lppmodule_2011

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INTRODUCTION

As we all know that business objectives involve optimal utilization of available resources besides other objectives. Linear programming techniques help in achieving such objectives. For example, a firm manufacturing different product with varying profit margins may face resource constraint in terms of available working hours or raw material. Now the firm may be interested in knowing as to how many units of each product shall be manufactured so that profit can be maximised with the given resources. The operation of the firm is to be confined within the given resources and the profit is to be maximized. Thus the use of resources is referred as constraint whereas achieving maximum profit is said to be objective of the firm. Linear Programming helps in dealing with such situations.

Linear programming is a mathematical technique to find value of objective function subject to certain constraints.

Structure of Linear Programming Problem

The structure of a Linear programming problem can be understood from the following problem:

Let x1 = quantity of product 1

x2= quantity of product 2

Then x1 and x2, are the decision variables.

Let profit on each unit of product 1 and product 2 be Rs 10 and Rs 15 respectively.

Aggregate profit = l0x1 + 15x2

Since the objective is to maximise the profit, it can be stated as Maximize l0x1 + 15x2

This is called the objective function. It is indicated by the symbol z’

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LINEAR PROGRAMMING

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Maximize z = 10x1 + l5x2

Let us assume that x1 requires 3 man-hours to produce each unit and x2 requires 4 man-hours and the total available man hours are 100. This requirement can be mathematically stated as:

3x1 + 4x2≤ 100

It is further assumed that product x1 need 5 Kg of material A and 3 Kg of material B and x2 need 6 Kg of material A and 2 Kg of material B. Now, if total material available is 70 Kg and material B available is 90 Kg, then we have following equations

5x1 + 6x2 ≤70

3x1+2x2 ≤90

The above problem can now be mathematically summarized in the form of a linear programming model as follows:

Objective function Maximize Z= 10x1 + l5x2 ---------(1)

Subject to constraints 3x1 + 4x2≤ 100 ---------(2)

5x1 + 6x2 ≤70 ---------(3)

3x1+2x2 ≤90 ---------(4)

Besides the above, the number of units of x1 and x2 can not to be negative

Therefore x1 ≥ 0, x2 ≥ 0 --------(5)

Here, equation (1) is objective functions

Equations (2) to (4) are constraint functions

Equation (5) is non negative constraint.

Thus, the structure of a Linear programming problem comprises of three main parts. which are:

(1) Objective function


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(2) Constraint function

(3) Non negative constraints.

Basic Assumptions of Linear Programming.

(1) Proportionality: According to this assumption, objective function is proportional to resource function. If the resources are doubled, the output would also get doubled.

(2) Additivity: This assumption means that the sum of resources used by different activities/product is equal to the total resources available for the complete activity. This implies that there is no interaction between the decision variables.

(3) Divisibility: According to this assumption, the solution given by linear programming may take any fractional value. The decision variable need not necessarily be a whole numbers.

(4) Certainty: It is assumed that the coefficients of the objective function and constraints do not change during the period under study. This implies that resource requirement and profit per unit remain constant.

(5) Finiteness: It is necessary that there should be finite number of constraints or resource restriction. In case, there would be infinite number of constraints, it would not be feasible to arrive at optimal solution.

(6) Optimality: In case of linear programming problem, the optimal solution always occurs at a corner point of the set of feasible solution.

Formulation of LP Problem

Formulation of LP problem means converting verbal description of the given problem into mathematical form in terms of objective function, constraints and non negative restriction. It basically involves the following steps:

(1) Identification of the decision variables whose values is to he determined.

(2) Formation of an objective functions as a linear function of the decision variable.


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(3) Identification of the set of constraints or restrictions. Express them as linear equations with appropriate equality or inequality sign against a given resource.

(4) Mention the non negative restrictions for the decision variables.

The technique of formulation of LP problems would become clearer from the following examples.

Example 1. A pharmaceutical company manufactures three type of medicines A, B and C. The processing is done through two plants, Plant I and Plant II. The number of medicines produced per day in each plant is given as under

The monthly demand for type A, B and C is 2000, 4000 and 6000 respectively. If plant I costs Rs 2000 per day and plant II costs Rs 3000 per day to operate, how many days should each be run per month to minimize cost. Formulate the problem as LPP.

