Download - Kxu Chap02 Solution
-
PROBLEM SUMMARY
1. Maximization
2. Maximization
3. Minimization
4. Sensitivity analysis (23)
5. Minimization
6. Maximization
7. Slack analysis (26)
8. Sensitivity analysis (26)
9. Maximization, graphical solution
10. Slack analysis (29)
11. Maximization, graphical solution
12. Minimization, graphical solution
13. Maximization, graphical solution
14. Sensitivity analysis (213)
*15. Sensitivity analysis (213)
16. Maximization, graphical solution
*17. Sensitivity analysis (216)
18. Maximization, graphical solution
19. Standard form
20. Maximization, graphical solution
21. Standard form
22. Maximization, graphical solution
23. Constraint analysis (222)
24. Minimization, graphical solution
25. Sensitivity analysis (224)
26. Sensitivity analysis (224)
27. Sensitivity analysis (224)
28. Minimization, graphical solution
29. Minimization, graphical solution
30. Sensitivity analysis (229)
31. Maximization, graphical solution
32. Maximization, graphical solution
33. Minimization, graphical solution
34. Maximization, graphical solution
5
Chapter Two: Linear Programming: Model Formulation and Graphical Solution
35. Sensitivity analysis (234)
36. Minimization, graphical solution
37. Maximization, graphical solution
38. Maximization, graphical solution
39. Sensitivity analysis (238)
40. Maximization, graphical solution
41. Sensitivity analysis (240)
42. Maximization, graphical solution
43. Sensitivity analysis (242)
44. Minimization, graphical solution
45. Sensitivity analysis (244)
46. Maximization, graphical solution
47. Sensitivity analysis (246)
48. Maximization, graphical solution
49. Sensitivity analysis (248)
50. Multiple optimal solutions
51. Infeasible problem
52. Unbounded problem
PROBLEM SOLUTIONS
1.
2.(a) maximize Z = 6x1 + 4x2 (profit, $)subject to
10x1 + 10x2 100 (line 1, hr)7x1 + 3x2 42 (line 2, hr)
x1,x2 0
0 2 4 6 8 10 12 14x
x
1
Point C is optimalZ
B
C
A2
4
6
8
10
12
2
A: x1 = 0x2 = 3Z = 6
B: x1 = 15/7x2 = 16/7Z = 152/7
*C: x1 = 5x2 = 0Z = 40
-
7
Wood
2x1 + 6x2 36 lb2(6) + 6(3.2) 36
12 + 19.2 3631.2 36
36 31.2 = 4.8
There is 4.8 lb of wood left unused.
8. The new objective function, Z = 400x1 + 500x2, is parallel to the constraint for labor, whichresults in multiple optimal solutions. Points B(x1 = 30/7, x2 = 32/7) and C (x1 = 6, x2 = 3.2)are the alternate optimal solutions, each with aprofit of $4,000.
9.(a) maximize Z = x1 + 5x2 (profit, $)subject to
5x1 + 5x2 25 (flour, lb)2x1 + 4x2 16 (sugar, lb)
x1 5 (demand for cakes)x1,x2 0
(b)
10. In order to solve this problem, you must substitute the optimal solution into the resourceconstraints for flour and sugar and determinehow much of each resource is left over.
Flour
5x1 + 5x2 25 lb5(0) + 5(4) 25
20 2525 20 = 5
There are 5 lb of flour left unused.
Sugar
2x1 + 4x2 162(0) + 4(4) 16
16 16
There is no sugar left unused.
11.
12.(a) minimize Z = 80x1 + 50x2 (cost, $)subject to
3x1 + x2 6 (antibiotic 1, units) x1 + x2 4 (antibiotic 2, units)
2x1 + 6x2 12 (antibiotic 3, units)x1,x2 0
(b)
13.(a) maximize Z = 300x1 + 400x2 (profit, $)subject to
3x1 + 2x2 18 (gold, oz)2x1 + 4x2 20 (platinum, oz)
x2 4 (demand, bracelets)x1,x2 0
12
10
8
6
4
2
B
C
A
x1
x2
Z
0 2 4 6 8 10 12 14
Point A is optimal
A : x1 = 0
x2 = 4
Z = 20
B : x1 = 2
x2 = 3
Z = 17
C : x1 = 5
x2 = 0
Z = 5
*
12
10
8
6
4
2
A
B
CZ
x1
x2
0 2 4 6 8 10 12
Point A is optimal
A : x1 = 0
x2 = 9
Z = 54
B : x1 = 4
x2 = 3
Z = 30
C : x1 = 4
x2 = 1
Z = 18
*
12
10
8
6
4
2
B
B
A :
A
C
C D
x1
x1
x2
x2
B :
Z
ZZ
Z
===
x1x2
===
x1x2
===0
6300
230
3
1290
13
D :
Z
x1x2
===
480
60
0 2 4 6 8 10 12 14
Point is optimal
*
:
-
9
21. maximize Z = 5x1 + 8x2 + 0s1 + 0s3 + 0s4subject to:
3x1 + 5x2 + s1 = 502x1 + 4x2 + s2 = 40x1 + s3 = 8x1, x2 0
A: s1 = 0, s2 = 0, s3 = 8, s4 = 0B: s1 = 0, s2 = 3.2, s3 = 0, s4 = 4.8C: s1 = 26, s2 = 24, s3 = 0, s4 = 10
22.
