Post on 24-Oct-2014
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.
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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.
Prepared by S Raju / Associate Professor / Dept. of CA / S.V.C.E.
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.
Prepared by S Raju / Associate Professor / Dept. of CA / S.V.C.E.
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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.
Prepared by S Raju / Associate Professor / Dept. of CA / S.V.C.E.
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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.
Prepared by S Raju / Associate Professor / Dept. of CA / S.V.C.E.
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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)
2. Maximize Z = 3x1 + 4x2
Subject to the constraints
4x1 + 2x2 80
2x1 + 5x2 180 and
x1, x2 ≥ 0
(Ans: x1= 2.5, x2 =3.5 and Max Z = 147.5)
3. Maximize Z = 2x1 + 3x2
Subject to the constraints
x1 + x2 400
2x1 + x2 600 and
x1, x2 ≥ 0
(Ans: x1= 0, x2 = 400 and Max Z = 1200)
4. Maximize Z = x1 + 5x2
Subject to the constraints
2x1 + 5x2 15
6x1 30 and
x1, x2 ≥ 0
(Ans: x1= 0, x2 = 3 and Max Z = 15)
Prepared by S Raju / Associate Professor / Dept. of CA / S.V.C.E.
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5. Minimize Z = 60x1 + 90x2
Subject to the constraints
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)
6. Minimize Z = 10x1 + 25x2
Subject to the constraints
x1 + x2 50
x1 ≥ 20
x2 40 and x1, x2 ≥ 0
(Ans: x1= 20, x2 = 0 and Min Z = -200)
7. Maximize Z = 20x1 + 50x2
Subject to the constraints
0.2x1 + 0.4x2 56
x1 400
x2 300 and
x1, x2 ≥ 0
(Ans: x1= 0, x2 = 140 and Max Z = 7000)
8. Maximize Z = 5x1 + 6x2
Subject to the constraints
3x1 + 4x2 24
9x1 + 4x2 36 and
x1, x2 ≥ 0
(Ans: x1= 2, x2 = 4.5 and Max Z = 37)
9. Maximize Z = 30x1 + 40x2
Subject to the constraints
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
Subject to the constraints
x1 + 2x2 720
5x1 + 4x2 1800
3x1 + x2 900 and x1, x2≥ 0
(Ans: x1= 0, x2 = 360 and Max Z = 36000)
11. Maximize Z = 2x1 + 10x2
Subject to the constraints
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
Subject to the constraints
2x1 + 2x2 160
x1 + 2x2 120
4x1 + 2x2 280 and x1, x2 ≥ 0
(Ans: x1= 40, x2 = 40 and Max Z = 100)
13. Maximize Z = 7x1 + 5x2
Subject to the constraints
x1 + 2x2 6
4x1 + 3x2 12 and
x1, x2 ≥ 0
(Ans: x1= 3, x2 = 0 and Max Z = 21)
14. Maximize Z = 4x1 + 10x2
Subject to the constraints
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
Subject to the constraints
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
Subject to the constraints
x + 2y 10
0 x 5
0 y 4
x, y ≥ 0
(Ans: x=2 , y =4 and Max Z =14 )
Prepared by S Raju / Associate Professor / Dept. of CA / S.V.C.E.
