SLM-Unit-02-MB0048

Operations Research Unit 2

Sikkim Manipal University Page No. 19

Unit 2 Linear Programming

Structure:

2.1 Introduction

Learning objectives

2.2 Requirements

Basic assumptions of linear programming problems

2.3 Linear Programming

Canonical forms

Case studies of linear programming problems

2.4 Graphical Analysis

Some basic definitions

2.5 Graphical Methods to Solve Linear Programming Problems

Working rule

Solved problems on mixed constraints LP problem

Solved problem for unbounded solution

Solved problem for inconsistent solution

Solved problem for redundant constraint

2.6 Summary

2.7 Terminal Questions

2.8 Answers to SAQs and TQs

Answers to Self Assessment Questions

Answers to Terminal Questions

2.9 References

2.1 Introduction

Welcome to the unit of Operations Research on Linear Programming. Linear

programming focuses on obtaining the best possible output (or a set of

outputs) from a given set of limited resources.

Minimal time and effort and maximum benefit coupled with the best possible

output or a set of outputs is the mantra of any decision-maker. Today,

decision-makers or managements have to tackle the issue of allocating

limited and scarce resources at various levels in an organisation in the best

possible manner. Man, money, machine, time and technology are some of

these common resources. The management’s task is to obtain the best

possible output (or a set of outputs) from these given resources.



You can measure the output from factors, such as the profits, the costs, the

social welfare, and the overall effectiveness. In several situations, you can

express the output (or a set of outputs) as a linear relationship among

several variables. You can also express the amount of available resources

as a linear relationship among various system variables. The management’s

dilemma is to optimise (maximise or minimise) the output or the objective

function subject to the set of constraints. Optimisation of resources in which

both the objective function and the constraints are represented by a linear

form is known as a linear programming problem (LPP).

Learning objectives

By the end of this unit, you should be able to:

Construct linear programming problem and analyse a feasible region

Evaluate and solve linear programming problems graphically

2.2 Requirements of LPP

The common requirements of a LPP are as follows.

i. Decision variables and their relationship

ii. Well-defined objective function

iii. Existence of alternative courses of action

iv. Non-negative conditions on decision variables

2.2.1 Basic Assumptions of LPP

1. Linearity: You need to express both the objective function and

constraints as linear inequalities.

2. Deterministic: All co-efficient of decision variables in the objective and

constraints expressions are known and finite.

3. Additivity: The value of the objective function and the total sum of

resources used must be equal to the 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 non-

negative values including fractions.

Self Assessment Questions

Fill in the blanks

1. Both the objective function and constraints are expressed in _____ forms.

2. LPP requires existence of _______, _______, ____ and _______.

3. Solution of decision variables can also be ____________.



2.3 Linear Programming

The LPP is a class of mathematical programming where the functions

representing the objectives and the constraints are linear. Optimisation

refers to the maximisation or minimisation of the objective functions.

You can define the general linear programming model as follows:

Maximise or Minimise:

Z = c1 x1 + c2 x2 + - - - - + cn xn

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 coefficients cj, bi aij

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

You can represent the general Linear Programming Problem (LPP)

mentioned above in the canonical form as follows:

Maximise Z = c1 x1+c2 x2 + ------ + cn

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 following are the characteristics of this form.

1. All decision variables are non-negative.

2. All constraints are of ≤ type.

3. The objective function is of the maximisation type.



You can represent any LPP in the canonical form by using five elementary

transformations, which are as follows:

1. The minimisation of a function is mathematically equivalent to the

maximisation of the negative expression of this function. That is,

Minimise Z = c1 x1 + c2x2 + ……. + cn xn

is equivalent to

Maximise – Z = – c1x1 – c2x2 – … – cn xn

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 (≥ 0, ≤ 0 or zero) is equivalent

to the difference between 2 non-negative variables. For example:

if x is unconstrained in sign then x = (x+ – x–) where x+ ≥ 0, x– ≤ 0

Caselet

An automobile company has two units X and Y which manufacture three

different models of cars - A, B and C. The company has to supply 1500,

2500, and 3000 cars of A, B and C respectively per week (6 days). It

costs the company Rs. 1,00,000 and Rs. 1,20,000 per day to run the units

X and Y respectively. On a day unit X manufactures 200, 250 and 400

cars and unit Y manufactures 180, 200 and 300 cars of A, B and C

respectively per day. The operations manager has to decide on how many

days per week should each unit be operated to meet the current demand

at minimum cost.

