2 linear programming problem

Operations Research Unit 2

Sikkim Manipal University 15

Unit 2 Linear Programming

Structure

2.1 Introduction

2.2. Requirements

2.2.1 Basic assumptions of L.P.P

2.3. Linear Programming

2.3.1 Canonical forms

2.3.2 Examples of a linear programming problem

2.4. Graphical analysis

2.4.1 Some basic definitions

2.5 Graphical Methods to solve L.P.P

2.5.1 Working Rule:

2.5.2 Examples 6 for mixed constraints LP problem

2.5.3 Examples 9 for Unbounded Solution

2.5.4 Examples 10 for Inconsistent:

2.5.5 Examples 11 for redundant Constraint:

2.6. Summary

Terminal Questions

Answers to SAQs to TQs

2.1 Introduction One of the most important problems in management decision is to allocate limited and scarce

resource among competing agencies in the best possible manner. Resources may represent man,

money, machine, time, technology on space. The task of the management is to derive the best

possible output (or set of outputs) under given restraints on resources. The output may be

measured in the form of profits, costs, social welfare, effectiveness, etc. In many situations the

output (or the set of outputs) can be expressed as a linear relationship among a number of

variables. The amount of available resources can also be expressed as a linear relationship among

some system variables. The management problem may be to optimize (maximize or minimize) the

output or the objective function subject to the set of constraints An optimization problem in which



both the objective function and the constraints are represented by linear forms is a problem in linear

programming.

Learning Objectives: After studying this unit, you should be able to understand the following

1. Formulate the LPP and observe the feasible region. 2. Graphically analyze and solve a L.P.P.

2.2Requirements of L.P.P i. Decisions variables and their relationship

ii. Well defined objective function

iii. Existence of alternative courses of action

iv. Nonnegative conditions on decision variables.

2.2.1 Basic assumptions of L.P.P 1. Linearity: Both objective function and constraints must be expressed as linear inequalities.

2. Deterministic: All coefficient of decision variables in the objective and constraints expressions

should be known and finite.

3. Additivity: The value of objective function for the given values of decision variables and the

total sum of resources used, must be equal to sum of the contributions earned from each

decision variable and the sum of resources used by decision variables respectively.

4. Divisibility: The solution of decision variables and resources can be any nonnegative values

including fractions.

Self Assessment Questions 1 Fill in the blanks

1. Both objective function and constraints are expressed in _________ forms. 2. L.P.P requires existence of _________ __________ _________ ________. 3. Solution of decision variables can also be ___________

2.3 Linear Programming The Linear Programming Problem (LPP) is a class of mathematical programming in which the

functions representing the objectives and the constraints are linear. Here, by optimization, we



mean either to maximize or minimize the objective functions. The general linear programming

model is usually defined as follows:

Maximize or Minimize

Z = c1 x1 + c2 x 2 + + cn x n subject to the constraints,

a11 x1 + a12 x2 + + a1n xn ∼ b1

a21 x1 + a22 x2 + + a2n xn ∼ b2

am1 x1 + am2 x2 + + amn xn ∼ bm

and x1 ≥ 0, x2 ≥ 0, xn ≥ 0.

Where cj, bi and aij (i = 1, 2, 3, ….. m, j = 1, 2, 3 n) are constants determined from the

technology of the problem and xj (j = 1, 2, 3 n) are the decision variables. Here ∼ is either ≤

(less than), ≥ (greater than) or = (equal). Note that, in terms of the above formulation the

coefficient cj, aij, bj are interpreted physically as follows. If bi is the available amount of resources

i, where aij is the amount of resource i, that must be allocated to each unit of activity j, the “worth”

per unit of activity is equal to cj.

