211310_634662513001162500

S r i V e n k a t e s w a r a C o l l e g e o f E n g i n e e r i n g , S r i p e r u m b u d u r

D e p a r t m e n t o f C o m p u t e r A p p l i c a t i o n s

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ANNA UNIVERSITY - M.C.A – IV SEMESTER

Problem Set – MC1752/MC9242/600416 – Resource Management Techniques

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UNIT I – LINEAR PROGRAMMING MODELS

1. Mathematical Formulation

1. Reddy Mikks Company produces two types of paints: Interior and Exterior paints from two

types of raw materials: M1 and M2. The following table provides the basic data of the

problem:

Raw

Material

Tons of raw Material per ton of Maximum Daily

Availability (tons) Exterior paint Interior paint

M1 6 4 24

M2 1 2 6

Profit / ton Rs. 5 Rs. 4

A market survey indicates that the daily demand for interior paint cannot exceed that for

exterior paint by more than 1 ton. Also the maximum daily demand for interior paint is 2

tons. Reddy Mikks wants to determine the optimum product mix of both interior and

exterior paints that maximizes the total daily profit.

2. A manufacturer produces two types of product: P1 and P2. Each model must go through

two processes: grinding and polishing. The Manufacturer has 2 grinders and 3 polishers.

The following table provides the information pertaining to this problem.

Product

Types

Processing time (in hours)

for

Profit /

unit

Grinding Polishing

P1 4 2 Rs. 3.00

P2 2 5 Rs. 4.00

Availability 40 Hrs. 60 Hrs.

How should the manufacturer allocate his production capacity to the two types of products

so that he may make the maximum profit in a week?

Prepared by S Raju / Associate Professor / Dept. of CA / S.V.C.E.

2

3. A Leather company produces two types of belts: A and B. Each belt type is processed on two

machines: G and H. The following table provides the information pertaining to this problem.

Determine how many belts of each type the company should produce each day in order to

get maximum profit.

Belt

Types

Processing time (in minutes) on Profit /

Belt Machine G Machine H

A 1 2 Rs. 2.00

B 1 1 Rs. 3.00

Availability 6 Hrs 40 Mts. 10 Hrs.

4. Old hens can be bought for Rs. 2 each but young ones cost Rs. 5 each. The old hens lay 3

eggs per week and the young ones, 5 eggs per week, each being worth 30 paise. A hen

costs Rs. 1 per week to feed. If I have only Rs. 80 to spend for hens, how many of each kind

should I buy to get a profit of more than Rs. 6 per week, assuming that I cannot house more

than 20 hens.

5. A manufacturer has three machines A, B and C with which he produces three different

articles P, Q and R. The following table shows the required time for processing. Determine

the number of different articles which should be made in order to maximize the profit.

Articles Processing time (in hours) on the Machine Profit /

Unit A B C

P 8 4 2 Rs. 20

Q 2 3 0 Rs. 6

R 3 0 1 Rs. 8

Availability 250 Hrs. 150 Hrs. 50 Hrs.

6. A small-scale manufacturer has producing facilities for producing two different products.

Each product requires three different operations: grinding, assembly and testing. Product I

requires 15, 20 and 10 minutes and Product II requires 7.5, 40 and 45 minutes respectively

for grinding, assembly and testing. The production run calls for at least 7.5 hours of

grinding, at least 20 hours of assembly time and at least 15 hours of testing time. If

product I costs Rs. 60 and product II costs Rs. 90 to manufacture, determine the number of

units of each product the firm should produce in order to minimize the total cost of

operations.


3

7. A manufacturer of furniture makes chairs and tables. They are processed on two machines:

A and B. The following table provides the information pertaining to this problem.

Determine how many chairs and tables, the company should produce each day in order to

get maximum profit.

Product Processing time (in hours) on Profit /

Unit Machine A Machine B

Chair 2 6 Rs. 1.00

Table 5 0 Rs. 5.00

Availability 15 30

8. Two grades of paper – M and N are produced on a paper machine. Due to the limitations on

the availability of raw material, not more than 400 tonnes of grade M and 300 tonnes of

grade N can be produced in a week. It requires 0.2 and 0.4 hours to produce a tonne of

products M and N respectively, with corresponding profits of Rs. 20 and Rs. 50 per tonne.

Determine the optimum product mix that maximizes the profit.

