Solution Manual For Introduction To Management Science, 11th Edition
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1-1
Chapter One: Management Science
PROBLEM SUMMARY
1. Total cost, revenue, profit, and
break-even
2. Total cost, revenue, profit, and
break-even
3. Total cost, revenue, profit, and
break-even
4. Break-even volume
5. Graphical analysis (1−2)
6. Graphical analysis (1−4)
7. Break-even sales volume
8. Break-even volume as a percentage
of capacity (1−2)
9. Break-even volume as a percentage
of capacity (1−3)
10. Break-even volume as a percentage
of capacity (1−4)
11. Effect of price change (1−2)
12. Effect of price change (1−4)
13. Effect of variable cost change (1−12)
14. Effect of fixed cost change (1−13)
15. Break-even analysis
16. Effect of fixed cost change (1−7)
17. Effect of variable cost change (1−7)
18. Break-even analysis
19. Break-even analysis
20. Break-even analysis
21. Break-even analysis; volume and
price analysis
22. Break-even analysis; profit analysis
23. Break-even analysis
24. Break-even analysis; profit analysis
25. Break-even analysis; price and volume analysis
26. Break-even analysis; profit analysis
27. Break-even analysis; profit analysis
28. Break-even analysis; profit analysis
29. Linear programming
30. Linear programming
31. Linear programming
32. Linear programming
33. Forecasting/statistics
34. Linear programming
35. Waiting lines
36. Shortest route
PROBLEM SOLUTIONS
1. a)= =
= =
= + = + =
= = =
= − =
f
v
f v
300, $8,000,
$65 per table, $180;
TC $8,000 (300)(65) $27,500;
TR (300)(180) $54,000;
$54,000 27,500 $26,500 per month
v c
c p
c vc
vp
Z
b)f
v
8,000 69.56 tables per month
180 65
c
v p c
= = =
− −
2. a)= = =
= = +
= +
=
= = =
= − =
f v
f v
12,000, $60,000, $9,
$25; TC
60,000 (12,000)(9)
$168,000;
TR (12,000)($25) $300,000;
$300,000 168,000 $132,000 per year
v c c
p c vc
vp
Z
b)= = =
− −
f
v
60,000 3,750 tires per year
25 9
c
v p c
3. a)= = =
=
= + = + =
= = =
= − = −
f v
f v
18,000, $21,000, $.45,
$1.30;
TC $21,000 (18,000)(.45) $29,100;
TR (18,000)(1.30) $23, 400;
$23, 400 29,100 $5,700 (loss)
v c c
p
c vc
vp
Z
b)
4.= = =
= = =
− −
f v
f
v
$25,000, $.40, $.15,
25,000 100,000 lb per month
.40 .15
c p c
c
v p cf
v
21,000 24,705.88 yd per month
1.30 .45
c
v p c
= = =
− −
Chapter One: Management Science
PROBLEM SUMMARY
1. Total cost, revenue, profit, and
break-even
2. Total cost, revenue, profit, and
break-even
3. Total cost, revenue, profit, and
break-even
4. Break-even volume
5. Graphical analysis (1−2)
6. Graphical analysis (1−4)
7. Break-even sales volume
8. Break-even volume as a percentage
of capacity (1−2)
9. Break-even volume as a percentage
of capacity (1−3)
10. Break-even volume as a percentage
of capacity (1−4)
11. Effect of price change (1−2)
12. Effect of price change (1−4)
13. Effect of variable cost change (1−12)
14. Effect of fixed cost change (1−13)
15. Break-even analysis
16. Effect of fixed cost change (1−7)
17. Effect of variable cost change (1−7)
18. Break-even analysis
19. Break-even analysis
20. Break-even analysis
21. Break-even analysis; volume and
price analysis
22. Break-even analysis; profit analysis
23. Break-even analysis
24. Break-even analysis; profit analysis
25. Break-even analysis; price and volume analysis
26. Break-even analysis; profit analysis
27. Break-even analysis; profit analysis
28. Break-even analysis; profit analysis
29. Linear programming
30. Linear programming
31. Linear programming
32. Linear programming
33. Forecasting/statistics
34. Linear programming
35. Waiting lines
36. Shortest route
PROBLEM SOLUTIONS
1. a)= =
= =
= + = + =
= = =
= − =
f
v
f v
300, $8,000,
$65 per table, $180;
TC $8,000 (300)(65) $27,500;
TR (300)(180) $54,000;
$54,000 27,500 $26,500 per month
v c
c p
c vc
vp
Z
b)f
v
8,000 69.56 tables per month
180 65
c
v p c
= = =
− −
2. a)= = =
= = +
= +
=
= = =
= − =
f v
f v
12,000, $60,000, $9,
$25; TC
60,000 (12,000)(9)
$168,000;
TR (12,000)($25) $300,000;
$300,000 168,000 $132,000 per year
v c c
p c vc
vp
Z
b)= = =
− −
f
v
60,000 3,750 tires per year
25 9
c
v p c
3. a)= = =
=
= + = + =
= = =
= − = −
f v
f v
18,000, $21,000, $.45,
$1.30;
TC $21,000 (18,000)(.45) $29,100;
TR (18,000)(1.30) $23, 400;
$23, 400 29,100 $5,700 (loss)
v c c
p
c vc
vp
Z
b)
4.= = =
= = =
− −
f v
f
v
$25,000, $.40, $.15,
25,000 100,000 lb per month
.40 .15
c p c
c
v p cf
v
