6 years.

Option 3Expected value of market research

Node (B): 0.95(100,000 × 6) + 0.05(25,000 × 6)= 570,000 + 7,500 = Shs. 577,500Deduct Shs. 35,000 for extensions

= 542,500.

Node (C): 0.95(90,000 × 6) + 0.05(30,000 × 6)= 513,000 + 9,000 = Shs. 522,000

Node 1: Compare B and CB is higher, thus = 542,000.

Node (D): 0.34(70,000 × 6) + 0.66(50,000 × 6)142,800 + 198,000 = Shs. 340,800

Node 2: Shs. 340,800 or 0 – no launch

Node (E): 0.59 × 542,500 + 0.41 × 340,800320,075 + 139,728 = Shs. 459,803Less market research expenditure459,803 – 30,000 = Shs. 429,803

Node 2: Final decision summaryOption 1 EMV = 253,022Option 2 EMV = 432,000Option 3 EMV = 429,803

Therefore we chose option 2 since it has the highest EMV.

Advantages of decision trees


Lesson Eight 282

1. it clearly brings out implicit assumptions and calculations for all to see question and revise

2. it is easy to understand

Disadvantages1. it assumes that the utility of money is linear with money 2. it is complicated by introduction of more variables and decision

alternatives3. it is complicated by presence of interdependent alternatives and

dependent variables

7.3 Game TheoryGame theory is used to determine the optimum strategy in a competitive situationWhen two or more competitors are engaged in making decisions, it may involve conflict of interest. In such a case the outcome depends not only upon an individuals action but also upon the action of others. Both competing sides face a similar problem. Hence game theory is a science of conflict Game theory does not concern itself with finding an optimum strategy but it helps to improve the decision process.Game theory has been used in business and industry to develop bidding tactics, pricing policies, advertising strategies, timing of the introduction of new models in the market e.t.c.

RULES OF GAME THEORYi. The number of competitors is finiteii. There is conflict of interests between the participantsiii. Each of these participants has available to him a finite set of

available courses of action i.e. choicesiv. The rules governing these choices are specified and known to all

playersWhile playing each player chooses a course of action from a list of choices available to him

v. the outcome of the game is affected by choices made by all of the players. The choices are to be made simultaneously so that no competitor knows his opponents choice until he is already committed to his own

vi. the outcome for all specific choices by all the players is known in advance and numerically definedWhen a competitive situation meets all these criteria above we call it a game

NOTE: only in a few real life competitive situation can game theory be applied because all the rules are difficult to apply at the same time to a given situation.

ExampleTwo players X and Y have two alternatives. They show their choices by pressing two types of buttons in front of them but they cannot see the


283 Decesion Theory

opponents move. It is assumed that both players have equal intelligence and both intend to win the game.This sort of simple game can be illustrated in tabular form as follows:

Player YButton R Button t

Player X Button m X wins 2 points X wins 3 pointsButton n Y wins 2 points X wins 1 point

The game is biased against Y because if player X presses button m he will always win. Hence Y will be forced to press button r to cut down his losses

Alternative examplePlayer Y

Button R Button tPlayer X Button m X wins 3 points Y wins 4 points

Button n Y wins 2 points X wins 1 point

In this case X will not be able to press button m all the time in order to win(or button n). similarly Y will not be able to press button r or button t all the time in order to win. In such a situation each player will exercise his choice for part of the time based on the probability


Lesson Eight 284

Standard conventions in game theoryConsider the following table

Y3 -4

X -2 1

X plays row I, Y plays columns I, X wins 3 pointsX plays row I, Y plays columns II, X looses 4 pointsX plays row II, Y plays columns I, X looses 2 pointsX plays row II, Y plays columns II, X wins 1 points

3, -4, -2, 1 are the known pay offs to X(X takes precedence over Y)here the game has been represented in the form of a matrix. When the games are expressed in this fashion the resulting matrix is commonly known as PAYOFF MATRIX

STRATEGYIt refers to a total pattern of choices employed by any player. Strategy could be pure or a mixed oneIn a pure strategy, player X will play one row all of the time or player Y will also play one of this columns all the time.In a mixed strategy, player X will play each of his rows a certain portion of the time and player Y will play each of his columns a certain portion of the time.

VALUE OF THE GAMEThe value of the game refers to the average pay off per play of the game over an extended period of time

Example

in this game player X will play his first row on each play of the game. Player y will have to play first column on each play of the game in order to minimize his loosesso this game is in favour of X and he wins 3 points on each play of the game.This game is a game of pure strategy and the value of the game is 3 points in favour of X

ExampleDetermine the optimum strategies for the two players X and Y and find the value of the game from the following pay off matrix


285 Decesion Theory

Strategy assume the worst and act accordinglyif X plays firstif X plays first with his row one then Y will play with his 2nd column to win 1 point similarly if X plays with his 2nd row then Y will play his 3rd column to win 7 points and if x plays with his 3rd row then Y will play his fourth column to win 9 pointsIn this game X cannot win so he should adopt first row strategy in order to minimize lossesThis decision rule is known as ‘maximum strategy’ i.e. X chooses the highest of these minimum pay offs

Using the same reasoning from the point of view of yIf Y plays with his 1st column, then X will play his 3rd row to win 4 pointsIf Y plays with his 2nd column, then X will play his 1st row to lose 1 pointIf Y plays with his 3rd column, then X will play his 1st row to win 4 pointsIf Y plays with his 4th column, then X will play his 1st row to win 2 points

Thus player Y will make the best of the situation by playing his 2nd

column which is a ‘Minimax strategy’This game is also a game of pure strategy and the value of the game is –1(win of 1 point per game to y) using matrix notation, the solution is shown below

In this case value of the game is –1Minimum of the column maximums is –1Maximum of the row is also –1i.e. X’s strategy is maximim strategy

Y’s strategy is Minimax strategy

Saddle PointThe saddle point in a pay off matrix is one which is the smallest value in its row and the largest value in its column. It is also known as equilibrium point in the theory of games.


Lesson Eight 286

Saddle point also gives the value of such a game. In a game having a saddle point, the optimum strategy for both players is to pay the row or column containing the saddle point. Note: if in a game there is no saddle point the players will resort to what is known as mixed strategies.

Mixed Strategies Example Find the optimum strategies and the value of the game from the following pay off matrix concerning two person game

In this game there is no saddle pointLet Q be the proportion of time player X spends playing his 1st row and 1-Q be the proportion of time player X spends playing his 2nd rowSimilarlyLet R be the proportion of time player Y spends playing his 1st column and 1-R be the proportion of time player Y spends playing his second rowThe following matrix shows this strategy

X’s strategy X will like to divide his play between his rows in such a way that his expected winning or loses when Y plays the 1st column will be equal to his expected winning or losses when y plays the second column

Column 1Points Proportion played Expected winnings1 Q Q5 1-Q 5(1-Q)

Total = Q + 5(1 –Q)Column 2

Points Proportion played Expected winnings4 Q 4Q3 1-Q 3(1-Q)

Total = 4Q + 3(1 –Q)Therefore Q + 5(1-Q) = 4Q +3(1-Q)Giving Q = and (1-Q) =


287 Decesion Theory

This means that player X should play his first row th of the time and his

second row th of the timeUsing the same reasoning 1×R + 4(1-R) = 5R +3(1-R)Giving R = and (1-R) = This means that player Y should divide his time between his first column and second column in the ratio 1:4

Short cut method of determining mixed matrices

Step ISubtract the smaller pay off in each row from the larger one and smaller pay off in each column from the larger one

Step IIInterchange each of these pairs of subtracted numbers found in step I

Thus player X plays his two rows in the ratio 2: 3And player Y plays his columns in the ratio 1:4This is the same result as calculated before

To determine the value of the game in mixed strategiesIn a simple 2 x 2 game without a saddle point, each players strategy consists of two probabilities denoting the portion of the time he spends on each of his rows or columns. Since each player plays a random pattern the probabilities are listed under

Pay off Strategies which produce this pay off

Joint probability


Lesson Eight 288

1 Row I column I

4 Row I column II

5 Row II column I

3 Row II column II

Expected value (or value of the game)Pay off Probability p(x) Expected value x

(p(x)1

4

5

3

Ƹx p(x) = 85/25 = 17/5 = 3.43.4 is the value of the game

DominanceDominated strategy is useful for reducing the size of the payoff tableRule of dominance

i. If all the elements in a column are greater than or equal to the corresponding elements in another column, then the column is dominated

ii. Similarly if all the elements in a row are less than or equal to the corresponding elements in another row, then the row is dominatedDominated rows and columns may be deleted which reduces the size of the gameNB always look for dominance and saddle points when solving a game

ExampleDetermine the optimum strategies and the value of the game from the following 2xm pay off matrix game for X and Y


289 Decesion Theory

In this columns I, II, and IV are dominated by columns III and V hence Y will not play these columnsSo the game is reduced to 2×2 matrix, hence this game can be solved using methods already discussed

GRAPHICAL METHODGraphical methods can be used in games with no saddle points and having pay off m×2 or 2×n matrixThe aim is to substitute a much simpler 2×2 matrix for the original m×2 or 2×m matrix

Example I Determine the optimum strategies and the value of the game from the following pay off matrix game.

Draw two vertical axes and plot two pay offs corresponding to each of the five columns. The pay off numbers in the first row are plotted on axis I and those in second row on axis II


Lesson Eight 290

Axis I Axis II 2 K 2

6 A 6 1 1

5 5 0 0

4 4 -1 L -1

3 B 3 -2 -2

2 2 -3 -3

1 1 -4 -4

0 0 -5 -5

-1 -1 -6 -6

-2 T -2 -7 -7

-3 K -3 -8 -8

-4 L -4 -9 -9Example I Example II

Thus the two pay off number 6 and 3 in the first column are shown respectively by point A on axis I and point B on axis IIOn the two intersecting lines at the very bottom thicken them from below to the point of intersection i.e. highest point on the boundary.The thick lines on the graph KT and LT meet at TThe two lines passing through T identify the two critical moves of Y which combined with X yield the following 2 × 2 matrix

The value of the game and the optimum strategies can be calculated using the methods described earlier

Example IIDetermine the optimum strategies and the value of the game from the following pay off matrix concerning two person 4 × 2 game


M

291 Decesion Theory

The method is similar to the previous example, except we thicken the line segments which binds the figure from the top and taken the lowest point on the boundaryThe segments KP, PM and ML drawn in thick lines bind the figure from the top and their lowest intersection M through which the two lines pass defines the following 2 × 2 matrix relevant to our purpose

The optimal strategies and the value of the game can now be calculated

Non Zero Sum GamesUntil recently there was no satisfactory theory either to explain how people should play non zero games or to describe how they actually play such gamesNigel Howard (1966) developed a method which describes how most people play non zero sum games involving any number of persons

ExampleEach individual farmer can maximize his own income by maximizing the amount of crops that he produces. When all farmers follow this policy the supply exceeds demand and the prices fall. On the other hand they can agree to reduce the production and keep the prices highThis creates a dilemma to the farmerThis is an example of a non zero sum gameSimilarly marketing problems are non zero sum games as elements of advertising come in. in such cases the market may be split in proportion to the money spent on advertising multiplied by an effectieness factor

Prisoners DilemmaIt is a type of non zero sum game and derives its name from the following storyThe district attorney has two bank robbers in separate cells and offers each a chance of confession. If one confesses and the other does not then the confessor gets two years and the other one ten years. If both confess they will get eight years each. If both refuse to confess there is only evidence to ensure convictions on a lesser charge and each will receive 5 years


Lesson Eight 292

Another exampleThe table below is a pay off matrix for two large companies A and B. initially they both have the same prices. Each consider cutting their prices to gain market share and hence improve profit

Corporation B

Corporation A

The entries in the pay off matrix indicate the order of preference of the players i.e. first A then B.We may suppose that if both player study the situation, they will both decide to play row I column I(3,3).

HoweverSuppose A’s reasoning is as followsIf B plays column I then I should play row 2 because I will increase my gain to 4In the same way B’s reasoning may be as followsIf A plays row I then I should play column 2 to get pay off 4 per play If both play 2(row 2 column 2) each two receives a pay off of 2 onlyIn the long run pay off forms a new equilibrium point because if either party departs from it without the other doing so he will be worse off before he departed from itGame theory seems to indicate that they should play (2,2) because it is an equilibrium point but this is not intuitively satisfying. On the other hand (3,3) is satisfying but does not appear to provide stability. Hence the dilemma.

Theory of MetagamesThis theory appears to describe how most people play non zero sum games involving a number of personsPrisoners dilemma is an example of this. The aim is to identify points at which players actually tend to stabilize their play in non zero sum games.This theory not only identifies equilibrium points missed by traditional game theory in games that have one or more such points but also does so in games in which traditional theory finds no such pointIts main aim is that each player is trying to maximize the minimum gain of his opponent

ADVANTAGES AND LIMITATIONS OF GAME THEORY


Maintain prices Decrease pricesmaintain prices 3,3 status quo 1 , 4 B gets market

share and profitDecrease prices

4, 1, A gains market share and profit

(2,2) Both retain market share but lose profit

293 Decesion Theory

Advantage Game theory helps us to learn how to approach and understand a conflict situation and to improve the decision making process

LIMITATIONS1. Businessmen do not have all the knowledge required by the theory of

games. Most often they do not know all the strategies available to them nor do they know all the strategies available to their rivals

2. there is a great deal of uncertainty. Hence we usually restrict ourselves to those games with known outcomes

3. The implications of the Minimax strategy is that the businessman minimizes the chance of maximum loss. For an ambitious business man, this strategy is very conservative

4. the techniques of solving games involving mixed strategies where pay off matrices are rather large is very complicated

5. in non zero sum games, mathematical solutions are not always possible. For example a reduction in the price of a commodity may increase overall demand. It is also not necessary that demand units will shift from one firm to another


Lesson Eight 294


QUESTION ONEAn Oil Company has recently acquired rights in a certain area to conduct surveys and test drillings to lead to lifting oil where it is found in commercially exploitable quantities. The area is already considered to have good potential for finding oil in potentially exploitable quantities. At the outset the company has the choice to conduct further geological tests or to carry out a drilling programme immediately. On the known conditions, the company estimates that there is a 70:30 chance of further tests showing a ‘success’.Whether the tests show the possibility of ultimate success or not, or even if no tests are undertaken at all, the company could still pursue its drilling programme or alternatively sell its rights to drill in the area. Thereafter, however, if it carries out its drilling programme, the likelihood of final success or failure is considered dependent on the foregoing stages. Thus:

If ‘successful’ tests have been carried out, the expectation of success in drilling is given as 80:20If test indicate ‘failure’ then the expectation of success in drilling is given as 20:80If no tests have been carried out at all then the expectation of success in drilling is given as 55:45Costs and revenues have been estimated for all possible outcomes and the net present value (NPV) of each is given below.

Outcome NPV

(£m)SUCCESS

With prior tests 100Without prior tests 120

FAILUREWith prior tests -50Without prior tests -40

SALE OF EXPLOITATION RIGHTS

Prior test show ‘success’ 65Prior test show ‘failure’ 15Without prior tests 45

Required.

a) Draw up a decision tree diagramb) Advise the company on the best course of action

QUESTION TWO


295 Decesion Theory

1. A construction company has a £1 million contract to complete a building by 31 March 1995, but is experiencing delays due to the complex design. The managers have to make a decision now whether to continue as at present, or to employ specialist-engineering consultants at a cost of £200000.

If the company continues as at present, it estimates there is only a 30% chance of completing the building on time, and that the delay could be one two or three months, with equal probability. If the building is late, there are penalties for each month’s delay (or part of a month).

The managers believe that if they employ specialist-engineering consultants, their chances of finishing the building on time will be trebled. But if the building is still late, it would only be one or two months late, with equal probability.

Requireda) To draw a tree diagram to represent this decision problem,

using squares for decision points, circles for random outcomes, and including probabilities revenues and penalties;

b) To analyse the tree using expected value techniques:c) To write a short report for the managers, with reasons and

comments, recommending which decision to make.

QUESTION THREEDefine minimax and maximax decision rules

QUESTION FOURA has two ammunition stores, one of which is twice as valuable as the other. B is an attacker who can destroy an undefended store but he can only attack one of them. A can only successfully defend one of them.

What would A do so as to maximize his return from the situation no matter what B may do?

QUESTION FIVEDetermine the optimum strategies and the value of the game for the following pay off matrix.

XY

1 2 -1-2 1 1 2 0 1

Compare your answers with those given in lesson 9 of the study pack


Lesson Eight 296

LESSON EIGHT

Operation Research- Linear programming- Transport and Assignments- Network Analysis

8.1 Linear programmingLinear programming is a technique of decision making used by managers to allocate limited resources eg machinery, raw materials and labor in order to minimize costs or maximize production. Decision variable are the amounts of each product to be made in a given time period. Linear programming assumes that the variable has a linear relationship.

Application of linear programming Production department to decide the quantity of pots to be produced subject to limited resources (constraints) eg labour, power, machine hours, raw materials etc.Marketing department: Allocation of salesmen to different sales regions subject to their expected performance.Human Resource: Scheduling personnel’s work hours and job description to either maximize production or minimize cost.

Steps in solving linear programming problems(problem formulation)1. Identify variables (eg product x and product y)2. Identify the objective (To maximize contribution or to minimize cost),

and write down its mathematical presentation in terms of variables.3. Identify the constraints (ie the limited resources shared among the

variables), and write down its mathematical representation in terms of variables.

4. write down the objectives and the constraints in terms of the variables.

These steps apply regardless of the number of the variables.

NOTE: If only two variables are involved, a graphical solution can be used otherwise for multivariable problems, an algebraic method is applied to find the solution.

Example 1:Long Castling Breweries manufactures two brands of beer, Benko lager and Benoni lager. Benko has a contribution of Sh.4 per unit and Benoni has a contribution of Sh.3 per unit. Benko requires 30 machine minutes and 30 labor minutes to manufacture a unit. Total available machine hours per day are 12hrs whereas total available labour hours per day are 14hrs.


297 Decesion Theory

Required:1. Formulate linear programming model.2. How much of each brand should Long Castling produce if it wishes to

maximize its daily contribution assuming that all the lager produced is sold.


Lesson Eight 298

Solution:1. Formulating a linear programming modelStep 1: Identifying variables:The variables here are the number of units of Benko and Benoni lager produced by Long castling breweries per day; we can represent them as:

X1= a unit of Benko lager.X2= a unit of Benoni lager.