2.3.1 Canonical forms: The general Linear Programming Problem (LPP) defined above can always be put in the following

form which is called as the canonical form:

Maximise Z = c1 x1+c2 x2 + + cn xn Subject to

a11 x1 + a12 x2 + + a1n xn ≤ b1

a21 x1 + a22 x2 + + a2n xn ≤ b2

am1 x1+am2 x2 + …… + amn xn ≤ bm

x1, x2, x3, … xn ≥ 0.

The characteristics of this form are: 1) all decision variables are nonnegative.

2) all constraints are of ≤ type.



3) the objective function is of the maximization type.

Any LPP can be put in the cannonical form by the use of five elementary transformations: 1. The minimization of a function is mathematically equivalent to the maximization of the

negative expression of this function. That is, Minimize Z = c1 x1 + c2x2 + ……. + cn xn is equivalent to Maximize – Z = – c1x1 – c2x2 – … – cnxn.

2. Any inequality in one direction (≤ or ≥) may be changed to an inequality in the opposite

direction (≥ or ≤) by multiplying both sides of the inequality by –1.

For example 2x1+3x2 ≥ 5 is equivalent to –2x1–3x2 ≤ –5. 3. An equation can be replaced by two inequalities in opposite direction. For example, 2x1+3x2 =

5 can be written as 2x1+3x2 ≤ 5 and 2x1+3x2 ≥ 5 or 2x1+3x2 ≤ 5 and – 2x1 – 3x2 ≤ – 5.

4. An inequality constraint with its left hand side in the absolute form can be changed into two

regular inequalities. For example: | 2x1+3x2 | ≤ 5 is equivalent to 2x1+3x2 ≤ 5 and 2x1+3x2 ≥ – 5

or – 2x1 – 3x2 ≤ 5.

5. The variable which is unconstrained in sign (i.e., ≥ 0, ≤ 0 or zero) is equivalent to the difference between 2 nonnegative variables. For example, if x is unconstrained in sign then x

= (x + – x – ) where x + ≥ 0, x – ≤ 0.

2.3.2 Examples Of A Linear Programming Problem: Example 1: A firm engaged in producing 2 models, viz., Model A and Model B, performs only 3 operations painting, assembly and testing. The relevant data are as follows:

Unit Sale Price Hours required for each unit

Assembly Painting Testing

Model A Rs. 50.00 Model B Rs. 80.00

1.0 1.5

0.2 0.2

0.0 0.1

Total number of hours available each week are as under assembly 600, painting 100, testing 30.

The firm wishes to determine the weekly productmix so as to maximize revenue.

Solution: Let us first write the notations as under: Z : Total revenue

x1 : Number of Units of Model A

x2 : Number of Units of Model B

X1, X2 : Are known as decision variables

b1 : Weekly hours available for assembly



b2 : Weekly hours available for painting

b3 : Weekly hours available for testing.

Since the objective (goal) of the firm is to maximize its revenue, the model can be stated as

follows:

The objective function, Z = 50x1 + 80x2 is to be maximized subject to the constraints

1.0 x1+1.5x2 ≤ 600 , (Assembly constraints)

0.2 x1+0.2x2 ≤ 100, ( Painting constaints)

0.0 x1+0.1x2 ≤ 30 , (Testing constraints)

and

x1 ≥ 0, x2 ≥ 0, The Nonnegativity conditions.

Example 2: A milk distributor supplier milk in bottles to houses in three areas A, B, C in a city. His delivery charges per bottle is 30 paise in area A, 40 paise in area B and 50 paise in area C. He has to spend on an average, 1 minute to supply one bottle in area A, 2 minutes per bottle in area B and 3 minutes per bottle in area C. He can spare only 2 hours 30 minutes for this milk distribution but not more than one hour 30 minutes for area A and B together. The maximum number of bottles he can deliver is 120. Find the number of bottles that he has to supply in each area so as to earn the maximum. Construct a mathematical model.

Solution: The decision variables of the model can be defined as follows: x1 : Number of bottles of milk which the distributor supplies in Area A.

x2 : Number of bottles of milk which the distributor supplies in Area B.

x3 : Number of bottles of milk which the distributor supplies in Area C.

