Narsee Monjee Institute of Management Studies

Operations Research MBA Banking Trimester III 2012-13

Linear programming

Problems

1. 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 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 10 AM-FM radios can be sold each week. (a) Formulate a linear programming model to determine the optimum production mix of AM

and FM radios that will maximize profits. (b) Solve the above problem using the graphic method.

2. Vitamins V and W are found in two different foods F 1 and F2. One unit of food F1 contains 2 units of vitamin V and 5 units of vitamin W. One unit of food F 2 contains

4 units of vitamin V and 2 units of vitamin W. One unit of food F 1 and F2 cost Rs. 30

and 25 respectively. The minimum daily requirements (for a person) of vitamin V and W is 40 and 50 units respectively. Assuming that anything in excess of daily minimum requirement of vitamin V and W is not harmful, find out the optimal

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mixture of food F1 and F2 at the minimum cost which meets the daily minimum

requirement of vitamins V and W. Formulate this as a linear programming problem.Ans Food 1 Food 2 Min. daily reqmt.

X1 X2

Vitamin V 2 4 40Vitamin W 5 2 50 Cost (Rs.) 30 25

Minimize Z = 30 X1 + 25 X2 Subject to : 2X1 + 4X3 ≥ 405X1 + 2X2 ≥ 50X1, X2, ≥ 0

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Dual: Max. Z = 40Y1 + 50Y2Sunject to: 2Y1 + 5Y2 ≤ 30

4Y1 + 2Y2 ≤ 25Yi ≥ 0

Dual Solution

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A (0,3)

1234X1

X2

3

2

1

0

X1 - X2 = 1

B (2,1)

X1 + X2 = 3

Unbounded Feasible Region

3. Use graphical method to solve the following LP problem

Maximise Z = 3x1 + 3x2

subject to the constraints

x1 - x2 ≤ 1

x1 + x2 ≥ 3

and x1, x2 ≥ 0

Solution:

The problem is depicted graphically in Fig. The solution space is shaded and is bound by A and B from below.

It is noted here that the shaded convex region (solution space) is unbounded. The two corners of the region are A = (0, 3) and B = (2, 1). The values of the objective function at these corners are:

Z(A) = 6 and Z(B) = 8.

But there exist number of points in the shaded region for which the value of the objective function is more than 8. For example the point (10,12) lies in the region and the function value at this point is 70 which is more than 8. Thus both the variables x 1 and x2 can be

made arbitrarily large and the value of Z also increased. Hence, the problem has an unbounded solution.

Remark: An unbounded solution does not mean that there is no solution to the given LP problem, but implies that there exist an infinite number of solutions.

4. (Problem with inconsistence system of constraints) Use graphical method to solve the following LP problem:

Maximise Z = 6x1 - 4x2

subject to the constraints

4

1234X1

X2

3

2

1

0

4X1 + 8X2 = 16

2X1 + 4X2 = 4

2x1 + 4x2 ≤ 4

4x1 + 8x2 ≥ 16

and x1, x2 ≥ 0

Solution:

The problem is shown graphically in Fig. The two inequalities that form the constraint set are inconsistent. Thus, there is no set of points that satisfies all the constraints. Hence, there is no feasible solution to this problem.

5. Solve graphically the following LPP:

Maximise Z = 8x1 + 16x2

Subject to x1 + x2 ≤ 200

x2 ≤ 125

3x1 + 6x2 ≤ 900 x1, x2 ≥ 0

Solution:

The constraints are shown plotted on the graph in Figure. Also, iso-profit lines have been graphed. We observe that iso-profit lines are parallel to the equation for third constraint 3x 1 + 6x2 = 900. As we move the iso-profit line farther from the origin, it coincides with the

portion BC of the constraint line that forms the boundary of the feasible region. It implies that there are an infinite number of optimal solutions represented by all points lying on the line segment BC, including the extreme points represented by B (50, 125) and C (100, 100). Since the extreme points are also included in the solutions, we may disregard all other solutions and consider only these ones to establish that the solution to a linear programming problem shall always lie at an extreme point of the feasible region.

The extreme points of the feasible region are given and evaluated here.