Solution. Let x1 be the number of days for which plant I should be run

and x2 be the number of days of which plant II should be run to minimize cost.

Minimize z : 2000x + 3000x2 (objective function)

It is further given that plant I can manufacture 100 units of A and plant II can

manufacture 150 units of 100X1 + 150x2≥ 2000 (Constraint)

Likewise, for product B and C, we have 75x1+ 90x2 ≥ 4000 (Constraint)

80x1 + 100x2 ≥ 6000 (constraint)

Since x1and x2, cannot be negative,

x1, x2 ≥ 0 (Non negative restrictions)


Medicine

A B C

Plant I 100 75 80

Plant II 150 90 100

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Example 2. Food I contains 10 units of vitamin A and 12 unit of vitamin B per gram and cost 20 paise per gram. Food II contains 7 unit of vitamin A and 15 units of vitamin B and costs 18 paise per gram. The daily requirement of vitamin A and B are atleast 80 units and 90 units respectively. Formulate the above as an LPP to minimize the cost.

Solution. Let x1 and x2 be units of Food I and Food II respectively to be consumed

Since cost per gram of A and B are 20 paise and 18 paise respectively.

Total Cost 20x1 + 18x2

Our objective is to minimize the cost Minimize z = 20x1 + 18x, (objective functions)

Further, at least 80 units of A is required

10x1 + 7x2 ≥ 80 (Constraint)

12x1 + 15x2 ≥ 90 (Constraint)

Further x1 and x2 cannot be negative

x1, x2 ≥ 0 (Non negative restriction)

Example 3. A firm manufactures three items A, B ,C. The time required for finishing and polishing along with the other necessary details are given below.

Product of Time Required in HoursAssembly Finishing Profit in Rs.

A 20 12 80B 14 15 60C 15 14 30Firms Capacity

2000 1500

Express the above data in the form of a Linear programming problem to maximize profit from the production.


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Solution. Let x1, x., and x3 be the number of units of A, B and C respectively to he

manufactured by a firm.

Since profit per unit is Rs 80, Rs 60 and Rs 30 for x1, x2 and x3 respectively

Total Profit = 80x1 + 60x2 + 30x3

our objective is to maximise total profit

Maximise = 80x1 + 60x2 + 30x3 (1)

It is further given that total available polishing hours are 2000 and the product A,

B and C require 20, 14 and 15 hours per unit. Since the total activity is required to

be confined within the given time.

we have 20x1+ 14x2 + l5x3≤ 2000 (2)

Here 20x1 represent total time of polishing x1 units of A. 14x2 represent total time of

polishing x2 units of B. 15x3 represent total time of polishing x units of C.

Similarly we have

20x1 - 15x2 + 14x3 ≤ 1500 (3)

Further x1, x2 and x3 cannot be negative

we have x1, x2, x3 ≥ 0 (4)

Thus equation (1) is objective function.

Equation (2) and (3) are constraint.

Equation (4) is non negative restriction.