The objective :

Maximize Z = 3 2 1 x 100 50 x

100 40 x

100 30 + + in rupees.

constraints:

1. Maximum number of milk bottles is 120, that is x1+x2+x3 ≤ 120.

2. Since he requires one minute per bottle in area A, 2 minutes per bottle in area B and 3

minutes per bottle in area C and he cannot spend more than 150 minutes for the work,

1.x1 + 2.x2 + 3.x3 ≤ 150.

3. Further, since he cannot spend more than 90 minutes for areas A and B. 1.x1+2.x2 ≤ 90.

4. Nonnegativity x1 ≥ 0, x2 ≥ 0.

The problem can now be stated in the standard L.P. form is



Maximize Z = 0.3x1 + 0.4x2 + 0.5x3

Subject to

x1 + x2 + x3 ≤ 120

x1 + 2x2 + 3x3 ≤ 150

x1 + 2x2 ≤ 90

and

x1 ≥ 0, x2 ≥ 0.

Example 3: An oil company has two units A and B which produce three different grades of oil super fine, medium and low grade oil. The company has to supply 12, 8, 24 barrels of super fine, medium and low grade oils respectively per week. It costs the company Rs. 1,000 and Rs. 800 per day to run the units A and B respectively. On a day Unit A produces 6, 2 and 4 barrels and the unit B produces 2, 2 and 12 barrels of super fine, medium and low grade oil per day. The manager has to decide on how many days per week should each unit be operated in order to meet the requirement at minimum cost. Formulate the LPP model.

Solution: The given data can be presented in summary as follows:

Product Capacity Requirements

Super fine Medium Low grade Cost

Unit A Unit B 12 8 24 –

6 2 4

Rs. 1,000

2 2 12

Rs. 800

Let x1 and x2 be the number of days the units A and B be operated per week respectively. Then

the objective of the manager is to,

Minimize the cost function

Z = 1000 x1 + 800 x2

Subject to the constraints 6x1+2x2 ≥ 12 (Super fine)

2x1+2x2 ≥ 8 (medium)

4x1+12x2 ≥ 24 (low grade)

and x1 ≥ 0, x2 ≥ 0.

Self Assessment Questions 2 State True / False

a. One of the characteristics of canonical form in the objective function must be of maximisation.

b. 2x – 3y ≤ 10 can be written as



2x + 3y ≥ 10

2.4 Graphical Analysis Linear programming with 2 decision variables can be analysed graphically. The graphical

analysis of a L.P.P. is illustrated with the help of the following example:

Maximize Z = 700 x1+500 x2

Subject to 4x1+3x2 ≤ 210

2x1+x2 ≤ 90

and

x1 ≥ 0, x2 ≥ 0.

Let the horizontal axis represent x1 and the vertical axis x2. First we draw the line 4x1 + 3x2 = 210.

(by replacing the inequality symbols by the equality) which meets the x1axis at the point A (52.50,

0) (put x2 = 0 and solve for x1 in 4x1 + 3x2 = 210) and the x2 – axis at the point B(0, 70) (put x1 = 0

in 4x1 + 3x2 = 210 and solve for x2).

Any point on the line 4x1+3x2 = 210 or inside the shaded portion will satisfy the restriction of the

inequality, 4x1+3x2 ≤ 210. Similarly the line 2x1+x2 = 90 meets the x1axis at the point C(45, 0) and

the x2 – axis at the point D(0, 90).



Combining we can sketch the area as follows:

The 3 constraints including nonnegativity are satisfied simultaneously in the shaded region

OCEB. This region is called feasible region.

2.4.1 Some Basic Definitions

E



Definition: Any nonnegative value of (x1, x2) (i.e.: x1 ≥ 0, x2 ≥ 0) is a feasible solution of the LPP

if it satisfies all the constraints. The collection of all feasible solutions is known as the feasible

region.

