Preview Extract

Chapter Two: Linear Programming: Model Formulation
and Graphical Solution
PROBLEM SUMMARY
1. Maximization (1โ40 continuation), graphical
solution
34. Maximization, graphical solution
35. Sensitivity analysis (2โ34)
2. Minimization, graphical solution
36. Maximization, graphical solution
3. Sensitivity analysis (2โ2)
37. Sensitivity analysis (2โ36)
4. Minimization, graphical solution
38. Maximization, graphical solution
5. Maximization, graphical solution
39. Sensitivity analysis (2โ38)
6. Slack analysis (2โ5), sensitivity analysis
40. Minimization, graphical solution
7. Maximization, graphical solution
41. Sensitivity analysis (2โ40)
8. Slack analysis (2โ7)
42. Maximization, graphical solution
9. Maximization, graphical solution
43. Sensitivity analysis (2โ42)
10. Minimization, graphical solution
44. Maximization, graphical solution
11. Maximization, graphical solution
45. Sensitivity analysis (2โ44)
12. Sensitivity analysis (2โ11)
46. Maximization, graphical solution
13. Sensitivity analysis (2โ11)
47. Minimization, graphical solution
14. Maximization, graphical solution
48. Sensitivity analysis (2โ47)
15. Sensitivity analysis (2โ14)
49. Minimization, graphical solution
16. Maximization, graphical solution
50. Sensitivity analysis (2โ49)
17. Sensitivity analysis (2โ16)
51. Maximization, graphical solution
18. Maximization, graphical solution
52. Minimization, graphical solution
19. Standard form (2โ18)
53. Sensitivity analysis (2โ52)
20. Maximization, graphical solution
54. Maximization, graphical solution
21. Constraint analysis (2โ20)
55. Sensitivity analysis (2โ54)
22. Minimization, graphical solution
56. Maximization, graphical solution
23. Sensitivity analysis (2โ22)
57. Sensitivity analysis (2โ56)
24. Minimization, graphical solution
58. Multiple optimal solutions
25. Minimization, graphical solution
59. Infeasible problem
26. Sensitivity analysis (2โ25)
60. Unbounded problem
27. Minimization, graphical solution
28. Maximization, graphical solution
29. Minimization, graphical solution
30. Maximization, graphical solution
31. Sensitivity analysis (2โ30)
32. Minimization, graphical solution
33. Maximization, graphical solution
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6×1 + 6×2 โฅ 36 (phosphate, oz)
x2 โฅ 2 (potassium, oz)
x1,x2 โฅ 0
PROBLEM SOLUTIONS
1. a) x1 = # cakes
x2 = # loaves of bread
maximize Z = $10×1 + 6×2
subject to
3×1 + 8×2 โค 20 cups of flour
45×1 + 30×2 โค 180 minutes
x1,x2 โฅ 0
b)
b)
5.
a) Maximize Z = 400×1 + 100×2 (profit, $)
subject to
8×1 + 10×2 โค 80 (labor, hr)
2×1 + 6×2 โค 36 (wood)
x1 โค 6 (demand, chairs)
x1,x2 โฅ 0
2.
a) Minimize Z = .05×1 + .03×2 (cost, $)
subject to
b)
8×1 + 6×2 โฅ 48 (vitamin A, mg)
x1 + 2×2 โฅ 12 (vitamin B, mg)
x1,x2 โฅ 0
b)
6.
a) In order to solve this problem, you must
substitute the optimal solution into the
resource constraint for wood and the
resource constraint for labor and determine
how much of each resource
is left over.
Labor
3.
The optimal solution point would change
from point A to point B, thus resulting in the
optimal solution
x1 = 12/5 x2 = 24/5 Z = .408
4.
a) Minimize Z = 3×1 + 5×2 (cost, $)
subject to
8×1 + 10×2 โค 80 hr
8(6) + 10(3.2) โค 80
48 + 32 โค 80
80 โค 80
There is no labor left unused.
10×1 + 2×2 โฅ 20 (nitrogen, oz)
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Sugar
Wood
2×1 + 6×2 โค 36
2(6) + 6(3.2) โค 36
12 + 19.2 โค 36
31.2 โค 36
36 โ 31.2 = 4.8
2×1 + 4×2 โค 16
2(0) + 4(4) โค 16
16 โค 16
There is no sugar left unused.
There is 4.8 lb of wood left unused.
9.
b) The new objective function, Z = 400×1 +
500×2, is parallel to the constraint for labor,
which results 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 a profit of $4,000.