Step 2: Identify the objective:Definition: An objective is the desired result i.e. optimization of a function dependent on decision variable and subject to some constraints.The objective of Long Castling Breweries is to maximize daily contribution. Objective function is the formula that will give us the total contribution in a day for both Benko lager and Benoni lager.The information above can thus be represented in a tabular form as:

PRODUCT Maximum available

(PER DAY) X1 X2 hours/dayMachine hours 0.5 0.33 12Labor hours 0.5 0.5 14Contribution 4 3

Objective function = 4X1 + 3X2

The objective is to maximize 4X1 + 3X2

Step 3: Identifying constraints (constraints formulation)Definition: Constraints are circumstances that govern achievement of an objective.Limitations must be quantified mathematically and they must be linear.For Long castling breweries we have limited machine hours (12hrs/day), which must be shared among production of Benko and Benoni lagers.Therefore production must be such that the numbers of machine hours required is less than or equal to 12 hours per day.

0.5X1+0.33X2 ≤ 12hrs

Similarly for labour hours we have:0.5X1+0.5X2 ≤ 14hrs

Non-Negativity: It is logical assumption to assume that the company cannot manufacture negative amounts of a product, thus it can only manufacture either zero product or more. Therefore we have: X1≥0

X2≥0 or X1, X2≥0

Thus the complete linear programming model is; Maximize 4X1 + 3X2

Subject to the constraints;


299 Decesion Theory

0.5X1 + 0.33X2 ≤ 120.5X1 + 0.5X2 ≤ 14X1, X2 ≥ 0

Solving linear programming problemsThe question requires us to optimize (in our case, maximize) the objective (the contribution function), or in simple terms we are required to solve the linear programming model.Solving linear programming model entails finding the values of variables that satisfy all inequalities simultaneously and optimize the objective.

Graphical solutionThis method is used to solve LP models in case where only two variables are involved. For more than two variables (multivariable) then the simplex technique (algebraic method may be used).Now in solving the problem above we first draw the axis, taking X1 to be the y axis and X2 to be the X axis.

X1

0X2

Next we plot the scales on each axis to approximate the scales to use them, we consider each constraint equation. We get the value of one of variables putting the other variable to be zero and by substituting the inequality ≤or ≥ with equality sign (=).

For: 0.5X1 + 0.33X2 ≤ 12When X1 = 0

0.5(0) + 0.33X2 = 120.33X2 = 12X2 = 12/0.33 = 36

Therefore point to plot is (36, 0) … Implying that when X2 = 36 then X1 = 0

When X2 = 00.5X1 + 0.33(0) = 120.5X1 = 12


Lesson Eight 300

X1 = 12/0.5 = 24Therefore the point is (0, 24)

For 0.5X1 + 0.5X2 ≤ 14When X1 = 0

0.5(0) + 0.5X2 = 140.5X2 = 14X2 = 14/0.5 = 28

Therefore the point is (28, 0)

When X2 = 00.5X1 + 0.5(0) = 140.5X1 = 14X1 = 14/0.5 = 28

Therefore the point is (0, 28)

Comparing these values we see that X2 ranges between 0 – 28, therefore we can have the graph plotted as:

Next draw each limitation (constraint) as separate line on the graph.For 0.5X1 + 0.33X2 ≤ 12The two points that represent this line are (36, 0) and (0, 24). This is plotted as a straight line from 36 on X2 axis to 24 on X1 axis.


301 Decesion Theory

Now including the Non-Negativity constraints since no negative product can be produced;X1 ≥ 0; x2 ≥ 0


Lesson Eight 302

We must now consider how to choose the production which will maximize contribution. This we do by plotting a line representing the objective function (4x1 + 3x2).First choose a convenient point inside the feasible region

eg X2(10) +3X1(20) = 40 +60 = Sh 100

All of the other product mixes that give a contribution of Sh.100 lies on the line:

100 = 4X1 + 3X2 .....................................................(i)<<This line is called a contribution line>>

Picking another point, say X2 =10 ad X1 = 20Its contribution value is SH 110, thus give a contribution line of

110 = 4X1 + 3X2.......................................................(ii)

Plotting these two contribution lines to our graph we get two parallel lines.

Until we reach the last feasible solution(s) before the line moves entirely out of the feasible region.


303 Decesion Theory

Point X is the last feasible solution. Coordinates of this point give a combination of the two lager’s production volumes that fetches the highest contribution.Coordinates of point X can be read from the graph, but for precision they are calculated by solving simultaneously the equations of the two lines that intersect at point X.The two constraints are called binding or limiting constraints. They are the resources being fully used thus preventing daily contribution from increasing further.Therefore to get point x we solve:

0.5X1 + 0.33X2 = 12 ….(i)0.5X1 + 0.5X2 = 14 ….(ii)

Since X is the intersection of these two constraints, solving by deducting (i) from (ii) we get

0.17X2=2X2 =11.76

And substituting X2 = 11.76 to equation (i) we get X1 = 16.24

Therefore 11.76 units of Benko lager and 16.24 units of Benoni lager need to be produced for maximum contribution.


Lesson Eight 304

Contribution = 4(16.24) + 3(11.76) = 100.24

Assumption made in linear programmingAssumptions that are made to solve these types of problems are that:. Proportionality: all activities in linear programming problems are

proportional to the level of decision variables. Divisibility: the solution to a linear programming problem does not

have to be an integer but for strictly whole number solutions, use integer programming.

Non-negativity: no decision variable can be negative. Additivity: the total of all activities in linear programming

problems are assumed to equal to the sum of individual activities.

Special cases in linear programming Infeasibility: This is when all constraints don’t satisfy a particular

point thus there is no feasible solution. Redundancy: A constraint is considered redundant if it does not

affect the feasible region. This happens in cases of excess resources since it does not limit attainment of the objective.

Multiple optimal solutions: This occurs when the objective function has the same slope as a binding constraint.

Minimizing problem

ExampleA manufacturing company has acquired new machine for producing product P at a rate of 25 units per hour with a 98% rate of efficiency. The company requires to produce atleast 1800 units of P per day. The 10 old machines that the company has, produce 15 units of P with a 95% efficiency.The cost of operating the new machine is Sh. 4 per hour and Sh. 3 per hour for the old ones. The cost incurred due to inefficiency is Sh. 2 per unit; It is government policy that at least 2 of the new machines must be indulged into production.The company wishes to optimally allocate the machines in order to minimize the total manufacturing cost if the total available hours for production in a day are 8 hours.

SolutionIn minimizing problems we use ≥ (greater or equal to) type inequality.

Step 1: Identifying variables.Since the problem requires us to appropriately allocate the machines in order to minimize the costs thus our variables are the new machines and the old machines, we can let;

X1 =new machinesX2=old machines

Step 2: Identify objectives:The objective is to minimize manufacturing costs.


305 Decesion Theory

Total manufacturing cost per machine=Operating cost + (inefficiency rate × number of units × cost of loss)Therefore cost for new machinesCost = 4+(0.02 × 25 × 2) = Sh. 5 per hour

=5 × 8= Sh. 40 per day]Similarly for old machines Cost = 3+(0.05×15×2) = Sh. 4.5 per hour =4.5×8=Sh. 36 per dayTherefore the objective function is to minimize 40X1 + 36X2

Step 3: Identifying constraint functions.X1 ≤ 8X2 ≤ 10

(25 × 8) X1 +(8 × 15) X2 ≥1800 (This can be simplified further as)200 X1 +120 X2 ≥1800 (dividing through by 40)5X1 +3 X2 ≥45X1 ≥2X1, X2 ≥0

Thus the LP model is:Minimize 40X1 + 36X2

Subject to: X1 ≤ 8X2 ≤ 105X1 +3 X2 ≥45X1 ≥2X1, X2 ≥0

Plotting this on a graph we get:


X1

X2

8

305

Lesson Eight 306

The line X1 ≥2 does not affect the feasible region (doesn’t cause reduction of the feasible region), this constraint doesn’t limit attainment of the objective, thus its known as a redundant constraint.Now picking a convenient point inside the feasible region, say (6, 10)

We get a total cost of 600 = (6(40)+10(36)) Thus the objective function line of 40X1 + 36X2 = 600

Moving this line parallel toward the origin to locate the last apex before the line completely fall off the feasible region, we get:

Point X is the point of optimal solution The binding constraints here are:

X1 ≤ 8And 5X1 +3 X2 ≥45Solving this to get coordinates of point X

5X1 +3 X2 = 45 (when X1 = 8)40 + 3 X2 = 45X2 = 5/3

Thus the solution is X1 = 8X2 = 5/3

Shadow or dual pricesDefinition: A shadow price or a dual price is the amount increase (or decrease) of the objective function when one more (or one less) of the binding constraints is made available.Consider example 1.


307 Decesion Theory

Maximize 4X1 + 3 X2

Subject to: 0.5X1 + 0.33X2 ≤ 12 (Machine hours)0.5X1 + 0.5X2 ≤ 14 (labor hours)

Starting with machine hours; let’s assume that one more machine hour is available (with labor hours remaining constant)We get:

0.5X1 + 0.33X2 = 130.5X1 + 0.5X2 = 14

Solving this simultaneously we get the values of X1 and X2 as0.17 X2 = 1X2 = 5.88X1 = 22.12

Thus the contribution is 4(22.12) + 3(5.88) = Sh.106.12

Comparing this with its original contribution of Sh.100.24 (see example 1) we see increasing machine hours by one unit has increased contribution by Sh.5.88, which is the shadow price per machine hour.Note: This figure is also arrived at if we assume that machine hours are reduced by 1 unit ie 12-1.Similarly assuming that one more labor hour is made available, then contribution change is:

0.5X1 + 0.33X2 = 12 0.5X1 + 0.5X2 = 15

Solving this simultaneously gives:0.17 X2 =3X2 = 17.65 X1 = 12.35

Which give a contribution of:4(12.35) + 3(17.65) = Sh.102.35

The contribution change is Sh.2.11 which is the shadow price per labor hour.

Note:The shadow prices apply in so far as the constraint is binding for example if more and more labor hours are available it will reach a point where labor hours are no longer scarce thus labor hours cease to be a binding constraint and its shadow price becomes a zero.(All non-binding constraints have zero shadow price). Logically its senseless to pay more to increase a resource, which is already abundant.

Interpretation of shadow pricesA shadow price of a binding constraint indicates to management how much extra contribution will be gained by increasing a unit of the scarce resource.In the example above Sh.2.11 is the shadow price for labor hours. This implies that management is ready to pay up to Sh.2.11 extra per hour for the extra hours i.e. say an employee is paid sh.5 per hour and one day he works for two hours extra (overtime), the management is prepared to pay up to sh.7.11 per hour for the two hours overtime worked.


Lesson Eight 308

Sensitivity AnalysisDefinition: Sensitivity analysis is the test of how certain changes in resources affect the optimal solution.

In sensitivity analysis we consider the effect of additional limiting or non-limiting constraints.We already know that adding more non-limiting constraints does not change the optimal solution.We also know that adding more binding constraints affects the objective function.It is very important for the management to know how much of a limit resource can be made available until it has no effect on the objective function (ie ceases to be a binding resource)

SIMPLEX METHODWhen analyzing linear programming problems with three or more variables the graphical method becomes enadequate, in such cases we employ simplex method . Simplex method is an algebraic procedure for solving systems of equations requiring optimization of the objective function..This method can be applied to any number of variables, the more they are the more complex it becomes to workout a solution on paper. Computer programs e.g. Tora are used to solve the most intricate problems. The first step in simplex method is conversion of inequalities to linear equations

ExampleConsider the linear problem.

MaximizeSubject to

Solution.1. to convert this problem to a system of linear equation, we

introduce slack variables to each constraint.

Subject to

where the structural variables are slack variables

2. we then place this information in a tabular form known as a tableu

Initial tableu


309 Decesion Theory

Solution

Variable

Products Slack Variables

Solution

Quantity

5 20 1 0 40010 15 0 1 450

Z 45 80 0 0 0

The tableu shows that

It also show that unused capacity is at maximum i.e. the value for the slack is 0.Afer several operations and when an optimal solution has been attained, these values will change to give an optimal feasible solution.

3. Select the column with the highest value of Z (i.e. 80), then devide the positive numbers in that column (i.e. the x2 column) into the quantity column.

chose the rowgiving the lowest result (in our case the row with x3 gives 20) and mark the element falling on the intersection of the selected row and selected column (i.e. 20: selected element)

4. Devide all the elements in the selected row by the value of the selected and change the solution variable to the heading of the identified column (from x3 to x2)

Thus the tableu will appear as follows.Solutio

n Variab

le


Solution

Quantity

1 0 2010 15 0 1 450

Z 45 80 0 0 0

5. next we conduct row operations that aim to reduce elements falling in the same column as the previously marked element to zero. These row operations may sometime necessitate multiplying or deviding the selected row with an arbitrary number.

Therefore:Row 2 – 15×Row 1


Lesson Eight 310

note that the aim was to attain the zero.

on replacing the new rows to the tableu we get2nd tableu.

Solution

Variable


Solution

Quantity

1 0 200 1 150

Z 25 0 -4 0 -1600

Since in the Z row under products column we still have values greator than zero, we conduct another operation.

Taking the column with a Z value of 25, we repeat the process in the same manner.

thus we pick the x4 row and mark the element , the row solution variable is changed to x1 and we devide the row by to convert the marked element to 1.Therefore; x1 1 0 = 24

Next we do the row operations


311 Decesion Theory

the tableu becomes, 3rd TableuSolutio

n Variab

le


Solution

Quantity

0 1 241 0 24

Z 0 0 -7 -4 -2200

This is the final tableu since the Z row has no values greator than zero thus we have the optimal solution.

Interpretation to maximise Z we need to produce 24 units x2 and 24 units of

x1, we obtain these values from the solutions quantity column

thus, Z = 24(45) + 24(80) = 3000

we have zero slack (unused quantities of constraints).Assume tableu 2 is the final tableu and let the constraint with variable x3 be labour hours and x4 be raw materials, the slack wouldhve been interpreted as:

o 150 units of raw materials were unusedo to maximize Z we produce 20units of x2 and none of x1.

The values represnts in Z row under slack vaiable column represents shadow prizes. Thus the shadow prize for the first constraint with x3 is 7 and the shadow prize for the second constraint with the vaiable x4 is 4.

INTERPRETATION OF COMPUTER GENERATED SOLUTION

ExampleMaximize

Subject to

The computer generated solution for this problem is as follows;Objective value = 71666.7Variable Value Obj. Coeff Obj Value

ContributionX1: Xtragrow 1666.7 25 4166.7X2: Youngrow 1500 20 30000


Lesson Eight 312

X3: Zupergrow 0 24 0Constraint RHS Slack-/

Surplus+1(<) 500 02(<) 1000 250-3(<) 800 316.67-4(<) 600 05(>) 1500 166.7+

Sensitivity AnalysisVariable Current obj

coeffMin obj

coeffMax Obj

CoeffReduced

costX1: Xtragrow

25 13.50 Infinity 0

X2: Youngrow

20 9.78 Infinity 0

X3: Zupergrow

24 -Infinity 31.67 7.67

Constraint Current RHS Min RHS Max RHS Dual price1(<) 500 450 975 83.332(<) 1000 750 Infinity 03(<) 800 483.3 Infinity 04(<) 600 0 800 505(>) 1500 -Infinity 1666.7 0

RequiredInterpret the data generated by the the computer.

Solution.Table 1:Objective value, is the solution to objective function (e.g the solution to this example is 71,436)The four columns of table 1 are to be interpreted as follows;

Variable: these are the variables of the model. In our example we have x1 = Xtragrow, x2 = Youngrow and x3 = Zupergrow

Value: this is value that the variables assume at optimal solution (to optimize the objective function one needs to produce this amounts of the variables). In our example we are required to produce 1,666.67 of x1 and 1,750 of x2 and none of x3

Objective coefficient: these are the coefficients of the objective function

Objective value contribution: this is the value contributed by each variable to the objective function (for x1=25×1,666.67), the total of this is equal to our objective value (i.e 41,666.67+35,000=76,666.67).

The 3 columns of the second part of table1 can be interpreted as follows; Constraints: this is constraints of the model representing the

limited resources.


313 Decesion Theory

RHS: the Right hand side value is the limiting value of the constraint. E.g for the first constraint the maximum amount of material A is 500 tons.

Slack/surplus: at optimal production not all the materials for some of the constrants will be fully utilized, slack is the amount of material that is left over after production. For constraint 1 and 4 no material remained, this also implies that these are the binding constraints i.e their adjustment directly affects the objective solution

Sensitivity AnalysisThis is the analysis of the effect of adjusting variables or constraint, whether te objective solution will be affected. How much of the objective coefficient (or the maximum available amount of a constraint) can be reduced or increased without affecting the objective solution.The columns of this table can be interpreted as follows;

Variable: as explained aboveCurrent objective coefficient: this is the value of coefficients of the

objective functionMinimum objective coefficient: this is how low the coefficient can be

reduced without affecting the optimal basis. The coefficient for x1 can be reduced from 25 to 13.50 (the prize for Xtragrow fall to as low as $13.50 from $25) but the optimal solution will remain the same

Maximum objective coefficient: this is how high the coefficient can be increased without affecting the optimal basis

Reduced cost: this is amount by which the coefficient of the variable has to be adjusted with for it to become a basic variable (included to the objective optimal solution). X1 and x2 have 0 reduced costs implying that they already make part of the optimal solution, x3 will require to be increased by 7.67 for it to make part of the basic variable.

The second part of the table is interpreted as follows; Constraints: as described above Current Rihgt Hand Side: the limiting value of the constraint. E.g

for the first constraint the maximum amount of material A is 500 tons.

Minimum RHS: the lowest the available amount of the constraint can be reduced without affecting the optimal basis

Maximum RHS: the highest the available amount of the constraint can be increased without affecting the optimal solution

Dual price: this is amount increase to the objective contribution due to a unit increase of the available constraint. Since there are


Lesson Eight 314

only 500 tons of material A if management decides to increase it by a unit to 501tons then the objective optimal solution will be 76,666.67+83.33.

TRANSPORTATIONA transportation problem deals with a number of sources of supply (e.g a manufacturing company, warehouse) and a number of destinations (e,g shops, houses). The usual objective is minimizing transportation costs of supplying items from a set of source points to a set of destinations. A major characteristic of this problem is the linearity requirement, i.e. transport cost fom one point to another must be clearly defined, if it will cost sh.50 to transport a bag from a warehouse to shop A then it will cost sh.250 to transport 5 bags.

Assumptions The model assumes a homogeneous commodity, one type of

commodity Total supply is equal to total demand

Example 164 chambers a computer support firm has three branches at different parts of the city, it receives orders for a total of 15 desktop computers from four customers. In total in the three branches there are 15 machines available. The management wish to minimise delivery costs by dispatching the computers from the appropriate branch for each customer.

Details of the availabilities, 'requirements, and transport costs per computer are given in the following table.

