Linear Programming ISolution Methods

Is a mathematical technique for finding the best uses of an organization’s resources.

LINEAR – is used to describe a relationship between two or more variables, a relationship which is directly and precisely proportional

PROGRAMMING – refers to the use of certain mathematical techniques to get the best possible solution to a problem involving limited resources

Linear Programming

1. Objective- The firm must have an objective to achieve which can be

expressed as a function.

Total selling price variable cost sales volume

Contribution per unit per unit in units

Four Major Characteristic

= _ x

2. Alternative courses of action- Should our firm allocate its manufacturing capacity to tables

& chair in the ratio of 50:50? 25:75? 70:30? And some other ratio.

3. Resources must be in limited supply.- Our furniture plant has limited number of machine hours

available per week; consequently, the more hrs it schedules for tables, the fewer chairs it can make.

4. We must be able to express the firm’s objective and its limitation as mathematical equations or inequalities, and these must be linear equations or inequalities.

P = 8 (number of tables + 6 ( number of chairs produced per week) produced per week)

Profit per table Profit per chair

EQUATION

INEQUALITIES

Example : The statement that the total cost of T tables (at a unit cost of $5

per table) & C chairs (at a unit cost of $4 per chair) must not exceed $120 is

5T + 4C ≤ 120

Profit per table Profit per chair

P = 8 (number of tables + 6 ( number of chairs produced per week) produced per week)

It is possible to solve linear programming problems graphically as long as the number of variables is no more than two.

Graphic method to solve linear programs

Dimensions, Ltd., makes two products, tables and chairs, which must be processed through assembly & finishing departments. Assembly has 60 hrs available per week; finishing can handle up to 48 hrs of work a week. Manufacturing one table requires 4 hrs in assembly and 2 hrs in finishing. Each chair requires 2 hrs in assembly & 4 hrs in finishing.

◦ If profit is $8 per table and $6 per chair the problem is to determine the best possible combination of tables & chairs to produce & sell in order to realize the maximum profit. There are two limitations (also called constraints) in the problem: the weekly time available in assembly & the weekly time available in finishing.

Example:

Let us use T to represent the number of table & C to represent the number of chairs

- 8T = total weekly profit from sale of tables- 6C = total weekly profit from sale of chairs

Objective function = 8T + 6C

Department Time Constraints:

Assembly: 4T + 2C ≤ 60 Finishing: 2T + 4C ≤ 48

FIRST STEP

In order to obtain meaningful answers, the values calculated for T and C can’t be negative; thus all element of the solution to a linear programming problem must be greater than or equal to 0. (T ≥ 0,C ≥ 0).

Maximize:

Profit = 8T + 6C

Subject to the constraints:

4T + 2C ≤ 60 2T + 4C ≤ 48 T ≥ 0

C ≥ 0

Mathematical Summary of the Problem

The inequality 4T + 2C ≤ 60 may be located on the graph by first locating its two terminal points & joining these points by a straight line. The two terminal points for inequality can be found in the ff: manner.

◦ 1. if we assume that none of the time available in assembly is used in making tables (the production of table is 0), then up to 30 chairs could be made. Thus if we let T = 0 , then C ≤ 30. If we make the maximum numbers of chairs, then C = 30. Our first point, thus, is (0, 30);

◦ 2. In order to find the second point, we assume that none of the time available in assembly is used in making chairs (the production of chairs is 0), under this assumption we could produce up to 15 tables. Thus if we let C = 0, then T ≤ 15. If we make the maximum numbers of tables, then T = 15. Our second point, thus, is (15, 0);

SECOND STEP

4T + 2C ≤ 60

4T + 2C = 60 0 0

4T = 60C = 0 ___ ___ 4 4

T = 0 ___ ___ 2 22C = 60

C

T

5

1

0

15

2

0

30

. .

.