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EXERCISE

(1) Maximize Z =30x1 + 80x2

Subject 3x1 + 4x2 ≤ 18

4x1 + 5x2 ≤ 21

x1, x2 ≥ 0

2) Maximize Z = .12x1 + .09x2 + .08x3 + .085x4 + .16x5 + .18 x6

Subject x1 + x2 + x3 + x4 + x5 + x6 ≤ 500000

x4 ≤ 175000

x5 + x6 ≤ 80000

x3 ≥ 100000

x5 + x6 ≤ x1 + x2

x1 ≤ x2

x1,x2,x3,x4,x5,x6 ≥0

3) Maximize Z =50000x1 + 40000x2 + 25000x3

Subject 1500x1 + 1250x2 +1000x3 ≤ 50000

x1 ≤ 12, x2 ≥ 5, 6 ≤ x3 ≥ 10

x1,x2,x3 ≥ 0

4) Maximize Z =0.8x1 + 0.8x2

Subject 0 .2x1 ≤ 18

0.6x1 + 0.5x2 ≤ 300,

0.3x2 ≤ 150 x1,x2 ≥ 0


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5) Maximize Z =x1 + 5x2

Subject 1000x1 + 4000x2 ≤ 16000

4x1 + 5x2 ≤ 21

X1 ≤ 4, X2 ≤ 4, X1, x2 ≥ 0

6) Maximize Z =24x1 + 8x2

Subject 2x1 + 5x2 ≤ 40

4x1 + x2 ≤ 20

10x1 + 5x2 ≤ 60

x1,x2 ≥ 0

7) Maximize Z =3x1 + 5x2

Subject x1 + 2x2 ≤ 2000

x1 + x2 ≤ 1500

x2 ≤ 600

x1,x2 ≥ 0

8) Maximize Z =.40x1 + .30x2

Subject 2x1 + x2 ≤ 1000

x1 + x2 ≤ 800

x1 ≤ 400 x2 ≤ 700

x1,x2 ≥ 0

9) Maximize Z =x1 + x2 + 3x3

Subject 3x1 + 2x2 + x3 ≤ 3


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2x1 + x2 + 2x3 ≤ 2

x1, x2, x3 ≥ 0

Q. 1 A Plastic company is producing two products A and B. The processing times arc 3 hours and 4 hours per unit .for A on operations one and two respectively and 4 hours and 5 hours per unit for B on operations on one and two respectively. The available times are 18 hours and 21 hours for operation one and two respectively. The product A and B can be sold. at a profit of Rs. 30 per unit and Rs. 80 per unit. Formulate this problem as s LP model so as to maximize the total profit.

Q. 2 A person is interested in investing Rs. 5,00,000 in a mix of investments. The investments choices and expected return on each one of them are:

InvestmentExpected Rate of Return

Mutual Fund A 0.12

Mutual Fund B 0.09

Money Market Fund 0.08

Government Bonds 0.085

Share A 0.16

Share B 0.18

The investor wants at least 35 per cent of his investment in government bonds. Because of the higher perceived risk of the two shares, he has specified that the combined investment in these not to exceed Rs. 80,000. The investor also has specified that at least 20 per cent of the investment should be in the money market fund and that the amount of money invested in shares should not exceed the amount invested in mutual funds. His final investment condition is that the amount invested in mutual fund A should be no more than the amount invested in mutual fund B. The problem is to decide the amount of money to be invested in each alternative so as to obtain the highest annual total return. Formulate the problem as a linear programming problem.

Q. 3 A media specialist plans to allocate advertising expenditure in three media where unit costs of a message are Rs. 1,500, Rs. 1,250 and Rs. 1,000 respectively.


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The total advertising budget available for the year is Rs. 50,000. The first medium is a monthly magazine and it is desired to advertise not more than once in one issue. At least five advertisements should appear in the second medium and the number of advertisements in the third medium should strictly lie between 6 and 10. The effective audience for unit advertisement in the three media is given below:

Medium 1 2 3

Expected effective audience

50,000 40,000 25,000

Formulate a linear programming problem to find the optimum allocation of advertisements in three media that would maximize the total effective audience.

Q. 4. The meat department of a super market deals in three types of meat, say, A, B and C. The manager of the department finds that he has 240 pounds of A, 900 pounds of B and 450 pounds of C on a Saturday morning. From past experience, he knows that he can sell two-thirds of these quantities as straight cuts. The remaining meat will have to be ground into hamburger patties and picnic patties for which there is a long weekend demand. Each pound of hamburger patties contains 20% A and 60% B. Each pound of picnic patties contains 50% B and 30% C. The remainder of each product consists of an inexpensive non-meat filler which is available in unlimited quantities. How many pounds of each product should be made so that the maximum amount of meat is used?

Q. 5 A firm is planning to advertise a special sale on radio and television during a particular week. A maximum budget of Rs. 16,000 is approved for this purpose. It is known that radio commercials cost Rs. 1,000 per 30-second spot with a minimum contract of four spots, which the firm intends to enter. Television commercials, on the other hand, cost Rs. 4,000 per spot. Because of heavy demand, at the most 4 television spots are available in the week. Also, it is believed that a TV spot is five times as effective as a radio spot in reaching consumers. How should the firm allocate its advertising to attract the largest number of buyers? How will the optimal solution be affected if the availability of TV spots is not constrained?

Q. 6 A steel manufacturer produces two types of steel. Type I requires 2 hours of melting, 4 hours of rolling and 10 hours of cutting. Type II requires 5 hours of


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melting, 1 hour of rolling and 5 hours of cutting. Forty hours are available for melting, 20 for rolling and 60 for cutting. The profit margin for type I is 24; for type TI it is 8. Formulate the above as LPP.