Table 1Custome

rCustome

rCustome

rCustome

r TotalA B C D

Computers 3 3 4 5 15transportatio

n costper unit

Branch X. 2 Sh.13 11 15 20Available Branch Y 6 Sh.17 14 12 13

Branch Z 7 Sh.18 18 15 12Total 15

SolutionStep 1 Make an initial feasible allocation of deliveries by selecting the

cheapest route first, and allocate as many as possible then the next cheapest and so on. The result of such an allocation is as follows.

Table 2Requirement

A B C D3 3 4 5

X 2 Units 2(1)


315 Decesion Theory

Available Y 6 Units 1(4) 1(3) 4(2)Z 7 Units 2(5) 5(2)

Note: the number in the table represent deliveries of computers and the number in the brackets (1), (2), etc represent the sequence in which they are inserted, lowest cost first i.e.

Sh.1. 2 units X → B sh.11/unit Total cost 222. 4 units Y → C sh.12/unit Total cost 48

5 units Z → D sh.12/unit Totals cost 603. The next lowest cost move which is feasible i.e. doesn’t exceed row

or column totals is 1 unit Y → B sh.14/unit 144. similarly the next lowest feasible allocation 1

unit Y→ A sh.17/unit 175. finally to fulfill the row /column totals 2 units Z →

A sh.18/unit __36197

Step 2. Check solution obtained to see if it represents the minimum cost possible. This is done by calculating ‘shadow costs’ (i.e. an imputed cost of not using a particular route) and comparing these with the real transport costs to see whether a change of allocation is desirable.

This is done as follows:Calculate a nominal 'dispatch' and 'reception' cost for each occupied cell by making an assumption that the transport cost per unit is capable of being split between dispatch and reception costs thus:


Lesson Eight 316

D(X) + R(B) = 11 D(Y) + R(A) = 17D(y) + R(B) = 14D(Y) + R(C) = 12D(Z) + R(A) = 18D(Z) + R(D) = 12

Where D(X), D(Y) and D(Z) represent Dispatch cost from depots X, Y and Z, and R(A) R(B), R(C) and R(D) represent Reception costs at customers A, B, C, D.By convention the first depot is assigned the value of zero i.e. D(X) = 0 and this value is substituted in the first equation and then all the other values can be obtained thus

R(A) = 14 D(X) = 0R(B) = 11 D(Y) = 3R(C) = 9 D(Z) = 4R(D) = 8

Using these values the shadow costs of the unoccupied cells can be calculated. The unoccupied cells are X : A, X : C, X : D, Y : D, Z : B, Z : C.

Shadowcosts

D(X) + R(A) = 0 + 14 = 14D(X) + R(C) = 0 + 9 = 9D(X) + R(D)

= 0 + 8 = 8D(Y) + R(D)

= 3 + 8 = 11D(Z) + R(B) = 4 + 11 = 15D(Z) + R(C) = 4 + 9 = 13

These computed 'shadow costs' are compared with the actual transport costs (from Tab- I), Where the actual costs are less than shadow costs, overall costs can be reduced by allocating units into that cell.

Actual Shadow + Cost increasecost - cost - Cost reduction

CellX:A 13 - 14 = -1X : C 15 - 9 = +6X : D 20 - 8 = + 12Y: D 13 - 11 = +2Z : B 18 - 15 = +3Z : C 15 - 13 = +2

The meaning of this is that total costs could be reduced by sh.1 for every unit that can be transferred into cell X : A. As there is a cost reduction that can be made the solution , Table 2 is not optimum.


317 Decesion Theory

Step 3: Make the maximum possible allocation of deliveries into the cell where actual costs are less than shadow costs using occupied cells i.e.

Cell X : A from Step 2, The number that can be allocated is governed by the need to keep within the row and column totals. This is done as follows:

Table 3Requirement

A B C D3 3 4 5

X 2 Units + 2 - Available Y 6 Units 1 - 1 + 4

Z 7 Units 2 5

Table 3 is a reproduction of Table 2 with a number of + and - inserted. These were inserted for the following reasons.Cell X : A + indicates a transfer in as indicated in Step 2Cell X : B - indicates a transfer out to maintain Row X total.Cell Y : B + indicates a transfer in to maintain Column B totalCell Y : A - indicates a transfer out to maintain Row Y and Column A totals.

The maximum number than can be transferred into Cell X : A is the lowest number in theMinus cells i.e. cells Y : A, and X : B which is 1 unit.Therefore 1 unit is transferred in the + and - sequence described above resulting in the following table

Table 4Requirement

A B C D3 3 4 5

X 2 Units 1 1Available Y 6 Units 2 4

Z 7 Units 2 5

The total cost of this solution is

Sh.Cell X:A 1 unit @ sh.13 = 13Cell X:B 1 Unit @ sh.11 = 11Cell Y:B 2 Units @ sh.14 = 28Cell Y:C 4 Units @ sh.12 = 48Cell Z:A 2 Units @ sh.18 = 36Cell Z:D 5 Units @ sh.12 = 60

196


Lesson Eight 318

The new total cost is sh.1 less than the total cost established in Step 1. This is the result expected because it was calculated in Step 2 that sh.1 would be saved for every unit we were able to transfer to Cell X : A and we were able to" transfer 1 unit only.

Notes: Always commence the + and - sequence with a + in the cell indicated by the (actual cost - shadow cost) calculation. Then put a - in the occupied cell in the same row which has an occupied cell in its column. Proceed until a - appears in the same column as the original +.

Step 4. Repeat Step 2 i.e. check that solution represents minimum cost. Each of the processes in Step 2 are repeated using the latest solution (Table 4) as a basis, thus: Nominal dispatch and reception costs for each occupied cell.

D(X) + R(A) = 13D(X) + R(B) = 11D(y) + R(B) = 14D(Y) + R(C) = 12DZ) + R(A) = 18D(Z) + R(D) = 12

On setting D(X) to be 0, the rest of the values are found to be

R(A) = 13 D(X) = 0R(B) = 11 D(Y) = 3R(C) = 9 D(Z) = 5R(D) = 7

Using these values the shadow costs of the unoccupied cells are calculated. The unoccupied cells are X:C , X:D, Y:A, Y:D, Z:B, and Z:C

Therefore;D(X) + R(C) = 9D(X) + R(D) = 7D(Y) + R(A) = 16D(Y) + R(D) = 10 D(Z) + R(B) = 16

D(Z) + R(C) = 14

The computed shadow costs are compared with actual costs to see if any reduction in cost is possible.

Actual Shadow + Cost increase

cost - cost - Cost reduction

Cell X :C 15 - 9= +6


319 Decesion Theory

X:D 20 - 7= +13Y:A 17 - 16 = +1Y:D 13 - 10 = +3Z:B 18 - 16 = +2Z:C 15 - 14 = +1

It will be seen that all the answers are positive, therefore no further cost reduction is possible and optimum solution has been reached.Thus the optimal solution is represented by table 4

UNEQUAL SUPPLY AND DEMAND QUANTITIESConsider the following example.Example 2Wanjiru books supplies in a firm dealing with import of books and it has three stores strategically situated around the country. Yesterday the company received orders to supply 100 books from 4 schools, of the books ordered the firm has 110 books in stock. The firm wishes to minimize cost and its seeking your advice, advise the firm. Below is a table of availability and requirement;

RequiredSch. A Sch. B Sch. C Sch. D Total

Books 25 25 42 8 100Store I 40 Sh.3 16 9 transport

costs perBook

Store II 20 Sh.1 9 3 8Available Store III 50 Sh.4 5 2 5

Total 110

SolutionStep 1: add a dummy destination to table 5 with zero transport costs

and requirements equal to the surplus availability.

Required

Sch. A Sch. B Sch. C Sch. DDummy TotalBooks 25 25 42 8 10 100

Store I 40 Sh.3 16 9 0 transportcosts perBook

Store II 20 Sh.1 9 3 8 0Available Store III 50 Sh.4 5 2 5 0

Total 110

Step 2. Now that the quantity available equals the quantity required (because of insertion of the dummy) the solution can proceed in exactly the same manner described in the first example. First set up an initial feasible solution

Requirement


Lesson Eight 320

A B C D Dummy25 25 42 8 10

I 40 5(4) 17(6) 8(3) 10(7)Available II 20 20(1)

III 50 8(5) 42(2)

The numbers in the table represent the allocations made and the numbers in brackets represent the sequence they were inserted based on lowest cost and the necessity to maintain row/column totals. The residue of 10 was allocated to the dummy. The cost of this allocation is

Sh. Sh.I→A 5 units @ 3 15I→B 17 units @ 16 272I→D 8units @ 2 16

I→Dummy 10 units @ zero costII→A 20 units @ 1 20III→B 8 units @ 5 40III→C 42 units @ 2 84

447

Step 3. Check solution to see if it represents the minimum cost possible in the same manner as previously described i.e.

Dispatch & Reception Costs of used routes:

D(I) + R(A) = 3

D(I) + R(B) = 16

D(I) + R(D) = 2D(I) + R(Dummy) = 12D(II) + R(A) = 1D(III) + R(B) = 5D(III) + R(C) = 2

Setting D(I) at zero the following values are be obtained

R(A) =3 D(I) =0R(B) =16 D(I) =-2R(C) =13 D(III) =-11R(D) =2R(Dummy) =0

Using these values the shadow costs of the unused routes can be calculated .The unused routes are I:C,II:B,II:C,II:D,II:Dummy,III:D,and Dummy

ShadowCosts


321 Decesion Theory

£D (I) + R(C) = 0+13 =13D (II). + R (B) = -2+16 =14D (II). + R(C) = -2+13 =11D (II) + R (D) = -2+ 2 =0D (II) + R (Dummy) = -2+0 =-2D (III) + R (A) = -11+3 =-8D (III) + R (D) = -11+2 =-9

D (III) + R (Dummy) = -11+0 =-11

The shadow costs are then deducted from actual costs

It will be seen that total cost can be reduced by £8 per unit for every unit that can be transferred into Cell II:C

Step4.Make the maximum possible allocation of deliveries into Cell II:C.This is done by inserting a sequence of +and -,maintaining row and column totals.


Lesson Eight 322

RequirementsA B C D Dummy25 25- 42 8 10

I 40 5+ 17- 8 10Available

II 20 20-

III 50 8+ 42-

The maximum transferable number is the lowest number in the minus cell, i.e. 17. after the transfer is made we get;

A B C D Dummy25 25- 42 8 10

I 40 22 0 8 10Available

II 20 3 17

III 50 25 25

Step 3 is repeated again to check if the cost is minimum after setting D(I) = 0.

In our case after deducting shadow costs from actual costs we find that there are no more negative numbers thus we deduce from the last table that the minimum transportation cost is,

(22×3) + (8×2) + (10×0) + (3×1) + (17×3) + (25×5) + (25×2) = Sh.311

Maximization using TransportationTransportation problems are usually minimizing problems, on occasions problems are framed so that the objective is to make the allocations from sources to destinations in a manner which maximizes contribution or profit. These problems are dealt with similar to minimizing problems but the reverse of it. i.e.a) Make initial feasible allocation on basis of maximum contribution

first, then next highest and so on.b) For optimum, the differences between actual and shadow

contributions for the unused routes should be all negative. If not, make allocation into cell with the largest positive difference.

c) In case there are more items available than are required, a dummy destination with zero contribution should be introduced and the maximizing procedure in a). followed

8.2 Assignment ModelsThe following example will be used as a basis of the step-by-step explanation.

Example 1


323 Decesion Theory

A company employs services engineers based at various locations throughout the country to service and repair their equipment installed in customer’s premises. Four requests for services have been received and the company finds that four engineers are available. The distances each of the engineers is from the various customers, is given in the following table and the company wishes to assign engineers to customers to minimise the total distances to be travelled.

CustomersW X Y Z

Alf 25 18 23 14

Bill 38 15 53 23

Charlie 15 17 41 30

Dave 26 28 36 29

Step 1. Reduce each column by the smallest figure in that column. The smallest figures are 15, 15, 23 and 14 and deducting these values from each element in the columns produces the following table.

Table 2

W X Y ZA 10 3 0 0B 23 0 30 9

C 0 2 18 16D 11 13 13 15

Step 2 Reduce each row by the smallest figure in that row.The smallest figures are 0, 0, 0 and 11 and deducting these values gives the following table.

Table 3 W X Y Z

A 10 3 0 0B 23 0 30 9

C 0 2 18 16D 0 2 2 4

Note: Where the smallest value in a row is zero (i.e. as in rows A, B and C above) the row is, of course, unchanged.

Step 3 Cover all the zero in the table 3 by the minimum possible number of lines. The lines may be horizontal or vertical.

Table 4

W X Y ZA 10 3 0 0


Lesson Eight 324

B 23 0 30 9C 0 2 18 16D 0 2 2 4

Note: Line 3, covering Row B, could equally well have been drawn covering column X.

Step 4.Compare the number of lines with the number of assignments to be made (in this example there are 3 lines and 4 assignments).If the number of line equals the number of assignments to be made go to step 6.

If the number of lines is less than the number of assignments to be made (i.e. as in this example which has three lines and four assignments) then

a) Find the smallest uncovered element from step 3, called X (in Table 4 this value is 2).

b) Subtract X to every element in the matrix.c) Add back to every element covered by a line. If an element is

covered by two lines, for example, cell A: W in Table 4, X is added twice.

Note: The effect of these steps is that X is subtracted from all covered by one line remain unchanged, and elements covered by two lines are increased by X.

Note: The effect of these steps is that X is subtracted from all uncovered elements, elements covered by one line remains unchanged, and elements covered by two lines are increased by X.

Carrying out this procedure on Table 4 produces the following results:In Table 4 the smallest elements is 2. New table is

Table 5

W X Y Z

A 12 3 0 0B 25 0 30 9C 0 0 16 14D 0 0 0 2

Note: It will be seen that cells A: W and B: W have been increased by 2; cells A : X, A : Y,A :Z, B :X,B:Y, B:Z, C:W and D:W are unchanged, and all other cells have been reduced by 2.

Step 5. Repeat steps 3 and step 4 until the number of lines covering the zero equals the number of assignments without any further repetition, thus:


325 Decesion Theory

Table 6

W X Y Z

A 12 3 0 0 Line 1B 25 0 30 9 Line 2C 0 0 16 14 Line 3D 0 0 0 2 Line 4

Step 6 when the number of lines equals the number of assignments to be made, use the following rules:

a) Assign to any zero which is unique to both a column and a row.

b) Assign to any zero which is unique to a column or a row.c) Ignoring assignments already made repeat rule (b) until all

assignments are made.

Carrying out this procedure for our example results in the following:a) (Zero unique to both a column and a row). None in this

example.b) (Zero unique column or row). Assign B to X and A to Z.

The position is now as follows.

Table 7

W X Y Z

A Row Satisfied Column satisfiedB Row Satisfied Column satisfiedC 0 Column Satisfied 16 Column SatisfiedD 0 Column Satisfied 0 Column Satisfied

c) Repeating rule (b) results in assigning D to Y and C to W.Notes:a) Should the final assignment not be to a zero, then more lines than

necessary were used in step 3.b) If a block of 4 or more zero’s is left for the final assignment, then a

choice of assignment exits with the same mileage.

Step 7 Calculate the total mileage of the final assignment.A to Z Mileage 14B to X 15C to W 15D toY 36

80 Miles

The assignment technique for maximising


Lesson Eight 326

A maximising assignment problem typically involves making assignments so as to maximise contribution. To maximise only one step 1 from above differs-the columns are reduced by the largest number in each column. From then on the same rules apply that are used for minimising.

Maximising example

Example 2The previous example No.1 will be used with the changed assumptions that the figures relate to contribution and not mileage and that it is required to maximise contribution .The solution would be reached as follows.(In each case the step number corresponds to the solution given for Example No 1.)

Original data

Table 8

W X Y ZA 25 18 23 14 ContributionsB 38 15 53 23 to be gainedC 15 17 41 30D 26 28 36 29

Step 1: Reduce each column by the largest figure in that column and ignore the resulting signs.

Table 9

W X Y ZA 13 10 30 16B 0 13 0 7C 23 11 12 0D 12 0 17 1


327 Decesion Theory

Step 2. Reduce each row by smallest figures in that row.

Table 10

W X Y ZA 3 0 20 6B 0 13 0 7C 23 11 12 0D 12 0 17 1

Step 3.Cover zeros by minimum possible number of lines.

Table 11

W X Y ZA 3 0 20 6B 0 13 0 7C 23 11 12 0D 12 0 17 1

Step 4. If a number of lines equals the number of assignments to be made go to step 6.If less, (as in this example), carry out the ‘uncovered element’ procedure previously described. This results in the following table:

Table 12

W X Y ZA 0 0 17 6B 0 16 0 10C 20 11 9 0D 9 0 14 1

Table 13

W X Y ZA 0 0 17 6B 0 16 0 10C 20 11 9 0D 9 0 14 1

Step 6. Make assignment in accordance with the rules previously described which result in the following assignment:

C to ZD to XA to WB to Y


Lesson Eight 328

Step 7.Calculate contribution to be gained from the assignments.

C to Z 30D to X 28A to W 25B to Y 53Total 136

Notes:a) It will be apparent that maximising assignment problems can be

solved in virtually the same manner as minimising problems.b) The solution methods given are suitable for any size of matrix. If a

problem is as small as the illustration used in this chapter, it can probably be solved merely by inspection.

Unequal sources and destinations5. To solve assignments problems in the manner described the matrix must be square, i.e. the supply must equal the requirements. Where the supply and requirements are not equal, an artificial source or destination must be created to square the matrix. The cost/mileage/contributions etc for the fictitious column or row should be zero throughout.

Solution method Having made the sources equal the destinations, the solutions method will be as normal, treating the fictitious elements as though they were real. The solution method will automatically assign a source or destination to the fictitious row or column and the resulting assignment will incur zero cost or gain zero contribution.

Points to notea) The assignment technique can be used for allocating type of

problems, e.g. taxis to customers, jobs to personnel.b) Most practical problems of size illustrated could be solved

fairly readily using nothing more than commonsense. However, the technique illustrated can be used to solve much larger problems.

Exercises with answers1. A foreman has four fitters and has been asked to deal with five jobs.

The times for each job are estimated as follows:

Fitters

Alf Bill Charlie Dave

Job 1 6 12 20 12Job 2 22 18 15 20Job 3 12 16 18 15Job 4 16 8 12 20


329 Decesion Theory

Job 5 18 14 10 17

Allocate the men to the jobs so as to minimise the total time taken and identify the job which will not be dealt with.

2. A company has four salesmen who have to visit four clients. The profits records from previous visits are shown in the table and it is required to maximise profits by the best assignments.