9. The ABC Company combines factors X and Y to form a product which must weigh 50 Kgs.

At least 20 Kgs. of X and no more than 40 Kgs. of Y can be used. The cost of X is Rs. 10 per

Kg. and that of Y is Rs. 25 per Kg. Determine the amount of factors X and Y which should

be used to minimize the total costs.

10. A production manager wants to determine the quantity to be produced per month of

products A snd B manufactured by his firm. The data on resources required and availability

of resources are given below. Find the product mix that would give maximum profit.

Resources

Requirements Maximum

per

month Product A Product B

Raw material (Kg.) 60 120 12000

Machine hours (piece) 8 5 600

Assembly man hours 3 4 500

Sale Price / piece Rs. 30 Rs. 40

11. Two spare parts X and Y are to be produced in a batch. Each one has to go through two

processes A and B. The time required in hours per unit and total time available is given

below. Find how many unit of spare parts of X and Y are to be produced in this batch to

maximize the profit.


4

Spare

Part

Process Time (in hours) for Profit /

Unit Process A Process B

X 3 9 Rs. 5.00

Y 4 4 Rs. 6.00

Availability 24 36

12. A company is manufacturing two products – A and B, involving three departments. The

process time, profit earned per unit and the total capacity of each department is given in the

following table. Determine the optimum product mix to maximize the profit.

Products Processing time (in hours) on the department Profit /

Unit Machining Fabrication Assembly

A 1 5 3 Rs. 80

B 2 4 1 Rs. 100

Availability 720 Hrs. 1800 Hrs. 900 Hrs.

13. A firm can produce three types of cloths – A, B and C. Three kinds of wool are required for

it – red, green and blue. The availability and requirement of raw material and profit details

are as follows. Determine the optimum production schedule.

Cloths

Type

Raw Material (wool) Required (in yards) Profit/Unit

of length Red wool Green wool Blue wool

A 2 0 3 Rs. 3

B 3 2 2 Rs. 5

C 0 5 4 Rs. 4

Availability 8 Yards 10 Yards 15 Yards

14. A manufacturer makes 2 products – P1 and P2 using two machines: M1 and M2. The

following table provides the information pertaining to this problem. What should be the

daily product mix to optimize the profit?

Product Processing time (in hours) on Profit /

Unit Machine M1 Machine M2

P1 2 6 Rs. 2

P2 5 0 Rs. 10

Availability 16 30

15. A company is manufacturing two products – A and B, involving three departments. The

process time, profit earned per unit and the total capacity of each department is given in the

following table. Determine the optimum product mix to maximize the profit.


5

Products Processing time (in hours) on the department Profit /

Unit Dept. 1 Dept. 2 Dept. 3

A 2 1 4 Rs. 1.00

B 2 2 2 Rs. 1.50

Availability 160 120 280 (Hrs. /week)

16. A company produces three types of food products – F1, F2 and F3 for which is uses three

types of vitamins – A, C and D. The daily requirement of vitamins with cost per unit is

shown below. Determine the best combination of food for minimum cost.

Vitamins Vitamin Required for food Type (in Mgs) Cost /

Unit F1 F2 F3

A 1 1 10 Rs. 10

C 100 10 10 Rs. 15

D 10 100 10 Rs. 5

Requirement 1 Mg. 50 Mgs. 10 Mgs.

17. A manufacturer makes 3 products – A, B and C using two machines: Cutting and Welding.

The following table provides the information pertaining to this problem. Determine how

much of each product must be produced to realize maximum profit.

Products Processing time (in hours) on Profit /

Unit Cutting M/C Welding M/C

A 9 11 Rs. 32

B 5 18 Rs. 20

C 20 6 Rs. 60

Availability 400 / week 750 / week

18. A firm makes two products X and Y, and has a total production capacity of 9 tonnes per day,

X and Y requiring the same production capacity. The firm has a permanent contract to

supply at least 2 tonnes of X and at least 3 tonnes of Y per day to another company. Each

tonne of X requires 20 machine hours production time and each tonne of Y requires 50

machine hours of production time. The daily maximum possible number of machine hours

is 360. All the firm’s output can be sold and the profit made is Rs. 80 per tonne of X and

Rs. 120 per tonne of Y. It is required to determine the production schedule for maximizing

profit and to calculate this profit.


6

19. One unit of product A contributes Rs. 7 and requires 3 units of raw material and 2 hours of

labour. One unit of product B contributes Rs. 5 and requires 1 unit of raw material and 1

hours of labour. Availability of raw material at present is 48 units and there are 40 hours of

labour. Formulate this problem as a Linear Programming Problem.