Q. 7 A toy manufacturer produces two types of dolls, a basic version doll A and a deluxe version doll B. Each doll of type B takes twice as long to produce as a doll of type A. The company has time to make a maximum of 2000 dolls of type A per day and each type requires equal amount of it. The deluxe version i.e. type B requires a fancy dress of which there are only 600 per day available. The company makes a profit of Rs. 3 and Rs. 5 per doll. respectively, on doll A and B. Formulate the above as LPP.

Q. 8 A company makes two kinds of leather belts. Belt A is a high quality belt and belt B is of lower quality. The respective profits are Rs. 0.40 and Rs. 0.30 per belt. Each belt of type A requires twice as much time as a belt of type B and if all belts were of types B, the company could make 1000 per day. The supply of leather is sufficient for only 800 belts per day (both A and B combined). Belt A requires a fancy buckle and only 400 per day are available. There are only 700 buckles a day available for belt B. Formulate the above as linear programming problem.

Q. 9 Three types of cakes are manufactured by a bakery. The different types C1, C2, and C3 give a contribution of Rs. 1, Rs. 1 and Rs. 3 respectively. The maximum time available at the two stages of manufacture is 3 and 2 hours respectively. The three cake varieties take 3, 2 and 1 hour at the first stage and 2, 1 and 2 hours respectively at the second stage respectively. It is required to find out the optimal product mix for maximum contribution. Formulate the above as linear programming problem.

Q. 10 Write a note on the business applications of linear programming.

Linear Programming Graphical approach

In order to understand graphical method of Linear programming the students must know how to draw graph of inequation, such as ax + by + c ≥ 0 or ax + by + c ≤ 0.

The following steps shall be followed in drawing graph of ax + by + c ≥ 0 or

ax + by + c ≤ 0

(a) Draw the graph of straight line equation ax + by + c = 0. (Ignoring inequality)


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(b) Substitute x = 0 and y = 0 is the given equation. If the inequality is satisfied, shade the portion from the line towards the origin (0,0), otherwise shade the portion from the line on the opposite side. The above concept would become clearer from the following examples.

Example 4. Given 2x + 4y ≥ 20, draw a line and shade the relevant area.

Ploting the above coordinates on X, Y graph, we have

Put x =0, y=0 in the given inequation we have

2(0) + 4(0) ≥ 20 0≥20

This is false, therefore, we have shaded opposite side of the line i.e. side away from ‘the origin.

Example 5. Given x + 2y ≤ 7, draw a line and shade the relevant area.

Solution :Assume x +2y ≤ 7 As x + 2y = 7 => Y= (7-x)/2


x 0 1 -1 10

y 5 4.5 5.5 0

X 0 1 -1 7

Y 3.5 1 4 0

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Put x = 0 and y = 0 in the above equation, we have

(0) + 2(0) ≤ 7, 0 ≤ 7

It is true, therefore the area from the line towards origin has been shaded.

When we have two or more linear inequalities in two variables, we should graph them and shade the relevant area. The area which is common to both graphs gives the required result.

The following example would help in understanding the concept properly:

Example 6. Exhibit graphically the following.

2x +2y ≤ 4

-x +y ≤ 1

Solution: Assume 2x + 2y ≤ 4

As 2x + 2y = 4 and find coordinates

X 0 2 0.5

Y 2 0 1.5

Put x = 0, y = 0 in the inequality

0< 4, true Shade towards the origin


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Likewise find coordinates for -x + y <1

Shade towards the origin.

Common Shaded Portion is the required area as shown in figure. It is observed that common area is OABC.

Using the above concept, graphic method of linear programming can be explained.

Following steps shall be followed in order to graphically solve the Linear

Programming problem.

(a) First of all formulate the given problem in terms of mathematical constraints and an objective function.

(b) The constraints would be inequations which shall be plotted and relevant area shall be shaded.

(c) The corner points of common shaded area shall be identified and the coordinates corresponding to these points shall be substituted in the objective function.

(d) The value which maximizes the objective function or minimizes the objective function, as the case may be, shall be optimal solution of the given problem.