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100200300X1

X2

200

150

100

50

0

X2 = 125A B

C

D E

Optimal Solutions

FEASIBLE REGION Iso-profit lines

Point x1 x2 Z = 8x1 + 16x2

0 0 0 0

A 0 125 2000

B 50 125 2400

C 100 100 2400 } Maximum

D 200 0 1600

The point B and C clearly represent the optima.

In this example, the constraint to which the objective function was parallel, was the one which formed a side of the boundary of the space of the feasible region. As mentioned in condition (a), if such a constraint (to which the objective function is parallel) does not form an edge or boundary of the feasible region, then multiple solutions would not exist.

6. A firm produces three products A, B, and C, each of which passes through three departments: Fabrication, Finishing and Packaging. Each unit of product A requires 3, 4 and 2; a unit of product B requires 5, 4 and 4, while each unit of product C requires 2, 4 and 5 hours respectively in the three departments. Everyday, 60 hours are available in the fabrication department, 72 hours are available in the finishing department, and 100 hours in the packaging department. The unit contribution of product A is Rs. 5, of product B is Rs. 10 and of product C is Rs. 8.Required:a. Formulate the problem as an LPP and determine the number of units of each of the products

that should be made each day to maximize the total contribution. Also determine if any capacity would remain unutilized.

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b. If the optimal solution obtained does not require the production of some product, explain as to why such product would not be produced. In this context indicate the quantity (quantities) of other product/s that would be foregone for producing such product.

c. What would be the effect on the solution of each of the following:i. Obtaining an order for 6 units of product A which has to be met.ii. An increase of 20% capacity in the fabrication department.

d. The firm is contemplating introduction of a new product D with a likely profit margin of Rs. 8. The product shall consume 3 hours each in the fabrication and finishing departments and 2 hours in the packaging department. Should the firm introduce this product?

Solution:

6.

a. X1 = 0, X2 = 8, X3 = 10. Unutilized capacity = 18 hours in the packaging department.b. X1 is not produced as producing 1 unit of X1 will reduce the objective function by 3.6667

If one were to produce 1 unit of X1, then X2 will reduce by .3333 units and X3 will reduce by .6667 units. Surplus packaging hours will increase by 2.6667.

c. If an order for 6 units of product were to be met, then X2 will become 8-6*.3333 = 6.0002; X3 will become 10 – 6*.6667 = 5.9998 and packaging hours left over will be 18+6*2.6667 = 34.0002; Z = 160 – 6*3.6667 = 138.

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ii. An increase of 20% capacity in the fabrication department will result in an increase in the objective function by 12*.6667 = 8.0004 (20% increase = .2*60 = 12);increase in production of X2 by 12*.3333 = 3.9996;decrease in X3 by 12*.3333 = 3.9996packaging hours will increase by 12*.3333 = 3.9996New solution will be X2 = 11.9996; X3 = 6.0004; S3 = 21.9996 and Z = 168.0002.

d. Contribution of D = 8. Worth of resources consumed by D = 3*.6667 + 3*1.6667 = 7.0002 ≤ 8.Therefore firm should produce product D.

7. A manufacturing company is engaged in producing three types of products: A, B and C. The production department produces, each day, components sufficient to make 50 units of A, 25 units of B and 30 units of C. The management is confronted with the problem of optimizing the daily production of products in assembly department where only 100 man-hours are available daily to assemble the products. The following additional information is available.

Type of Product Profit Contribution per Unit of Product (Rs)

Assembly Time per Product (hrs)

A 12 0.8

B 20 1.7

C 45 2.5

The company has a daily order commitment for 20 units of products A and a total of 15 units of products B and C. Formulate this problem as an LP model so as to maximize the total profit.

Plant A Plant B Plant CX1 X2 X3

Maximise Z = 12 X1 + 20 X2 + 45 X3

Subject to : X1 <= 50X2 <=25X1 >=20X3 <=30 X2 + X3 >=15 0.8 X1 + 1.7 X2 + 2.5 X3 <=100 X1, X2, X3 >=0

8. A company has two plants, each of which produces and supplies two products: A and B. The

plants can each work up to 16 hours a day. In plant 1, it takes three hours to prepare and pack 1,000 gallons of A and one hour to prepare and pack one quintal of B. In plant 2, it takes two hours to prepare and pack 1,000 gallons of A and 1.5 hours to prepare and pack a quintal of B. In plant 1, it costs Rs 15,000 to prepare and pack 1,000 gallons of A and Rs 28,000 to prepare and pack a quintal of B, whereas these costs are Rs 18,000 and Rs 26,000, respectively in plant 2. The company is obliged to produce daily at least 10 thousand gallons of A and 8 quintals of B.

