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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
Solution: Chapter 5: LP Graph
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
Therefore, the LP formulation is
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
Solution: Chapter 5: LP Graph
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
Solution: Chapter 5: LP Graph
Therefore, the LP formulation is
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
Solution: Chapter 5: LP Graph
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
Solution: Chapter 5: LP Graph
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
Solution: Chapter 5: LP Graph
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.
7
Solution: Chapter 5: LP Graph
x3 = number of PCs for finance dept.
Therefore, the LP formulation is
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.
8
Solution: Chapter 5: LP Graph
(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
Solution: Chapter 5: LP Graph
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
Solution: Chapter 5: LP Graph
For C
For D
Now we calculate the objective function values at the four corner points:
11
Solution: Chapter 5: LP Graph
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
Solution: Chapter 5: LP Graph
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
Solution: Chapter 5: LP Graph
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
Solution: Chapter 5: LP Graph
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
Solution: Chapter 5: LP Graph
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
Solution: Chapter 5: LP Graph
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
Solution: Chapter 5: LP Graph
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
Solution: Chapter 5: LP Graph
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
________
19