Definition: A set X is convex if for any points x1, x2 in X, the line segment joining these points is

also in X.

(That is, x1, x2 ∈ X, 0 ≤ λ ≤ 1 ⇒ λx2 + (1λ)x1 ∈ X). By convention, a set containing only a single

point is also a convex set.

λx2 + (1λ)x1 (where 0 ≤ λ ≤ 1) is called a convex combination of x1 and x2.

A point x of a convex set X is said to be an extreme point if there do not exist x1, x2 ∈X (x1 ≠ x2)

such that x = λx2 + (1λ)x1 for some λ with 0 < λ < 1.

Definition: A linear inequality in two variables is known as a half plane. The corresponding equality or

the line is known as the boundary of the half plane.

Definition: A convex polygon is a convex set formed by the intersection of finite number of closed halfplanes.

Convex regions

Nonconvex regions

Note: The objective function is maximized or minimized at one of the extreme points which is the

Optimum solution. Extreme points are referred to as vertices or corner points of the convex

regions.

Definition: A redundant constraint is a constraint which does not affect the feasible region.

Definition: A basic solution of a system of m equations and n variables (m < n) is a solution where at least nm variables are zero.

E E

E E

E

E E

E

E

E E



Definition: A basic feasible solution of a system of m equations and n variables (m < n) is a solution where m variables are nonnegative (≥ 0) and nm variables are zero.

Definition: Any feasible solution that optimizes the objective function is called an optimal feasible solution.

Example: Find all basic solutions for the system x1 + 2x2 + x3 = 4, 2x1 + x2 + 5x3 = 5.

Solution: Here A =

5 1 2 1 2 1

, X =

3

2

1

x x x

and b =

5 4

.

i) If x1 = 0, then the basis matrix is B =

5 1 1 2

. In this case 2x2 + x3 = 4, x2 + 5x3 = 5.

If we solve this, then x2 = 3 5 and x3 =

3 2 . Therefore x2 =

3 5 , x3 =

3 2 is a basic feasible

solution.

ii) If x2 = 0, then the basis matrix is B =

5 2 1 1

. In this case, x1 + x3 = 4, 2x1 + 5x3 = 5.If we

solve this, then x1 = 5 and x3 = 1. Therefore x1 = 5, x3 = 1 is a basic solution. (Note that this solution is not feasible, because x3 = 1 < 0).

iii) If x3 = 0, then the basis matrix is B =

1 2 2 1

. In this case, x1 + 2x2 = 4.

2x1 + x2 = 5. If we solve this, then x1 = 2, and x2 = 1. Therefore x1 = 2, x2 = 1 is a basic feasible solution.

Therefore (i) (x2, x3) = (5/3, 2/3), (ii) (x1, x3) = (5, 1), and

(iii) (x1, x2) = (2, 1) are only the collection of all basic solutions.

Self Assessment Questions 3 a. The collection of all feasible solutions is known as the _________ region.

b. A linear inequality in two variables is known as a _________.



2.5 Graphical Methods To Solve The Linear Programming Problems

A LPP with 2 decision variables x1 and x2 can be solved easily by graphical method. We consider the x1 x2 – plane where we plot the solution space, which is the space enclosed by the constraints. Usually the solution space is a convex set which is bounded by a polygon; since a linear function attains extreme (maximum or minimum) values only on boundary of the region, it is sufficient to consider the vertices of the polygon and find the value of the objective function in these vertices. By comparing the vertices of the objective function at these vertices, we obtain the optimal solution of the problem.

The method of solving a LPP on the basis of the above analysis is known as the graphical method. The working rule for the method is as follows:

2.5.1 Working Rule: Step I:Write down the equations by replacing the inequality symbols by the equality symbol in the given constraints.

Step II: Plot the straight lines represented by the equations obtained in step I.