A B C D

Customer 1 6 12 20 1222 18 15 2012 16 18 1516 8 12 20

Answers to exercises

1. Dummy fitter inserted to square matrixA B C D DUMMY

1 6 12 20 12 02 22 18 15 20 03 12 16 18 15 04 16 8 12 15 05 18 14 10 17 0

Reduce columns by the smallest element and cover by lines

0 4 10 0 016 10 5 8 06 8 8 3 010 0 2 8 012 6 0 5 0

4 lines so not optimum, smallest element 3Therefore reduce uncovered elements by 3 and increase elements crossed by 2 lines by 3

0 4 10 0 313 7 2 5 03 5 5 0 010 0 2 8 312 6 0 5 3

5 Lines so optimum.

AssignmentsB to 4C to 5


Lesson Eight 330

A to 1Dummy to 2

8.3 NETWORK ANALYSISThis is a system of interrelationship between jobs and tasks for planning and control of resources of a project by identifying critical path of the project.


Dangling activity

331 Decesion Theory

TerminologyActivity. Task or job of work, which takes time and resources e.g building

a bridge. It is represented by an arrow which indicates where the task begins and ends

Event (node). This is a point in time and it indicates the start or finish of

an activity e.g in building a bridge, rails installed. It is represented by a circle.

Dummy activity. An activity that doesn’t consume time or resources, it is merely to show logical dependencies between activities so as abide by rules of drawing a network, it is represented by a dotted arrow

Network. This is a combination of activities and events (including dummy activities)

Rules for Drawing a Networka) A network should only have one start point and one finish point

(start event and finish event )b) All activities must have at least one preceding event (tail event)

and at least one succeeding event (head event), but an activity may not share the same tail event and head event.

c) An activity can only start after its tail event has been reachedd) An event is only complete after all activities leading to it are

complete.e) Activities are identified by alphabetical or numeric codes i.e.

A,B,C; 1,2,3 or identification by head or tail events 1-2, 2-4, 3-4,1-4…

f) Loops (a series of activities leading back to the same event) and danglers (activities which do not link to the overall project) are not allowed

Dummy EventsThis is an event that does not consume time or resources, it is represented by dotted arrow. Dummies are applied when two or more events occur concurrently and they share the same head and tail events


Loop

331

B

A CRCA

Lesson Eight 332

e.g. when a car goes to a garage tires are changed and break pads as well, instead of representing this as;

These events are represented as;

Example of a network.

Activities1-2 - where 1 is the preceding event where as 2 is the succeeding event of the activity1-32-42-53-54-54-65-66-7

8.4 Network Analysis-Time AnalysisAssessing the time

a) After drawing the outline of the network time durations of the activities are then inserted.a) Time estimates. The analysis of the projects time can be

achieved by using :i. Single time estimates for each activity. These estimates

would be based on the judgment of the individual


A- Tires Changed

B- Break pads Changed

Car Arrives (CA) Car ready (CR)

1

2

3

4

5

6 7

B 2

D4

333 Decesion Theory

responsible or by technical calculations using data from similar projects

ii. Multiple time estimates for each activity. the most usual multiple time estimates are three estimates for each activity , i.e. optimistic (O), Most Likely (ML), and Pessimistic (P). These three estimates are combined to give an expected time and the accepted time formula is:

Expected time = For example assume that the three estimates for an activity are Optimistic 11 days Most likely 15 days Pessimistic 18 days

Expected time =

= 14.8 daysb) Use of time estimates. as three time estimates are converted

to a single time estimate. There is no fundamental difference between the two methods as regards the basic time analysis of a network. However, on completion of the basic time analysis, projects with multiple time estimates can be further analyzed to give an estimate of the probability of completing the project by a scheduled date.

c) Time units. Time estimates may be given in any unit, i.e. minutes , hours, days depending on the project. All times estimates within a project must be in the same units otherwise confusion is bound to occur.

Basic time analysis – critical path b) The critical path of a network gives the shortest time in which

the whole project can be completed. It is the chain of activities with the longest duration times. There may be more than one critical path which may run through a dummy.

Earliest start times (EST) – Forward pass, Once the activities have been timed we can assess the total project time by calculating the ESTs for each activity. The EST is the earliest possible time at which a succeeding activity can start.Assume the following network has been drawn and the activity times estimated in days.


2

0 1 3 4 5

333

A1

E 1

C3

F2

B 2

A1

E 1

C3

F2

D4

B 2

A1

E 1

C3

F2

D4

Lesson Eight 334

The ESTs can be inserted as follows.

EST

The method used to insert the ESTs is also known as the forward pass, this is obtained by;

EST = The greater of [EST (tail event) + Activity duration]

a) Start from the start event giving it 0 values, b) For the rest of the events EST is obtained by summing the EST of

the tail event and the activity durationc) Where two or more routes converge into an activity, calculate

individual EST per route and then select the longest route (time)d) The EST of the finish event is the shortest time the whole project

can be completed.

Latest Start Times (LST) – Backward pass. this is the latest possible time with which a preceding activity can finish without increasing the project duration. After this operation the critical path will be clearly defined.

From our example this is done as follows;

LST

LST = Lowest of [LST (head event) – activity duration]


23

00

11

34

47

59

23 3

00 0

11 1

34 6

47 7

59 9

B 2

A1

E 1

C3

F2

D4

A B10

C

Section of the network

335 Decesion Theory

a) Starting at the finish event, insert the LST (i.e. 9 for our example) ,and work backwards through the network.

b) deduct each activity duration from the previously calculated LST (i.e. head LST).

c) Where the tails of activities join an event, the lowest number is taken as the LST for that event

Critical Path. . This is the chain of activities in a network with the longest duration Assessment of the resultant network shows that one path through the network (A, B, D, F) has EST's and LST's which are identical this is the critical path.

The critical path can be indicated on the network either by a different colour or by two small transverse lines across the arrows along the path, thus in our example we have;

Activities along the critical path are vital activities which must be completed by their EST's/LST's otherwise the project will be delayed.

Non critical activities (in the example above, C and E) have spare time or float available. C and/ or E could take up to an additional 2 days in total without delaying the project duration. If it is required to reduce the overall project duration then the time of one or more of the activities on the critical path must be reduced perhaps by using more labour, or better equipment to reducing job times.

FLOATFloat or spare time can only be associated with activities which are non-critical. By definition, activities on the critical path cannot have float. There are three types of float, Total Float, Free Float and Independent Float. To illustrate these types of float we use the following example.


23 3

00 0

11 1

34 6

47 7

59 9

510 20

640 50

335

Lesson Eight 336

a) Total float. Amount of time by which a path of activities could be delayed without affecting the overall project duration. The path in this example consists of one activity only i.e. B

Total Float = Latest Finish time (LFT) - Earliest Start time(EST) time – Activity Duration

Total Float = 50 - 10 - 10= 30 days

b) Free float Amount of time an activity can be delayed without affecting the commencement of a subsequent activity at its earliest start time, but may affect float of a previous activity.

Free Float = Earliest Finish Time(EFT) - EST - Activity Duration

Free Float = 40-10-10= 20 days

c) Independent float. Amount of time an activity can be delayed when all preceding activities are completed as late as possible and all succeeding activities commenced as early a possible. Independent float therefore does not affect the float of either preceding or subsequent activities.

Independent float = EFT- Latest Start time (EST) - Activity Duration

Independent float = 40 - 20 - 10= 10 days

Note: for examination purposes, float always refers to total float The total float can be calculated separately for each activity but it is

often useful to find the total float over chains of non-critical activities between critical events

Example.The following represents activities of a network.

Activity Preceding Activity Duration DaysA - 4B A 7C A 5D A 6E B 2F C 3


A4 C

5

B 7

D6

E 2

F 3

G5

H 11

I 7

J4

K 3

L4

E 2

G5

H 11

337 Decesion Theory

G E 5H B,F 11I G,H 7J C 4K D 3L I,J,K 4

Required:a) Draw the network diagram and find the critical pathb) Calculate the floats of the network in question

Solution. (a)

First we draw the network structure ensuring it fits the data above We then label all activities from 1 to 12 and indicate activity

duration Conduct a forward pass operation (to obtain the diagram above) Operate backward pass to establish the critical path, thus we

have…


10 2

4

311

6 9

4 13

930

823

512

1034

7 10

11 15

13 18

23 23

12 12 337

A4 C

5

B 7

D6

F 3

I 7

J4

K 3

L4

Lesson Eight 338

Therefore we get the critical path to be, A- C- F- H- I- Lb) The floats of the network,

Activity

Duration

Total Float Free Float

Independent Float

Activity EST LST EFT LFT D LFT -EST- D

EFT-EST-D

EFT-LST-D

*A 0 0 4 4 4 - - - B 4 4 11 15 7 4 - -*C 4 4 9 9 5 - - - D 4 4 10 22 6 12 - - E 11 15 13 21 2 8 - -*F 9 9 15 15 3 - - -G 13 21 23 23 5 5 5 -*H 12 12 23 23 11 - - -*I 23 23 30 30 7 - - -J 9 9 30 30 4 17 17 17K 10 22 30 30 3 17 17 5*L 30 30 34 34 4 - - -

The total float on the non-critical chains are;

Non-critical chain

Time required(sum of

duration)

Time available(LFT of last activity-EST of 1st

activity)

Total Float over chain

B,E,G 14 19 5B,Dummy 7 8 1

D,K 9 26 17


0 0

4 4 9 9

30 30

34 34

10 27

339 Decesion Theory

J 4 21 17

SlackThis is the difference between the EST and LST for each event. Strictly it does not apply to activities but on occasions the terms are confused in examination questions and unless the context makes it abundantly clear that event slack is required, it is likely that some form of activity float is required. Events on the critical path have zero slack.

8.5 Cost SchedulingThis is done by calculating the cost of various project durations, cost analysis seeks to find the cheapest way of reducing the overall cost duration of a project by increasing labour hours, equipment e.t.c.

TerminologiesNormal cost. The costs associated with a normal time estimate for an

activity. Often the normal time estimate is set at the point where resources (labour, equipment, etc.) are used in the most efficient manner.

Crash cost. The costs associated with the minimum possible time for an activity. Crash costs, because of extra wages, overtime premiums, extra facility costs are always higher than normal costs.

Crash time. The minimum possible time that an activity is planned to take. . The minimum time is invariably brought about by the application of extra resources, e.g. more labour or machinery.

Cost slope. This is the average cost of shortening an activity by one time unit (day, week, month as appropriate). The cost slope is generally assumed to be linear and is calculated as follows:

Cost slope = Crash cost – Normal cost Normal time – Crash time

ExampleA project has the following activities and costs. You are required to prepare the least cost schedules for all possible durations from normal time – normal cost to crash time – crash cost.


A 4

B 8

D 9

C 5 E

5

Lesson Eight 340

Activity Preceding

Activity

Duration

days

Crash time

Cost (Shs).

Crash cost

Cost slope

A - 4 3 360 420 60B - 8 5 300 510 70C A 5 3 170 270 50D A 9 7 220 300 40E B,C 5 3 200 360 80

Project duration and costs

(a) Normal duration = 14 daysCritical path = A,C,EProject cost (cost of all activities in normal time) = Shs. 1,250.

(b) Reduce by 1 day the activity on the critical path with the lowest cost slope. Thus we reduce C at extra cost of Shs. 50.

NowProject duration = 13 daysProject cost = Shs. 1,300

Note: that all activities are now critical.

(c) Further reducing the critical path by 1 day will require that more than one activity is affected because there exist several critical paths.

Reduce by 1 day

Extra cost Activities critical

A and B 60 + 70 = 130 AllD and E 40 + 80 = 120 AllB, C and D 70 + 50 + 40 =

160All

A and E 60 + 80 = 140 A, D, B, E


0 0 0

1 4 4

2 9 9

3 14 14

341 Decesion Theory

From this we realize that reducing D and E is the cheapest.

However closer examination of the fourth alternative reveals that C is now non-critical and has 1 day float. Since we earlier reduced C for Shs. 50, if we reduce A and E and increase C by a day which will save Shs. 50.

Then the net cost for 12 day duration = 1,300 + (140 – 50) = 1,390.

The network becomes………

(d) Next we reduce D & EProject duration = 11 daysProject cost = 1,510Critical activities = All

(e) Final reduction possible is by reducing B, C & D for Shs. 160 the network then becomes.


13 3

00 0

27 7

310

10

3 (crash)

C4

B7

D

7 (Crash)

E3 (crash)

A

13 3

00 0

27 7

312

12

3 (crash) C

5

B8

D9

E4

A

341

Lesson Eight 342

Duration = 10 daysCost = Shs. 1,670Critical activities = All.

Note: only critical activities affect project duration. : Always look for a possibility of increasing the duration of a previously crashed activity.

SCHEDULING RESOURCES AND GANTT CHART

Apart from time, cost network analysis also help in controlling and planning of resources.

ExampleA project has the following activity durations and resource requirements.

Activity

Preceding activity

Duration (days)

Resource requirement (man power)

A - 6 3B - 3 2C - 2 2D C 2 1E B 1 2F D 1 1 Requiredi) What is the networks critical pathii) Draw a gantt chart diagram indicating activity times, using their

estimate.iii) Show resource requirement on a day to day basis assuming all

events commence at their estimates.iv) Assuming that only six employees are available, how will the

activities be planned for?

Solutioni)Activities Duration EST LST Man powerA 6 0 0 3B 3 0 0 2C 2 0 0 2D 2 2 3 1E 1 3 5 2F 1 4 5 1


343 Decesion Theory


Lesson Eight 344

ii) A gantt chart or a bar chart. This is a diagram indicating a resource scaled network.

iii) Resource requirements on a day to day basis.


345 Decesion Theory

iv) When on 6 manpower resources are available then we adjust the activities to accommodate this and still end at the given critical time duration i.e.

Node NetworksThis network also known as a procedure diagram is represented with the same information as a network diagram.Its characteristics are;

i) Activities are shown in boxes instead of arrowsii) Events are not represented.iii) The arrows linking boxes indicate the sequence precedence

of activities.iv) Dummies aren’t necessary.

E.g.

Would appear as


Lesson Eight 346


347 Decesion Theory

A full activity node network is represented as;

This is represented as;

Note:i) EST and LST are calculated by the same process we learnt earlier.ii) EFT and LFT are calculated by adding the activity time duration to

EST and LST respectively.iii) Critical path is similarly identified by identifying equal EST and

LST throughout the path.


Lesson Eight 348


QUESTION ONERegal Investments has just received instructions from a client to invest in two shares; one an airline share, the other an insurance share. The total maximum appreciation in share value over the next year is to be maximized subject to the following restrictions:- the total investment shall not exceed Sh.100,000- at most Sh.40,000 is to be invested in the insurance sharesquarterly dividends must total at least Sh.2,600

The airline share is currently selling for Sh.40 per share and its quarterly dividend is Sh.1per share. The insurance share is currently selling for Sh.50 per share and the quarterly dividend is Sh.1.50 per share. Regal’s analysts predict that over the next year, the value of the airline share will increase by Sh.2 per share and the value of the insurance share will increase by Sh.3 per share.A computer software provided the following part solution output:

Objective Function Value = 5,400

Variable Number Reduced cost Airline shares 1,500 0.000Insurance shares 800 0.000

Constraint Slack/Surplus Dual pricesTotal investment 0.000 0.050Investment in insurance 0.000 0.010Dividends 100.000 0.000


349 Decesion Theory

Objective Coefficient Ranges

Variable Lower limit Current value Upper limitAirline share 2.500 3.000 No upper limitInsurance share 0.000 2.000 2.400

Right-hand Side Ranges

Constraint Lower limit Current value Upper limitTotal investment 96,000.00 100,000 No upper limitInvestment in insurance20,000.00 40,000 100,000.00Dividends No lower limit 2,600 2,700.00

Required:a) Formulate the above problem.b) Explain what reduced cost and dual prices columns above mean.c) How should the client’s money be invested to satisfy the restrictions?d) Suppose Regal’s estimate of the airline shares appreciation is an

error, within what limits must the actual appreciation lie for the answer in (c) above to remain optimal?

(Q 6 Dec 2001)

QUESTION TWOa) A baker makes two products; large loaves and small round loaves. He

can sell up to 280 of the large loaves and up to 400 small round loaves per day. Each large loaf occupies 0.01m3 of shelf space, each small loaf occupies 0.008m3 of space, and there is 4m3 of shelf space available. There are 8 hours available each night for baking, and he can produce large loaves at the rate of 40 per hour, and small loaves at the rate of 80 per hour. The profit on each large loaf is Sh.5.00 and Sh.3.00 profit on the small round loaf.

Required:In order to maximize profits, how many large and small round loaves

should he produce?

b) Summarize the procedure for solving the kind of quantitative technique you have used to solve part (a) above.

(Q 6 June 2001)

QUESTION THREEa) A small company will be introducing a new line of lightweight bicycle

frames to be made from special aluminium alloy and steel alloy. The frames will be produced in two models, deluxe and professional. The anticipated unit profits are currently Sh.1,000 for a deluxe frame and Sh.1,500 for a professional frame. The number of kilogrammes of each alloy needed per frame is summarized in the table below. A


Lesson Eight 350

supplier delivers 100 kilogrammes of the aluminium alloy and 80 kilogrammes of the steel alloy weekly.

Aluminium alloy Steel alloyDeluxe 2 3Professional 4 2

Required:i) Determine the optimal weekly production schedule.ii) Within what limits must the unit profits lie for each of the frames

for this solution to remain optimal?

b) Explain the limitations of the technique you have used to solve part (a) above. (Q 6 Dec 2000)

QUESTION FOURa) Define the following terms as used in linear programming:

i) Feasible solutionii) Transportation problemiii) Assignment problem

b) The TamuTamu products company ltd is considering an expansion into five new sales districts. The company has been able to hire four new experienced salespersons. Upon analysing the new salesperson’s past experience in combination with a personality test which was given to them, the company assigned a rating to each of the salespersons for each of the districts .These ratings are as follows:

c)Districts

Salespersons

1 2 3 4 5A 92 90 94 91 83B 84 88 96 82 81C 90 90 93 86 93D 78 94 89 84 88

The company knows that with four salespersons, only four of the five potential districts can be covered.

Required:i) The four districts that the salespersons should be assigned to in

order to maximize the total of the ratingsii) Maximum total rating. (Q 6 June

2002)

QUESTION FIVE


351 Decesion Theory

a) Explain the value of sensitivity analysis in linear programming problems and show how dual values are useful in identifying the price worth paying to relax constraints.

b) J.A Computers is a small manufacturer of personal computers. It concentrates on production of three models- a Desktop 386, a Desktop 286, and a Laptop 486, each containing one CPU Chip. Due to its limited assembly facilities JA Computers are unable to produce more than 500 desktop models or more than 250 Laptop models per month. It has one hundred and twenty 80386 chips (these are used in Desktop-386) and four hundred 80286 chips (used in desktop 286 and Laptop 486) for the month. The Desktop 386 model requires five hours of production time, the Desktop 286 model requires four hours of production time, and the Laptop 486 requires three hours of production time. J.A Computers have 2000 hours of production time available for the coming month. The company estimates that the profit on Desktop 386 is Sh. 5,000. for a desktop 286 the profit is Sh.3,400 and Sh.3,000 profit for a laptop 486.