7. a) Maximize Z = x1 + 5×2 (profit, $)
subject to
5×1 + 5×2 โค 25 (flour, lb)
2×1 + 4×2 โค 16 (sugar, lb)
x1 โค 5 (demand for cakes)
x1,x2 โฅ 0
b)
10. a) Minimize Z = 80×1 + 50×2 (cost, $)
subject to
3×1 + x2 โฅ 6 (antibiotic 1, units)
x1 + x2 โฅ 4 (antibiotic 2, units)
2×1 + 6×2 โฅ 12 (antibiotic 3, units)
x1,x2 โฅ 0
b)
8.
In order to solve this problem, you must
substitute the optimal solution into the
resource constraints for flour and sugar and
determine how much of each resource is left
over.
11. a)
Maximize Z = 300×1 + 400×2 (profit, $)
subject to
3×1 + 2×2 โค 18 (gold, oz)
2×1 + 4×2 โค 20 (platinum, oz)
x2 โค 4 (demand, bracelets)
x1,x2 โฅ 0
Flour
5×1 + 5×2 โค 25 lb
5(0) + 5(4) โค 25
20 โค 25
25 โ 20 = 5
There are 5 lb of flour left unused.
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b)
The profit for a necklace would have to
increase to $600 to result in a slope of โ3/2:
400×2 = Z โ 600×1
x2 = Z/400 โ 3/2×1
However, this creates a situation where both
points C and D are optimal, ie., multiple
optimal solutions, as are all
points on the line segment between
C and D.
14. a) Maximize Z = 50×1 + 40×2 (profit, $) subject
to
12.
3×1 + 5×2 โค 150 (wool, yd2)
10×1 + 4×2 โค 200 (labor, hr)
x1,x2 โฅ 0
The new objective function, Z = 300×1 +
600×2, is parallel to the constraint line for
platinum, which results in multiple optimal
solutions. Points B (x1 = 2, x2 = 4) and C (x1
= 4, x2 = 3) are the alternate optimal
solutions, each with a profit of $3,000.
b)
The feasible solution space will change. The
new constraint line, 3×1 + 4×2 = 20, is
parallel to the existing objective function.
Thus, multiple optimal solutions will also be
present in this scenario. The alternate
optimal solutions are at x1 = 1.33, x2 = 4 and
x1 = 2.4, x2 = 3.2, each with a profit of
$2,000.
13. a) Optimal solution: x1 = 4 necklaces, x2 = 3
bracelets. The maximum demand is not
achieved by the amount of one bracelet.
15.
b) The solution point on the graph which
corresponds to no bracelets being produced
must be on the x1 axis where x2 = 0. This is
point D on the graph. In order for point D to
be optimal, the objective function โslopeโ
must change such that it is equal to or greater
than the slope of the constraint line, 3×1 + 2×2
= 18. Transforming this constraint into the
form y = a + bx enables us to compute the
slope:
The feasible solution space changes from the
area 0ABC to 0AB’C’, as shown on the
following graph.
2×2 = 18 โ 3×1
x2 = 9 โ 3/2×1
From this equation the slope is โ3/2. Thus,
the slope of the objective function must be at
least โ3/2. Presently, the slope of the
objective function is โ3/4:
400×2 = Z โ 300×1
x2 = Z/400 โ 3/4×1
The extreme points to evaluate are now A,
B’, and C’.
A:
*B’:
x1 = 0
x2 = 30
Z = 1,200
x1 = 15.8
x2 = 20.5
Z = 1,610
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C’:
18.
x1 = 24
x2 = 0
Z = 1,200
Point B’ is optimal
16. a) Maximize Z = 23×1 + 73×2
subject to
x1 โค 40
x2 โค 25
x1 + 4×2 โค 120
x1,x2 โฅ 0
19.
b)
Maximize Z = 5×1 + 8×2 + 0s1 + 0s3 + 0s4
subject to
3×1 + 5×2 + s1 = 50
2×1 + 4×2 + s2 = 40
x1 + s3 = 8
x2 + s4 = 10
x1,x2 โฅ 0
A: s1 = 0, s2 = 0, s3 = 8, s4 = 0
B: s1 = 0, s2 = 3.2, s3 = 0, s4 = 4.8
C: s1 = 26, s2 = 24, s3 = 0, s4 = 10
20.
17. a) No, not this winter, but they might after they
recover equipment costs, which should be
after the 2nd winter.
b) x1 = 55
x2 = 16.25
Z = 1,851
21.
No, profit will go down
c)
x1 = 40
x2 = 25
Z = 2,435
Profit will increase slightly
d) x1 = 55
x2 = 27.72
Z = $2,073
It changes the optimal solution to point A
(x1 = 8, x2 = 6, Z = 112), and the constraint,
x1 + x2 โค 15, is no longer part of the solution
space boundary.
