Chapter Two: Linear Programming: Model Formulation and Graphical Solution40p6zu91z1c3x7lz71846qd1-wpengine.netdna-ssl.com/wp-content/... · 6 PROBLEM SUMMARY 1. Maximization (1–28
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7. In order to solve this problem, you must substitute the optimal solution into the resourceconstraint for wood and the resource constraint forlabor and determine how much of each resource isleft over.
12
10
8
6
4
2
A
B
C
D x1
x2
Z
0 2 4 6 8 10 12 14 16 18
Point C is optimal
A : x1 = 0
x2 = 6
Z = 600
B : x1 = 30/7
x2 = 32/7
Z = 2,171
D : x1 = 6
x2 = 0
Z = 2,400
C : x1 = 6
x2 = 3.2
Z = 2,720
*
12
10
8
6
4
2
A
B
C
x1
x2
Z0 2 4 6 8 10 12 14
Point C is optimal
A : x1 = 0
x2 = 10
Z = 50
B : x1 = 1
x2 = 5
Z = 28
*C : x1 = 4
x2 = 2
Z = 22
TAYLMC02_0131961381.QXD 4/14/09 8:32 AM Page 7
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Labor
8x1 + 10x2 ≤ 80 hr8(6) + 10(3.2) ≤ 80
48 + 32 ≤ 8080 ≤ 80
There is no labor left unused.
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.
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.
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
*
Sugar
2x1 + 4x2 ≤ 16
2(0) + 4(4) ≤ 16
16 ≤ 16
There is no sugar left unused.
11.
12. a) minimize Z = 80x1 + 50x2 (cost, $)subject to
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
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
*
:
TAYLMC02_0131961381.QXD 4/14/09 8:32 AM Page 8
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However, this creates a situation where both points C and D are optimal, ie., multipleoptimal solutions, as are all points on the linesegment between C and D.
16. a) maximize Z = 50x1 + 40x2 (profit, $)subject to
14. The new objective function, Z = 300x1 +600x2, is parallel to the constraint line forplatinum, which results in multiple optimalsolutions. Points B (x1 = 2, x2 = 4) and C (x1 = 4, x2 = 3) are the alternate optimalsolutions, each with a profit of $3,000.
The feasible solution space will change. The new constraint line, 3x1 + 4x2 = 20, isparallel to the existing objective function. Thus,multiple optimal solutions will also be presentin this scenario. The alternate optimal solutionsare at x1 = 1.33, x2 = 4 and x1 = 2.4, x2 = 3.2,each with a profit of $2,000.
15. a) Optimal solution: x1 = 4 necklaces, x2 = 3 bracelets. The maximum demand is notachieved by the amount of one bracelet.
b) The solution point on the graph which corresponds to no bracelets beingproduced must be on the x1 axis where x2 = 0.This is point D on the graph. In order for pointD to be optimal, the objective function “slope”must change such that it is equal to or greaterthan the slope of the constraint line, 3x1 + 2x2 =18. Transforming this constraint into the form y = a + bx enables us to compute the slope:
2x2 = 18 – 3x1x2 = 9 – 3/2x1
From this equation the slope is –3/2. Thus, the slope of the objective function must be at least–3/2. Presently, the slope of the objectivefunction is –3/4:
400x2 = Z – 300x1x2 = Z/400 – 3/4x1
The profit for a necklace would have to increase to $600 to result in a slope of –3/2:
400x2 = Z – 600x1x2 = Z/400 – 3/2x1
b)
17. The feasible solution space changes from the area 0ABC to 0AB'C', as shown on thefollowing graph.
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
60
40
50
30
20
10
A
C C
B B
x1
x2
Z
0 10 20 30 40 50 60 70
′
′
60
40
50
30
20
10
A
C
B
x1
x2
Z
0 10 20 30 40 50 60 70
Point B is optimal
A : x1 = 0
x2 = 30
Z = 1,200
B : x1 = 10.5
x2 = 23.7
Z = 1,473
C : x1 = 20
x2 = 0
Z = 1,000
*
TAYLMC02_0131961381.QXD 4/14/09 8:32 AM Page 9
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C': x1 = 24x2 = 0Z = 1,200
Point B' is optimal
18.