Formulate this problem as an LP model to find out as to how the company should organize its production so that the required amounts of the two products be obtained at minimum cost.

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AnsPlant 1 Plant 2

Product A X1 X2

Product B X3 X4 Minimise Z = 15000 X1 + 18000 X2 + 28000X3 + 26000 X4 Subject to : 3X1 + X3 <= 162X2 + 1.5X4 <=16X1 + X2 >=10X3 + X4 >=8 X1, X2, X3,X4>=0

9. An electronic company is engaged in the production of two components C1 and C2 used in radio sets. Each unit of C1 costs the company Rs 5 in wages and Rs 5 in material, while each of C2 costs the company Rs 25 in wages and Rs 15 in material. The company sells both products on one-period credit terms, but the company’s labour and material expenses must be paid in cash. The selling price of C1 is Rs 30 per unit and of C2 it is Rs 70 per unit. Because of the strong monopoly of the company for these components, it is assumed that the company can sell at the prevailing prices as many units as it produces. The company’s production capacity is, however, limited by two considerations. First, at the beginning of period 1, the company has an initial balance of Rs 4,000 (cash plus bank credit plus collections from past credit sales). Second, the company has available in each period 2,000 hours of machine time and 1,400 hours of assembly time. The production of each C1 requires 3 hours of machine time and 2 hours of assembly time, whereas the production of each C2 requires 2 hours of machine time and 3 hours of assembly time. Formulate this problem as an LP model so as to maximize the total profit to the company.C1 C2

X1 X2

Maximise Z = 20 X1 + 30 X2 Subject to : 10X1 + 40X2 <= 40003X1 + 2X2 <= 20002X1 + 3X2 <= 1400X1, X2 >=0

10. A company has two grades of inspectors 1 and 2, who are to be assigned for a quality control inspection. It is required that at least 2,000 pieces be inspected per 8 – hour day. Grade 1 inspector can check pieces at the rate of 40 per hour, with an accuracy of 97 per cent. Grade 2 inspector checks at the rate of 30 pieces per hour with an accuracy of 95 per cent. The wage rate of a Grade 1 inspector is Rs 5 per hour while that of a Grade 2 inspector is Rs 4 per hour. An error made by an inspector costs Rs 3 to the company. There are only nine Grade 1 inspectors and eleven Grade 2 inspectors available in the company. The company wishes to assign work to the available inspectors so as to minimize the total cost of the inspection. Formulate this problem as an LP model so as to minimize daily inspection cost.Grade 1 Grade 2X1 X2

Minimize 68.8 X1 + 68 X2

Subject to:320X1 + 240 X2 >=2000X1<=9

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X2<=11X1, X2 >=0

11. Let us assume that you have inherited Rs. 1,00,000 from your father-in-law that can be invested in a combination of only two stock portfolios, with the maximum investment allowed in either portfolio set at Rs. 75,000. The first portfolio has an average rate of return of 10%, whereas the second has 20%. In terms of risk factors associated with these portfolios, the first has a risk rating of 4 (on a scale from 0 to 10), and the second has 9. Since you wish to maximise your return, you will not accept an average rate of return below 12% or a risk factor above 6. Hence, you then face the important question. How much should you invest in each portfolio?