The operations manager along with his team uses a LPP model to arrive

at the minimum cost solution.



2.3.2 Case Studies of linear programming problems

Case Study 1

A firm engaged in producing 2 models namely, Model A and Model B,

performs only 3 operations - painting, assembly and testing. The relevant

data are as follows:

Table 2.1: Unit sale price and hours required for each unit

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 numbers of hours available each week are as under assembly 600,

painting 100, and testing 30. The firm wishes to determine the weekly

product-mix so as to maximise 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 maximise its revenue, the model

can be stated as follows:

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

constraints

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

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

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

x1 ≥0, x2 ≥ 0, (the non-negativity conditions)



Case Study 2

A milk distributor supplies milk in bottles to houses in three areas A, B, C

in a city. His delivery charge 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:

Maximise Z = 321 x10050x

10040x

10030 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. Non-negativity x1 ≥ 0, x2 ≥ 0.

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

Maximise 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



Case Study 3

An oil company has two units A and B which produce three different

grades of oil - super fine, medium and low grade. The company has to

supply 12, 8, 24 barrels of super fine, medium and low grade oil

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 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:

Table 2.2: Capacity and requirements details of an oil company

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:

Minimise 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




State True/False

4. One of the characteristics of canonical form in the objective function

must be of maximisation.

5. 2x – 3y ≤ 10 can be written as -2x + 3y ≥-10

2.4 Graphical Analysis

You can analyse linear programming with 2 decision variables graphically.

Example

Let’s look at the following illustration.

Maximise 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, draw the

line 4x1 + 3x2 = 210, (by replacing the inequality symbols by the equality)

which meets the x1-axis 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).

Figure 2.1: Linear programming with 2 decision variables



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 x1-axis at the point C(45, 0) and the x2 – axis at the point D(0, 90).

Figure 2.2: Linear programming with 2 decision variables

Combining the two graphs, you can sketch the area as follows:

Figure 2.3: Feasible region

The 3 constraints including non-negativity are satisfied simultaneously in the

shaded region OCEB. This region is called feasible region.



2.4.1 Some basic definitions

Sr. No.

Terms Definitions

1. Feasible Region

Any non-negative 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.

2. Convex

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 1 x2 + (1-)x1 X By convention, a set containing only a single point is also a convex set.

x2 + (1-)x1 (where 0 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 does not exist x1, x2 X (x1 x2) such

that x = x2 + (1-)x1 for some with 0 < < 1.

3. Half Plane 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.

4. Convex Polygon A convex polygon is a convex set formed by the inter-section of finite number of closed half-planes. Refer figures 1.3 and 1.4.

5. Redundant constraint

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

6. Basic Solution A basic solution of a system of m equations and n variables (m < n) is a solution where at least n-m variables are zero.

7. Basic Feasible Solution

A basic feasible solution of a system of m equations and n variables (m < n) is a solution

where m variables are non-negative ( 0) and n-m variables are zero.

8. Optimal Feasible Solution

Any feasible solution that optimises the objective function is called an optimal feasible solution.

E E E E

E

E E E E E E

Figure 2.4: Convex regions



Figure 2.5: Non – convex regions

Note: The objective function is maximised or minimised at one of the

extreme points referred to as optimum solution. Extreme points are referred

to as vertices or corner points of the convex regions.

Solved Problem 1

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

Solution: Here:

A =

512

121, X =

3

2

1

x

x

x

and b =

5

4.