X 0 -1 0.5

Y 1 0 1.5

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This method is based on a theorem, called extreme point theorem, which is stated as under.

Extreme Point Theorem

An optimal solution to a linear programming problem, if it exists, occurs at one of

the corner or extreme points of the area of feasible region.

Put x = 0, y = 0 in the inequality we have 0, 1which is true

Example 6. Maximize the following objective function Z= 2x+3y

Subject x + y ≤ 20

x + 10y ≤ 110

x, y ≥ 0

Solution. First of all, plot the graph of the constraints

x + y ≤ 20 ; x + 10y ≤ 110

and shade the relevant area , as under

Given x + y ≤ 20

x + y =20 => y = 20 - x

X 0 10 -10 20

Y 20 10 30 0


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Put x = 0 and y = 0 in the given inequation, we have 0 ≤ 20 which is true, therefore, area covering the origin shall be shaded.

Further we have

X + 10y ≤ 110

Put x + 10 y =110 => y = (110 – x ) / 10

X 0 10 -10 20

Y 11 10 12 9

Put x = 0, y = 0 in the above equation, we have

(0) + 10(0) = 0 ≤ 110

which is true, therefore, shade the area covering origin.

The inequalities are satisfied by the common shaded area; ABCD as shown in figure

The extreme points of ABCD are

A(0,l1), B(10,10), C(20,0) and D(0,0)


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Subsituting these values in the objective function

Z = 2x + 3y we have

Z(0,l1)= 2(0) + 3(11) =33

Z(10,10) = 2(10) + 3(10) = 50

Z(20,0) = 2(20) + 3(0) = 40

Z(0,0)= 2(0) + 3(0) = 0

Z acquires maximum value at x = 10, y = 10.

Example 7 . Solve the following graphically

Maximise z = 4x + 3y

Subject to x + 2y ≤ 45

x ≤ 20

y ≤ 15

Solution. First of all plot the graphs of following equations.

x+2y ≤ 45

X≤ 20

y≤ 15

Assume x+2y ≤ 45

as x+2y= 45

we have following coordinates satisfying the above equation.

x 0 45 20 15

y 22.5 0 12.5 15


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Further, the equations x ≤ 20 and y ≤ 15 would be represented by straight lines parallel to y-axis and x-axis respectively and the required area shall be shaded, as explained previously

The above equations are shown graphically in figure

Required Area = OABCD.

The points corresponding to OABCD are (0, 0); (20, 0); (20, 12.5); (15, 15) and (0.15)

Substituting the value of corner points in the given objective function, we have

Z at (0,0)= 4(0)+3(0) =0

Z at (20,0) = 4(20) + 3(0) = 80

Z at (20,12.5) = 4(20) + 3(12.5) = 117.5

Z at (15,15) = 4(15) + 3(15) = 105

Z at (0,15)= 4(0)+3(15) =45

Z achieves maximum value when x 20 and y = 12.5.

Example 8. Solve the following graphically.

Maximize Z = 2x + 5y (objective function)


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Subject to x + 3y ≤ 90

Solution First of all plot the given constraint functions and shade the relevant area, as under

Given x+3y ≤ 90

3x+y ≤ 120

.x ≤ 30

As me .x + 3y = 90 and find set of coordinates

x 0 90 30 33.5

Y 30 0 20 18.75

Similarity for 3x + y ≤ 120

Assume 3x + y = 120 and find the set of coordinates as under.

X 0 40 30 33.5

Y 120 0 30 18.75

x ≤ 30 would be represented by a straight line parallel to Y-axis. The graph of the above equation and relevant area is shown in Figure 1.6.

Corner points of the required area O, A, B, C are (0, 0); (30, 0); (30, 20) and (0, 30).

Substituting these points in the objective function

(0,0) = 2(0) + 5(0) = 0 (300) = 2(30) + 5(0) = 60

(30.20) = 2(30) + 5(20) = 160

= 2(0) + 5(30) = 150


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Thus the maximum point of the function exist at x = 30 and y = 20.