20. A furniture company can produce four types of chairs. Each chair must go through two

types of processes: Carpentry shop and finishing shop. Man hours required for each

process, available man hours and profit per chair are shown in the table below. Determine

the number of chairs in each type to be produced to get maximum profit.

Type of

Chair

Processing time (in hours) on Profit /

Chair Carpentry Shop Finishing Shop

C1 4 1 Rs. 12

C2 9 1 Rs. 20

C3 7 3 Rs. 18

C4 10 40 Rs. 40

Availability of

Man Hours 6000 / week 4000 / week

2. Graphical Solution of Linear Programming Models

Solve the Following Linear Programming Problems by graphical method.

1. Maximize Z = 5x1 + 4x2

Subject to the constraints

6x1 + 4x2 24

x1 + 2x2 6

- x1 + x2 1

x2 2 and x1, x2 ≥ 0

(Ans: x1= 3, x2 =1.5 and Max Z = 21)



4x1 + 2x2 80

2x1 + 5x2 180 and

x1, x2 ≥ 0

(Ans: x1= 2.5, x2 =3.5 and Max Z = 147.5)



x1 + x2 400

2x1 + x2 600 and

x1, x2 ≥ 0

(Ans: x1= 0, x2 = 400 and Max Z = 1200)

4. Maximize Z = x1 + 5x2


2x1 + 5x2 15

6x1 30 and

x1, x2 ≥ 0

(Ans: x1= 0, x2 = 3 and Max Z = 15)


7

5. Minimize Z = 60x1 + 90x2


15x1 + 7.5x2 ≥ 7.5

20x1 + 40x2 ≥ 20

10x1 + 45x2 ≥ 15 and x1, x2 ≥ 0

(Ans: x1= 0.33, x2 = 0.33 and Min Z = -50)



x1 + x2 50

x1 ≥ 20

x2 40 and x1, x2 ≥ 0

(Ans: x1= 20, x2 = 0 and Min Z = -200)



0.2x1 + 0.4x2 56

x1 400

x2 300 and

x1, x2 ≥ 0

(Ans: x1= 0, x2 = 140 and Max Z = 7000)



3x1 + 4x2 24

9x1 + 4x2 36 and

x1, x2 ≥ 0

(Ans: x1= 2, x2 = 4.5 and Max Z = 37)



60x1 + 120x2 12000

8x1 + 5x2 600

3x1 + 4x2 500 and x1, x2 ≥ 0

(Ans: x1= 18.18, x2 = 90.91 and Max Z = 4181.82)

10. Maximize Z = 8x1 + 100x2


x1 + 2x2 720

5x1 + 4x2 1800

3x1 + x2 900 and x1, x2≥ 0

(Ans: x1= 0, x2 = 360 and Max Z = 36000)



2x1 + 5x2 16

6x1 30 and

x1, x2 ≥ 0

(Ans: x1= 0, x2 = 3.2 and Max Z = 32)

12. Maximize Z = x1 + 1.5x2


2x1 + 2x2 160

x1 + 2x2 120

4x1 + 2x2 280 and x1, x2 ≥ 0

(Ans: x1= 40, x2 = 40 and Max Z = 100)



x1 + 2x2 6

4x1 + 3x2 12 and

x1, x2 ≥ 0

(Ans: x1= 3, x2 = 0 and Max Z = 21)



2x1 + x2 50

2x1 + 5x2 100

2x1 + 3x2 90 and

x1, x2 ≥ 0

(Ans: x1= 0, x2 = 20 and Max Z = 200)

15. Maximize Z = 4x + 10y


2x + y 50

2x + 5y 100

2x + 3y 90 and

x, y ≥ 0

(Ans: x=0 , y = 20 and Max Z =200 )

16. Maximize Z = x + 3y


x + 2y 10

0 x 5

0 y 4

x, y ≥ 0

(Ans: x=2 , y =4 and Max Z =14 )


8



4x1 + 5x2 10

3x1 + 2x2 9

8x1 + 3x2 12 and x1, x2 ≥ 0

(Ans: x1= 1.07, x2 =1.14 and Max Z = 9.93)



-2x1 + x2 2

x1 - x2 2

3x1 + 2x2 9 and x1, x2 ≥ 0

(Ans: x1=2.6 , x2 = 0.6 and Max Z =18 )