Formulate this as a Linear Programming Problem.Ans: Maximise Z = 0.1X1 + 0.2X2

Subject to : 4X1 + 9X2 ≤ 6(X1 + X2)0.1X1 + 0.2X2 ≥ 0.12(X1 + X2)X1 + X2 ≤ 1,00,000X1 ≤ 75,000X2 ≤ 75,000X1 , X2 >=0

12. An electronic company produces three types of parts for automatic washing machines. It purchases casting of the parts from a local foundry and then finishes the part on drilling, shaping and polishing machines. The selling prices of parts A, B and C, respectively are Rs 8, Rs 10 and Rs 14. All parts made can be sold. Castings for parts A, B and C, respectively cost Rs 5, Rs 6 and Rs 10. The shop possesses only one of each type of machine. Costs per hour to run each of the three machines are Rs 20 for drilling, Rs 30 for shaping and Rs 30 for polishing. The capacities (parts per hour) for each part on each machine are shown in the following table:

Machine Capacity Per Hour

Part A Part B Part C

Drilling 25 40 25

Shaping 25 20 20

Polishing 40 30 40

The management of the shop wants to know how many parts of each type it should produce per hour in order to maximize profit for an hour’s run. Formulate this problem as an LP model so as to maximize total profit to the company.Ans A B CX1 X2 X3

Maximize Z = 0.25 X1 + X2 + 0.95 X3

Subject to : X1/25 + X2/40 + X3/25 <=1X1/25 + X2/20 + X3/20 <=1X1/40 + X2/30 + X3/40 <=1X1, X2, X3 >=0

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13. A media campaign has to be designed for a new footwear company with an ad budget of Rs. 20 lakhs. The media, cost and its effectiveness are displayed in the table below:

Advertising Media

Cost per unit (RS)

Estimated no. of

exposureNewspaper 20,000 100,000Radio 40,000 500,000Television 100,000 1,000,000

The client wishes that at least 5,000,000 exposure must be achieved. Also newspaper advertising should not exceed 500,000. Design a media plan.Ans: X1 : No.of releases in newspaper

X2 : No. of releases on radioX3 : No. of releases on Television

Maximise Z = 100,000 X1 + 500,000X2 + 1000,000X3Subject to: 20,000 X1 + 40,000X2 + 100,000X3 ≤ 2000,000

100,000 X1 + 500,000 X2 + 1000,000 X3 ≥ 5000,00020,000 X1 ≤ 500,000X1, X2, X3 ≥ 0

14. A company engaged in producing tinned food, has 300 trained employees on the rolls, each of whom can produce one can of food in a week. Due to the developing taste of the public for this kind of food, the company plans to add existing labour force by employing 150 people, in a phased manner, over the next 5 weeks. The newcomers would have to undergo a two week training program before being put to work. The training to be given by employees from among existing ones and it is known that one employee can train three trainees. Assume that there will be no production from trainers and the trainees during the training period as the training is off-the-job. However the trainees would be remunerated at the rate of Rs 300 per week, the same rate as for the trainers.The company has booked following orders to supply during the next five weeks :

Week 1 2 3 4 5No of cans 280 298 305 360 400

Assume that The production in any week would not be more than the number of cans ordered for , so that every delivery of the food would be ‘fresh’Formulate this problem as an LP model to develop a training schedule that minimizes the labour costs over the five week period.Ans:Minimize Z= 5X1 + 4X2 + 3X3 + 2X4 + X5

Subject to 300 – X1/3 >=280300 – X1/3 – X2/3 >=298300 + X1 – X2/3 – X3/3 >=305300 + X1 + X2 – X3/3 – X4/3 >=360300 + X1 + X2 + X3 – X4/3 – X5/3 >=400X1 + X2 + X3 + X4 + X5 =150

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X1,X2,X3,X4 & X5>=0

15. A firm plans to purchase at least 200 tons of scrap containing high grade Metal H and Low grade metal L. The firm wants to end up with at least 100 tons of H and not more than 35 tons of L. The contents of scrap from two dealers A and B is given in the table

Metal Supplier A Supplier BH 25% 75%L 10% 20%

Cost (per ton)

200 400

Formulate the buying plan to minimise the cost.Let X1 tons from Supplier A and X2 tons from Supplier BMinimise Z = 200X1 + 400X2 Subject to : X1+X2>=200

0.25X1+0.75X2>=1000.1X1+0.2X2 <=35X1, X2, >=0

16. A public limited company is planning its capital structure that will consist of equity capital, 15% debentures and term-loan. Debentures are to be repaid on face value, interest rate is payable half yearly and annualised cost of issue of debenture is 1/2%. Interest on term-loan is 18% p.a. to be paid annually while the cost of equity is estimated at 20%. It is decided not to have outsiders funds more than 2 times of equity funds; also the amount of term-loan must be at least 50% of the debenture amount. Formulate a suitable linear programming model so as to minimise average cost of capital of the company.