Step III: Identify the convex polygon region relevant to the problem. We must decide on which side of the line, the halfplane is located.

Step IV: Determine the vertices of the polygon and find the values of the given objective function Z at each of these vertices. Identify the greatest and least of these values. These are respectively the maximum and minimum value of Z.

Step V: Identify the values of (x1, x2) which correspond to the desired extreme value of Z. This is an optimal solution of the problem.

Example 4: We can solve the L.P.P. discussed in Example I. Maximize Z = 50x1 + 80x2 Subject to the constraints

1.0x1 + 1.5x2 ≤ 600

0.2 x1 + 0.2x2 ≤ 100

0.0x1 + 0.1x2 ≤ 30

and x1 ≥ 0, x2 ≥ 0

Let the horizontal axis represent x1 and the vertical axis x2. Plot the constraint lines and mark the feasibility region as has been shown in the figure.



Feasible region of the two dimensional LPP

Any point on the thick line or inside the shaded portion will satisfy all the restrictions of the problem. Then ABCDE is the feasibility region carried out by the constraints operating on the objective function. This depicts the limits within which the values of the decision variables are permissible. The intersection points C and D can be solved by the linear equations x2 = 30; x1 + 1.5 x2 = 600, and 0.2x1 + 0.2x2 = 100 and x1 + 1.5x2 = 600 i.e. C (150, 300) and D (300, 180).

After doing this, the next step is to maximise revenues subject to the above shaded area. We work out the revenues at different corner points as tabulated below:

At point

Feasible solution of the productmix

Corresponding revenue Total

revenue x 1 x2 From x1 From x2

A B C D E

0 0

150 300 500

0 300 300 180 0

0 0

7500 15000 25000

0 2400 24000 14,400

0

0 24000 31500 29400 25,000

From the above table we find that revenue is maximum at Rs. 31,500 when 150 units of x1 and

300 units of x2 are produced.

Example 5: For conducting a practical examination, the chemistry department of a college requires 10, 12 and 7 units of three chemicals X, Y, Z respectively. The chemicals are available in two types of boxes: Box A, Box B. Box A contains 3, 2 and 1 units of X, Y, Z respectively and costs Rs. 300. Box B contains 1, 2 and 2 units of X, Y, Z respectively and costs Rs. 200. Find



how many boxes of each type should be bought by the department so that the total cost is minimum.

Solution: First, we summarize the given data in the following table:

Units Units in Box A Units in Box B Units required X Y Z

3 2 1

1 2 2

10 12 7

Cost Rs. 300 Rs. 200 ––

Let x1 be the number of boxes of Atype to be bought and x2 be the number of boxes of Btype. Then the total cost is, Z = 300x1 + 200x2.

Obiviously x1 ≥ 0, x2 ≥ 0.

From the details tabulated in the table, we find that x1 and x2 are subject to the following constraints:

3x1 + x2 ≥ 10

2x1 + 2x2 ≥ 12

x1 + 2x2 ≥ 7

Now, we consider the lines L1: 3x1 + x2 = 10, L2: 2x1 + 2x2 = 12 L3: x1 + 2x2 = 7. These lines are shown in fig.

We note that for the coordinates (x1, x2) of a point satisfy the inequalities. The convex region

bounded by these lines and the coordinate axes is an unbounded region, this is shaded in fig.



We check that a point (x1, x2) that lies inside or on the boundary lines of this region satisfies the

conditions x1 ≥ 0, x2 ≥ 0 and the constraints.