19. Maximize Z = 2x1 + x2


x1 + 2x2 10

x1 + x2 6

x1 - x2 2

x1 - 2x2 1 and x1, x2 ≥ 0

(Ans: x1=4 , x2 =2 and Max Z =10 )



x1 - 2x2 2

2x1 + x2 6

x1 + 2x2 5

-x1 + x2 2 and x1, x2 ≥ 0

(Ans: x1=2.33 , x2 =1.33 and Max Z =16 )

21. Maximize Z = 400x1 + 100x2


2x1 + x2 800

5x1 + 2x2 2400

9x1 + 3x2 3200 and x1, x2 ≥ 0

(Ans: x1=355.56 , x2 =0 and Max Z =142222.22 )



x1 + x2 30

x1 - x2 ≥ 0

x2 ≥ 3

0 x1 20

0 x2 12 and x1, x2 ≥ 0

(Ans: x1=18 , x2 =12 and Max Z =72 )

23. Minimize Z = 20x1 + 10x2


x1 + 2x2 40

3x1 + x2 ≥ 30

4x1 + 3x2 ≥ 60 and x1, x2 ≥ 0

(Ans: x1= 6, x2 =12 and Min Z = 240)



x1 + 2x2 ≥ 3

x1 + x2 4

0 x1 5/2

0 x2 3/2 and x1, x2 ≥ 0

(Ans: x1=5/2 , x2 =3/2 and Max Z =22 )



5x1 + x2 ≥ 10

x1 + x2 ≥ 6

x1 + 4x2 ≥ 12 and x1, x2 ≥ 0

(Ans: x1= 1, x2 =5 and Min Z = 13)



x1 + 2x2 10

x1 + x2 6

x1 - x2 2

x1 - 2x2 1 and x1, x2 ≥ 0

(Ans: x1=4 , x2 =2 and Max Z =10 )

27. Maximize Z = 15x1 + 10x2


4x1 + 6x2 360

3x1 180

5x2 200 and x1, x2 ≥ 0

(Ans: x1=60, x2 = 20 and Max Z =1100 )



3x1 + x2 48

2x1 + x2 40 and

x1, x2 ≥ 0

(Ans: x1=0 , x2 =40 and Max Z =200 )


9

3. Simplex Method

Solve the following LPP by Simplex Method:



6x1 + 4x2 24

x1 + 2x2 6

- x1 + x2 1

x2 2 and x1, x2 ≥ 0

(Ans: x1= 3, x2 =1.5 and Max Z = 21)



4x1 + 2x2 80

2x1 + 5x2 180 and

x1, x2 ≥ 0

(Ans: x1= 2.5, x2 =3.5 and Max Z = 147.5)



x1 + x2 400

2x1 + x2 600 and

x1, x2 ≥ 0

(Ans: x1= 0, x2 = 400 and Max Z = 1200)

4. Maximize Z = x1 + 5x2


2x1 + 5x2 15

6x1 30 and

x1, x2 ≥ 0

(Ans: x1= 0, x2 = 3 and Max Z = 15)

5. Maximize Z = x1 - x2 + 3x3


x1 + x2 + x3 10

2x1 – x3 2

2x1 - 2x2 + 3x3 0

x1, x2, x3 ≥ 0

(Ans: x1= , x2 = , x3 = and Max Z = )

6. Maximize Z = 2x + 4y + 3z


3x + 4y + 2z 60

2x + y + 2z 40

x + 3y + 2z 80

x, y, z ≥ 0

(Ans: x= , y= , z= and Max Z = )



0.2x1 + 0.4x2 56

x1 400

x2 300 and x1, x2 ≥ 0

(Ans: x1= 0, x2 = 140 and Max Z = 7000)



3x1 + 4x2 24

9x1 + 4x2 36 and

x1, x2 ≥ 0

(Ans: x1= 2, x2 = 4.5 and Max Z = 37)



60x1 + 120x2 12000

8x1 + 5x2 600

3x1 + 4x2 500 and x1, x2 ≥ 0

(Ans: x1= 18.18, x2 = 90.91 and Max Z =

4181.82)

10. Maximize Z = 8x1 + 100x2


x1 + 2x2 720

5x1 + 4x2 1800

3x1 + x2 900 and x1, x2 ≥ 0

(Ans: x1= 0, x2 = 360 and Max Z = 36000)


10



2x1 + 5x2 16

6x1 30 and

x1, x2 ≥ 0

(Ans: x1= 0, x2 = 3.2 and Max Z = 32)