Lesson Eight 352

Required:Formulate this problem as a profit maximization problem and mention the basic assumptions that are inherent in such models.

c) An extract of the output from a computer package for this problem is given below:

Output solutionX1=120, X2 = 200, X3 = 200Dual values Constraints 3 150

Constraints 4 90Constraints 5 20

Sensitivity analysis of objective function coefficients:

Variable

Lower limit

Original value

Upper limit

X1 100 250 No limitX2 150 170 200X3 127.5 150 170

Sensitivity analysis on R.H.S ranges.

Constraints

Lower limit

Original value

Upper limit

1 320 500 No limit2 200 250 No limit3 80 120 1304 350 400 412.55 1950 2000 2180

X1=Monthly production level for Desktop 386.X2 =Monthly production level for Desktop 286.X3=Monthly production level for Laptop 486.

Required:i) Interpret the output clearly, including optimum product mix,

monthly profit, unused resources and dual valuesii) Explain the purpose of upper limits and lower limits for the

variables X1,X2,X3 and constraints 1 to 5.iii) Calculate the increase in profit if the company is able to produce

a further 10 CPU 80386 chips.(Q7 July 2000 Pilot paper)

QUESTION SIXPreface Retailers is a high-technology retailer and mail order business. In order to improve its process the company decides to install a new microcomputer system to manage its entire operation (i.e. payroll, accounts, inventory).


353 Decesion Theory

Terminals for each of its many stores will be networked for fast, dependable service. The specific activities that Preface will need to accomplish before the system is up and running are listed below. The table also includes the necessary increased staffing to undertake the project.


Lesson Eight 354

Activity PrecedingActivities

Duration

(Days)

Increased

StaffA. Build insulated enclosureB. Decide on computer systemC. Electrical wiring of roomD. Order and collect computerE. Install air conditioningF. Install computerG. Staff testingH. Install softwareI. Staff training

--ABA

D, EB

C, FG, H

413242523

132122111

Required:a) Draw a network diagram for the project and determine the critical

path and its duration.b) Assuming that all activities start as soon as possible, draw a

progress chart for the project, showing the times at which each activity takes place and the manpower requirements.

c) The union has decided that any staff employed on the project must be paid for the duration of the project whether they work or not, at a rate of £500 per day.

Assuming that the same staff is employed on the different activities, determine the work schedule that will minimise labour costs though not necessarily the project time. What is the cost associated with this schedule?

Comment on the validity of the assumption.


355 Decesion Theory

COMPREHENSIVE ASSIGNMENT FOURWork out these question for three hours (exam condition) then hand




QUESTION ONEa) Define the following terms as used in game theory:i) Dominance. (2

marks)ii) Saddle point. (2

marks)iii) Mixed strategy. (2

marks)iv) Value of the game (2

marks)

b) Consider the two person zero sum game between players A and B given the following pay-off table:

Player B Strategies1 2 3 4

Player A Strategies

1 2 2 3 -1

2 4 3 2 6

Required:i) Using the maximin and minimax values, is it possible to determine the

value of the game? Give reasons. (3

marks)ii) Use graphical methods to determine optimal mixed strategy for

player A and determine the value of the game. (9

marks)(Total: 20

marks)

QUESTION TWOCentral and Eastern Industries is planning to introduce a new mobile phone service. To do so, the following activities are necessary:


Lesson Eight 356

Preceding Expected StandardActivity Activity Time

(weeks)Deviation (weeks)

A - 6 1.0B - 3 0.5C A 5 1.0D A 4 1.0E A 3 0.5F C 3 0.5G D 5 1.0H B.D.E 5 1.0I H 2 0.5J F.G.I 3 1.0

The costs of the project are estimated to be Sh.10 million. If the projects is completed within 24 weeks the expected net revenue will be about Sh.100 million but if the deadline of 24 weeks is not met, the service will fail to penetrate the market and a net revenue of Sh.2 million is expected.

Required:a) Determine how long the project would take.

(8 marks)b) If the start of activities B, E and G are respectively delayed by 3,2 and

2 weeks, how would this affect the total project time?

(6 marks)c) Determine a 95% confidence interval for the expected time of the

project and explain your answer. Ignore the delays referred to in (b) above.

(3 marks)d) Determine the expected profit on this project. Again ignore the

delays referred in (b) above (3

marks)(Total: 20

marks)

QUESTION THREEa) Explain the value of sensitivity analysis in linear programming

problems and show how dual values are useful in identifying the price worth paying to relax constrains. (4marks)

b) J.A Computers is a small manufacturer of personal computers. It concentrates production on three models, a Desktop 286,and a laptop 486,each containing one CPU Chip. Due to its limited assembly facilities J.A Computers are unable to produce more than 500 desktop model or more than 250 Laptop models per month. It has one hundred and twenty 80386 chips (these are used in Desktop – 386) and four hundred 80286 chips(used in desktop 286 model requires


357 Decesion Theory

four hours of production time, and the laptop 486 requires three hours of production time. J.A Computers have 2000hours of production time available for the coming month. The company estimates that the profit on a desktop 386 is Sh.5000, for a desktop 286 the profit is Sh.3400 and Sh.3000 profit for a Laptop 486.

Required:Formulate this problem as a profit maximization problem and mention the basic assumptions that are inherent in such models.

(7 marks)

a) An extract of the output from a computer package for this problem is given below:

Output solution

X1 = 120, X2 = 200, X3 = 200

Dual values Constraints 3 150Constraints 4 90

Constraints 5 20

Sensitivity analysis of objective function coefficientsVariable Lower limit Original value Upper limitX1 100 250 No limitX2 150 170 200X3 127.5 150 170

Sensitivity analysis on R. H. S rangesConstraints Lower limit Original value Upper limit1 320 500 No limit2 200 250 No limit3 80 120 1304 350 400 412.55 1950 2000 2180

X1 = Monthly level for Desktop 386.X2 = Monthly production level for Desktop 286X3 = monthly production level for Laptop 486

Required:i) Interpret the output clearly, including optimum product mix, monthly

profit, unused resources and dual values. (3

marks)ii) Explain the purpose of upper limits for the variables X1, X2 ,X3 and

constraints 1 to 5. (3 marks)


Lesson Eight 358

iii) Calculate the increase in profit if the company is able to produce a further 10 CPU 80836 chips. (3 marks)

(Total: 20 marks)

QUESTION FOURa) Give two applications of simulation in business. (2

marks)

b) Collins Simiyu recently acquired a piece of land in Kitale. A property development company has offered him Sh.300,000 for the piece of land. He has to make a decision on whether to cultivate the piece of land or to sell it to the property development company. if he decides to cultivate the land, there is a probability of getting a high, medium or low harvest. The expected net income for each of the above states of harvest is shown below:

State of harvest

Net income (Sh.)

High 500,000Medium 100,000Low (200,000)

From past experience, there is a 10 per cent probability that the harvest will be low, a 30 pr cent probability that the harvest will be medium and a 60 per cent probability that the harvest will be high. Collins Simiyu can engage an agricultural expert to carry out a survey on the productivity of the land, which will cost him Sh.30,000. The agricultural expert gives the following information as to the reliability of such surveys (prior probabilities).

Results of survey

State of harvest

High Medium Low TotalAccurate 0.35 0.10 0.05 0.5Not accurate 0.25 0.10 0.15 0.5

0.60 0.20 0.20 1.0

Required:i) Construct a decision tree for the above problem.

(6 marks)ii) The expected monetary value for each decision

(10 marks)iii) The decision that you would recommend (2

marks)(Total: 20

marks)


359 Decesion Theory

QUESTION FIVEa) Explain the difference between assignment and transportation

problems. (4 marks)

b) State the assumptions made in solving a transportation problem.(4 marks)

c) Umoja Engineering Works Ltd. Has a network of branches all over Kenya. The branches are used to service, repair and install equipment for their clients. Currently, the Nairobi branch has four clients who require installation of equipment. Each client requires the services of one engineer.

There are four engineers who are not engaged at the moment and can be assigned any one of the tasks. However, these engineers have to travel from different locations and the Nairobi branch has to meet their travel and subsistence allowances. The allowances vary from one engineer to another and according to the client the engineer has been assigned to work for.

The table below shows the costs (in thousands of shillings) associated with each engineer.

ClientEngineer 1 2 3 4A 37.0 27.0 34.0 21.0B 57.0 22.0 79.0 34.0C 22.0 25.0 61.0 45.0D 39.0 42.0 54.0 43.0

Required:i) The assignments to be made in order to minimize the total cost of the

engineers.(10 marks)

ii) The minimum cost of using engineers.(12 marks)

(Total: 20 marks)

SECTION IIQUESTION SIXa) Define the following terms as used in the network analysis:

i) Crash time (2 marks)

ii) Optimistic time (2 marks)

iii) Forward pass (2 marks)


Lesson Eight 360

iv) Dummy activity (2 marks)

v) Slack (2 marks)

b) James Mutiso is a computer engineer in an information technology firm. The firm has decided to install a new system to be used by the firm’s help desk. James Mutiso has identified some activities required to complete the installation.

The table below provides a summary of the activities’ durations and the required number of technicians:

Activity Duration (Weeks)

Required number of technicians

1-2 3 21-3 1 42-4 3 42-5 2 23-4 2 43-6 4 44-5 2 25-6 2 26-7 2 2

Required:i) Draw a gantt chart for the project. (6

marks)ii) Mr. Mutiso would like to reschedule activities so that not more than 6

technicians are required each week.

Determine if this is possible and how it can be achieved by rescheduling

the activities. (4 marks)

(Total: 20 marks)

QUESTION SEVENRegal investments has just received instructions from a client to invest in two shares; one an airline share, the other an insurance share. The total maximum appreciation in share value over the next year is to be maximized subject to the following restrictions:

The total investment shall not exceed Sh.100,000At most Sh.40,000 is to be invested in the insurance sharesQuarterly dividends must total at least Sh.2,600


361 Decesion Theory

The airline share is currently selling for Sh.40 per share and its quarterly dividend is Sh.1 per share. The insurance share is currently selling for Sh.50 per share and the quarterly dividend is Sh.1.50 per share. Regal’s analysts predict that over the next year, the value o f the airline share will increase by Sh.2 per share and the value of the insurance share will increase by Sh.3 per share.

A computer software provided the following part solution output:Objective Function Value = 5,400

Variable Number Reduced cost

Airline shares 1,500 0.000Insurance shares 800 0.000

Constraint Slack/Surplus

Dual prices

Total investment 0.000 0.050Investment in insurance 0.000 0.010Dividends 100,000 0.000

Objective coefficient RangesVariable Lower limit Current

valueUpper limit

Insurance share 2.5000 3.000 No upper limit

Airline share 0.000 2.000 2.400

Right-hand Side RangesConstraint Lower limit Current

valueUpper limit

Total investment 96,000.00 100,000 No upper limit

Investment in insurance 20,000.00 40,000 100,000.00Dividends No lower limit 2,600 2,700.00

Required:i) Formulate the above problem. (7

marks)ii) Explain what reduced cost and dual prices columns above mean.

(5 marks)iii) How should the client’s money be invested to satisfy the restrictions?

(4 marks)iv) Suppose Regal’s estimate of the airline shares appreciation is in

error, within what limits must the actual appreciation lie for the answer in (c)

above toremain optimal? (4

marks)


Lesson Eight 362

(Total: 20 marks)

QUESTION EIGHTThe linear programming model and output model below represent a problem whose solution will tell a road side kiosk owner how many of the four different types of household goods to stock in order to maximize profits. It is assumed that every item stocked will be sold. The variables measure the packets of Unga, Spaghetti, Rice and Sugar to stock respectively. The constraints measure storage space in units, special display racks, demand and a marketing restriction, respectively.

Maximize Z = 4X1+ 6X2+ 5X3+ 3.5X4

Subject to :2X1 + 3X2+ 3X3+ X4 ≤ 120 (1)1.5X1 + 2X2 ≥ 54 (2)

2X2 + X3 + X4 ≤ 72 (3) X2 + X3 ≥ 12 (4)

Where X1 = packets of UngaX2 = packets of SpaghettiX3 = packets of RiceX4 = packets of Sugar

Optimal solutionVariable Value Reduced costX1 12.00 -X2 0.00 0.50X3 12.00 -X4 60.00 -

Constraint Stack/surplus

Dual/shadow price

1 - 2.002 - -3 - 1.504 - -2.50

Objective Coefficient RangesVariable Lower limit Current

valueUpper limit

X1 1.50 - 5.00X2 No limit - 6.50X3 4.50 - 7.50X4 3.00 - No limit


363 Decesion Theory

Right Hand Side RangesConstraint

Lower limit Current value Upper limit

1 96 - 1682 18 - No limit3 24 - 964 0 - 24

Required:a) Determine the retailer’s optimal profit level.

(2 marks)b) Determine and interpret the missing values under:

i) Reduced cost column. (2 marks)

ii) Slack/surplus column, indicate whether the value is a slack or a surplus.

(2 marks)

iii) Dual/shadow price column. (2 marks)

iv) Current value column under objective coefficient ranges.(2 marks)

v) Current value column under right hand side ranges.(2 marks)

c) Interpret the value 0.50 under the reduced cost column and values; 2.00, 1.50 and -2.50 under the dual/shadow price column (2

marks)d) Determine whether the current, optimal solution would change if the

current profit of packets of Unga is increased by Sh.2.00.(2 marks)

e) Determine by how much the amount of space would increase before thereis a change in the dual/shadow price.(2 marks)

f) The above problem could have been solved manually. Explain how theoptimal solution can be determined using the manual approach.(2 marks)

(Total: 20 marks)


Revision Aid 364

LESSON NINE

Revision Aid

SOLUTIONS

LESSON ONE

QUESTION ONE

1. (–3, 12) 2. (12, 0) 3. (10, 4) 4. ( , )

QUESTION TWO

QUESTION THREE Equillibrium will be attained

Population

QUESTION FOUR

1. The vector for final demand =

The input/output coefficient matrix (also called the matrix for technical coefficients). Call it A

let X be the total output vector.

Thus; X = AX + YX = (I – A )-1Y


365 Lesson Nine

Inverse (I - A)-1=

Therefore X =

= in Shs

QUESTION FIVEi) Express each percentage as a decimal and the matrix equation

becomes,

ii) Put t1= 400 and t2= 700 and the matrix equation becomes.

That is b1 = = 580 kilos

b2 = = 375 kilos

iii) To establish the value of t1 and t2 which correspond to b1 and b2 it is necessary to form the inverse of the matrix.

which is


Revision Aid 366

the equation to determine t1 and t2 now becomes,

Thus;t1 = - 1.25×600 + 3.0 × 450

= 600 kilosand

t2= - 2.5×600 - 2 .0 × 450= 600 kilos

QUESTION SIXi)

2 2 2 2 10 -2A3 = =

3 -3 3 -3 -3 15

2 2 10 -2 14 26A2 = =

3 -3 -3 15 39 -51

ii) F(A) = A3 – 3A2 – 2A + 41

14 26 10 -2 2 2 1 0= -3 -2 +4

39 -51 -3 15 3 -3 0 1

14 26 30 6 4 4 4 0= - - +

39 -51 -9 45 6 -6 0 4

-16 28=

42 -86

iii)

-3 -2 1 1


367 Lesson Nine

1_ = 4 6A-1 = -12 -3 2 1 -1

4 6

QUESTION SEVENi)

12

10 15 6 0 8 348

15 20 6 4 18 = 536

16 24 8 6 22 660

Cost of Standard set = 348 pence = £3.48Cost of Deluxe set = 536 pence = £5.36Cost of super set = 660 pence = £6.60

ii)10 15 16 3,000

100 10 15 6 015 20 24 4,300

80 = or (100 80 50)1520 6 46 6 8 1,480

50 16 24 8 60 4 6 620

Squares 3,000, triangles 4,300 hexagons 1,480, octagons 620

iii)12

8(3,000 4,300 1,480 620)

18

20= 110,680 in pence

= £1,106.80

QUESTION EIGHTa)

A B C storage maintenance

10 12 50 2 0.5 156 48NP = 60 0 20 3 1.5 = 160 40

2 0.5

Cost of storing in warehouse Y = 156p = £1.56


Revision Aid 368

Cost of storing in warehouse W =160p = £1.60Cost of maintaining warehouse Y = 48p = £0.48Cost of maintaining warehouse W = 40p = £0.40

b)A B C storage

maintenance

Day I Y 10 10 50 2 0.5 150 45cost in

W 40 0 20 3 1.5 = 120 30 pence2 0.5

A B C storage maintenance

Day II 10 10 60 2 0.5 191 60.5 cost in

40 15 20 3 1.5 = 165 52.5 pence2 0.5

For total cost add Day I and Day II values

c)A B C

10 150 45 191 60.5 +3 × Cost in pence

120 30 165 52.5


369 Lesson Nine

LESSON TWO

QUESTION ONEa) 6

b)c)

d)

QUESTION TWO

Therefore Q = 15 at minimum

Note: which is positive indicating a minimum value

QUESTION THREEn(є) = 250

P + 12 + 59 = 147 giving P = 76Q + 59 + 37 = 102 giving Q = 6

i) Those who did not vote= 250 – (76 + 12 + 14 + 59 + 6 + 37)= 250 – 204 = 46

ii) x = 76 + 12 + 14 = 102


Revision Aid 370

x = 12 + 59 + 6 = 77

z = 37 + 14 + 6 = 57

iii) x won the election.

QUESTION FOURi) To find maximum or minimum value we use differential calculus as

follows:

i.e. 81 – x = 0 Which gives x = 18 or = -18 t hus x =18 or –18

Check for a maximum or a minimum

=

when x = 18, = -9 which is negative

Therefore at x = 18, the value of R is maximumSimilarly at x = -18, the value of R is minimum.

Therefore, the number of units that maximize the revenue = 18 units

ii) The maximum revenue is given by

R = 14 + 81 × 18 –

= Ksh.986

iii) The price per unit to maximize the revenue is

= 45.78 i.e. Ksh.54.78

QUESTION FIVEi) Profit is maximized when


371 Lesson Nine

Marginal Revenue = Marginal cost

Note: Profit = Revenue – Costπ = R – C

grad of π = Grad of R – grad of C= (marginal Revenue – marginal Cost)

Grad of π= 0 for maximum and minimum.