22. a) Minimize Z = 64×1 + 42×2 (labor cost, $)
subject to
16×1 + 12×2 โฅ 450 (claims)
x1 + x2 โค 40 (workstations)
0.5×1 + 1.4×2 โค 25 (defective claims)
x1,x2 โฅ 0
Profit will go down from (c)
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25.
b)
23. a) Changing the pay for a full-time claims
solution to point A in the graphical solution
where x1 = 28.125 and x2 = 0, i.e., there will
be no part-time operators.
b) Changing the pay for a part-time operator
from $42 to $36 has no effect on the number
of full-time and part-time operators hired,
although the total cost will be reduced to
$1,671.95.
c)
26.
The problem becomes infeasible.
27.
Eliminating the constraint for defective
claims would result in a new solution,
x1 = 0 and x2 = 37.5, where only part-time
operators would be hired.
d) The solution becomes infeasible; there are
not enough workstations to handle the
increase in the volume of claims.
28.
24.
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b)
29.
30. a) Maximize Z = $4.15×1 + 3.60×2 (profit, $)
subject to
33. a) Maximize Z = 800×1 + 900×2 (profit, $)
subject to
2×1 + 4×2 โค 30 (stamping, days)
4×1 + 2×2 โค 30 (coating, days)
x1 + x2 โฅ 9 (lots)
x1,x2 โฅ 0
x1 + x2 โค 115 (freezer space, gals.)
0.93 x1 + 0.75 x2 โค 90 (budget, $)
x1 2
โฅ or x1 โ 2 x2 โฅ 0 (demand)
x2 1
b)
x1 ,x2 โฅ 0
34. a) Maximize Z = 30×1 + 70×2 (profit, $) subject
to
4×1 + 10×2 โค 80 (assembly, hr)
14×1 + 8×2 โค 112 (finishing, hr)
x1 + x2 โค 10 (inventory, units)
x1,x2 โฅ 0
b)
31.
No additional profit, freezer space is not a
binding constraint.
32. a) Minimize Z = 200×1 + 160×2 (cost, $)
subject to
6×1 + 2×2 โฅ 12 (high-grade ore, tons)
2×1 + 2×2 โฅ 8 (medium-grade ore, tons)
4×1 + 12×2 โฅ 24 (low-grade ore, tons)
x1,x2 โฅ 0
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b)
b)
35.
The slope of the original objective function
is computed as follows:
Z = 30×1 + 70×2
70×2 = Z โ 30×1
x2 = Z/70 โ 3/7×1
slope = โ3/7
The slope of the new objective function is
computed as follows:
Z = 90×1 + 70×2
70×2 = Z โ 90×1
x2 = Z/70 โ 9/7×1
slope = โ9/7
37. a) 15(4) + 8(6) โค 120 hr
60 + 48 โค 120
108 โค 120
120 โ 108 = 12 hr left unused
b) Points C and D would be eliminated and a
new optimal solution point at x1 = 5.09,
x2 = 5.45, and Z = 111.27 would result.
38. a) Maximize Z = .28×1 + .19×2
x1 + x2 โค 96 cans
x2
โฅ2
x1
The change in the objective function not
only changes the Z values but also results in
a new solution point, C. The slope of the
new objective function is steeper and thus
changes the solution point.
A: x1 = 0
x2 = 8
Z = 560
C: x1 = 5.3
x2 = 4.7
Z = 806
B:
D: x1 = 8
x2 = 0
Z = 720
x1 = 3.3
x2 = 6.7
Z = 766
x1 ,x2 โฅ 0
b)
36. a) Maximize Z = 9×1 + 12×2 (profit, $1,000s)
subject to
4×1 + 8×2 โค 64 (grapes, tons)
5×1 + 5×2 โค 50 (storage space, yd3)
15×1 + 8×2 โค 120 (processing time, hr)
x1 โค 7 (demand, Nectar)
x2 โค 7 (demand, Red)
x1,x2 โฅ 0
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39.
The model formulation would become,
maximize Z = $0.23×1 + 0.19×2
subject to
x1 + x2 โค 96
โ1.5×1 + x2 โฅ 0
x1,x2 โฅ 0
The solution is x1 = 38.4, x2 = 57.6, and
Z = $19.78
The discount would reduce profit.