8
17. Maximize Z = 5x1 + 4x2
Subject to the constraints
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)
18. Maximize Z = 6x1 + 4x2
Subject to the constraints
-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
Subject to the constraints
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 )
20. Maximize Z = 4x1 + 5x2
Subject to the constraints
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
Subject to the constraints
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 = 2x1 + 3x2
Subject to the constraints
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
Subject to the constraints
x1 + 2x2 40
3x1 + x2 ≥ 30
4x1 + 3x2 ≥ 60 and x1, x2 ≥ 0
(Ans: x1= 6, x2 =12 and Min Z = 240)
24. Maximize Z = 7x1 + 3x2
Subject to the constraints
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 )
25. Minimize Z = 3x1 + 2x2
Subject to the constraints
5x1 + x2 ≥ 10
x1 + x2 ≥ 6
x1 + 4x2 ≥ 12 and x1, x2 ≥ 0
(Ans: x1= 1, x2 =5 and Min Z = 13)
26. Maximize Z = 2x1 + x2
Subject to the constraints
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
Subject to the constraints
4x1 + 6x2 360
3x1 180
5x2 200 and x1, x2 ≥ 0
(Ans: x1=60, x2 = 20 and Max Z =1100 )
28. Maximize Z = 7x1 + 5x2
Subject to the constraints
3x1 + x2 48
2x1 + x2 40 and
x1, x2 ≥ 0
(Ans: x1=0 , x2 =40 and Max Z =200 )
Prepared by S Raju / Associate Professor / Dept. of CA / S.V.C.E.
9
3. Simplex Method
Solve the following LPP by Simplex 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)
2. Maximize Z = 3x1 + 4x2
Subject to the constraints
4x1 + 2x2 80
2x1 + 5x2 180 and
x1, x2 ≥ 0
(Ans: x1= 2.5, x2 =3.5 and Max Z = 147.5)
3. Maximize Z = 2x1 + 3x2
Subject to the constraints
x1 + x2 400
2x1 + x2 600 and
x1, x2 ≥ 0
(Ans: x1= 0, x2 = 400 and Max Z = 1200)
4. Maximize Z = x1 + 5x2
Subject to the constraints
2x1 + 5x2 15
6x1 30 and
x1, x2 ≥ 0
(Ans: x1= 0, x2 = 3 and Max Z = 15)
5. Maximize Z = x1 - x2 + 3x3
Subject to the constraints
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
Subject to the constraints
3x + 4y + 2z 60
2x + y + 2z 40
x + 3y + 2z 80
x, y, z ≥ 0
(Ans: x= , y= , z= and Max Z = )
7. Maximize Z = 20x1 + 50x2
Subject to the constraints
0.2x1 + 0.4x2 56
x1 400
x2 300 and x1, x2 ≥ 0
(Ans: x1= 0, x2 = 140 and Max Z = 7000)
8. Maximize Z = 5x1 + 6x2
Subject to the constraints
3x1 + 4x2 24
9x1 + 4x2 36 and
x1, x2 ≥ 0
(Ans: x1= 2, x2 = 4.5 and Max Z = 37)
9. Maximize Z = 30x1 + 40x2
Subject to the constraints
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
Subject to the constraints
x1 + 2x2 720
5x1 + 4x2 1800
3x1 + x2 900 and x1, x2 ≥ 0
(Ans: x1= 0, x2 = 360 and Max Z = 36000)
Prepared by S Raju / Associate Professor / Dept. of CA / S.V.C.E.
10
11. Maximize Z = 2x1 + 10x2
Subject to the constraints
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
Subject to the constraints
2x1 + 2x2 160
x1 + 2x2 120
4x1 + 2x2 280 and x1, x2 ≥ 0
(Ans: x1= 40, x2 = 40 and Max Z = 100)
13. Maximize Z = 7x1 + 5x2
Subject to the constraints
x1 + 2x2 6
4x1 + 3x2 12 and
x1, x2 ≥ 0
(Ans: x1= 3, x2 = 0 and Max Z = 21)
14. Maximize Z = 4x1 + 10x2
Subject to the constraints
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
Subject to the constraints
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
Subject to the constraints
x + 2y 10
0 x 5
0 y 4
x, y ≥ 0
(Ans: x=2 , y =4 and Max Z =14 )
17. Maximize Z = 5x1 + 4x2
Subject to the constraints
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)
18. Maximize Z = 6x1 + 4x2
Subject to the constraints
-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
Subject to the constraints
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 )
20. Maximize Z = 4x1 + 5x2
Subject to the constraints
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
Subject to the constraints
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
Subject to the constraints
x1 + 2x2 + 2x3 10
2x1 + 4x2 + 3x3 15
x1, x2 , x3 ≥ 0
Prepared by S Raju / Associate Professor / Dept. of CA / S.V.C.E.