21,000 24,705.88 yd per month
1.30 .45
c
v p c
= = =
− −
1-1
Chapter One: Management Science
PROBLEM SUMMARY
1. Total cost, revenue, profit, and
break-even
2. Total cost, revenue, profit, and
break-even
3. Total cost, revenue, profit, and
break-even
4. Break-even volume
5. Graphical analysis (1−2)
6. Graphical analysis (1−4)
7. Break-even sales volume
8. Break-even volume as a percentage
of capacity (1−2)
9. Break-even volume as a percentage
of capacity (1−3)
10. Break-even volume as a percentage
of capacity (1−4)
11. Effect of price change (1−2)
12. Effect of price change (1−4)
13. Effect of variable cost change (1−12)
14. Effect of fixed cost change (1−13)
15. Break-even analysis
16. Effect of fixed cost change (1−7)
17. Effect of variable cost change (1−7)
18. Break-even analysis
19. Break-even analysis
20. Break-even analysis
21. Break-even analysis; volume and
price analysis
22. Break-even analysis; profit analysis
23. Break-even analysis
24. Break-even analysis; profit analysis
25. Break-even analysis; price and volume analysis
26. Break-even analysis; profit analysis
27. Break-even analysis; profit analysis
28. Break-even analysis; profit analysis
29. Linear programming
30. Linear programming
31. Linear programming
32. Linear programming
33. Forecasting/statistics
34. Linear programming
35. Waiting lines
36. Shortest route
PROBLEM SOLUTIONS
1. a)= =
= =
= + = + =
= = =
= − =
f
v
f v
300, $8,000,
$65 per table, $180;
TC $8,000 (300)(65) $27,500;
TR (300)(180) $54,000;
$54,000 27,500 $26,500 per month
v c
c p
c vc
vp
Z
b)f
v
8,000 69.56 tables per month
180 65
c
v p c
= = =
− −
2. a)= = =
= = +
= +
=
= = =
= − =
f v
f v
12,000, $60,000, $9,
$25; TC
60,000 (12,000)(9)
$168,000;
TR (12,000)($25) $300,000;
$300,000 168,000 $132,000 per year
v c c
p c vc
vp
Z
b)= = =
− −
f
v
60,000 3,750 tires per year
25 9
c
v p c
3. a)= = =
=
= + = + =
= = =
= − = −
f v
f v
18,000, $21,000, $.45,
$1.30;
TC $21,000 (18,000)(.45) $29,100;
TR (18,000)(1.30) $23, 400;
$23, 400 29,100 $5,700 (loss)
v c c
p
c vc
vp
Z
b)
4.= = =
= = =
− −
f v
f
v
$25,000, $.40, $.15,
25,000 100,000 lb per month
.40 .15
c p c
c
v p cf
v
21,000 24,705.88 yd per month
1.30 .45
c
v p c
= = =
− −
Chapter One: Management Science
PROBLEM SUMMARY
1. Total cost, revenue, profit, and
break-even
2. Total cost, revenue, profit, and
break-even
3. Total cost, revenue, profit, and
break-even
4. Break-even volume
5. Graphical analysis (1−2)
6. Graphical analysis (1−4)
7. Break-even sales volume
8. Break-even volume as a percentage
of capacity (1−2)
9. Break-even volume as a percentage
of capacity (1−3)
10. Break-even volume as a percentage
of capacity (1−4)
11. Effect of price change (1−2)
12. Effect of price change (1−4)
13. Effect of variable cost change (1−12)
14. Effect of fixed cost change (1−13)
15. Break-even analysis
16. Effect of fixed cost change (1−7)
17. Effect of variable cost change (1−7)
18. Break-even analysis
19. Break-even analysis
20. Break-even analysis
21. Break-even analysis; volume and
price analysis
22. Break-even analysis; profit analysis
23. Break-even analysis
24. Break-even analysis; profit analysis
25. Break-even analysis; price and volume analysis
26. Break-even analysis; profit analysis
27. Break-even analysis; profit analysis
28. Break-even analysis; profit analysis
29. Linear programming
30. Linear programming
31. Linear programming
32. Linear programming
33. Forecasting/statistics
34. Linear programming
35. Waiting lines
36. Shortest route
PROBLEM SOLUTIONS
1. a)= =
= =
= + = + =
= = =
= − =
f
v
f v
300, $8,000,
$65 per table, $180;
TC $8,000 (300)(65) $27,500;
TR (300)(180) $54,000;
$54,000 27,500 $26,500 per month
v c
c p
c vc
vp
Z
b)f
v
8,000 69.56 tables per month
180 65
c
v p c
= = =
− −
2. a)= = =
= = +
= +
=
= = =
= − =
f v
f v
12,000, $60,000, $9,
$25; TC
60,000 (12,000)(9)
$168,000;
TR (12,000)($25) $300,000;
$300,000 168,000 $132,000 per year
v c c
p c vc
vp
Z
b)= = =
− −
f
v
60,000 3,750 tires per year
25 9
c
v p c
3. a)= = =
=
= + = + =
= = =
= − = −
f v
f v
18,000, $21,000, $.45,
$1.30;
TC $21,000 (18,000)(.45) $29,100;
TR (18,000)(1.30) $23, 400;
$23, 400 29,100 $5,700 (loss)
v c c
p
c vc
vp
Z
b)
4.= = =
= = =
− −
f v
f
v
$25,000, $.40, $.15,
25,000 100,000 lb per month
.40 .15
c p c
c
v p cf
v
21,000 24,705.88 yd per month
1.30 .45
c
v p c
= = =
− −
1-2
5.