19. maximize Z = 1.5x1 + x2 + 0s1 + 0s2 + 0s3subject to:
x1 + s1 = 4
x2 + s2 = 6
x1 + x2 + s3 = 5
x1, x2 ≥ 0
A: s1 = 4, s2 = 1, s3 = 0
B: s1 = 0, s2 = 5, s3 = 0
C: s1 = 0, s2 = 6, s3 = 1
20.
21. maximize Z = 5x1 + 8x2 + 0s1 + 0s3 + 0s4subject to:
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
b)
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
TAYLMC02_0131961381.QXD 4/14/09 8:32 AM Page 10
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30. The problem becomes infeasible.
31.
32.
33.
12
10
8
6
4
2BA
D
Cx1
x2
(5)
(2)
(3)
(4)
(1)
0 2 4 6 8 10 12
Point C is optimal
A : x1 = 2.67x2 = 2.33
Z = 22
B : x1 = 4x2 = 3
Z = 30
D : x1 = 3.36x2 = 3.96
Z = 33.84
C : x1 = 4x2 = 1
Z = 18
*
12
10
8
6
4
2B
A
x1
x2
0 2 4 6 8 10 12 14
Feasible space
Point A 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
0–2
–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 = 4x2 = 3.5
Z = 19
B : x1 = 5x2 = 3
Z = 21
C : x1 = 4x2 = 1
Z = 14
*
12
10
8
6
4
2
–2
B
A
x1
x2
02 6 8 10 12 14
C
Point A is optimal
A : x1 = 3.2
x2 = 6
Z = 37.6
B : x1 = 5.33
x2 = 3.33
Z = 49.3
C : x1 = 9.6
x2 = 1.2
Z = 79.2
*
4
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.
29.
x1
x2
0–2–4–6 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
TAYLMC02_0131961381.QXD 4/14/09 8:32 AM Page 11
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34. a) Maximize Z = $4.15x1 + 3.60x2 (profit, $)subject to
b)
35. No additional profit, freezer space is not a bindingconstraint.
36. a) minimize Z = 200x1 + 160x2 (cost, $)subject to
39. The slope of the original objective function is computed as follows:
Z = 30x1 + 70x270x2 = Z – 30x1
x2 = Z/70 – 3/7x1slope = –3/7
12
14
10
8
6
4
2
BA
C
x1
x2
0 2 4 6 8 10 12 14 16
Point B is optimal
A : x1 = 3
x2 = 6
Z = 7,800
B : x1 = 5
x2 = 5
Z = 8,500
C : x1 = 6
x2 = 3
Z = 7,500
*
12
14
10
8
6
4
2
B
A
C D
x1
x2
0 2 4 6 8 10 12 14
Point B is optimal
A : x1 = 0
x2 = 6
Z = 960
B : x1 = 1
x2 = 3
Z = 680
C : x1 = 3
x2 = 1
Z = 760
D : x1 = 6
x2 = 0
Z = 1,200
*
x2
x1
B
A
0 20 40 60 80 100 120
20
40
60
80
100
120
Point A is optimal
*A : x1 = 68.96
x2 = 34.48
Z = 410.35
B : x1 = 96.77
x2 = 0
Z = 47.59Z = 401.6
12
14
10
8
6
4
2
A
BC
D x1
x2
0 2 4 6 8 10 12 14 16 2018
A : x1 = 0
x2 = 8
Z = 560
B : x1 = 3.3
x2 = 6.7
Z = 568
C : x1 = 5.3
x2 = 4.7
Z = 488
D : x1 = 8
x2 = 0
Z = 240
*
Point B is optimal
TAYLMC02_0131961381.QXD 4/14/09 8:32 AM Page 12
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The slope of the new objective function is computed as follows:
Z = 90x1 + 70x270x2 = Z – 90x1
x2 = Z/70 – 9/7x1slope = –9/7
The change in the objective function not only changes the Z values but also results in a new solutionpoint, C. The slope of the new objective function issteeper and thus changes the solution point.
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.