40. a) Minimize Z = $0.46×1 + 0.35×2
subject to
.91×1 + .82×2 = 3,500
x1 โฅ 1,000
x2 โฅ 1,000
.03×1 โ .06×2 โฅ 0
x1,x2 โฅ 0
b) 477 โ 445 = 32 fewer defective items
b)
42. a) Maximize Z = $2.25×1 + 1.95×2
subject to
8×1 + 6×2 โค 1,920
3×1 + 6×2 โค 1,440
3×1 + 2×2 โค 720
x1 + x2 โค 288
x1,x2 โฅ 0
b)
41. a) Minimize Z = .09×1 + .18×2
subject to
.46×1 + .35×2 โค 2,000
x1 โฅ 1,000
x2 โฅ 1,000
.91×1 โ .82×2 = 3,500
x1,x2 โฅ 0
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43.
A new constraint is added to the model in
45.
The feasible solution space changes if the
fertilizer constraint changes to 20×1 + 20×2 โค
800 tons. The new solution space is
A’B’C’D’. Two of the constraints now have
no effect.
x1
โฅ 1.5
x2
The solution is x1 = 160, x2 = 106.67,
Z = $568
The new optimal solution is point C’:
A’: x1 = 0
x2 = 37
Z = 11,100
B’: x1 = 3
x2 = 37
Z = 12,300
44. a) Maximize Z = 400×1 + 300×2 (profit, $)
subject to
x1 + x2 โค 50 (available land, acres)
10×1 + 3×2 โค 300 (labor, hr)
8×1 + 20×2 โค 800 (fertilizer, tons)
*C’: x1 = 25.71
x2 = 14.29
Z = 14,571
D’: x1 = 26
x2 = 0
Z = 10,400
46. a) Maximize Z = $7,600×1 + 22,500×2
subject to
x1 + x2 โค 3,500
x2/(x1 + x2) โค .40
.12×1 + .24×2 โค 600
x1,x2 โฅ 0
x1 โค 26 (shipping space, acres)
x2 โค 37 (shipping space, acres)
x1,x2 โฅ 0
b)
b)
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wine, then there would logically be no waste
for wine but only for beer. This amount
โlogicallyโ would be the waste from 266.67
bottles, or $20, and the amount from the
additional 53 bottles, $3.98, for a total of
$23.98.
47. a) Minimize Z = $(.05)(8)x1 + (.10)(.75)x2
subject to
5×1 + x2 โฅ 800
5 x1
= 1.5
x2
8×1 + .75×2 โค 1,200
x1, x2 โฅ 0
x1 = 96
x2 = 320
Z = $62.40
49. a) Minimize Z = 3700×1 + 5100×2
subject to
x1 + x2 = 45
(32×1 + 14×2) / (x1 + x2) โค 21
.10×1 + .04×2 โค 6
b)
x1
โฅ .25
( x1 + x2 )
x2
โฅ .25
( x1 + x2 )
x1, x2 โฅ 0
b)
48.
The new solution is
50. a) No, the solution would not change
x1 = 106.67
x2 = 266.67
Z = $62.67
b) No, the solution would not change
c)
If twice as many guests prefer wine to beer,
then the Robinsons would be approximately
10 bottles of wine short and they would have
approximately 53 more bottles of beer than
they need. The waste is more difficult to
compute. The model in problem 53 assumes
that the Robinsons are ordering more wine
and beer than they need, i.e., a buffer, and
thus there logically would be some waste,
i.e., 5% of the wine and 10% of the beer.
However, if twice as many guests prefer
Yes, the solution would change to China (x1)
= 22.5, Brazil (x2) = 22.5, and
Z = $198,000.
51. a) x1 = $ invested in stocks
x2 = $ invested in bonds
maximize Z = $0.18×1 + 0.06×2 (average
annual return)
subject to
x1 + x2 โค $720,000 (available funds)
x1/(x1 + x2) โค .65 (% of stocks)
.22×1 + .05×2 โค 100,000 (total possible loss)
x1,x2 โฅ 0
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One more hour of Sarahโs time would
reduce the number of regraded exams from
10 to 9.8, whereas increasing Brad by one
hour would have no effect on the solution.
This is actually the marginal (or dual) value
of one additional hour of labor, for Sarah,
which is 0.20 fewer regraded exams,
whereas the marginal value of Bradโs is
zero.
b)
54. a) x1 = # cups of Pomona
x2 = # cups of Coastal
Maximize Z = $2.05×1 + 1.85×2
subject to
16×1 + 16×2 โค 3,840 oz or (30 gal. ร
128 oz)
(.20)(.0625)x1 + (.60)(.0625)x2 โค 6 lbs.
Colombian
(.35)(.0625)x1 + (.10)(.0625)x2 โค 6 lbs.
Kenyan
(.45)(.0625)x1 + (.30)(.0625)x2 โค 6 lbs.