The extreme points to evaluate are now A, B',and C'.
A: x1 = 0x2 = 30Z = 1,200
*B': x1 = 15.8x2 = 20.5Z = 1,610
C': x1 = 24x2 = 0Z = 1,200
Point B' is optimal
18.
19. maximize Z = 1.5x1 + x2 + 0s1 + 0s3subject to:
x1 + s2 = 4x2 + s2 = 6x1 + x2 + s3 = 5x1, x2 0
A: s1 = 4, s2 = 1, s3 = 0B: s1 = 0, s2 = 5, s3 = 0C: s1 = 0, s2 = 6, s3 = 1
20.
12
10
8
6
4
2
A
BC x1
x2
Z
0 2 4 6 8 10 12 14
Point B is optimal
A : x1 = 0
x2 = 5
Z = 5
B : x1 = 4
x2 = 1
Z = 7
C : x1 = 4
x2 = 0
Z = 6
*
12
10
8
6
4
2
A
B
Cx1
x2
Z
0 2 4 6 8 10 12 14 16 2018
Point B is optimal
A : x1 = 0
x2 = 10
Z = 80
B : x1 = 8
x2 = 5.2
Z = 81.6
C : x1 = 8
x2 = 0
Z = 40
*
12
14
16
10
8
6
4
2
A B
Cx1
x2
Z
0 2 4 6 8 10 12 14 16 18 20
Point B is optimal
A : x1 = 8
x2 = 6
Z = 112
B : x1 = 10
x2 = 5
Z = 115
C : x1 = 15
x2 = 0
Z = 97.5
*
23. It changes the optimal solution to point A(x1 = 8, x2 = 6, Z = 112), and the constraint,x1 + x2 15, is no longer part of the solutionspace boundary.
24.(a) Minimize Z = 64x1 + 42x2 (labor cost, $)subject to
16x1 + 12x2 450 (claims)x1 + x2 40 (workstations)
0.5x1 + 1.4x2 25 (defective claims)x1, x2 0
-
10
29.
30. The problem becomes infeasible.
31.
32.
12
10
8
6
4
2BA
Cx1
x2
Z
0 2 4 6 8 10 12
Point C is optimal
A : x1 = 2.67
x2 = 2.33
Z = 22
B : x1 = 4
x2 = 3
Z = 30
C : x1 = 4
x2 = 1
Z = 18
*
12
10
8
6
4
2B
A
x1
x2
0 2 4 6 8 10 12 14
Feasible space
Point B is optimal
A : x1 = 4.8
x2 = 2.4
Z = 26.4
B : x1 = 6
x2 = 1.5
Z = 31.5
*
A
C
B
x1
x2
02
2
4
4
6
6
8
8
10 2 4 6 8 10 12
12
10
8
6
4
2
Point B is optimal
A : x1 = 4
x2 = 3.5
Z = 19
B : x1 = 5
x2 = 3
Z = 21
C : x1 = 4
x2 = 1
Z = 14
*
21.(b)
25. Changing the pay for a full-time claimsprocessor from $64 to $54 will change thesolution to point A in the graphical solutionwhere x1 = 28.125 and x2 = 0, i.e., there will beno part-time operators. Changing the pay for apart-time operator from $42 to $36 has noeffect on the number of full-time and part-timeoperators hired, although the total cost will bereduced to $1,671.95
26. Eliminating the constraint for defective claimswould result in a new solution, x1 = 0 and x2 =37.5, where only part-time operators would behired.
