chapter 5_lp graph

Solution: Chapter 5: LP Graph

Chapter 5: Linear Programming

1. Let x1 = number of HCC-1 computers to be produced x2 = number of HCC-2 computers to be produced

Therefore, the LP formulation is

Maximize Z = 4,000x1 + 6,000x2

subject to25x1 + 30x2 1500

x1 20 x2 30

x1, x2 0

Graphical solution:

25x1 + 30x1 = 1500 -----------------(1)

Two points are (0, 50), (60, 0)

x1 = 20 ----------------(2)x2 = 30 ----------------(3)

For A

x1 = 20, x2 = 30, hence A = (20, 30).

1

10 20 30 40 50 60 70

10

20

30

40

50

60

C

B

A

Feasible region

x1

x2


For B

25x1 + 30x2 = 1500x2 = 30Or, 25x1 = 1500 – 30 30 = 600 x1 = 600/25 = 24

B = (24, 30)

For C

25x1 + 30x2 = 1500x1 = 2030x2 = 1500-2520 = 1000 x2 = 1000/30 = 33.33

C = (20, 33.33)

Z at A = 4000x1 + 6000x2 = (400020)+(600030) = 260,000 Z at B = (400024)+(600030) = 276,000Z at C = (400020)+(600033.33) = 279,980

Optimal solution is

C: x1 = 20, x2 = 33.33; Zmax = Rs. 279,9802. x1 = number of undergraduate courses to be offered

x2 = number of postgraduate courses to be offered


Minimize Z = 4,200x1 + 6,000x2

subject to x1 + x2 65

x1 35 x2 20

x1, x2 0

Graphical solution:

x1 + x2 = 65 -----------------(1)

Two points are (0, 65), (65, 0)

x1 = 35 ----------------(2)x2 = 20 ----------------(3)

2

10 20 30 40 50 60 70

10

20

30

40

50

60

B

A

x1

x2

80

70

80

Feasible region


For Ax1 + x2 = 65Since x2 = 20x1 = 65-20=45, hence A = (45, 20).

For B

x1 + x2 = 65x1 = 35x2 = 65– 35= 30

B = (35, 30)

Z at A = 4200x1 + 6000x2 = (420045)+(600020) = 309,000 Z at B = (420035)+(600030) = 327,000Optimal solution is

A: x1 = 45, x2 = 20

Zmin = $ 309,000

3. Let x1 = number of tables to be produced x2 = number of chairs to be produced

3



Maximize Z = 120x1 + 80x2

subject to 3x1 + 2x2 200

2x1 +2x2 180 x1 + x2 40

x2 – 2x1 0

Graphical solution:

3x1 + 2x2 = 200 -----------------(1)

Two points are (0, 100), (66.67, 0)

x1 +x2 = 90 ----------------(2)

Two points are (0, 90), (90, 0)

x1 + x2 = 40 ----------------(3)

Two points are (0, 40), (40,0)

x2 = 2x1

Two points are (50,100), (25,50)

B

A

For A

x1 + x2 = 40

4

10 20 30 40 50 60 70 80 90 100

10

20

30

4050

60

70

80

90

100

Feasible region

x1

x2


x2 = 2x1

x1 + 2x2 = 40 x1 = 40/3 = 13.33x2 = 26.66, Hence A = (13.33, 26.66).

B = (0, 40)

Objective function value calculation: Z at A = 120x1 + 80x2 = (12013.33)+(8026.66) = RM 3732.4 Z at B = (1200)+(8040) = RM 3200

Hence optimal solution is

A: x1 = 13.33, x2 = 26.66

Zmax = RM 3732.4

Remark: Constraints (1) and (2) are redundant.

4. Let x1 = number of acres to be allocated for tomatoes x2 = number of acres to be allocated for lettuce x3 = number of acres to be allocated for radishes

Revenue from tomatoes per acre of land = 20001.00 = Rs. 2000Per acre fertilizer expenditure for tomatoes = 1000.50 = Rs. 50Per acre expenditure due to labor for tomatoes = 520 = Rs. 100.

Hence profit from 1 acre of tomatoes = Rs (2000-50-100) = Rs. 1850

Similarly profit from 1 acre of letture and 1 acre of radishes are Rs 2080 and Rs. 1875, respectively. Hence the LP formulation is

Maximize Z = 1850x1 + 2080x2 + 1875x3

subject to x1 + x2 + x3 = 100

5x1 +6x2 + 5x3 400 x1, x2, x3 0

5. Let x1 = number of newspaper ads x2 = number of radios ads

Therefore, the LP formulation is the following:

Maximize Z = 6,000x1 + 2,000x2

subject to 600x1 + 400x2 7200

x1 + x2 15 x1 2

x2 2

5


x1, x2 0

Graphical solution:

600x1 + 400x1 = 7200 -----------------(1)

Two points are (0, 18), (12, 0)

x1 + x2 = 15----------------(2)

Two points are (0, 15), (15, 0)

x1 = 2 --------------------(3)x2 = 2 --------------------(4)

A

B

C

For A

600x1 + 400x2 = 7200x1 = 2, 400x2 = 7200-1200 = 6000 x2 = 6000/400 = 15Hence A = (2, 15).