We find that the vertices for the region of interest here are P, Q, R, S. Where P is the point at

which L meets the x2 – axis, Q is the point of intersection of L1 and L2, R is the point of inter

section of L2 and L3 and S is the point at which L3 meets the x1 – axis. We find that P(0, 10), Q(2,

4), R(5, 1) and S(7, 0).:

At P (0, 10), Z = 300 × 0 + 200 × 10 = 2000

At Q (2, 4), Z = 300 × 2 + 200 × 4 = 1400

At R (5, 1), Z = 300 × 5 + 200 × 1 = 1700

At S (7, 0), Z = 300 × 7 + 200 × 0 = 2100

Evidently, Z is minimum at the vertices Q (2, 4) for which x1 = 2, x2 = 4. Thus the cost is minimum

if 2 boxes of type A and 4 boxes of type B are bought. The minimum cost is Rs. 1400.

2.5.2 Examples 6 on mixed constraints LP problem: By using graphical method, find the maximum and minimum values of the function Z = x – 3y where x and y are nonnegative and are subject to the following conditions:

3x + 4y ≥ 19,

2x – y ≤ 9

2x + y ≤ 15

x – y ≥ – 3

Solution: First, we write the constraints (conditions) to be satisfied by x, y in the following standard (less than or equal) form:

– 3x – 4y ≤ – 19

2x – y ≤ 9

2x + y ≤ 15

– x + y ≤ 3

Now, consider the equations: – 3x – 4y = – 19, 2x – y = 9, 2x + y = 15, – x + y = 3 which represents straight lines in the xy – plane. Let us denote them by L1, L2, L3 and L4 respectively. These are shown in fig.:



From the figure, we note that the lines L1, L2, L3 and L4 form a quadrilateral ABCD that lies in the

first quadrant of the xy – plane. We readily see that the region bounded by this quadrilateral is

convex. As such, the points (x, y) that lie within or on the boundary lines of this quadrilateral

satisfy the inequalities x ≥ 0, y ≥ 0 and the constraints. The coordinates of the vertices A, B, C, D

of the quadrilateral are obtained by solving equations taken two of them at a time, we find that A

(1, 4), B (5, 1), C (6, 3), D (4, 7)

we get the solution

Zat A(1, 4) = 1 – 3×4 = – 11

Zat B(5, 1) = 5 – 3×1 = 2

Zat C(6, 3) = 6 – 3×3 = – 3

Zat D(4, 7) = 4 – 3×7 = – 17

Evidently, Z is maximum at the vertex B and minimum at the vertex D. The maximum value of Z is

Zat B(5, 1) = 2, which corresponds to x = 5, y = 1, and the minimum values of Z is –17 at D(4, 7)

which corresponds to x = 4, y = 7.

Examples 7: Use the graphical method to solve the following LP problem: Maximize Z = 7x1+3x2



Subject to the constraints

x1+2x2 ≥ 3

x1+x2 ≤ 4

0 ≤ x1 ≤

0 ≤ x2 ≤

and x1, x2 ≥ 0

Solution: Rewriting the given constraints as follows:

1 x

; 1 x

1 4 x

4 x

1 x

3 x

2 3 2

2 5 1

2 1

2 3 2 1

≤ ≤

≤ +

≥ +

Note: The equation 1 b y

a x

= + is called intercept form of the straight line. Here a and b are the

distance from orgin to the intersection points on the coordinate axes.

Graph each constraint by first treating it as a linear equation. Then use the inequality condition of each constraint to make the feasible region as shown in fig.:

2 5

2 3



The coordinates of the extreme points of the feasible region are

4 1 ,

2 5 A

2 3 ,

2 5 B and

2 3 , 0 C . The value of the objective function at each of these extreme points is as follows:

Extreme point Coordinates (x1, x2) Objective function

value Z= 7x1 + 3x2

A

B

C

4 1 ,

2 5

2 3 ,

2 5

2 3 , 0

7 ¾

22

9/2

The maximum value of the objective function Z= 22 occurs at the extreme points

2 3 ,

2 5 B .

Hence the optimal solution to the given LP problem is 2 3 x ,

2 5 x 2 1 = = and Max. Z = 22.

In linear programming problems may have: i) a unique optimal solution or ii) many number of optimal solutions or



iii) an unbounded solution or iv) no solutions.