Ans: Equity Debentures Term LoanX1 X2 X3

Minimize Z= 0.2X1 + 0.155X2 + 0.18X3 Subject to X2 + X3 ≤ 2X1 X3 ≥ 0.5X2

X1,X2,X3, ≥ 0

17. Smart Ltd. makes two types of adhesives – Quik and Tuff. The table gives the raw material consumed by the two brands.:

Raw material Price(Rs / ton)

Quik Tuff

N 600 350 200A 400 50 100P 400 50 100I 200 550 600

1000kg 1000kg

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Quik can be blended at 1000 Kg per hour, whereas Tuff at 1250 Kg per hour. The respective selling prices are Rs. 1010 and Rs. 845. (per Ton)Variable costs are Rs. 500 per hour of plant production time. The Maximum availability of raw materials is given below.

Raw material Max. available(kg)

N 1000A 300P 250I 1800

Find the optimum product mix to maximise profits.Ans

Maximize Z = 150 X1+ 125X2

Subject to 350X1+200X2<=100050X1+100X2<=30050X1+100X2<=250550X1+600X2<=1800X1,X2>=0

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18. A company has four machines on which three jobs have to be done. Each job can be assigned to one and only one machine. The cost of each job on each machine is given in the following Table.

Formulate this problem as an LP model to find out the job assignments which will minimize the cost. Ans: Let the allocation of jobs to machines be as under:

Minimse Z = 18 X11 + 24X12 + 28 X13 + 32 X14 + 8 X21+ 13 X22 + 17X23 + 19X24 + 10X31 + 15 X32 + 19X33 + 22X34

Subject to:X11 + X12 + X13 + X14 = 1X21 + X22+ X23 + X24 = 1X31 + X32 + X33 + X34 = 1X11 + X21 + X31 ≤ 1X12 + X22 + X32 ≤ 1X13 + X23+ X33 ≤ 1X14 + X24 + X34 ≤ 1X11, X12, X13, X14, , X21, X22, X23, X24, X31, X32, X33, X34, ≥ 0

19. A manufacturer has distribution centers at X , Y, & Z. these centers have availability 40, 20 and 40 units of his product. His retail outlets at A, B, C, D & E require 25, 10, 20, 30 & 15 units respectively. The transportation cost ( in Rs) per unit between each center and outlet are given below.

Distribution Centers

Retail outlets

AB C D E

X 55 30 40 50 50

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P Q R S

A 18 24 28 32

Job B 8 13 17 19

C 10 15 19 22

P Q R S Supply

A X11 X12 X13 X14 1

Job B X21 X22 X23 X24 1

C X31 X32 X33 X34 1

Demand 1 1 1 1

Y 35 30 100 45 60Z 40 60 95 35 30

Formulate this problem as an LP model to find out how many units from each distributers should be transported to which retail outlets so as to minimize cost.Ans: Let the allocation of units from the distributor to the retail outlets be as under:

Distribution Centers

Retail outlets

AB C D E Availability

X X11 X12 X13 X14 X15 40Y X21 X22 X23 X24 X25 20Z X31 X32 X33 X34 X35 40

Requirement 25 10 20 30 15

Minimse Z = 55 X11 + 30 X12 + 40 X13 + 50 X14 + 50X15 + 35 X21+ 30 X22 + 100 X23 + 45X24 +60 X25 + 40X31 + 60 X32 + 95X33 + 35 X34 + 30X35

Subject to: X11 + X12 + X13 + X14 + X15 = 40X21 + X22+ X23 + X24 + X25 = 20X31 + X32 + X33 + X34 + X35 = 40X11 + X21 + X31= 25X12 + X22 + X32 = 10X13 + X23+ X33 = 20X14 + X24 + X34 = 30X15 + X25 + X35 = 15

X11, X12, X13, X14, X15, X21, X22, X23, X24, X25, X31, X32, X33, X34, X35 ≥ 0

20. Manager of an oil refinery has to decide the optimum mix of two possible blending processes of which inputs and outputs per production run are given in the table.