Indonesian
x2/x1 = 3/2
x1,x2 โฅ 0
b) Solution:
x1 = 87.3 cups
x2 = 130.9 cups
Z = $421.09
52.
x1 = exams assigned to Brad
x2 = exams assigned to Sarah
minimize Z = .10×1 + .06×2
subject to
x1 + x2 = 120
x1 โค (720/7.2) or 100
x2 โค 50(600/12)
x1,x2 โฅ 0
53.
If the constraint for Sarahโs time became x2
โค 55 with an additional hour then the
solution point at A would move to
x1 = 65, x2 = 55 and Z = 9.8. If the constraint
for Bradโs time became x1 โค 108.33 with an
additional hour then the solution point (A)
would not change. All of Bradโs time is not
being used anyway so assigning him more
time would not have an effect.
55. a) The only binding constraint is for
Colombian; the constraints for Kenyan and
Indonesian are nonbinding and there are
already extra, or slack, pounds of these
coffees available. Thus, only getting more
Colombian would affect the solution.
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One more pound of Colombian would
increase sales from $421.09 to $463.20.
58.
Increasing the brewing capacity to 40
gallons would have no effect since there is
already unused brewing capacity with the
optimal solution.
b) If the shop increased the demand ratio of
Pomona to Coastal from 1.5 to 1 to 2 to 1 it
would increase daily sales to $460.00, so the
shop should spend extra on advertising to
achieve this result.
56. a) x1 = 16 in. pizzas
x2 = hot dogs
Maximize Z = $22×1 + 2.35×2
Subject to
$10×1 + 0.65×2 โค $1,000
324 in2 x1 + 16 in2 x2 โค 27,648 in2
x2 โค 1,000
x1, x2 โฅ 0
b)
Multiple optimal solutions; A and B alternate
optimal
59.
60.
57. a) x1 = 35, x2 = 1,000, Z = $3,120
Profit would remain the same ($3,120) so
the increase in the oven cost would decrease
the seasonโs profit from $10,120 to $8,120.
b) x1 = 35.95, x2 = 1,000, Z = $3,140
Profit would increase slightly from $10,120
to $10, 245.46.
c) x1 = 55.7, x2 = 600, Z = $3,235.48
Profit per game would increase slightly.
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The graphical solution is shown as follows.
CASE SOLUTION: METROPOLITAN
POLICE PATROL
The linear programming model for this case
problem is
Minimize Z = x/60 + y/45
subject to
2x + 2y โฅ 5
2x + 2y โค 12
y โฅ 1.5x
x, y โฅ 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 รท 20, or x/60. In
the first two constraints,
2x + 2y represents the formula for the
perimeter of a rectangle.
Changing the objective function to
Z = $16×1 + 16×2 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
.90×2 โ .10×1 โฅ 0 to .80×2 โ.20×1 โฅ 0
has no effect on the solution.
The graphical solution is displayed as
follows.
CASE SOLUTION: ANNABELLE
INVESTS IN THE MARKET
x1 = no. of shares of index fund
x2 = no. of shares of internet stock fund
Maximize Z = (.17)(175)x1 + (.28)(208)x2
= 29.75×1 + 58.24×2
subject to
175×1 + 208×2 = $120, 000
x1
โฅ .33
x2
The optimal solution is x = 1, y = 1.5, and Z
= 0.05. This means that a patrol sector is 1.5
miles by 1 mile and the response time is
0.05 hr, or 3 min.
x2
โค2
x1
x1, x2 > 0
CASE SOLUTION: โTHE
POSSIBILITYโ RESTAURANT
The linear programming model formulation
is
Maximize = Z = $12×1 + 16×2
subject to
x1 + x2 โค 60
.25×1 + .50×2 โค 20
x1/x2 โฅ 3/2 or 2×1 โ 3×2 โฅ 0
x2/(x1 + x2) โฅ .10 or .90×2 โ .10×1 โฅ 0
x1x2 โฅ 0
x1 = 203
x2 = 406
Z = $29,691.37
x2
โฅ .33
x1
will have no effect on the solution.
Eliminating the constraint
x1
โค2
x2
will change the solution to x1 = 149,
x2 = 451.55, Z = $30,731.52.
Eliminating the constraint
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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 or approximately $0.25.
Increasing by another dollar will increase
profit by another $0.25, and increasing the
amount available by one more dollar
will again increase profit by $0.25. This
indicates that for each extra dollar invested a
return of $0.25 might be expected with this
investment strategy.
Thus, the marginal value of an extra dollar
to invest is $0.25, which is also referred to
as the โshadowโ or โdualโ price as described
in Chapter 3.
2-15
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