PRODUCTION DEPARTMENT

Marginal Revenue = Revenue per unit = 6 + 6 × 0.8= Shs.10.8 per unit

Cost per unit = + 6 + 0.0002 (marginal cost)

Therefore + 6.0002 = 10.8

Giving Q = = 4,166.84

= 4,167 units

ii) SALES DEPARTMENT

Profit = Revenue – CostRevenue = Sales price x Quantity = p(40,000 – 2,000p)

= 40,000p – 2,000p2

Cost = 2 X q + 6,000 + 6q + 80% of 6q = 2q + 6p + 4.8q + 6,000

= 12.8q + 6,000= 12.8 (40,000 – 2,000p) + 6,000

Profit = (40,000p – 2,000p2) – [6,000 + 12.8(40,000 – 2,000p)]

π = 40,000p – 2,000p2 – 6,000 – 512,000 + 25,600p

π = -2,000p2 + 65,600p – 518,000

= -4,000p + 65,600

= -4,000p + 65,600 When = 0


Revision Aid 372

P = = 16.4

Selling price = Shs.16.4

iii) Quantity 4,167 (as in i)Selling price Shs 16.4 (as in ii)

Firms Profit = Total Revenue – Total CostRevenue = 4,167 × 16.4 = Shs.68,338.8

Cost of production

Production Department 20,000 + 6.0002 × 4,167Sales Department 6,000 + 2 × 4,167

Total = 26,000 + 25,002.833 + 8,334

= 59,336.83

= Shs.59,336.85

Profit = 68,338.8 – 59,336.85

= Shs.9,002

iv) Quantity and sales price that maximizes the shop’s profit.

Revenue = (40,000 – 2,000p)p [Q = 40,000 – 2,000p]

Cost = (40,000 – 2,000p) (6 + 0.0002 + 2) + 20,000 + 6,000

Profit = (40,000p – 2,000p2) – [(40,000 – 2,000p) (8.0002) + 26,000]

π = 40,000p – 2,000p2 – 320,008 + 16,000.4p - 26,000

= -2,000p2 + 56,000.4p – 340,008

For maximum profit

= 0 and is –ve

= -4,000p + 56,000.4

Therefore -4,000p + 56,000.4 = 0 giving p = 1.0001 = Shs.14


373 Lesson Nine

= -4,000 which is –ve

for maximum profit = -2,000 (14) + 56,000.4 × 14 – 340,008

= -392,000 + 784,005.6 – 340,008

= 47,998

Quantity Q = 40,000 – 2,000 × 14

= 12,000 units.

QUESTION SIXa) Quadratic functions in decision making.

Due to economies of scale, the cost of production is usuall dependent upon volume of sales.

Total revenue = sale price per unit × quantity sold (say x) Sales price is usually a function of x; f(x).Hence total revenue = f(x) × x ; which may be a

quadratic functionUsing calculus techniques (i.e. maxima and minima) we can calculate the optimum value of x, which might give maximum profits or minimum costs.

b) Demand function p = 400 – q (price p in shillings and q in Kg)

Average total cost of prodcing the quantity =

Hence total cost

Revenue = Price × Quantity = p× q = (400 – q) q = 400q –q2

Profit π = (400q – q2) – (1000 + 100q – 5q2 + q3)= 300q + 4q2 – q3 – 1000

i.

= average fixed cost/unitas quantity sold gradually increases, average fixed cost per unit decreases

ii. for maximum profit


Revision Aid 374

iii.


375 Lesson Nine

QUESTION SEVENi.

ii.1. Marginal productivity


Revision Aid 376

2. Marginal productivity is the increase in output of electric furnaces per week if the capitalization is increased by Sh.1 million.


377 Lesson Nine

LESSON THREE

QUESTION ONEa) Discrete data have distinct values with no intermediate points, whereas

continuous data can have any values over a range either a whole number or any fraction.

b) Dispersion is the variation or scatter of a set of values.

Standard deviation is represented by;

Where s: is for a sample and σ: is for a population

c) This is the coefficient of dispersion of a distribution that is used in comparing dispersion between distributions. It is given by;

.

d) See Text

QUESTION TWOa)

Payment in daysMid-point Number of customers

x f f fx2

5 – 9 7 4 28 19610 – 14 12 10 120 1,4405 – 19 17 17 289 4,91320 – 24 22 20 440 9,68025 – 29 27 22 594 16,03830 -34 32 16 512 16,38435 – 39 37 8 296 10,95240 – 44 42 3 123 5,292

100 2,405 64,895


Revision Aid 378

b)

c) Histogram to show payment record of 100 customers

Number of Customers 25

22 20 20

17 15 16

10 10 8

5 4 3

4.5 9.5 14.5 19.5 24.5 29.5 34.5. 39.5 44.5

Days taken to settle debt

d) Out of 100, 16 i.e. in the class 30 to 34 days and 8 lie in the class 35 to 39 days. Therefore, the best estimate that an unpaid invoice chosen at random will be between 30 and 39 days old is

= 0.24

QUESTION THREEa) The smallest value in the distribution is 105, the largest in the

distribution is 142. The range to be spanned is 142 – 105, i.e. 37. The following grouping is a suggestion.

The classes should be of equal width.Group Tally Frequency105 but less than 110 I I 2110 115 I I I I 5115 120 I I I I 4120 125 I I I I I I I 8


379 Lesson Nine

125 130 I I I I I I I I 10130 135 I I I I 5135 140 I I I I 4140 145 II 2

40

b) Calculate the cumulative frequency.

Group FrequencyCumulativeFrequency

105 but less than 110 2 2110 115 5 7115 120 4 11120 125 8 19125 130 10 29130 135 5 34135 140 4 38140 145 2 40

SummaryMedian value = 125 pFirst quartile = 119 pThird quartile = 131 p

The semi-interquartile range is given by the formula ½ (third quartile value – first quartile value)

Thus the semi-interquartile range = ½(131 – 119)= ½ × 12= 6p

c)

Group Mid-point f fx105 but less than 110 107.5 2 215.0110 115 112.5 5 562.5115 120 117.5 4 470.0120 125 122.5 8 980.0125 130 127.5 10 1,275.0130 135 132.5 5 662.5135 140 137.5 4 550.0140 145 142.5 2 285.0

40 5,000

Calculate the mean value first using the

Formula = Σfx Σf


Revision Aid 380

Thus = 5,000 ÷ 40

= 125p

Continuing now with = 125 we can calculate the standard deviation

Group x (x - ) f f (x- )2

105 but less than 110 107.5 -17.5 2 612.50110 115 112.5 -12 5 781.25115 120 117.5 .-7.55 4 225.00120 125 122.5 -2.5 8 50.000125 130 127.5 2.5 10 62.50130 135 132.5 7.5 5 281.25135 140 137.5 12.5 4 625.00140 145 1425 17.5 2 612.50

40 3,250.00The standard deviation σ is given by the formula

d)

i) This distribution is very nearly normal and so consequently the mean at 125 and the median at just over 125 are close to one another.

ii) The semi-interquartile range and the standard deviation both measure dispersion. The semi-interquartile range, in this case 6p, gives the dispersion around the median. The standard deviation measures the dispersion around the mean, in this case 9p, for the whole distribution

QUESTION FOURUse the same method as in question 77 to find mean and standard deviation or use the formulae

Mean =

Supermarket A


381 Lesson Nine

Mean = = 49.92

Standard Dev = 30.49

= 30.5

Supermarket B

Mean = = 51.08

= 51.1

Standard Dev = 34.25

Coefficient of Variation Supermarket A

= × 100

= 61.1%

Coefficient of Variation Supermarket B

= 34.25 × 10051.1

= 67.02Hence variability of supermarket B is relatively greater than supermarket A

QUESTION FIVEThe objective of this question is to test the candidate’s knowledge of:Use of base and current weighting index numbers and contrast their construction.

a) To establish the base weighted indices, the weights are the quantities used in 1981. the following tabulation leads to the solution.

For 1981Competent Weight Price (£) Price X weight

(£)A 3 3.63 10.89B 4 2.11 8.44C 1 10.03 10.03D 7 4.01 28.07

57.4


Revision Aid 382

For 1982Competent Weight Price (£) Price X weight

(£)A 3 4.00 12.00B 4 3.10 12.40C 1 10.36 10.36D 7 5.23 36.61

71.37For 1983Competent Weight Price (£) Price X weight

(£)A 3 4.49 13.47B 4 3.26 13.04C 1 12.05 12.05D 7 5.21 36.47

75.03

The base weighted price indices are therefore for

1981, 100

1982, × 100 = 124.27

1983 × 100 = 130.65

b) The current weighted indices use the weights of the components in 1982 to establish the 1982 index, then the weights of the components in 1983 for the 1983 index. The following tabulation leads to the solution.

The 1981 index as before 100

For 1982

Component

Weight1982

1981Price (£)

1981Price ×Weight

(£)

1982Price (£)

1982Price ×Weight

(£)A 2 3.63 7.26 4.00 8.00B 5 2.11 10.55 3.10 15.50C 1 10.03 10.03 10.36 10.36D 6 4.01 24.06 5.23 31.38

51.90 65.24

The 1982 current weighted price index is therefore


383 Lesson Nine

× 100 that is 125.70

Component

Weight 1981Price (£)

1981Price ×Weight

(£)

1982Price (£)

1982Price ×Weight

(£)A 2 3.63 7.26 4.49 8.98B 6 2.11 12.66 3.26 19.56C 1 10.03 10.03 12.05 12.05D 5 4.01 20.05 5.21 26.05

50.00 66.64

The 1983 current weighted price index is therefore

× 100 that is 133.28

c) Laspeyres price indices use weights at the base period, whereas Paasche price indices use weights from the current period.

The weights at the base period will always be available whereas the current weights may not always be available. In application like Retail or Consumer Price Index establishing the current weights will be much more difficult and expensive than establishing the current prices. These arguments give a preference for the Laspeyres type of index number.

It can be argued that the use of current weights reflects the present situation more accurately, giving a preference for the Paasche type of index number. However, in computing the series of index numbers, it is clearly demonstrated in part (a) that the denominator need only be calculated once in the series for base weighted indices, whereas a recomputation is needed for current weighted indices. This leads to a favouring of the Laspeyres type index.

QUESTION SIXa)i) The Laspeyres Price index is often summarized by the formula

That is the weighting factor in the calculation is the quantity at the base period.

For 1983

Compon Quantit Price Quantity Price Quantity1


Revision Aid 384

ent y (1982) 1982× Price1982

1983 982×

Price1982

A 10 3.12 31.20 3.17 31.70B 6 11.49 68.94 11.58 69.48C 5 1.40 7.00 1.35 6.75D 9 2.15 19.35 2.14 19.26E 50 0.32 16.00 0.32 16.00

142.49 143.19

Thus the Laspeyres Price Index for 1983

=

= 100.49

For 1984

Component

Quantity Price Quantity 1982

1982 1984 × Price 1984A 10 3.2 32.00B 6 11.67 70.20C 5 1.31 6.55D 9 2.63 23.67E 50 0.32 16.00

148.24

Thus the Laspeyres Price Index for 1984

= × 100

= 104.04

ii) The Paasche Price Index is often summarized by the formula

That is the weighting factor in the calculation is the quantity at the current year.

For 1983

Compone Quantit Price Quantity Price Quantity19


385 Lesson Nine

nt y (1982) 1982× Price1982

1983 82×

Price1982

A 12 3.12 37.44 3.17 38.04B 7 11.49 80.43 11.58 81.06C 8 1.40 11.20 1.35 10.80D 9 2.15 19.35 2.14 19.26E 53 0.32 16.96 0.32 16.96

165.38 166.12

Thus the Paasche Price Index for 1983

=

= 100.45

For 1984

Component

Quantity

Price (1982)

Quantity 1982

× Price1982

Price 1983 Quantity1

982×

Price1982

A 14 3.12 43.68 3.20 44.80B 5 11.49 57.45 11.67 58.35C 9 1.40 12.60 1.31 11.79D 10 2.15 21.50 2.63 26.30E 57 0.32 18.24 0.32 18.24

153.47 159.48

Thus the Paasche Price Index for 1984= × 100= 103.92

iii) The following is a comparison between the index numbers.

Year Laspeyres

Paasche

1983 100.49 100.451984 104.03 103.92

There is little to choose between the two measures in 1983 as the weightings, that is, the quantity for 1982 and 1983 are close. The situation is different in 1984 where the weightings have increased in three cases by a significant amount over the 1982 figures resulting in a large increase in the Laspeyres index than the Paasche index.


Revision Aid 386

b) An index of industrial production would probably by calculated on a month by month basis by central government, indicating in percentage terms by how much production in the industrial sector has either grown or declined over the previous month or year. The full index would be made up from components which apply to particular sectors and so it is possible for an employer to measure increase or decrease in production in that sector and increase in production may thus be rewarded and a decrease would be looked upon with less favour.

An index of retail prices would also probably be calculated on a month by month basis by central government, and is an indication in percentage terms of the increase or decrease in retail prices. This measure is often used to quantify inflation. A trade union may therefore argue its case for an increase in pay to be greater than or equal to the rate of inflation to keep up with the cost of living and not decrease the living standard s of its members. An index of wages may also be published by central government, and may be available for different industrial sectors.

QUESTION SEVENNote: The production for 1980 and 1990 is given in 1,000 boxes. As long as units are kept the same throughout the problem, the rates will not change.

Produce

Production1,000 boxes

Price per box

Po(Shs)Pn

PnPo

Po Qo

1980 1990 1980 1990

Cabbages

48,600 62,000 100 150 1.5 4,860,000

7,290,000

Tomatoes

22,000 37,440 220 310 1.4091 4,840,000

6,820,000

Onions 47,040 61,430 180 200 1.1111 8467,200 9,408,000

Spinach 43,110 55,720 130 170 1.3077 5,604,300

7,328,700

Σ 5.3279 23,71,500

30,846,700

a) Mean Relative Index of prices = 133.20

Hence there is a 33.2% increase in average price of the four horticultural products from 1980 to 1990.


387 Lesson Nine

b) Laspeyre’s Price Index = × 100 = 129.76

% increase 29.76%

c) Paasche’s Price Index = Σpnqn × 100ΣpOqn

And

d) Marshall Hedge-worth index = ΣPn(qn + qn) × 100ΣPO (qn + q n )

pnqn poqn pn + qn pn (qn + pn) pn(pn + qn)Cabbages 9,300,000 6,200 110,600 16,590,00

011,060,00

0Tomatoes 11,606,40

08,236,800 59,400 18,426,40

013,076,80

0Onions 12,286,00

011,057,40

0108,470 21,694,00

019,524,60

0Spinach 9,472,400 7,43,600 98,830 16,801,10

012,847,90

0Σ 42,664,80

032,737,80

073,511,50

056,509,30

0

e) Paasche’s index= × 100 = 130.32

% increase 30.32%

f) Marshall Hedge-worth Index= × 100 = 130.09

% increase = 30%

g) Fisher’s Price Index = (129.76 × 130.32)1/2 =130.4

This gives a percentage increase = 30%


Revision Aid 388

LESSON FOURQuestion 1 and 2See text

QUESTION THREEr2

x1 = 0.78

This is the coefficient of determination of miles traveled to cost and means that 78% of total cost is attributable to mileage

r2x1 = 0.16 i.e. 16% of cost is accounted for by the type of journey

R2 = 0.88 is the overall coefficient of determination and indicates that the multiple regression equation accounts for 88% of the total variation in costs.The coefficient in the equation are:

a = 86 = fixed costsb1 = 0.37 = amount per mileb2 = 0.08 = influence of the type of journey

QUESTION FOURForecasts produced by3 monthlyMoving average

6 monthly moving average

12 monthly moving average

450437417397 423383 410377 397380 388407 395443 410470 425 424


389 Lesson Nine

QUESTION FIVEa)Year Qtr Sum of

four qtrs

Sum of two qtrs

Trend Actual

minus trend

Actual/

trend

1990/91

1 49

2 37 2113 58 212 423 52.875 5.125 1.0974 67 213 425 53.125 13.875 1.2611 50 214 427 53.375 -3.375 0.9372 38 215 429 53.625 -

15.6250.709

3 59 216 431 53.875 5.125 1.0954 68 218 434 54.25 13.75 1.2531 51 219 437 54.625 -3.625 -.9342 40 221 440 55 -15 0.7273 60 220 441 55.125 4.875 1.0884 70 222 442 55.25 14.75 1.2671 50 223 445 55.625 -5.625 0.8992 423 61

b) EitherAdditive model

QuarterYear Q1 Q2 Q3 Q41990/91 5.125 13.8751991/1992 -3.375 -15.625 5.125 13.751992/93 -3.625 -15 4.875 14.751993/94 -5.625 ______ _____ ______Total -12.625 -30.625 15.125 42.375Average -4.208 -15.313 5.042 14.125 = -

0.354Seasonal variation

-4 -15 5 14 = 0

Or

Multiplicative modelQuarter

Year Q1 Q2 Q3 Q41990/91 1.097 1.2611991/1992 0.937 0.709 1.095 1.2531992/93 0.934 0.727 1.088 1.2671993/94 0.899Total 2.770 0.718 3.280 3.781Average 0.923 0.718 1.093 1.260 = 3.994


Revision Aid 390

Adjustment factor*

1.0015 1.0015 1.0015 1.0015

Seasonal variation

0.924 0.719 1.095 1.262 = 4.000

* adjustment factor = 4/(3.994) = 1.0015

c) An explanation of the forecasting method:i) Plot a graph of the trend.

ii) By eye, establish an appropriate forecast of the trend for the last qurter of 1993/94 and the first three quarters of 1994/5. (Note: linear regression or the high/low method may be appropriate methods to establish the forecast.)

iii) Adjust the forecast trend for these quarters for the seasonal variations:

Additive modelEstimated data value = forecast trend value + appropriate seasonal variation value.

Multiplicative methodMultiply each point by the appropriate seasonal factor.

QUESTION SIXa) A = y – bx

b =

A B A × Bx x - y y - (x -

)2

2 6 -4 60 104.8 -44.8 179.2 168 6 2 132 104.8 27.2 54.4 46 6 0 100 104.8 -4.8 0 08 6 2 120 104.8 15.2 30.4 4

10 6 4 150 104.8 45.2 180.8 164 6 -2 84 104.8 -20.8 41.6 44 6 -2 90 104.8 -14.8 29.6 42 6 -4 68 104.8 -36.8 147.2 06 6 0 104 104.8 -0.8 0 16

10 6 4 140 104.8 35.2 140.8 1660 1,048 804.0 80

=


391 Lesson Nine

Therefore b = = 10.05

a = 104.8 – (10.05 × 6) = 44.5

where y = total cost (x 10)x = age in years

b) Maintenance costs using formula from (a):

Age of vehicles

Cost

(years) (£ = 10)1 54.552 64.603 74.654 84.705 94.756 104.807 114.858 124.909 134.95

10 145.00

c) 12-year-old vehicle would have an estimated maintenance cost of:

44.5 + (10.05 × 12) = 165.1 (£ x 10), or £ x 10), or 1,651.00

This forecast is an extrapolation beyond the data and consequently is less sound.