Example 8: Maximize Z = 100x1 + 40x2 Subject to

10x1 + 4x2 ≤ 2000

3x1 + 2x2 ≤ 900

6x1 + 12x2 ≤ 3000

and x1, x2 ≥ 0

Solutions: The given constraints can be rewritten as

1 250 x

500 x

1 450 x

300 x

1 500 x

200 x

2 1

2 1

2 1

≤ +

≤ +

≤ +

The values of (x1 x2) at the points are 0(0, 0), A(200, 0) B(125, 187.5) and C(0, 250). The feasible region is OABC. The values of Z at the points are

Z at O(00) = 0 Z at A(200, 0) = 20000 Z at B(125, 187.5) = 20000 Z at C(0, 250) = 10,000



Thus the maximum value of Z occurs at 2 vertices at A and B. Any point on the line joining A and B will also give the same maximum value of Z. Therefore, there are infinite number of feasible solutions which yield the same maximum value of Z.

Suppose a linear programming problem has an unbounded feasible solution space.

If the set of all values of the objective function at different feasible solutions is not bounded above (respectively, bounded below), and if the problem is a maximization (respectively, minimization) problem, then we say that the given problem has an unbounded solution.

In the following, we present an example with unbounded solution.

2.5.3 Example 9 for Unbounded Solution: Maximize Z = 2x1+3x2 Subject to

x1 – x2 ≤ 2

x1 + x2 ≥ 4

and x1, x2 ≥ 0

The intersection point A of the straight lines x1 – x2 = 2 and x1+x2 = 4 is A(3, 1). Here the

solution space is unbounded. The vertices of the feasible region are A(3, 1) and B (0, 4). Value of

objective at these vertices are

Z atA(31) = 2×3+3×1 = 9



Z at B(0, 4) = 2×0+4×3 = 12.

But there are points in the convex region for which Z will have much higher values. For example E

(10, 9) lies in the shaded region and the value of Z there at 47. In fact, the maximum values of Z

occurs at infinity. Thus the problem has an unbounded solutions.

2.5.4 Example 10 for Inconsistent: Maximize Z = 4x1+3x2 Subject to

x1 – x2 ≤ – 1

– x1 + x2 ≤ 0

and x1, x2 ≥ 0.

There being no point (x1, x2) common to both the shaded regions, the LPP cannot be solved.

Hence the solution does not exist, since the constraints are inconsistent.

2.5.5 Example 11 for redundant Constraint: A company making cold drinks has 2 bottling plants located at towns T1 and T2. Each plant

produces three drinks A, B and C and their production capacity per day is shown below:

Cold drinks Plant at

T1 T2

A B C

6000 1000 3000

2000 2500 3000



The marketing department of the company forecasts a demand of 80,000 bottles of A, 22,000

botles of B and 40,000 bottles of C during the month of June. The operating costs per day of

plants at T1 and T2 are Rs. 6,000 and Rs. 4,000 respectively. Find the number of days for which

each plant must be run in June so as to minimize the operating costs while meeting the market

demand.

Solution: Let the plants at T1 and T2 be run for x1 and x2 days.

Then the objective is to minimize the operation costs.

Minimum of Z = 6000 x1 + 4000 x2.

Constraints on the demand for the 3 cold drinks are

6000 x1 + 2000 x2 ≥ 80,000 – (i)

1000 x1 + 2500 x2 ≥ 22000 – (ii)

3000 x1 + 3000 x2 ≥ 40000 – (iii)

also x1, x2 ≥ 0

Thus the LPP is to minimize the objective function subject to the constraints (i), (ii) and (iii). The

solution space is unbounded. The constraint (iii) is dominated by the constraints (i) and (ii) and

hence does not affect the solution space. Such a constraint 3000 x1 + 3000 x2 ≥ 40000 is called

the redundant constraint.