5 10 15 20 30

C T15

30

Table of Values Sample

Assembly (4T + 2C ≤ 60) 1st point (0,30) – this denotes the production of 0 tables & 30 chairs per week. 2nd point is (15,0) – this denotes the production of 15 tables & 0 chairs per week.

30

25

20

15

10

5

00 5 10 15 20 25 30

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Numbers of tables

T

C

a

b (0, 30)

c (15, 0)

Capacity constraint in assembly department.

Finishing (2T + 4C ≤ 48)

1st point (0,12) – this denotes the production of 0 tables & 12 chairs per week. 2nd point is (24,0) – this denotes the production of 24 tables & 0 chairs per week.

24

20

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8

4

00 4 8 12 16 20 24

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Numbers of tables

Ta

e (0,12)

(24,0) f

Capacity constraint in finishing department.

C

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Ta

C

b (0,30)

c (15,0) f (24,0)

e (0,12)

d

←Assembly department

Finishing department

Graphic representation of problem constraints.

Plot combination of Assembly and Finishing department.

The combinations of tables and chairs that fall within “aedc” is the feasible region . Combinations outside “aedc” are called infeasible.

Example 1. For 10 tables and 5 chairs per week.

Assembly:4T + 2C ≤ 60 hr available

4(10)+ 2(5) = 50 hr required

Finishing:2T + 4C ≤ 48 hr available

2(10)+ 4(5) = 40 hr required

The time required to make 10 tables and 5 chairs per week falls within the time available in both departments (see example figure 3) and so 10 tables and 5 chairs is a feasible region solution.

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b

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Example 1. For 10 tables and 5 chairs per week.

Assembly: 4T + 2C ≤ 60 hr available 4(10)+ 2(5) = 50 hr required

Finishing: 2T + 4C ≤ 48 hr available

2(10)+ 4(5) = 40 hr required

[Feasible]

Example3

(10,5)

.

(8,12). .(11,10)

Example 2. For 11 tables and 10 chairs per week.

Assembly: 4T + 2C ≤ 60 hr available 4(11)+ 2(10) = 64 hr required

Finishing: 2T + 4C ≤ 48 hr available

2(11)+ 4(10) = 62 hr required

[Infeasible]

Example 3. For 8 tables and 12 chairs per week.

Assembly: 4T + 2C ≤ 60 hr available 4(8)+ 2(12) = 56 hr required

Finishing: 2T + 4C ≤ 48 hr available

2(8)+ 4(12) = 64 hr required

[Infeasible]

Locate point D, because once that point is known, all the points defining the shaded area “aedc” will have been delineated precisely. This is because we already have three points. a (0,0) , e (0,12) and c (15,0)◦ How can d be located? One possibility is to read its location from an

accurately drawn graph below.

THIRD STEP

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b (0,30)

c (15,0) f (24,0)

e (0,12)

d

←Assembly department

Finishing department

Another method is to solve simultaneously the equation of the two lines which intersect to from point d, the only point common to both equations. The equations to be solved are:

4T + 2C = 602T + 4C = 48

First, multiply the first equation by -2:

-2 (4T + 2C=60) = -8T – 4C = -120 + 2T – 4C = 48 Add the second equation. -6T = -72 (divide both side by -6) T = 12

And now substitute 12 for T in the second equation: 2T + 4C = 48 2(12) + 4C = 48 24 + 4C = 48 = 24 + 4C = 48 so … 4C = (48-24)

4C = 24 (divide both side by 4) C = 6

So point d = (12,6)

Solving d algebraically

Test the four corners of the shaded area to see which yields the greatest weekly dollar profit;

Objective function/Profit = 8(T ) + 6(C)

Point a (0, 0) : 8 (0) + 6 (0) = 0Point e (0,12) : 8 (0) + 6 (12) = 72Point c (15,0) : 8 (15) + 6 (0) = 120Point d (12,6) : 8 (12) + 6 (6) = 132

FOURTH STEP

The point which yields the greatest weekly profit is point d ($132).