11
23. Maximize Z = 30x1 + 23x2 + 29x3
Subject to the constraints
6x1 + 5x2 + 3x3 26
4x1 + 2x2 + 5x3 7
24. Maximize Z = 12x1 + 20x2 + 18x3 +
40x4
Subject to the constraints
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
Subject to the constraints
x1 + 4x2 8
x1 + 2x2 4
x1, x2 ≥ 0 (Degenerate Solution)
**26. Maximize Z = 2x1 + 4x2
Subject to the constraints
x1 + 2x2 5
x1 + x2 4
x1, x2 ≥ 0 (Alternative or infinite
Solutions)
**27. Maximize Z = 2x1 + x2
Subject to the constraints
x1 - x2 10
2x1 40
x1, x2 ≥ 0 (Unbounded Solution)
**28. Maximize Z = 3x1 + 2x2
Subject to the constraints
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
Subject to the constraints
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
Subject to the constraints
-2x1 + 3x2 + 3x3 = 2
2x1 + 3x2 + 4x3 = 1 and x1,
x2, x3 ≥ 0
(Ans: No feasible Solution exists)
3. Minimize Z = 5x1 + 3x2
Subject to the constraints
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
Subject to the constraints
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)
Prepared by S Raju / Associate Professor / Dept. of CA / S.V.C.E.
12
5. Minimize Z = 600x1 + 500x2
Subject to the constraints
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
Subject to the constraints
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
Subject to the constraints
2x1 + x2 + x3 2
3x1 + 4x2 + 2x3 8
x1, x2, x3 ≥ 0
(Ans: x1=0, x2 =2, x3 =0 and Max Z = 4)
8. Minimize Z = 4x1 + 2x2
Subject to the constraints
3x1 + x2 ≥ 27
x1 + x2 ≥ 21
x1 + 2x2 ≥ 30 and x1, x2 ≥
0
(Ans: x1= , x2 = and Min Z = )
6. Minimize Z = 4x + y
Subject to the constraints
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
Subject to the constraints
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)
2. Minimize Z = 4x + y
Subject to the constraints
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
Subject to the constraints
3x1 + 10x2 + 5x3 15
33x1 - 10x2 + 9x3 33
x1 + x2 + x3 ≥ 4 and x1, x2, x2
≥ 0
4. Minimize Z = 6x1 + 21x2
Subject to the constraints
x1 + 2x2 ≥ 3
x1 + 4x2 ≥ 4 and
x1, x2 ≥ 0
Prepared by S Raju / Associate Professor / Dept. of CA / S.V.C.E.
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.
1. Maximize Z = 2x1 + x2
Subject to the constraints
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
Subject to the constraints
2w1 + w2 + 2w3 2
4w1 + 2w2 + w3 2
w1, w2, w3 ≥ 0
3. Maximize Z = 3x1 + 4x2
Subject to the constraints
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
Subject to the constraints
6x1 + 5x2 + 3x3 26
4x1 + 2x2 + 5x3 7 and x1, x2,
x3 ≥ 0
5. Maximize Z = 15x1 + 10x2
Subject to the constraints
3x1 + 5x2 ≥ 5
5x1 + 2x2 ≥ 3 and
x1, x2 ≥ 0
(Ans: x1=, x2 = and Max Z =)
6. Minimize Z = 2x1 + 9x2 +x3
Subject to the constraints
x1 + 4x2 + 2x3 ≥ 5
3x1 + x2 + 2x3 ≥ 4 and x1, x2,
x3 ≥ 0
7. Maximize f(x) = 6x + 5y + 2z
Subject to the constraints
x + 3y + 2z ≥ 5
2x + 2y + z ≥ 2
4x – 2y + 3z ≥ -1 and x, y, z ≥ 0
8. Maximize Z = 3x + 5y + 4z
Subject to the constraints
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.