The values of the convex region A, B, C are A (22, 0), B (12, 4) and C (0, 40). The values

of the objective function Z at the vertices are

Zat A = 132000

Zat B = 88,000

Zat C = 1,60,000



Thus the minimum value of Z is Rs. 80,000 and it occurs at B. Hence the optimal solution to the

problem is x1 = 12 days, x2 = 4 days. Example 12: Final the maximum and minimum value of Z = 2x + 3y

Subject to x + y ≤ 30

x – y ≥ 0

y ≥ 3

0 ≤ x ≤ 20

0 ≤ y ≤ 12.

Solution: Any point (x, y) satisfies the conditions x ≥ 0, y ≥ 0 lies in the first quadrant only.

The desired point (x, y) lies with in the feasible convex region ABCDE.

Its vertices are A (3, 3) B (10, 3) C (20, 10), D (18, 12) and B (12, 12). The values of Z at the

five vertices are

Zat A (3, 3) = 2 × 3 + 3 × 3 =15

Z at B (20, 3) = 49

Z at C (20, 10) = 70

Z at D (18, 12) = 72

Z zt E (12,12) = 60

Since the maximum value of Z is 72 which occurs at the vertix D (18, 12). Therefore the solution



of the LPP is x = 18, y = 12 and the minimum value of z is 15 at x = 3, y = 3.

Self Assessment Questions 4

State True / False

1. The feasible region is a convex set.

2. The optimum value occurs anywhere in feasible region.

2.6 Summary

In LPP we first identify the decision variables which are some economic or physical quantities

whose values are of interest to the management. The problems must have a welldefined

objective function expressed in terms of the decision variable. The objective function may have to

be maximized when it expresses the profit or contribution. In case the objective function indicates

a cost, it has to be minimized. The decision variables interact with each other through some

constraints. These constraints occur due to limited resources, stipulation on quality, technical,

legal or variety of other reasons. The objective function and the constraints are linear functions of

the decision variables. A LPP with two decision variables can be solved graphically. Any non

negative solution which satisfies all the constraints is known as a feasible solution of the problem.

The collection of all feasible solutions is known as a feasible region. The feasible region of a LPP

is a convex set. The value of the decision variables which maximise or minimize the objectives

function is located on the extreme point of the convex set formed by the feasible solutions.

Sometimes the problem may be infeasible indicating that no feasible solution of the problem

exists.

Terminal Questions.

1. Use graphical method and solve the L.P.P.

Maximize Z= 5x1 + 3x2

subject to: 3x1 + 5x2 ≤ 15

5x1 + 2x2 ≤ 10

x1, x2 ≥ 0

2. Mathematically formulate the problem. A firm manufactures two products; the net profit on

product 1 is Rs. 3 per unit and the net profit on product 2 is Rs. 5 per unit. The manufacturing



process is such that each product has to be processed in two departments D1 and D2. Each

unit of product 1 requires processing for 1 minute at D1 and 3 minutes at D2; Each unit of

product 2 requires processing for 2 minute at D1 and 2 minutes at D2.

Machine time available per day is 860 minutes at D1 and 1200 minutes at D2. How much of

products 1 and 2 should be produced every day so that total profit is maximum. Formulate this

as a problem in L.P.P.

Answers To Self Assessment Questions


1. Linear 2. Alternate course of actions 3. Fractious


1. True 2. True Self Assessment Questions 3

1. Feasible 2. Half – plan Self Assessment Questions 4

1. True 2. False

Answer for Terminal Questions

1. 19 5 Z max

19 45 x

19 20 x 2 1 = = =

2. Maximize 3x1 + 5x2

Subject to x1 + 2x2 ≤ 800 (minutes)

3x1 + 2x2 ≤ 1200 (minutes)

x1, x2 ≥ 0

2 linear programming problem

Documents

xn 0

inequality

set

sikkimmanipaluniversity

maximize

unit

maximizeorminimize

area