Example 9. Solve the following linear problem graphically,

Minimize z = 2x + 3y subject to

2x+y ≥ 40

2x + 2y ≥ 50

x, y ≥ 0

Solution: First of all plot the given constraint equation and shade the relevant area. as under. 2x+y ≥ 40 ---(1)

X 0 20 15

Y 40 0 10

Put x =0, y = 0 in equation (I), we have 2(0) + (0) ≥ 40

0 ≥ 40, this is false therefore shade the area against the origin.


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Similarly, we can solve the other equation and shade the relevant area. Corresponding to 2x + 2y ≥ 50, we have following coordinates

X 0 25 15

Y 25 0 10

The corner points are (0,40) (15,10) and (20,0) The value of objective function at these points would be

Z(0,40) = 2(0) ÷ 3(40) = 120

Z(15,10)= 2(15)+3(10) = 60

Z(20.0) = 2(20) + 3(0) = 40

Thus, Minima of objective function occurs at (20,0)

Example 10. Maximize Z =10x + 6y

subject to

3x + y ≤ 12

2x + 5y ≤ 34

x, y ≥ 0 by graphic method.

Solution. We first plot the lines corresponding to the given constraints

3x+y ≤ 12 , 2x+5y ≤ 34


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Corresponding to 3x + y = 12, and 2x + 5y = 34, we have following coordinates

3x+Y=l2 2x+5y=34

x 0 4 2 x 0 17 02

y 12 0 06 y 34/5 0 06

Note that the line (1) meets the axis at the points (0, 12) and (4, 0), and the line (2) at (0, 34/5) and (17, 0). The shaded area corresponding to the above constraints is shown in Fig.

The corner points of the feasible region are 0 (0,0), A (4,0), B (2,6) and C(0, 34/5). The values of the objective function at these corner points are summarized as follows:

Since objective function is maximum at B, the optimum solution occurs at x = 2 and = 6, yielding the maximum value of Z as 56.


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EXERCISE

Q. 1 Show the solution zone of the following inequalities on a graph paper:

5X+Y ≥10; x + y ≥ 6; x+4y ≥ 12 x ≥ 0 , y ≥ 0

Find x and y for which 3x + 2y is minimum subject to these inequalities.

Use graphic method. [x=l, y =5 and Z = 13]

Q. 2 Solve the following L.P.P. by graphic method:

Maximize Z = 2.75x1 + 4.15x2

Subject to

2x1 + 2.5x2≤ 100

4x1+8x2≥ 160

7.5x1 + 5x2 ≥ 150

x1,x2 ≥ 0 [x = 0, x2 = 40 and Z = 1661

Q. 3 Use graphical method to solve the following linear problems.

(a) Maximize Z= 10x + 6y

Subject to 3x+y ≤12

2x + 5y ≤ 34

x, y ≥ 0 [x=2,y=6,Z=56]

(b) Minimize Z =3x + 2y

Subject to 5x+y ≥ 10

x + y ≥ 6

x + 4y ≥ 12


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x, y ≥ 0 [x= 1, y=5, Z= 13]

(c) Maximize Z = 4x +6y

Subject to x + y = 5

X ≥ 2

y ≤ 4

x, y ≥ 0 (x= 2, y= 3, Z— 26)

(d) Minimize Z = 18x + l0y

Subject to 4x+y ≥ 20

2x + 3y ≥ 30

x, y ≥ 0 [x=.3,y=8,Z=.134]

(e) Maximize Z = 7x + 10y

Subject to x + y ≤ 30000 [(x = 3, ‘ = 8. Z 134]

y≤ 12000

x ≥ 6000

x, y ≥ 0 [x =18000, y = 12000, Z = 246000]

Q. 4 consider a firm that produces two products X1 and X2 with three fixed resources Z1, Z2, Z3. The profit contribution is Rs 20 per unit of X1 and Rs 25 per Unit of X2 sold. Production of a unit of X1 requires 10 units of Z1 and 15 units of Z2, while that of X2 requires 10 units of Z2 and 30 Units of Z1, Z2 and Z3 are available in amounts of 5000, 9000 and 12000 respectively. Find the profit maximizing product mix.

[x1 = 200, x2 = 400]

Q. 5 A carpenters has 90, 80 and 50 running feet respectively of teak, plywood and rosewood. The product A requires 2, 1 and 1 running feet and product B requires 1,


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2 and I running feet of teak, plywood and rosewood respectively. If A would sell for Rs 48 and B would sell for Rs 40 per unit, how much of each should he make and sell in order to obtain the maximum gross income out of his stock of wood? Use graphical method to solve the problem.