The concept that the most profitable combination of tables and chairs is found at point d (12,6) can be seen more clearly by first plotting the objective function 8T + 6C (given in the first step) directly on a graph of the feasible region.

To accomplish this, we first let profits equal some minimum dollar figure we know we can attain without violating a constraint. In this case we have elected to let weekly profits equal to $48, a profit easily attainable.

Objective function 48 = 8T + 6C

First locate two terminal points and join them with a straight line. (when T=0, C=8 and when C=0, T=6)

This line represent all the possible

combinations of tables & chairs which

would yield a total profit of $48. You

might want to check one such

combination.

For example,

point “X” represents the

manufacture of 4 tables

and 2 ⅔ chairs per week.

4(8) + 2 ⅔(6) = 48

Ta

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4 8 12 16 20

e

d

c

48 = 8T +

6C (X)

Suppose we now graph another isoprofit line representing all combinations of tables and chairs which would produce a $96 weekly profit.

When T = 0, C = 16 and when C = 0, T = 12. …

96 = 8T + 6C

Both profit equations (48 = 8T + 6C and 96 = 8T + 6C) are illustrated on the graph.

It is obvious, then, that

the isoprofit line which can be located

farthest from the origin (a) will contain

all the combination of tables & chairs

which will generate the greatest

possible profit.

Ta

C

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Numbers of tables

20

16

12

8

4

4 8 12 16 20

e

d

c

48 = 8T +

6C

96 = 8T +

6C0

It is also true that one parallel isoprofit line will pass through point “d”. This particular profit line (line 3), together with the first two profit lines. Although most of the combinations of tables and chairs on profit line 3 do not fall within feasible region (aedc), one point does, point d.

That point represent the most profitable combinations of products. Point d lies on isoprofit line 3 and is still within the feasible region;

thus it represent the most profitable

combination of tables(12) and

chairs (6) for Dimensions, Ltd.,

to manufacture each week.

Ta

C

Nu

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Numbers of tables

20

16

12

8

4

4 8 12 16 20

e

d

c

48 = 8T +

6C

96 = 8T +

6C0

CheckingLet say T=12, and C=6.

Assembly: 4T + 2C ≤ 60 4(12) + 2(6) ≤ 60 48 + 12 = 60

Finishing: 2T + 4C ≤ 48 2(12) + 4(6) ≤ 48 24 + 16 = 48

Isoprofit:144 = 8T + 6C

144 = + 6C

144 = 8T + 6C 144 = 8T +

T C

0

0 6 6

24 = C

24

8 8

18 = T

18

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20

16

12

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4

4 8 12 16 20

e

d

c

48 = 8T +

6C

96 = 8T +

6C0

( Chapter 13)

Significance of an Integer Solution

The objective in our Dimension, Ltd., problem was to maximize profits. However, we can also consider linear programming problems in which the objective is to be minimized.

Example: (Consider a modification of the Dimension, Ltd., problem)

Jeff Smith, the Director of Marketing at Dimensions, has promised customers the firm will make at least 2 tables (T ≥ 2) and at least 4 chairs (C ≥ 4) per week.

Jeff has determined that it costs $20 per unit to manufacture a table & $8 per unit to manufacture a chair. In order to see a simple example where the objective function is to be minimized, let’s suppose that Jeff wants to minimize the total manufacturing cost per chairs and tables.

Graphic procedure for a minimization problem

Ta

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Numbers of tables

20

16

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4 8 12 16 20

d

g

h

T ≥ 2 C ≥ 4

Manufacturing Cost = 20T + 8C Objective Quantity:

Constraints:

When we add these two new constraints to the problem, the feasible region is now the shaded area “ghid” .

i

Ta

CN

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bers

of

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Numbers of tables

20

16

12

8

4

4 8 12 16 20

d

g

ih

20T +

8C

=

180

20T +

8C

=

288

Line 3 passes through point “h”

Example (in order to do this, we have plotted three isocost lines)

These 3 lines are;

20T + 8C = 288 (passing through point d)

20T + 8C = 180

20T + 8C = 72 (passing through point h)

In this case, point h represents the least costly combination of tables (2) and chairs (4) for Dimensions to manufacture per week.