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

51

12. 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 =

52

11. 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 =

12

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

2x1 + x2 = 5. After solving, 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

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




Fill in the blanks

6. The collection of all feasible solutions is known as the ________ region.

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

2.5 Graphical Methods to Solve LPP

Solving a LPP with 2 decision variables x1 and x2 through graphical

representation is easy. Consider x1 x2 – the plane, where you plot the

solution space enclosed by the constraints. The solution space is a convex

set bounded by a polygon; since a linear function attains extreme (maximum

or minimum) values only on the boundary of the region. You can consider

the vertices of the polygon and find the value of the objective function in

these vertices. Compare the vertices of the objective function at these

vertices to obtain the optimal solution of the problem.

2.5.1 Working rule

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.

Step 1: Write down the equations by replacing the inequality symbols by

the equality symbols in the given constraints.

Step 2: Plot the straight lines represented by the equations obtained in

step I.

Step 3: Identify the convex polygon region relevant to the problem. Decide

on which side of the line, the half-plane is located.

Step 4: 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 5: Identify the values of (x1, x2) which correspond to the desired

extreme value of Z. This is an optimal solution of the problem



Solved Problem 2

Solve the given LPP using the graphical method. (Solve the LPP as

discussed in example 1) Maximise 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

Solution:

The horizontal axis represents x1 and the vertical axis x2. Plot the

constraint lines and mark the feasibility region as shown in the figure.

Figure 2.6: 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. The 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 inter-section 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

That is C (150, 300) and D (300, 180).

The next step is to maximise revenues subject to the above shaded area.

You can work out the revenues at different corner points as tabulated

below:



Table 2.3: Revenues at different corner points

At point

Feasible solution of the product-mix

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 maximum revenue is at Rs. 31,500

when 150 units of x1 and 300 units of x2 are produced.

Solved Problem 3

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 minimal.

Solution: First, you need to summarise the given data in the following table:

Table 2.4: Representation of chemical units, number of units in box types and units required

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 A-type to be bought and x2 be the number of boxes of B-type. Then the total cost is,

Z = 300x1 + 200x2.

Obviously,

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, you consider the lines L1: 3x1 + x2 = 10, L2: 2x1 + 2x2 = 12 L3: x1 + 2x2 = 7 as shown in figure 2.7


You can see that the co-ordinates (x1, x2) of a point satisfy the inequalities. The convex region bounded by these lines and the co-ordinate axes is the shaded region in the unbounded region. Check the point (x1, x2) that lies inside or on the boundary lines of this region

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

You will find that the vertices for the region are P, Q, R, S.

Where P is the point at which L meets the x2 – axis, Q is the point of inter-section 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 minimal, if 2 boxes of type A and 4 boxes of type B are bought. The minimum cost is Rs. 1400.



2.5.2 Solved problems on mixed constraints LP problem

Solved Problem 4

1. By using graphical method, find the maximum and minimum values of

the function Z = x – 3y where x and y are non-negative and subject to

the following conditions:

3x + 4y 19,

2x – y 9

2x + y 15

x – y – 3

Solution: You can start by writing 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

Consider the equations:

– 3x – 4y = – 19, 2x – y = 9,

2x + y = 15, – x + y = 3,

Both the above equations represent straight lines in the xy – plane.

Denote the straight lines by L1, L2, L3 and L4 respectively.

You can see the lines L1, L2, L3 and L4 form a quadrilateral ABCD lying in

the first quadrant of the xy – plane. You can 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

co-ordinates of the vertices A, B, C, D of the quadrilateral are obtained by

solving equations taking two at a time, you will find that A (1, 4), B (5, 1),

C (6, 3), D (4, 7), hence the solution is

Zat A(1, 4) = 1 – 34 = – 11

Zat B(5, 1) = 5 – 31 = 2

Zat C(6, 3) = 6 – 33 = – 3

Zat D(4, 7) = 4 – 37 = – 17

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.

Solved Problem 5

Use the graphical method to solve the following LPP.

Maximise Z = 7x1+3x2

Subject to the constraints

x1+2x2 ≥ 3

x1+x2 ≤ 4

0 ≤ x1 ≤

25

0 ≤ x2 ≤

23

and x1, x2 0



Solution:

Rewriting the given constraints as follows:

1x

;1x

14

x

4

x

1x

3

x

23

2

25

1

21

23

21

Note: The equation 1b

y

a

x is called intercept form of the straight line.