For B

x1 + x2 = 15

6

2 4 6 8 10 12 14 16 18

2

4

6

8

10

12

14

16

18

Feasible region

x1

x2

20

20


x1 = 2x2 = 13

B = (2, 13)

For C

600x1 + 400x2 = 7200x1 + x2 = 15Or, 600x1 = 7200-400(15-x1) = 7200-6000+400x1

Or, 200x1 = 1200 x1 = 6, x2 = 9

C = (6, 9)

Objective function value calculation:

Z at A = 6,000x1 + 2000x2 = (60002)+(200015) = 42,000 Z at B = (60002)+(200013) = 38,000Z at C = (60006)+(20003) = 54,000

Optimal solution is: C (6, 9); Zmax = 54,000.

6. Let xDE = number of enamel paint cans produced at Dubai plant xDL = number of latex paint cans produced at Dubai plant xAE = number of enamel paint cans produced at Abu Dhabi plant xAL = number of Latex paint cans produced at Abu Dhabi plant

Profit from one can of enamel paint produced at Dubai plant is 8-5 = $3, and the profit from one can of latex paint produced at Dubai plant is 7-4.50 = $2.50. Similarly, profits from one can of enamel and one can of latex paint produced at Abu Dhabi plant are: $2, and $4, respectively.

The LP formulation is the following:

Maximize Z = 3xDE + 2.5xDL +2xAE + 4xAL

subject to xDE 500

xDL 500 xDE + xDL 500 xAE + xAL 800 5xDE + 4.5xDL 20,000 6xAE + 3xAL 30,000 xDE + xAE 600 xDL + xAL 800xDE , xDL , xAE , xAL 0

7. Let x1 = number of PCs for production dept. x2 = number of PCs for marketing dept.

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x3 = number of PCs for finance dept.


Maximize Z = 5x1 + 3x2 + 2x3

subject tox1 + x2 + x3 20

x1 5 x3 x2/2 x2 x1/3

x1, x2, x3 0

Or,

Maximize Z = 5x1 + 3x2 + 2x3

subject tox1 + x2 + x3 20

x1 5 -x2 + 2x3 0 -x1 + 3x3 0

x1, x2, x3 0

8. Let x1 = number of untrained workers x2 = number of semi-trained workers x3 = number of highly trained workers

Total cost of providing training to an untrained workers is (28+35)5 = $315Similarly, the costs of providing training to a semi-trained and highly-trained workers are $450.5 and $367.5, respectively.

The required LP formulation is:

Minimize Z = 315x1 + 450.5x2 + 367.5x3

subject to x1 + x2 + x3 25 28x1 +23x2 + 15x3 700

35x1 +30x2 + 20x3 775 x1, x2, x3 0

9. Let x1= the quantity of phosphate used x2 = the quantity of potassium used

Then the problem can be formulated as a linear programming as follows:

Minimize Z = 5x1+6x2

subject to: x1+ x2 =1000 x1 300 x2 150 x1, x2 0.

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(b) x1 + x2 = 1000

If x1 = 0, x2 = 1000If x2 = 0, x1 = 1000

Hence the two points are (0,1000), (1000,0).

The graph is the following:

100 200 300 400 500 600 700 800 900 1000

The feasible region is on the line segment AB.

A = (300,700), B = (0,1000)Z at A = (5300) + (6700) = Z at B = (50) + (61000) = 6000Hence the best solution is to mix 300 pounds of phosphate and 700 pounds of potassium and minimum cost will be RM 5,700.

10. The given LPP is:

x1+ x2 = 20

9

100

200

300

400

500

600

700

800

900

1000

A

B

570

x1

x2

maximize Z = 12x1+8x2

subject to: x1 + x2 ≤ 20 (1) 3x1 + x2 ≥ 30 (2) 2x1 + 6x2 ≥ 60 (3) 2x1 – x2 ≥ 0 (4) x1, x2 ≥ 0


Two points are (0,20), (20,0).

3x1+x2 = 30Put x1 = 0, x2 = 30, (0, 30)Put x2 = 0, x1 = 10, (10, 0)

2x1+6x2 = 60Put x1 = 0, x2 = 10, (0, 10)Put x2 = 0, x1 = 30, (30, 0)

2x1- x2 = 0 or 2x1 = x2; Two points are: (5, 10), (10, 20).