12. Maximize Z = x1 + 1.5x2


2x1 + 2x2 160

x1 + 2x2 120

4x1 + 2x2 280 and x1, x2 ≥ 0

(Ans: x1= 40, x2 = 40 and Max Z = 100)



x1 + 2x2 6

4x1 + 3x2 12 and

x1, x2 ≥ 0

(Ans: x1= 3, x2 = 0 and Max Z = 21)



2x1 + x2 50

2x1 + 5x2 100

2x1 + 3x2 90 and

x1, x2 ≥ 0

(Ans: x1= 0, x2 = 20 and Max Z = 200)

15. Maximize Z = 4x + 10y


2x + y 50

2x + 5y 100

2x + 3y 90 and

x, y ≥ 0

(Ans: x=0 , y = 20 and Max Z =200 )

16. Maximize Z = x + 3y


x + 2y 10

0 x 5

0 y 4

x, y ≥ 0

(Ans: x=2 , y =4 and Max Z =14 )



4x1 + 5x2 10

3x1 + 2x2 9

8x1 + 3x2 12 and

x1, x2 ≥ 0

(Ans: x1= 1.07, x2 =1.14 and Max Z = 9.93)



-2x1 + x2 2

x1 - x2 2

3x1 + 2x2 9 and

x1, x2 ≥ 0

(Ans: x1=2.6 , x2 = 0.6 and Max Z =18 )



x1 + 2x2 10

x1 + x2 6

x1 - x2 2

x1 - 2x2 1 and x1, x2 ≥ 0

(Ans: x1=4 , x2 =2 and Max Z =10 )



x1 - 2x2 2

2x1 + x2 6

x1 + 2x2 5

-x1 + x2 2 and x1, x2 ≥ 0

(Ans: x1=2.33 , x2 =1.33 and Max Z =16 )

21. Maximize Z = 400x1 + 100x2


2x1 + x2 800

5x1 + 2x2 2400

9x1 + 3x2 3200 and x1, x2 ≥ 0

(Ans: x1=355.56 , x2 =0 and Max Z =142222.22 )

22. Maximize Z = 3x1 + 2x2 - 2x3


x1 + 2x2 + 2x3 10

2x1 + 4x2 + 3x3 15

x1, x2 , x3 ≥ 0


11

23. Maximize Z = 30x1 + 23x2 + 29x3


6x1 + 5x2 + 3x3 26

4x1 + 2x2 + 5x3 7

24. Maximize Z = 12x1 + 20x2 + 18x3 +

40x4


4x1 + 9x2 + 7x3 + 10x4 6000

x1 + x2 + 3x3 + 40x4 4000 and

x1, x2, x3, x4 ≥ 0

(Ans: x1=4000/3, x2= x3=0, x4=200/3 and Max Z

=56000/3 )

**25. Maximize Z = 3x1 + 9x2


x1 + 4x2 8

x1 + 2x2 4

x1, x2 ≥ 0 (Degenerate Solution)

**26. Maximize Z = 2x1 + 4x2


x1 + 2x2 5

x1 + x2 4

x1, x2 ≥ 0 (Alternative or infinite

Solutions)

**27. Maximize Z = 2x1 + x2


x1 - x2 10

2x1 40

x1, x2 ≥ 0 (Unbounded Solution)

**28. Maximize Z = 3x1 + 2x2


2x1 + x2 2

3x1 + 4x2 12

x1, x2 ≥ 0 (Infeasible Solution)

4. Simplex Method – Artificial Variable Technique

Solve the following LPP by Big-M Method or Artificial Variable Method or Cost- Penalty

Method or Charnes’ penalty method:

1. Minimize Z = x1 + x2


2x1 + 4x2 ≥ 4

x1 + 7x2 ≥ 7 and x1, x2 ≥

0

(Ans: x1= 21/13, x2 =10/13 and Min Z =

31/13)

2. Minimize Z = x1 - 2x2 - 3x3


-2x1 + 3x2 + 3x3 = 2

2x1 + 3x2 + 4x3 = 1 and x1,

x2, x3 ≥ 0

(Ans: No feasible Solution exists)



2x1 + 4x2 12

2x1 + 2x2 = 10

5x1 + 2x2 ≥ 10 and x1, x2 ≥

0

(Ans: x1= 4, x2 = 1 and Min Z = 23)

4. Maximize Z = x1 + 2x2 +3x3 – x4


x1 + 2x2 + 3x3 = 15

2x1 + x2 + 5x3 = 20

x1 + 2x2 + x3 + x4 = 10 and x1, x2,

x3, x4 ≥ 0

(Ans: x1=15/6, x2 =15/6, x3 =15/6, x4=0 and

Max Z = 15)