A and B are the distance from the origin to the intersection points on the

co-ordinate axes.

Figure 2.9: Plot of line AB

You have to 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 figure 2.11




The co-ordinates of the extreme points of the feasible region are

41,

25A

2

3,

2

5B and

2

3,0C . The value of the objective function at

each of these extreme points is as follows:

Table 2.5: Extreme points and their coordinates

Extreme point Co-ordinates (x1, x2) Objective function value

Z= 7x1 + 3x2

A

B

C

41,

25

23,

25

23,0

7 ¾

22

9/2

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

points

2

3,

2

5B . Hence the optimal solution to the given LP problem is

2

3x,

2

5x 21 and Max. Z = 22.

In linear programming problems, you may have:

i) a unique optimal solution or

ii) many number of optimal solutions or

iii) an unbounded solution or

iv) no solutions.



Solved Problem 6

Solve the given LPP in the graphical method.

Maximise Z = 100x1 + 40x2 Subject to

10x1 + 4x2 2000

3x1 + 2x2 900

6x1 + 12x2 3000

and x1, x2 0

Solution: The given constraints can be rewritten as

1250

x

500

x

1450

x

300

x

1500

x

200

x

21

21

21


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

The maximum value of Z occurs at 2 vertices - A and B. Join any point on the line joining A and B to get the same maximum value of Z. Therefore, there are infinite numbers 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 maximisation (respectively, minimisation) problem, then we say that the given problem has an unbounded solution.



2.5.3 Solved problem for unbounded solution

Solved Problem 7



Subject to

x1 – x2 2

x1 + x2 4

and x1, x2 0

Solution:


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). Values of objective at these vertices are:

Z at A (31) = 23+31 = 9

Z at B (0, 4) = 20+43 = 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 is at 47. In fact, the maximum value of Z occurs at infinity. Thus the

problem has unbounded solutions.



2.5.4 Solved problem for inconsistent solution

Solved Problem 8



Subject to

x1 – x2 – 1

– x1 + x2 0

and x1, x2 0.

Solution: Due to lack of point (x1, x2) common to both the shaded regions,

the LPP cannot be solved. Consequently the constraints are inconsistent

to arrive at the solution.




2.5.5 Solved problem for redundant constraint

Solved Problem 9

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:

Table 2.6: Production capacity at different bottling plants

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 bottles 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 minimise the operating costs while meeting

the market demand.

Solution: Let the plants at T1 and T2 work on x1 and x2 days. Then the

objective is to minimise 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 minimise 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.



Solution: Any point (x, y) satisfies the conditions x 0, y 0 lies in the

first quadrant only. The desired point (x, y) lies within 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

The maximum value of Z is 72 which occur at the vertex 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.


State True/False

8. The feasible region is a convex set

9. The optimum value occurs anywhere in feasible region



2.6 Summary

In a LPP, you first identify the decision variables with economic or physical

quantities whose values are of interest to the management. The problems

must have a well-defined objective function expressed in terms of the

decision variable.

The objective function is to maximise the resources when it expresses profit

or contribution. Here, the objective function indicates that cost has to be

minimised. The decision variables interact with each other through some

constraints. These constraints arise 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 satisfying 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 minimise the

objectives function is located on the extreme point of the convex set formed

by the feasible solutions. Sometimes the problem may be unfeasible

indicating that no solution exists for the problem.

2.7 Terminal Questions

1. Use the graphical method to solve the LPP.

Maximise 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.

2.8 Answers to SAQs and TQs

Answers to Self Assessment Questions

1. Linear

2. Alternate course of action

3. Fractious

4. True

5. True

6. Feasible

7. Half-plan

8. True

9. False

Answers to Terminal Questions

1. 19

5Zmax

19

45x

19

20x 21

2. Maximise 3x1 + 5x2, subject to x1 + 2x2 800 (minutes)

3x1 + 2x2 1200 (minutes) x1, x2 0

2.9 References

No external sources have been referred.

SLM-Unit-02-MB0048

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