ProcessInput Output

Crude A

Crude B

Gasoline X

Gasoline Y

I 5 3 5 8II 4 5 4 4

The maximum amounts of Crude A and B available are 200 units and 150 units respectively. Market requirements show that at least 100 units of Gasoline X and 80 units of Gasoline Y must be produced. Profits per production run from Process I and Process II are Rs. 4000 and Rs. 5000 respectively. Determine the optimum production runs of each process to maximise profits.

Maximise Z = 4000X1 + 5000X2

Subject to : 5 X1 + 4X2 <=2003X1 + 5X2 <=1505 X1 + 4X2 >=100

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8X1 + 4X2 >=80 X1 , X2 >=0

21. A cooperative farm owns 100 acres of land and has Rs 25,000 in funds available for investment. The farm members can produce a total of 3500 man-hours worth of labour during the months September-May and 4000 man-hours during June-August. Cash income can be obtained from the three main crops and two types of livestock: dairy cows and laying hens. No investment funds are needed for the crops. However, each cow will require an investment outlay of Rs 3200 and each hen will require Rs 15.

Moreover, each cow will require 1.5 acres of land, 100 man-hours during the summer and 50 man-hours during June-Aug. Each cow will produce a net annual cash income of Rs 3500 for the farm. The corresponding figures for each hen are: no acreage, 0.6 man-hours during September-May; 0.4 man-hours during June-August, and an annual net cash income of Rs 200. The chicken house can accommodate a maximum of 4000 hens and the size of the cattle-shed limits the members to a maximum of 32 cows.Estimated man-hours and income per acre planted in each of the three crops are:

Paddy Bajra JowarMan-hoursSeptember-May 40 20 25June-August 50 35 40Net annual cash income (Rs) 1200 800 850The cooperative farm wishes to determine how much acreage should be planted in each of the crops and how many cows and hens should be kept to maximise its net cash income.

LP model formulation: The data of the problem is summarised as follows:

Constraints Crop Total

Cow Hens Paddy Bajra Jowar availability

Man-hours

Sept-May 100 0.6 40 20 25 3500

June-Aug 50 0.4 50 35 40 4000

Land 1.5 - 1 1 1 100

Cow 1 - - - - 32

Hens - 1 - - - 4000

Investment Outlay 3200 15 25000

Net annual

Cash income (Rs) 3500 200 1200 800 850

Decision variables: Let

x1 and x

2= number of dairy cows and laying hens respectively

x3, x

4 and x

5= average of paddy crop, bajra crop and jowar crop respectively

The LP model

Maximise (net cash income) Z = 3500x1 + 200x

2 + 1200x

3 + 800x

4 + 850x

5

subject to the constraints

(i) Man-hours constraints

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100x1 + 0.6x

2 + 40x

3 + 20x

4 + 25x

5 3500 (Sept.-May duration)

50x1 + 0.4x

2 + 50x

3 + 35x

4 + 40x

5 4000 (June-Aug. duration)

(ii) Land availability constraints

1.5x1 + x

3 + x

4 + x

5 100

(ii i) Livestock constraints

x1

32 (dairy cows)

x2

4000 (laying hens)

(iv) Funds availability constraint

3200x1 + 15x

2 25000

and x1, x

2, x

3, x

4, x

5 0

22. A firm is engaged in producing two products : P1 and P2. The relevant data are given here:Per Unit Product P1 Product P2

(i) Selling price Rs 200 Rs 240(ii) Direct materials Rs. 45 Rs. 50(iii) Direct wages

Dept A 8 hrs @ Rs. 2 / hr 10 hrs @ Rs 2 /hrDept B 10 hrs @ Rs 2.25 / hr 6 hrs @ Rs 2.25 /hrDept C 4 hrs @ Rs 2.5 / hr 12 hrs @ Rs 2.5 / hr

(iv) Variable overheads Rs. 6.50 Rs 11.50 Fixed overhead = Rs 2,85,000 per annum No of employees in the three departments: Dept A = 20

Dept B = 15Dept C = 18

No. of hours / employee / week = 40 in each departmentNo. of weeks per annum = 50

(a) Formulate the given problem as a linear programming problem and solve graphically to determine (i) the product mix as will maximize the contribution margin of the firm (ii) the amount of contribution margin and profit obtainable per year

(a) From the graph, do you observe any constraint that is redundant? Which one, if yes?