QUESTION SEVENa) Two possible reasons for the large variation in output each month

are:

Seasonal variation Production problems in some months

b) Graph showing relationship between output and costs.


Revision Aid 392

The graph shows there is a strong positive relationship between output and costs. This means that output may be used to predict costs.

x y xy x2 y2

16 170 2,720 256 28,90020 240 4,800 400 57,60023 260 5,980 529 67,60025 200 7,500 625 90,00025 280 7,000 625 78,40019 230 4,370 361 52,90016 200 3,200 256 40,00012 160 1,920 144 25,60019 240 4,560 361 57,60025 290 7,250 625 84,10028 350 9,800 784 122,50012 200 2,400 144 40,000

240 2,920 61,500 5,110 745,200

b = = = 10

a = – 10 × = 43.333

y = 43.333 = 10x


393 Lesson Nine

This means that when output is zero, costs are zero £43.333, and that for every one unit increase in output, costs will rise by £10. This assumes linearity.

QUESTION EIGHTi. The tabulation of the trend pattern is as follows and has been

computed using the formula Trend = Sales – Seasonal deviation

Year Quarter Sales Seasonal deviation

Trend

£ 000 £ 000 £ 0001983 2 360

3 5304 354 -42 396

1984 1 304 -128 4322 430 -37 4673 750 276 4744 395 -93 488

1985 1 340 -145 4852 500 12 4883 660 153 5074 509 -15 524

1986 1 374 -165 5392 590 43 5473 710 153 5574 521

1987 1 440

ii. The seasonal variations are established from the following table

Quarter Year 1 2 3 41983 -421984 -128 -37 276 -931985 -145 12 153 -151986 -165 43 153Total -438 18 582 -150Average -146 6 194 -50 Total 4Adjusted average

-147 5 193 -51 Total 0

Adjustment to each average –4/4 that is – 1To the nearest integer the seasonal variations are


Revision Aid 394

Quarter 1 -1472 53 1934 -51

iii. Forecast sales 1987 quarter II = 590 + 5 (£ 000)= 595 (£ 000)

QUESTION NINEThe object of this question was to test the candidates knowledge of the time series and the ability to present data on a labeled diagram

Year Quarter

Costs £ Four quarter total

Four quarter total in pairs

Trend Deviation

1980 IV 15601981 I 1730 6348

II 1554 6418 12766 1595.77 -41.75III 1504 6638 13056 1632.00 -128.00IV 1630 6679 13317 1664.63 -34.63

1982 I 1950 6715 13394 1674.25 275.75II 1595 6785 13500 1687.50 -92.50III 1540 6695 13480 1685.00 -145.00IV 1700 6809 13504 1688.00 12.00

1983 I 1860 6843 13652 1706.50 153.50II 1709 6933 13776 1722.00 -13.00III 1574 6983 13916 1739.50 -165.50IV 1790 6995 13978 1747.25 42.75

1984 I 1910 7061 14056 1757.00 153.00II 1721III 1640

a) The calculations for the trend figures and the deviations are summarized in the above table The seasonal effect is removed from the data by first totaling four quarter figures, then totaling the four quarter figures in pairs and finally dividing by eight and centering the trend figure at the middle pointDeviation is the difference between the costs and the corresponding

trend figureb) The seasonal deviation is another calculation produced from a table


395 Lesson Nine

SeasonalQuarter

Deviations

Year I II III IV1983 -41.75 -128.00 -34.631984 275.75 -92.50 -145.00 12.001985 153.50 -13.00 -165.50 42.751986 153.00Total 582.25 -147.25 -438.50 20.12Average 194.08 -49.08 -146.17 6.71 Total = 5.54Adjusted average

192.70 -50.46 -147.55 5.33 Total = 0.02

Seasonal deviation

193.00 -50.00 -148.00 5

The adjustment is obtained by reducing each figure by 1.38 (that is 5.54 ∸4)c) From the trend the forecast trend value of Quarter IV of 1984 is £

1790. the seasonal deviation is £ 5, hence the forecast heating costs for quarter IV for 1984 are £ 1795

It can be seen that the trend is a little variable and the seasonal deviations are not very regular, possibly due to the weather.Another possible reason for the irregular fluctuation in the heating cost trend line is an uneven increase in the price of the heating medium (for example, oil or gas or electricity.)


Revision Aid 396

LESSON FIVE

QUESTION ONEa) The Poisson distribution is appropriate because there is a small

probability of an event occurring these are discrete values and the average of these events (i.e. m = np) is below

10 (50 × 0.1 = 5)

b) The Poisson formula is

In this exmple m = np = 50(0.02) =1

P ( x 2) = P(x=0) +P ( x=1 ) + p (x=2 )

c) Similarly , the following can be drawn up

P Pa0 1.000

0.02 0.9200.05 0.5440.10 0.1250.15 0.020

QUESTION TWO(a)

(i) Probability = 0.7 × 0.3× 0.7 = 0.147

(ii) Probability =3C2 (0.7)2 (0.3) = 0.441;

(iii) The answer to (ii) is 3× the answer to (i)


397 Lesson Nine

This is because either A or B could be detected, A and C, or B and C. Each of these situations is equally probable, with the probability given in part (i).

(b) (i) The probability that the defect is undetected by the inspection procedure is 0.3.

The probability that it is undetected by the secondary check is 0.4.

The probability that it is undetected by both is 0.3 × 0.4 = 0.12

(ii) Call the two faults A and B.

We can have the possibilities:

A is found by inspection procedure, B not found by either procedure.

B is found by inspection procedure, A not found by either procedure.

A is found by secondary check, A and B not found by inspection procedure.

B is found by secondary check, A and B not found by inspection procedure.

Adding together the probabilities corresponding these four possibilities, the required probability is

(0.7×0.3×0.4) + (0.3×0.7×0.4) + (0.3×0.3×0.6×0.4) + (0.3×0.3×0.4×0.6)

=0.0840 + 0.0840 + 0.0216 + 0.0216 + 0.1680 + 0.0432

=0.2112

(iii) Probability that a fault is detected by the inspection procedure is 0.7. Probability that a fault is detected by secondary check is 0.3 × 0.6 =0.18

Therefore the proportion of faults detected by the inspection procedure and secondary check, respectively, is

0.7 0.18 and 0.7 0.18 0.7 0.18

70 18that is and 88 88

QUESTION THREELet F represent a unit which has been found to be faulty.


Revision Aid 398

Let P(S1) = probability that a unit chosen at random comes from S1



P(S1) = 0.40P(S2) = 0.35

P(S3) =

The percentages of faulty unit are as follows:

The required probability may be expressed as;

the unknown probability is P(F) to be slotted into the formula

Note that

the faulty part can only have come from S1 or S2 or S3 and so

since P(F) is a denominator and the sum equals unity then the expression

must be equal to P(F)

thus P(F)=

= 0.0285

Substitution into

Gives

= 0.2807


399 Lesson Nine

QUESTION FOUR

QUESTION FIVEP = 0.6 q = (1-p) = 1-.06 = 0.4

0.65 0.60 1.430.035

z

Which gives 0.4236 (42.36%)

This means there is a (0.5-0.4236) 0.0764(7.64%) chance of 65% or more passing the first attempt. This is graphically shown below.

Probability density

42.36%

7.64%

0.60 0.65 Xproportions


Revision Aid 400

i.e. the mean

QUESTION SIXa) The left-hand tail of the distribution below 900 hours represents the

number of lamps that will fail before 900 hours. Accordingly, if the probability of the distribution above 900 hours is found and deducted from 0.5, the required number can be found thus:

The probability of which is 0.4082

Lamps failing before 900 hours =5000(0.5- 0.4082)=459

b)1000 950 0.67

75z


Lamps failing between 1000 and 950 hours = 5000 (0.2486)=1243

c)

1000 925 175

z


Proportion failing between before 925 hours = 0.5 – 0.3413 = 15.87%

d) A probability of 0.3(0.5-02) is found in the tables with a Z score of 0.84

Thus 0.84=1000-916 84z= =1 s.d.= =100

s.d. 0.84


1000 900 1.3375

z

401 Lesson Nine

LESSON SIX

QUESTION ONEThe sample mean is 150 grams so that the estimate of the population mean is 150 grams.

i.e. Where means ̀ best estimate

of ́.

When n= 625Standard error of the mean

s 30= = =1.2grams625n

When n =1225

x30s = =0.857 grams1225

QUESTION TWOCorrection factor =

n 80Approximation to correction factor= 1- 1 0.9486N 800

It will be seen that to three significant figures, it is accurate enough for all practical purposes, the two formulae give the same result, i.e. 0.949.

Standard deviation error of the means =

Note: The standard error without correction is 6 0.671 grams80


N-n 800 80 0.9493N-1 800 1

Revision Aid 402

Thus the precision of the sample estimate, measured by the standard error, is determined not only by the absolute size of the sample but also to some extent by the proportion of the population sampled.


403 Lesson Nine

QUESTION THREEThe Central Limit Theorem states that the means of samples (and the medians and standard deviations) tend to be normally distributed almost regardless of the shape of the original population.

QUESTION FOURSee text

QUESTION FIVEH0 :2 =1

H1 : 2 1 (one-tail test)

Pooled sample proportion

and

The critical value of one-tailed test t the 5% level is 1.64 so that as the calculated value is lower than this value we conclude there is insufficient evidence to reject the null hypothesis.

QUESTION SIXIt will be seen that this is a two-tailed test.The common standard deviation is calculated first.


Revision Aid 404

and, finally the t score can be calculated

At 5% level with (n1+n2-2)= (12+9-2) = 19 degrees of freedom value is 2.093.

Since the calculated value is greater than 1.96, the null hypothesis can be rejected at the 5% level, i.e. we conclude that there is a significant difference between the mean monthly incomes.

QUESTION SEVENWe are testing whether the observed number of defects fits a binomial distribution, thus;

H0: the observed number of defects conforms to a binomial distribution of the form(p+q) 5 where p=0.18

H1 :that the observations do not conform.

The observed frequencies are already given so its only necessary to calculate the frequencies expected from a binomial distribution to the power 5 i.e. (p+q)5, where p=0.18 and q=1-0.18= 0.82.

The probabilities of the various values of p and q can be found from binomial probability tables if available. Alternatively they can be calculated from the binomial expansion. I.e.

P5 +5 (p4 q)+ 10(p3 q2)+ 10(p2 q3)+ 5(p q4)+q5

This shows the probabilities for 5, 4, 3, 2, 1 and 0 defectives and when p = 0.18 (the probability of a bulb being defective) and q=0.82 (the probability of not being defective) the probabilities range from 0.0002 (i.e. 0.185) for five defectives to 0.3711 (0.825) for no defectives. When the probabilities are known they are multiplied by 100 boxes to find the expected frequencies which are used in the normal X2 procedures.

The table below summarises the calculations:


405 Lesson Nine

Defectives No. of boxesObserved

Binomial Probabilities

Expected Frequency

(O - E) 2 (O - E) 2

E0 40 0.3711 37.11 8.35 0.221 37 0.4069 40.69 13.62 0.332 17 0.1786 17.86 0.74 0.043 5

10 } 0.0392 3.92

0.40.02 }4 0.0040 2.76 0.64

5 0.0002 1.23

Note: Because of the very small values of the expected frequencies for 3, 4 and 5 defectives they have been combined into one but it makes little difference to the results if they are not combined.

The calculated value is compared with the value for the appropriate degrees of freedom. Because the last three classes have been combined there are four classes remaining, i.e. for 1, 2 and the combined class for 3-5 rejects thus V =n – 2 = 4 - 2=2.

The X2 value for two degrees of freedom at the 5% level is 5.991 and, as the calculated value of 1.23 is well below this, we accept the null hypothesis and conclude that the observed values fit a binomial distribution to the power 5 when p = 0.18.

Note: If the last three classes had not been combined, the calculated X2 score would have been 1.8 and there would have been 4 degrees of freedom. At 4 degrees of freedom the score is 9.488, so the conclusion would be the same, i.e. we accept the Null Hypothesis.

QUESTION EIGHTSee textQUESTION NINE1. a)

S.e. =

Where p =proportion late

Actual p = = 0.075


Revision Aid 406

This is less than the standard errors at the 5% level ± 1.96 so we conclude there is no significant improvement in deliveries.

b) 0.75 standard errors from the mean would cover ± 0.2734 i.e. 54.68%, say 55% of the population, so the MD’s claim could be accepted at any level of confidence

2. Calculated mean of sample = 4.68 gmsCalculated sample s.d. = 1.968

There are 10-1 =9 d.f and it is a one-tailed test. The 5%value for a one-tailed test is 1.833.

The sample mean should be within 1.833*0.66 gms = 1.21 gms.

The actual difference is 4.68-3.8 gms= 0.88 gms so the figures do not support the Purchasing Manager’s assertion.


407 Lesson Nine

LESSON SEVEN

QUESTION ONEa)

b). Drilling Immediately:

Tests:Successful and drill, Fail and drill, There fore EV =

Overall EV = £53.5

Thus course of action is: first carry out tests, if successful proceed to drill if tests fail, sell the exploitation rights.


Fail 0.45

Succeed 0.55

Fail 0.2

Succeed 0.8

Sell 70

15

Fail 0.8

NPV

45

120

-40

65

100

-50

15

100

-50

Succeed 0.2 Drill

Fail 0.3

Succeed 0.7 53.5Tests

Drill immediately 53.5

1

2

2

3

3

4

1

Sell immediately

Sell

-20

70

48

Drill

Revision Aid 408

QUESTION TWOa) tree diagram

b) EV calculationsEV(present arrangement) = £(0.3×1000) + (0.23×900) + (0.23×800) + (0.23×700)

= 860

EV (with consultants) = 785

c) on strict EV calculations the firm should not use consultants, however management may consider to use consultants in order to improve the chances of completing on time thus safeguarding their reputation.

QUESTION THREESee text

QUESTION FOURLet the value of the small store be = 1

And let the value of the large store be = 2

If both survive, A loses nothing, if only larger store survives, A loses 1 and if smaller store survives, A will lose 2.


Continue as present

Employ consultants(Cost = £200,000)

On time (p = 0.3)

1 month delay (p = 0.23)

2 months delay (p = 0.23)


1 month delay (p = 0.05)


On time

1,000

1,000 – 100 = 900

1,000 – 200 = 800

1,000 – 300 = 700

1,000 – 200 = 800

1,000 – 200 - 100 = 700

1,000 – 200 - 200 = 600

Profits before construction costs (£’000)

409 Lesson Nine

GAIN MATRIX FOR A

DEFENDERA

BI II Row

Minimum

I

II

0 -2

-1 0

-2

-1

Column Maximum

0 0

There is no saddle point.

Hence this is a problem of mixed strategy.

Using method given I II

I

II

0 -2

-1 0

0-(-2)=2

0-(-1)=1

0-(-1)=10-(-2)=2

The final strategy is given by the matrix

A

A plays his first row 1/3 rd of the time (randomly)

A plays his second row 2/3 rd of the time.

Similarly


Attack the smaller store

I

Attack the larger store

IIDefend the smaller

storeI

Both survive0

The larger store destroyed

-2Defend the larger

storeII

The smaller store destroyed

-1

Both survive0

I III

II

0 -2

-1 0

1/3

2/3

2/3 1/3

Revision Aid 410

B plays his first column 2/3 rd of the time

B plays his second column 1/3 rd of the time.

The values of the game

=0×1/3×2/3 +(-2)×(1/3×1/3) +(-1)×2/3×2/3 +0×1/3×2/3

=0-2/9-4/9+0

= -6/9

= -2/3

QUESTION FIVE

YRow Minimum

X1 2 -1 -1-2 1 -2 -22 2 1 0

Column Maximum

2 2 1

There is no saddle effect.

Let the probabilities of y be p, q, r.

Then the three payoffs to y corresponding to each of the three counter moves of his opponent x must all be equal to the optimal value v of the game.

Y’s payoffs against the three moves of x are

1p+2q+(-1r) -2p+1q+1r 2p+0q+1r

We obtain three equations by equating each of the three payoffs to v

1p+2q-1r = -2p+1q+1r = 2p+0q+1r = v Also p+q+r = 1 (Total probability)

1p+2q+1r= -2p+1q+1r….. (1)1p+2q+1r= 2p+0q+1r……(2)p+q+r = 1 …..(3)

Solving these equations simultaneously, we get

P=2/17, q=8/17, and r=7/17


411 Lesson Nine

Similarly using the same reasoning as before

Let the three probabilities of x be p’, q’, r’

We get 1p’+2q’+1r’ = 2p’+1q’+0r’ = -1p’+1q’+1r’

Also p+q+r = 1

Solving them simultaneously, we get

P’=3/17, q’=5/17, and r’= 9/17

Hence x should play his rows in the ratio 3: 5: 9And y should play his columns in the ratio 2: 8: 7

(Note: Rows and columns should be played at random)

Σ [(payoff * joint probability of the payoff]

+6 other values calculated in the same way as above which amount to 11/17

Alternatively

Value of the game is


Revision Aid 412

LESSON EIGHT

QUESTION ONEa) Let x1 and x2 be the number of shares invested in airline and

insurance shares respectively. Then the objective function will be as follows:Objective function (maximize)

z = 2x1 + 3x2 Share appreciation.

Subject to the following restrictions:1. 40x1 + 50x2 ≤ 100,000 sh. Total investment.2. 50x1 ≤ 40,000 sh. Investment in insurance.3. x1 + 1.5x2 ≥ 2600 sh. Dividends.4. x1, x2 ≥ 0 Non-negativity of number.

b) Reduced cost represents the amount that objective function coefficient of a non-basic decision variable must improve (increase in this case) to be put into the basic. In this case since reduced cost is equal to zero for both variables, it means they are in the basic.

Dual prices on the other hand mean the amount that share appreciation will improve in case any of the limiting constraints is increased by one unit. This occurs for constraints that the slack is exactly zero. In this case total investment and investment in insurance do have positive dual prices while Dividends does have zero dual price (can not increase share appreciation if increased by one unit).

c) From the computer solution, the client’s money should be invested as follows, to satisfy the restrictions:1500 shares to be invested in Airline shares, and 800 shares to be invested in insurance shares, to give an optimum quarterly share appreciation of sh. 5400.

d) For the optimal decision to remain the airline shares appreciation should not be lower than 2.5, but can be any higher amount. That is, the optimum solution is insensitive to increase in the share appreciation.