Let us plot the points one by one and obtain the following graph:

For A

For B

10

5 10 15 20 25 30 35 40

5

10

40

15

20

25

30

35

O

Feasible region

A B

CD

④

①

② ③


For C

For D

Now we calculate the objective function values at the four corner points:

11


11. Minimize Z = 2x1 + 3x2

subject to:

x1 125 x1 + x2 350 2x1 + x2 600

x1, x2 0

Graphical solution:

x1 = 125 ----------------(1)x1 + x2 = 350 -----------------(2)

(0, 350), (350, 0)

2x1 + x2 = 600 -----------------(3) (0, 600), (300, 0)

12

50 100 150 200 250 300 350 400 450

50

100

150

200

250

300

350

400

450

Feasible region

x2

500

500

550

600

550 600

B

A

C


For A

x1 = 125, x1+x2 = 350x2 = 350-125 = 225 Hence A = (125, 225).

For B

2x1 + x2 = 600x1+x2 = 350Or, 2x1 = 600 – (350-x1) = 600 – 350 + x1

x1 = 600-350 = 250x2 = 350-250 = 100

B = (250, 100)

For C

x1 = 1252x1 + x2 = 600

x2 = 600-(2125) = 600 – 250 = 350

C = (125, 350)

Calculation of objective function values:

Z at A = 2x1 + 3x2 = (2125)+(3225) = 925

13

x1


Z at B = (2250)+(3100) = 800Z at C = (2125)+(3350) = 1300

Optimal solution is:

B = (250, 100)Zmin = 800

12. Given LPP is:

Minimize Z = 2x1 + 3x2

subject to:

2x1 + 7x2 22 x1 + x2 6 5x1 + x2 10 x1, x2 0

Graphical solution:

2x1 + 7x2 = 22 -----------------(1)

(0, 3.1), (11, 0)

x1 + x2 = 6 -----------------(2) (0, 6), (6, 0)

5x1 + x2 = 10 -----------------(3) (0, 10), (2, 0)

14

1 2 3 4 5 6 7 8 9

1

2

3

4

5

6

7

8

9

Feasible region

x1

x2

10

10

11

12

11 12

B

A

C

D

O


For A

A = (11, 0), D = (0, 10)

For B

2x1 + 7x2 = 22x1+x2 = 6Or, 2x1 = 22 – 7(6-x1) = 22 – 42 + 7x1

5x1 = 20x1 = 4x2 = 2

B = (4, 2)

For C

5x1 + x2 = 10x1 + x2 = 65x1 = 10-(6-x1) = 10 – 6 + x1 or, 4x1 = 4or, x1 = 1x2 = 5 C = (1, 5)

Calculation of objective function values:

15


Z at A = 2x1 + 3x2 = (211)+(30) = 22 Z at B = (24)+(32) = 14Z at C = (21)+(35) = 17Z at D = (20)+(310) = 30

Optimal solution is:

B = (4, 2)Zmin = 14

13. a) Given LPP is:

Maximize Z = 2x1 + x2

subject to:

x1 + x2 6 3x1 + 2x2 16

x2 9 x1, x2 0

Graphical solution:

x1 + x2 = 6 -----------------(1)

(0, 6), (6, 0)

3x1 + 2x2 = 16 -----------------(2) (0, 8), (5.33, 0)

x2 = 9 -----------------(3)

16

2 4 6 8 10 12

2

4

6

8

10

12

A

B

C

Feasible region

x1

x2

D

0


Feasible region is not bounded and given LPP is of maximization type. Hence the given problem has no finite solution.

b) Minimize Z = -2x1 + 5x2

subject to:

x1 + x2 7 10x1 + 7x2 40 x1, x2 0

Graphical solution:

x1 + x2 = 7 -----------------(1)

(0, 7), (7, 0)

10x1 + 7x2 = 40 -----------------(2) (0, 5.7), (4, 0)

17

1 2 3 4 5 6 7 8 9

1

2

3

4

5

6

7

8

9

x1

x2

10

10


The linear programming problem does not have feasible solution.

c) Maximize Z = 3x1 + 2x2

subject to 6x1 + 4x2 24

x1 3 x1, x2 0

Graphical solution:

6x1 + 4x1 = 24 -----------------(1)

(0, 6), (4, 0)

x1 = 3 ----------------(2)

18

1 2 3 4 5 6 7 8 9

1

2

3

4

5

6

7

8

9

x1

x2

10

10

Feasible region

A

B

C


A = (3, 0)

For B

6x1 + 4x2 = 24x1 = 3Or, 4x2 = 24 – 18 = 6 x2 = 6/4 = 3/2

B = (3, 3/2)

Further, C = (0, 6)

Objective function value calculation:

Z at A = 3x1 + 2x2 = (33)+(20) = 9 Z at B = (33)+(23/2) = 12Z at C = (30)+(26) = 12

Multiple Optimal solutions exist. Two optimal solutions are: B(3, 3/2), C (0, 6) and Zmax = 12

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chapter 5_lp graph

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