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Assemblyconstraint

Finishingconstraint

Marketing constraint(1)T ≥ 16

Marketing constraint(2)C ≥ 12

- The problem of Dimentions, Ltd., from earlier in this chapter with two additional constraints: (1) the marketing manager must have at least 16 tables a week, and (2) the marketing manager must have at least 12 chairs a week.

InfeasibilityMeans there is no solution which satisfies all the constraints

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Isoprofit line1

Isoprofit line2

Isoprofit line3

Isoprofit line4

Isoprofit line5MarketingConstraint (1)T ≥ 16

MarketingConstraint (2)C ≥ 12

Unboundedness- If the objective can be made infinitely large without violating any of the constraints.

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Assemblyconstraint

Finishingconstraint

Marketing constraint ≤ 20

Redundancy- Additional Constraint:

he cannot sell more than 20 chairs a week.

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12

8

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c

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d

Isoprofit line 1

Isoprofit line 2

Isoprofit line 3

Line 4 concludes withConstraint line ed

Alternative Optima- In this situation below is the example of Alternative Optima; here the isoprofit line

farthest from the origin coincides with one of the constraint lines, specially line “ed”

Linear Programming II

The simplex method

Introduce the simplex method of linear programming Examine in detail all the steps of the simplex method Define and discuss all the elements in the simplex

tableau in terms of their significance to the problem statement and problem solution

Demonstrate the use of the simplex method on minimization problems and how to handle all three kinds of constraints

Show how the simplex method deals with infeasibility, unboundedness, alternative optima and degeneracy

Chapter Objectives

1. Problem formulation: gathering the relevant information, learning what questions need to be answered, and setting the problem up as a linear program.

2. Problem solution: finding the optimal solution to this linear programming.

3. Solution interpretation and implementation: checking that the solution to the linear program is indeed a solution to the original real problem

Linear Programming is a three-stage process

Maximize: Profit = 8T + 6C

Subject to:Assembly: 4T + 2C ≤ 60Finishing: 2T + 4C ≤ 48

All variables ≥ 0

Recall that the variables in the problem were T and C, the numbers of tables and chairs that Dimensions Ltd., should manufacture per week.

The algebraic statement of the problem is:

Sᴀ = slack variable (unused weekly time) in assemblySғ = slack variable (unused weekly time) in finishing

Sᴀ = 60hrs (assembly) Sғ = 48hrs (finishing)

We can express these two statements in mathematical form by writing equations for slack variables Sᴀ and Sғ as follows:

Assembly = Sᴀ = 60 – 4T – 2TFinishing = Sғ = 48 – 2T – 4T

Using slack variables to generate equations

Assume that in assembly we process 5 tables and 3 chairs per week:

Sᴀ = 60 – 4(5) – 2(3) = 34hr unused time in assembly

Assume that in assembly we process 4 tables and 6 chairs per week

Sғ = 48 – 2T – 4T = 16hr unused time in finishing

example

By adding a slack variable to each inequality constraint, we convert them into these equations:

4T + 2C + Sᴀ = 602T + 4C + Sғ = 48

(final form)

Maximize:Profit = 8T + 6C + 0Sᴀ + 0Sғ

Subject to:4T + 2C + Sᴀ + 0Sғ = 602T + 4C + 0Sᴀ + Sғ = 48

All variables ≥ 0

Constraint Equation

Cj column (profits per unit) Product-mix column

Constant column (quantities of product in the mix)

Variable columns

Cj Product 8 6 0 0 mix Quantity T C Sᴀ Sғ

0 Sᴀ 60 4 2 1 00 Sғ 48 2 4 0 1

Parts of the simplex tableau and their functions

Realproducts

Slacktime

Cj row

Variable row

2 rows illustratingConstraint equations

(coefficients only)