[A = 40 units, B = 10 units, Z = 2320]

Q. 6 A firm makes two types of furniture: chairs and tables. The contribution for each product as calculated by the accounting department is Rs. 20 per chair and Rs. 30 per table. Both products are processed on three machines M1, M2 and M3. The time required in hours by each product and the total time available in hours per week on each machine is as follows:

How should the manufacturer schedule his production in order to maximize contribution? (Use graphical method only).

[x = 3, y = 9, Z = 330]

Q. 7 A retired person wants to invest up to an amount of Rs. 30,000 in the fixed income securities. His broker recommends investing in two bonds — bond A yielding 7% per annum and bond B yielding 10% per annum. After some consideration he decides to invest at the most Rs. 12,000 in bond B and at least Rs. 6,000 in bond A. He also wants that the amount invested in bond A must be at least equal to the amount invested in bond B. What should the broker recommend if the investor wants to maximize his return on investment? Solve graphically.

[A = 18000, B= 12000, Z= 2460]

Q. 8 The ABC Company has been a producer of picture tubes for television sets and certain printed circuits for radios. The company has just expanded into full scale production and marketing of AM and AM-FM radios. It has built a new plant that can operate 48 hours per week. Production of an AM radio in the new plant will require 2 hours and production of an AM-FM radio will require 3 hours. Each


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AM radio will contribute Rs. 40/= to profits while an AM-FM radio will contribute Rs. 80/= to profits.

The marketing department, after extensive research has determined that a maximum of 15 AM radios and 11) AM-FM radios can he sold each week.

i) Formulate a linear programming model to determine the optimal production mix of AM-FM radios that will maximize profits. A Solve the above problem using the graphical method.

[x1 = 9, x, = 10, Z = Rs 11601

Q. 9 A firm is engaged in breeding Pigs. The pigs are fed on various products grown on the farm. In view of the need to ensure certain nutrient constituents, it is necessary to buy two additional products, say A and B.

The content of the various products (per unit) in nutrient constituent (e.g. Vitamins, proteins, etc.) are given in the following table:

The last column of the above table gives the minimum amounts of nutrient constituents 4’I. M, and M3 which must be given to the pigs. If products A and B cost Rs. 20 and Rs. 40 per unit respectively, how much of each of these two products should be bought so that the total cost is minimized? (Use graphical method).

[x = 4, y 2, Z = 160]

Q. 10 The manager of an oil refinery must decide on the optimal mix of two possible blending processes of which the inputs and outputs per production run are as follows:


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The maximum quantities available of crudes A and B are 200 units and 150 units respectively. Market requirements show that at least 100 units of gasoline X and 80 units of gasoline )‘ must be produced. The profit per production run from process I and process 2 are Rs. 300 and Rs. 400 respectively. Formulate the problem as linear programming problem and solve it by graphical method.

Q. 11 A firm is planning to advertise a special sale on radio and television during a particular week. A maximum budget of Rs. 16000 is approved for this purpose. It is

known that radio commercials cost Rs. 1000 per 30-second spot with a minimum contract of four spots, which the firm intends to enter. Television commercials, on the other hand, cost Rs. 4000 per spot. Because of heavy demand, at the most 4 television spots are available in the week. Also, it is believed that a TV spot is five times as effective as a radio spot in reaching consumers. How should the firm allocate its advertising to attract the largest dumber of buyers? How will be optimal solution be affected if the availability of TV spot is not constrained?

Q. 12 A firm has two kinds of fruit juices available viz, pineapple and orange juice. These are mixed and two types of mixtures obtained which are sold as soft drinks A and

B. 1 tin of A needs 4 pounds of pine apple juice and 1 pound of orange juice. 1 tin of B needs 2 pounds of pine apple and 3 pounds of orange juice. The firm has only 46 pounds of pine apple juice and 24 pounds of orange juice. Each tin of A and B is sold at a profit of Rs. 4 and Rs. 3 respectively. How many tins of A and B should the firm produce to maximize profit?

Formulate the problem mathematically. Solve it graphically.


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[x=9, y=5, Z= 51]

[x=4, y=3, Z= 19] (Not affected)


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