52. x1 = exams assigned to Bradx2 = exams assigned to Sarah
minimize Z = .10x1 + .06x2
subject to
x1 + x2 = 120x1 ≤ (720/7.2) or 100x2 ≤ 50(600/12)
x1,x2 ≥ 0
53. If the constraint for Sarah’s time became x2 ≤55 with an additional hour then the solutionpoint at A would move to x1 = 65, x2 = 55 andZ = 9.8. If the constraint for Brad’s timebecame x1 ≤ 108.33 with an additional hourthen the solution point (A) would not change.All of Brad’s time is not being used anywayso assigning him more time would not have aneffect.
One more hour of Sarah’s time would reducethe number of regraded exams from 10 to 9.8,whereas increasing Brad by one hour wouldhave no effect on the solution. This is actuallythe marginal (or dual) value of one additionalhour of labor, for Sarah, which is 0.20 fewerregraded exams, whereas the marginal valueof Brad’s is zero.
54. a) x1 = # cups of Pomonax2 = # cups of Coastal
maximize Z = $2.05x1 + 1.85x2
subject to
16x1 + 16x2 ≤ 3,840 oz or (30 gal. � 128 oz)(.20)(.0625)x1 + (.60)(.0625)x2 ≤ 6 lbs. Colombian(.35)(.0625)x1 + (.10)(.0625)x2 ≤ 6 lbs. Kenyan(.45)(.0625)x1 + (.30)(.0625)x2 ≤ 6 lbs. Indonesian
x2/x1 = 3/2x1, x2 ≥ 0
b) Solution:
x1 = 87.3 cupsx2 = 130.9 cupsZ = $421.09
55. a) The only binding constraint is for Colombian;the constraints for Kenyan and Indonesian arenonbinding and there is already extra, or slack, pounds of these coffees available. Thus, only getting more Colombian wouldaffect the solution.
One more pound of Colombian wouldincrease sales from $421.09 to $463.20.
Increasing the brewing capacity to 40 gallonswould have no effect since there is alreadyunused brewing capacity with the optimalsolution.
b) If the shop increased the demand ratio ofPomona to Coastal from 1.5 to 1 to 2 to 1 itwould increase daily sales to $460.00, so theshop should spend extra on advertising toachieve this result.
50
500
A
B
100
100
150
150 200 X1
X2
200
*A : x1 = 70
x2 = 50
Z = 10
B : x1 = 100
x2 = 20
Z = 11.2
optimal
200
A
400
600
800
1000
2000 400 600 800 1000X1
X2
A : x1 = 87.3
x2 = 130.9
Z = 421.09
TAYLMC02_0131961381.QXD 4/14/09 8:32 AM Page 16
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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,i.e., x/3, by the travel speed, i.e., 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.
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
x1x2 ≥ 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
56.
Multiple optimal solutions; A and B alternate optimal
57.
58.
60
70
80
50
40
30
20
10
x1
x2
0 10 20 30 40 50 60 70 80
Infeasible Problem
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
60
70
80
50
40
30
20
10
x1
x2
0 10–10 20–20 30 40 50 60 70 80
Unbounded Problem
TAYLMC02_0131961381.QXD 4/14/09 8:32 AM Page 17
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Changing the objective function to Z = $16x1 + 16x2 would result in multiple optimal solutions, the end points being B and C. The profit in each case would be $960.
Changing the constraint from.90x2 – .10x1 ≥ 0 to .80x2 –.20x1 ≥ 0 has no effect on the solution.
CASE SOLUTION:ANNABELLE INVESTS IN THEMARKET
x1 = no. of shares of index fundx2 = no. of shares of internet stock fund
Maximize Z = (.17)(175)x1 + (.28)(208)x2= 29.75x1 + 58.24x2
subject to
x1 = 203x2 = 406Z = $29,691.37
Eliminating the will have no
effect on the solution.
Eliminating the will change
the solution to x1 = 149, x2 = 451.55,Z = $30,731.52.
Increasing the amount available to invest (i.e.,$120,000 to $120,001) will increase profit from Z = $29,691.37 to Z = $29,691.62 orapproximately $0.25. Increasing by anotherdollar will increase profit by another $0.25, andincreasing the amount available by one moredollar will again increase profit by $0.25. Thisindicates that for each extra dollar invested areturn of $0.25 might be expected with thisinvestment strategy. Thus, the marginal value ofan extra dollar to invest is $0.25, which is alsoreferred to as the “shadow” or “dual” price asdescribed in Chapter 3.