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Yes the constraint 8X1 + 10 X2 <=40000 is redundant.

23. Solve graphically:

Minimise Z = 6x1 + 14x2

Subject to

5x1 + 4x2 ≥ 60

3x1 + 7x2 ≤ 84

x1 + 2x2 ≥ 18

x1, x2 ≥ 0

Solution:

The restrictions in respect of the given problem are depicted graphically in Figure 4.6. The feasible area has been shown shaded. It may be observed here that although the iso-cost line is parallel to the second constraint line represented by 3x 1 + 7x2 = 84, and this constraint

does provide a side of the area of feasible solutions, yet the problem has a unique optimal solution, given by the point D. Here condition (b) mentioned earlier, is not satisfied. This is because, being a minimisation problem, the optimal movement of the objective function would be towards the origin and the constraint forms a boundary on the opposite side. Since the constraint is not a binding one, the problem does not have multiple optima.

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612182430 X1

X2

15

12

9

6

3

0

A

B

CD

FEASIBLE REGION

We can show the uniqueness of the solution by evaluating various extreme points as done here.

Point x1 x2 Z = 6x1 + 14x2

A 8 5 118

B 84/23 240/23 168

C 28 0 148

D 18 0 108 Minimum

The optimum solution is 108 units for x1 = 18 and x2 = 0.

24. Cashewco has two grades of cashew nuts: Grade I – 750 kg and Grade II – 1,200 kg. These are to be mixed in two types of packages of one kilogram each- economy and special. The economy pack consists of grade I and grade II cashews in the proportion of 1:3, while the special pack combines the two in equal proportion. The profit margin on the economy and special packs is, respectively, Rs. 5 and Rs. 8 a pack.

(a) Formulate this as a linear programming problem. (b) Ascertain graphically the number of packages of economy and special types to be made that will

maximize the profits.

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Would your answer be different if the profit margin on a special pack be Rs. 10? Yes. It would change. X1 = 0, X2 = 1500, Z = 15000. Upper bound is 10 Therefore solution changes for 10 and above.

25. A local travel agent is planning a charter trip to a major sea resort. The eight day/seven-night package includes the fare for round-trip travel, surface transportation, and boarding and loading and selected tour options. The charter trip is restricted to 200 persons and past experience indicates that there will not be any problem for getting 200 persons. The problem for the travel agent is to determine the number of Deluxe, Standard, and Economy tour packages to offer for this charter. These three plans differ according to seating and service for the fight, quality of accommodations, meal plans and tour options. The following table summarizes the estimated prices for the three packages and the corresponding expenses for the travel agent. The travel agent has hired an aircraft for the flat fee of Rs 2,00,000 for the entire trip.

Prices and Costs for Tour Packages per Person

Tour Plan Price(Rs)

Hotel Costs (Rs)

Meals & Other Expenses (Rs)

Deluxe 10,000 3,000 4,750Standard 7,000 2,200 2,500Economy 6,500 1,900 2,200

In planning the trip, the following considerations must be taken into account: (i) At least 10 per cent of the packages must be of the deluxe type.(ii) At least 35 per cent but no more than 70 per cent must be of the standard type.(iii) At least 30 per cent must be of the economy type. (iv) The maximum number of deluxe packages available in any aircraft is restricted to 60.

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(v) The hotel desires that at least 120 of the tourists should be on the deluxe and standard packages together.

The travel agent wishes to determine the number of packages to offer in each type so as to maximize the total profit. (a) Formulate this problem as a linear programming problem.(b) Restate the above linear programming problem in terms of two decision variables, taking

advantage of the fact that 200 packages will be sold. (c) Find the optimum solution using graphical method for the restated linear programming

problem and interpret your results.

Ans : 20 – D , 100 – S, 80 – E Profit = 280000 – 13000 = 267000.

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