27. The solution becomes infeasible; there are notenough workstations to handle the increase inthe volume of claims.
28.
x2
x1
BC
DA
0 5 10 15 20 25 30 35 40 45 50
5
10
15
20
25
30
35
40
45
50
Point B is optimal
A : x1 = 28.125
x2 = 0
Z = 1,800
B : x1 = 20.121
x2 = 10.670
Z = 1,735.97
C : x1 = 5.55
x2 = 34.45
Z = 2,437.9
* D : x1 = 40
x2 = 0
Z = 2,560
x1
x2
0246 2 4 6 8 10 12
12
10
8
6
4
2
Point B is optimal
A : x1 = 2
x2 = 6
Z = 52
B : x1 = 4
x2 = 2
Z = 44
C : x1 = 6
x2 = 0
Z = 48
*
A
B
CZ
C : x1 = 34.45
x2 = 5.55
-
14
49. The feasible solution space changes if the fertilizer constraint changes to 20x1 + 20x2 800 tons. The new solution space is A'B'C'D'.Two of the constraints now have no effect.
The new optimal solution is point C':
A': x1 = 0 *C': x1 = 26x2 = 37 x2 = 14Z = 11,100 Z = 14,600
B': x1 = 3 D': x1 = 26x2 = 37 x2 = 0Z = 12,300 Z = 10,400
50.
Multiple optimal solutions; A and B alternate optimal
60
70
80
50
40
30
20
10
A
B
C
D x1
x2
0 10 20 30 40 50 60 70 80
Multiple optimal solutions; Aand B alternate optimal.
A : x1 = 0
x2 = 60
Z = 60,000
B : x1 = 10
x2 = 30
Z = 60,000
C : x1 = 33.33
x2 = 6.67
Z = 106,669
D : x1 = 60
x2 = 0
Z = 180,000
120
100
80
60
40
20
A
x1
x2
0 20 40 60 80 100 120 140
B
C
D
47. A new constraint is added to the model in
1.5
The solution is, x1 = 160, x2 = 106.67, Z = $568
48.(a) maximize Z = 400x1 + 300x2 (profit, $)subject to
x1 + x2 50 (available land, acres)10x1 + 3x2 300 (labor, hr)8x1 + 20x2 800 (fertilizer, tons)
x1 26 (shipping space, acres)x2 37 (shipping space, acres)
x1,x2 0
(b)
120
100
80
60
40
20BA
DEF
C
x1
x2
0 20 40 60 80 100 120 140
A : x1 = 0
x2 = 37
Z = 11,100
B : x1 = 7.5
x2 = 37
Z = 14,100
D : x1 = 21.4
x2 = 28.6
Z = 17,140
E : x1 = 26
x2 = 13.3
Z = 14,390
C : x1 = 16.7
x2 = 33.3
Z = 16,680
F : x1 = 26
x2 = 0
Z = 10,400
*
Point D is optimal
x1x2
500
450
400
350
300
250
200
150
100
50
0 50 100 150 200 250 300 350 400 450 500
X1
X2
A
B
*A: X1=106.67A: X2=160A: Z=568
B: X1=240*A: X2=0*A: Z=540
Point A is optional
-
15
The graphical solution is displayed as follows.
The optimal solution is x = 1, y = 1.5, and Z = 0.05. This means that a patrol sector is 1.5miles by 1 mile and the response time is 0.05hr, or 3 min.
CASE SOLUTION:THE POSSIBILITY RESTAURANT
The linear programming model formulation is
Maximize = Z = $12x1 + 16x2
subject to
x1 + x2 60.25x1 + .50x2 20
x1/x2 3/2 or 2x1 3x2 0x2/(x1 + x2) .10 or .90x2 .10x1 0
x1,x2 0
The graphical solution is shown as follows.
6
5
4
3
2
1
A
B
C
D
x
y
0 1 2 3 4 5 6 7
point Optimal
51.
52.
CASE SOLUTION:METROPOLITAN POLICE PATROL
The linear programming model for this case problem is
minimize Z = x/60 + y/45
subject to
2x + 2y 52x + 2y 12
y 1.5xx,y 0
The objective function coefficients are determined by dividing the distance traveled, ie., x/3, by the travel speed, ie., 20 mph. Thus, the x coefficient is x/3 20, or x/60. In the first two constraints, 2x + 2y represents the formula for the perimeter of a rectangle.
60
70
80
50
40
30
20
10
x1
x2
0 10 20 30 40 50 60 70 80
Infeasible Problem
60
70
80
100
50
40
30
20
10
A
B
Cx1
x2
x1 x2
0 10 20 30 40 50 60 70 80 100
Optimal point
2 3 0
x1x2.90 .10 0
x1 x2+ 20.25 .50
x1 x2+ 60
A : x1 = 34.3
x2 = 22.8
Z = $776.23
B : x1 = 40
x2 = 20
Z = $800optimal
C : x1 = 6
x2 = 54
Z = $744
60
70
80
50
40
30
20
10
x1
x2
0 1010 2020 30 40 50 60 70 80
Unbounded Problem