12

5. Minimize Z = 600x1 + 500x2


2x1 + x2 ≥ 80

x1 + 2x2 ≥ 60 and

x1, x2 ≥ 0

(Ans: x1= 100/3, x2 =40/3 and Min Z =

80,000/3)

6. Minimize Z = 2x + y


3x + y 3

4x + 3y ≥ 6

x + 2y 3 and x, y ≥ 0

(Ans: x=3/5, y=6/5 and Min Z = 12/5)

7. Maximize Z = 3x1 + 2x2 +3x3


2x1 + x2 + x3 2

3x1 + 4x2 + 2x3 8

x1, x2, x3 ≥ 0

(Ans: x1=0, x2 =2, x3 =0 and Max Z = 4)



3x1 + x2 ≥ 27

x1 + x2 ≥ 21

x1 + 2x2 ≥ 30 and x1, x2 ≥

0

(Ans: x1= , x2 = and Min Z = )



3x + y = 3

4x + 3y ≥ 6

x + 2y 4 and x, y ≥ 0

(Ans: x=2/5, y=9/5, and Min Z = 17/5)

5. Variants of Simplex Method – Two Phase Method

Solve the following LPP by Two-Phase Method:

1. Minimize Z = 7.5x1 - 3x2


3x1 - x2 - x3 ≥ 3

x1 - x2 + x3 ≥ 2 and x1, x2,

x3 ≥ 0

(Ans: x1= 5/4, x2 = x3 = 0 and Min Z = 75/8)



3x + y = 3

4x + 3y ≥ 6

x + 2y 4 and x, y ≥ 0

(Ans: x=2/5, y=9/5, and Min Z = 17/5)

3. Maximize Z = 2x1 + 3x2+ 5x3


3x1 + 10x2 + 5x3 15

33x1 - 10x2 + 9x3 33

x1 + x2 + x3 ≥ 4 and x1, x2, x2

≥ 0



x1 + 2x2 ≥ 3

x1 + 4x2 ≥ 4 and

x1, x2 ≥ 0


13

6. Variants of Simplex Method – Duality

Write the dual of the following primal LPP: Try to solve the dual problem and obtain the solution

of the given primal from the Dual problem by simplex method.



x1 + 2x2 10

x1 + x2 6

x1 - x2 2

x1 - 2x2 1 and x1, x2 ≥ 0

(Ans: w1= 0, w2 =3/2, w3 =1/2, w4 =0 and

Min Z* = 10)

(Ans: x1= 4, x2 =2 and Max Z = 10)

2. Maximize Z = w1 + w2 + w3


2w1 + w2 + 2w3 2

4w1 + 2w2 + w3 2

w1, w2, w3 ≥ 0



x1 - x2 1

x1 + x2 ≥ 4

x1 - 3x2 3 and x1, x2 ≥ 0

(Ans: x1= , x2 = and Max Z = )

4. Maximize Z = 30x1 + 23x2 +29x3


6x1 + 5x2 + 3x3 26

4x1 + 2x2 + 5x3 7 and x1, x2,

x3 ≥ 0



3x1 + 5x2 ≥ 5

5x1 + 2x2 ≥ 3 and

x1, x2 ≥ 0

(Ans: x1=, x2 = and Max Z =)

6. Minimize Z = 2x1 + 9x2 +x3


x1 + 4x2 + 2x3 ≥ 5

3x1 + x2 + 2x3 ≥ 4 and x1, x2,

x3 ≥ 0

7. Maximize f(x) = 6x + 5y + 2z


x + 3y + 2z ≥ 5

2x + 2y + z ≥ 2

4x – 2y + 3z ≥ -1 and x, y, z ≥ 0

8. Maximize Z = 3x + 5y + 4z


2x + 3z ≤ 8

5x + 2y + 2z ≤ 10

5y + 4z ≤ 15 and x, y, z ≥ 0

9. A company produces three products P, Q and R from three raw materials: A, B and C. One

unit of P requires 2 units of A and 3 units of B. One unit of Q requires 2 units if B and 5 units of

C and one unit of product R require 3 units of A, 2 units of B and 4 units of C. The company has

8 units of material A, 10 units of material B and 15 units of material C available to it. Profit per

unit of products P, Q and R are Rs. 3, Rs. 5, and Rs. 4 Respectively.

(a) Formulate the problem as a mathematical model.

(b) How many units of each product should be produced so as to maximize the profit?

(c) Write the dual problem.

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