QUESTION TWOa) Formulation of the problem

Take x1 to be the number of large loaves and x2 to be the number of small loaves.

Objective function

z = 5x1 + 3x2 Profit


413 Lesson Nine

Constraints.Line 1 x1 ≤ 280 Maximum number of large loaves.Line 2 x2 ≤ 400 Maximum number of small loaves.Line 3 10x1 + 8x2 ≤ 4000 Space (1000-3) m3

Line 4 25x1 + 12.5x2 ≤ 8000 Hours (1000-3)5 x1, x2 ≥ 0 Non-negativity.


Revision Aid 414

Using graphical method.

The feasible area is that enclosed by corner points ABCDEFCorner points are where the optimum feasible solution exists. The upper points are to be considered for maximization problem.

At corner point A, Line 2 and x1=0 intersectSo x2= 400Putting these value in the objective function gives the followingz = 5x1 + 3x2=50 + 3400=1200 ProfitA

At corner point B, Line 2 and Line 3 intersect

x2= 400 and x1=

Putting these value in the objective function gives the followingz = 5x1 + 3x2=580 + 3400=1600 ProfitB

At corner point C, Line 3 and Line 4 intersect

The equations for the lines are:Line 310x1 + 8x2 = 4000 (1)Line 425x1 + 12.5x2 = 8000 (2)Multiplying equation (1) by 25 and equation (2) by 10 gives the following


BA

C

D

EF

415 Lesson Nine

(10x1 + 8x2 = 4000) 25 (25x1 + 12.5x2 = 8000) 10

250x1 + 200x2 = 100,000 (3)250x1 + 125x2 = 80,000 (4)Deducting equation (4) from Equation (3) gives:

75x2 =20,000Þx2 =266.7

Þx1 =

Putting these values in the objective function gives the followingz = 5x1 + 3x2=5 186.7 + 3 266.7=1733 ProfitC

At point D, Line 1 and Line 4 intersect.x1 = 280

Þx2 =

The profit at this point is then equal to:z = 5x1 + 3x2=5280 + 380=1640 ProfitD

At point E, Line 1 and x2 = 0 intersectx2 = 0 Þ x1 = 280z = 5x1 + 3x2=5280 + 30=1400 ProfitE

Comparing these profits, it is at point C that profit is maximized.So the solution is that:

x1 = 186 No. of large loaves produced.x2 = 266 No. of small loaves produced.

And the maximum profit is Profit c = Shs. 1,733 where x1 = 186.7, x2 = 266.7

NOTE: Two methods can be used to solve the problem. It is easily solved using the graphical rather than the simplex method, since it is just two variables and sensitivity analysis is not required.

b) To solve this kind of problem (linear programming problem) the following procedure is followed:- First, the problem has to be formulated. That is, the objective

function and constraints are determined.Objective function is that which is to be optimised.Constraints are the limitations in resources.

- Secondly, the method of solving is determined. In this case, of a two-variable problem, the better method to use is graphical method, rather than simplex method.

- Thirdly, the constraints are taken as equalities and a line graph drawn. The unwanted regions are shaded out. Resulting region indicates the feasible region. The optimum point exists where there are corner points, which show extreme amounts. For


Revision Aid 416

maximization it is the outer ones to the right and up. For minimization it is the lower side.

- Lastly, the profit is determined at those points where there is maximum profit, is the point to be used.

NOTE: This part simply asks for the procedure followed.

QUESTION THREEa)

i) Simplex method will be appropriate.Formulation of problem.

Objective function.Let x1 and x2 be the number of Deluxe and Professional bicycle frames produced respectively per week.

z = 1000x1 + 1500x2 Profit sh.

Constraints:2x1 + 4x2 ≤ 100 Aluminum alloy3x1 + 2x2 ≤ 80 Steel alloyx1, x2 ≥ 0

In standard form:0 = z – 1000x1 – 1500x2 + 0s1 + 0s2

100 = 2x1 + 4x2 + s1 + 0s2 80 = 3x1 + 2x2 + 0s1 + s2

Table 1

x1 x2 s1 s2 Solution Ratios1 2 4 1 0 100 25 Üs2 3 2 0 1 80 40z -1000 -1500 0 0 0

ÝTable

2x2 1/4 1 1/4 0 25 50s2 2 0 -1/2 1 30 15 Üz -250 0 375 0 37,500

ÝTable

3x2 0 1 3/8 -1/4 17.5x1 1 0 -1/4 1/2 15z 0 0 312.5 125 41,250

Stop hereThe optimal weekly production schedule is as follows:Deluxe bicycle Frame = 17.5 ≈17Professional bicycle Frame = 15


417 Lesson Nine

ii) Let Δ1 be the change in profit from Deluxe bicycle frame.Δ2 be the change in profit from Professional bicycle frame. So C1 = 1000 + Δ1 and C2 = 1500 + Δ2 limit of profit.

From the final table:To avoid entry of s1 Þ312.5 – 1/4Δ1 > 0 Þ Δ1 < 1250s2 Þ125 + 1/2Δ1 > 0 Þ Δ1 > -250

From the two conditions:-250 < ∆1 < 1250 and750 < C1 < 2250

To avoid entry of s1 Þ312.5 + 3/8Δ2 > 0 Þ Δ2 > -833.33s2 Þ125 – 1/4Δ2 > 0 Þ-Δ2 > -500 ÞΔ2 < 500

So from the two conditions:-833.33 < ∆2 < 500And C2 varies as follows666.7 < C2 < 2000

NOTE: This problem could be solved graphically with part (i) Easily determined. Part (ii) Limits will be determined from equating slopes of the objective function which has coefficients with constraints nearest to it.For part (ii), accurate drawings will be required. Intuition will have to be followed and there will be an assumption that fractions are possible.

b) The technique is really involving.Assumes fractions are possible, which is not really the case like here where we cannot make ½ a bicycle frame.

QUESTION FOURa)

i) A feasible solution is one that satisfies the objective function and given constraints

ii) Transportation problem is a special linear programming problem where there a number of sources and destinations and an optimum allocation plan is required. Total demand equal total supply

iii) Assignment problem is a special kind of transportation problem where the number of sources equals the number of destinations. That means for every demand there is one supply.

b) This is a case of assignment problem.Assignment problems usually require that the number of sources equal the number of supply. Here there are 5 districts and only 4


Revision Aid 418

salespersons. A dummy salesperson E is introduced with zero ratings.

Districts1 2 3 4 5

A 92 90 94 91 83Sales persons B 84 88 96 82 81

C 90 90 93 86 93D 78 94 89 84 88E 0 0 0 0 0

By following the Hungarian method:Firstly:For each row, the lowest rating is reduced from each rating in the particular row. This results to a row reduced rating table. Then all the zeroes are to be crossed by the least number of vertical and horizontal lines. If the number of lines equal the number of rows (or columns = 5 in this case) then the final assignment has been determined. Otherwise the following steps are followed.


419 Lesson Nine

1 2 3 4 5A 9 7 11 8 0B 3 7 15 1 0C 4 4 7 0 7D 0 16 11 6 10E 0 0 0 0 0

Secondly, for each column, the lowest rating is reduced from every rating in the particular column. In this case the table will remain the same since the dummy salesperson has ratings of zero for every district.Thirdly a revision of the opportunity-rating table is done.The smallest rating in the table not covered by the lines is taken (in this case it is one). This is reduced from all the uncrossed ratings and added to the ratings at the intersection of the crossings. Then all the zeroes are to be crossed by the least number of vertical and horizontal lines. If the number of lines equal the number of rows (or columns = 5 in this case) then the final assignment has been determined. Otherwise the following steps are followed.

1 2 3 4 5A 8 6 10 8 0B 2 6 14 0 0C 4 4 7 0 8D 0 16 11 6 11E 0 0 0 0 1

Third step is repeated as follows:1 2 3 4 5

A 6 4 8 8 0B 0 4 12 0 0C 2 2 5 0 8D 0 16 11 8 13E 0 0 0 2 3

Still the optimal solution has not been reached. Third step is again repeated to give the following table:

1 2 3 4 5A 6 2 6 8 0B 0 2 10 0 0C 2 0 3 0 8D 0 14 9 8 13E 0 0 0 4 5

An optimal assignment can now be determined since the number of lines crossing the ratings is equal to 5.Lastly, the assignment procedure is that a row or column with only one zero is identified and assigned. This row or column is now eliminated. The other zeroes are then assigned until the last zero is


Revision Aid 420

assigned. This step-by-step assignment is shown on the following table from the first one to the fifth one.


421 Lesson Nine

District

1 2 3 4 5

A 6 2 6 8

Sales person B 0 2 10 0

C 2 3 0 8

D 14 9 8 13

E 0 0 4 5

The assignment is as follows

Salesperson

District Rating

A 5 83B 4 82C 2 90D 1 78Total rating 333

The total rating is 333.

QUESTION FIVEa) Sensitivity analysis measures how sensitive a linear programming

solution is to changes in the values of parameters. These parameters include the coefficients of objective function, limiting resources and non-limiting resources.

So sensitivity analysis involves changing any of these parameters and showing how the linear programming problem is affected.Dual values indicate the additional improvement of the solution due to additional unit of limiting resource. In that way, the additional improvement of solution is the price worth paying to release a constraint

b) Let x1, x2 and x3 be the units of desktop 386, Desktop 286 and laptop 486Maximize profitZ = 5000x1 + 3400x2 + 3000x3

Subject tox1+x2500 limit of desktop modelsx3 250 limit of laptop model


01

02

03

05

04

Revision Aid 422

x3 120 limit of 80386 chipsx2+x3400 limit of 80286 chips5x1+4x2+3x22000 hours availableAssumptions x1, x2, x3 ³ 0

Linearity/proportionality Divisibility Deterministic Additive

c)i) The optimum product mix is that the numbers of units to produce

are Desktop 386-120Desktop 286-200Laptop 486-200Maximum profit is Z=5000120+3400200+3000200=Sh1,880,000Unused resources include the followingJK computers can still produce 180 more desktop models (500-120-200) and 50 laptop models (250-200)For used up resources the prices to pay for any additional unit are as followsSh150 for 80386 chipSh90 for 80286 chipSh20 for any hour

ii) The range for the variables x1, x2 and x3 are to indicate where the number of units can change without affecting the basic solution

The range for the constraints indicate the extent the resources can be changed without altering the basic solution of the linear programming problem

iii) The dual value of 80386 chip is Sh 150. That is the addition increase in profit due to increase of one chip. So if the company increases the number of chips by 10, the additional profit will be 10150=Sh1,500.

QUESTION SEVENa) Network


423 Lesson Nine

The critical path is A, E, F, H, I.The duration of the critical path is 15 days.

b) Assuming all the activities start as soon as possible, the following chart shows when activities will start and finish.

Consequently the following resource allocation is required on a day to day basis.

Day Activities Number of increased staff

1 A, B 1+3 = 42, 3 A, D, G 1+1+1 = 34 A, G 1+1 = 25, 6 C, E, G 2+2+1 = 57 C, E 2+2 = 4 8 E 2 = 29, 10 F 2 = 2 11, 12 H 1 = 113, 14, 15 I 1 = 1

Current costs with 5 staff = 15 days 5 £500 = £37500If activity C is delayed to start on day 7 only 4 staff are required and the project duration unchanged. The cost is

15 days 4 £500 = £30000

However if the duration is increased by starting B on Day 1 and delaying activities A, E, F, H, I by one day and C until day 12 only a maximum of 3 staff are required.


1 6

4 4

8 8 10 10 12 12 15 15

EST LST

Critical activities

Time scale(days)

G

D B

A E F H I

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

C

Other activities

Revision Aid 424


425 Lesson Nine

MOCK EXAMINATION

Work out these question for three hours (exam condition) then hand them in to DLC for marking



QUESTION ONEa) Explain the importance of set theory in business

(4marks)b) By use of matrix algebra, develop the leontief inverse matrix.

(8marks)c) Digital ltd manufactures and sells floppy disks at Nairobi industrial

area. The average revenue (AR)(in thousands of shillings )of producing x floppy disks are given by the following functions

AndAR=800-2X2

Where: x is the number of floppy disks produced

Required:i. The profit function

(3marks)ii. The number of floppy disks required to maximize profit

(3marks)iii. The maximum profit

(2marks)(Total: 20marks)

QUESTION TWOa) State any five problems encountered in the construction of the

consumer price index(5marks)

b) An investment analyst gathered the following data on the 91-day Treasury bill rates for the years 2003and 2004

Month Treasury bill rates (%)2003 2004

January 3.2 5.5February 3.0 5.2March 2.8 4.3


Revision Aid 426

April 2.5 3.6May 2.9 3.3June 3.4 2.7July 3.7 2.4August 4.0 2.0September 3.8 2.3October 4.2 2.8November 4.5 3.1December 5.1 3.7

The analyst would like to determine if on average there was a significant change in the Treasury bill rates over the two years.

Required:i. The mean and variance of the Treasury bill rates for each year

(10marks)ii. Determine if there is a significant difference in the average

Treasury bill rates (use a significant level of 1%).(5marks)

Note: S2=

QUESTION THREEa) Describe the characteristics of the following statistical distributions

i. Binomial distribution (3marks)ii. Poisson distribution

(3marks)

b) High Grade Meat Ltd produces beef sausages And sells them to various supermarkets .In order to satisfy the industry’s requirements ,the firm may only produce 0.2percent of sausages below a weight of 80 grammes .The sausage producing machine operates with a standard deviation of 0.5 grammes .The weights of the sausages are normally distributed

The firm’s weekly output is 300,000sausages and the sausage ingredients cost shs5.00 per 100 grammes ,sausages with weights in excess of 82 grammes require additional ingredients costing sh 2.50 per sausage

Required i. The mean weight at which the machine should be set

(4marks)ii. The firm’s weekly cost of production

(10marks)(Total: 20marks)

QUESTION FOUR


427 Lesson Nine

a) A survey of undergraduate students at High Fliers University (HFU)showed the following results regarding gender and the fields of specialization in their studies

Field of specializationGender

Business

Science

Arts Total

Male 100 250 100 450Female

200 50 100 350

Total 300 300 200 800

Required i. Determine if the field of specialization in the studies is dependent

on gender (use significance level of 5%)(10marks)

ii. An earlier survey showed that the proportion of female students taking science was only 10%of the total student population taking science .Does the data above show any significant improvement in the proportion of female students taking science (use a significance level of 5%)

(6marks)

b) Charles Nzioka who is a barber has found out that he can shave on average 4 customers per hour .The arrival rate of customers averages 3customers per hour

Requiredi. The proportion of time that Charles Nzioka is idle

(1 mark)ii. The probability that a customer receives immediate service

upon arrival (1mark)iii. Average number of customers in the queuing system

(1 mark)iv. Average time a customer spends in the queuing system

(1 mark)(Total:

20marks)

QUESTION FIVEa. Differentiate between the additive model and the multiplicative model

as used in time series analysis (4marks)

b. The sales data of XYZ Ltd (in millions of shillings) for the year 2001 to 2004 inclusive are given below.

QuarterYear 1 2 3 42001 40 64 124 582002 42 84 150 62


Revision Aid 428

2003 46 78 154 962004 54 78 184 106

Requiredi. The trend in the data using the least square method

(8marks)ii. The estimated sales for each quarter of the year 2004

(4marks)iii. The percentage variation of each quarter’s actual sales for the

year 2004 (4marks)(Total:

20marks)

SECTION IIQUESTION SIXa. Give two applications of simulation in business

(2marks)b. Collins Simiyu recently acquired a piece of land in Kitale .A property

development company has offered him 300,000 for the piece of land .He has to make a decision on whether to cultivate the piece of land or to sell it to the property development company If he decides to cultivate the land ,there is a probability of getting a high ,medium ,or low harvest .The expected net income for each of the above states of harvest is shown below:

State of harvest

Net income (sh)

High 500,000Medium 100,000Low (20,000)

From past experience there is a 10percent probability that the harvest will be low, a 30 per cent probability that the harvest will be medium and a 60percent probability that the harvest will be high .Colins Simiyu can engage an agricultural expert to carry out a survey on the productivity of the land which will cost him sh30, 000. The agricultural expert gives the following information as to the reliability of such surveys (prior probabilities)

Results of survey state of harvestHigh Medium Low Total

Accurate 0.35 0.10 0.05 0.5Not accurate 0.25 0.10 0.15 0.5

0.60 0.20 0.20 1.0

Required i. Construct a decision tree for the above problem

(6marks)ii. The expected monetary value for each decision

(10marks)


429 Lesson Nine

iii. The decision that you would recommend(2marks)

(Total: 20marks)

QUESTION SEVENa. Explain the difference between assignment and transportation

problems (4marks)b. State the assumptions made in solving a transportation

problem (4marks)c. Umoja Engineering Works Ltd has a network of branches all

over Kenya .The branches are used to service, repair and install equipment for their clients .Currently, the Nairobi branch has four clients who require installation of equipment .Each client requires the services of one engineer There are four engineers who are not engaged at the moment and can be assigned any one of the tasks .However, these engineers have to travel from different locations and the Nairobi branch has to meet their travel and subsistence allowances. The allowances vary from one engineer to another and according to the client the engineer has been assigned to work for.The table below shows the costs (in thousand of shillings) associated with each engineer

ClientEngineer

1 2 3 4

A 37.0 27.0 34.0 21.0B 57.0 22.0 79.0 34.0C 22.0 25.0 61.0 45.0D 39.0 42.0 54.0 43.0

Required i. The assignments to be made in order to minimise the

total cost of engineers (10marks)

ii. The minimum cost of using the engineers(2marks)

(Total 20marks)

QUESTION EIGHTa. Define the following terms as used in network analysis:

i. Crash time(2marks)

ii. Optimistic time (2marks)iii. Forward pass (2marks)iv. Dummy activity (2marks)v. Slack (2marks)


Revision Aid 430

b. James Mutiso is a computer engineer in an information technology firm .The firm has decided to install a new computer system to be used by the firm’s helpdesk .James Mutiso has identified nine activities required to complete the installation.The table below provides a summary of the activities durations and the required number of technicians

Activity Duration(weeks) Required number of technicians

1-2 3 21-3 1 42-4 3 42-5 2 23-4 2 43-6 4 44-5 2 25-6 2 26-7 2 2

Required:i. Draw a gantt chart for the project (6

marks)ii. Mr. Mutiso would like to reschedule the activities so that not more

than 6 technicians are required each weekDetermine if this is possible and how it can be achieved by

rescheduling the activities. (4marks)

(Total 20marks)


quantitative techniques study pack 2006 wip

Documents