6.
7.−
−
f
v
$25,000
= = = 1,250 dolls
30 10
c
v p c
8.= = = =
Break-even volume as percentage of capacity
3,750 .469 46.9%
8,000
v
k
9.= = = =
Break-even volume as percentage of capacity
24,750.88 .988 98.8%
25,000
v
k
10.= = = =
Break-even volume as percentage of
100,000
capacity .833 83.3%
120,000
v
k
11.f
v
60,000 2,727.3tires
31 9
per year; it reduces the break-even
volume from 3,750 tires to 2,727.3
tires per year.
c
v p c
= = =
− −
12.= = =
− −
f
v
25,000 55,555.55 lb
.60 .15
per month; it reduces the break-even
volume from 100,000 lb per month
to 55,555.55 lb.
c
v p c
13.f
v
25,000 65,789.47 lb
.60 .22
per month; it increases the break-even
volume from 55,555.55 lb per month
to 65,789.47 lb per month.
c
v p c
= = =
− −
14.= = =
− −
f
v
39,000 102,613.57 lb
.60 .22
per month; it increases the break-even
volume from 65,789.47 lb per month
to 102,631.57 lb per month.
c
v p c
15.f v
f
v
Initial profit: (9,000)(.75)
4,000 (9,000)(.21) 6,750 4,000 1,890
$860 per month; increase in price:
(5,700)(.95) 4,000 (5,700)(.21) 5, 415
4,000 1,197 $218 per month; the dair
Z vp c vc
Z vp c
vc
= − − = −
− = − − =
= − −
= − − = −
− = y should not
raise its price.
16.−
f
v
35,000
= = = 1,750
30–10
c
v p c
The increase in fixed cost from $25,000 to
$35,000 will increase the break-even point from
1,250 to 1,750 or 500 dolls; thus, he should not
spend the extra $10,000 for advertising.
17. Original break-even point (from problem 7) = 1,250
New break-even point:= = =
− −
f
v
17,000 1,062.5
30 14
c
v p c
Reduces BE point by 187.5 dolls.
18. a)= = =
− −
f
v
$27,000 5,192.30 pizzas
8.95 3.75
c
v p c
b)=
5,192.3 259.6 days
20
c) Revenue for the first 30 days = 30(pv − vcv)
= 30[(8.95)(20) −
(20)(3.75)]
= $3,120
$27,000 − 3,120 = $23,880, portion of fixed cost
not recouped after 30 days.= = =
− −
f
v
$23,880
New 5,685.7 pizzas
7.95 3.75
c
v p c
Total break-even volume = 600 + 5,685.7 =
6,285.7 pizzas
5.
6.
7.−
−
f
v
$25,000
= = = 1,250 dolls
30 10
c
v p c
8.= = = =
Break-even volume as percentage of capacity
3,750 .469 46.9%
8,000
v
k
9.= = = =
Break-even volume as percentage of capacity
24,750.88 .988 98.8%
25,000
v
k
10.= = = =
Break-even volume as percentage of
100,000
capacity .833 83.3%
120,000
v
k
11.f
v
60,000 2,727.3tires
31 9
per year; it reduces the break-even
volume from 3,750 tires to 2,727.3
tires per year.
c
v p c
= = =
− −
12.= = =
− −
f
v
25,000 55,555.55 lb
.60 .15
per month; it reduces the break-even
volume from 100,000 lb per month
to 55,555.55 lb.
c
v p c
13.f
v
25,000 65,789.47 lb
.60 .22
per month; it increases the break-even
volume from 55,555.55 lb per month
to 65,789.47 lb per month.
c
v p c
= = =
− −
14.= = =
− −
f
v
39,000 102,613.57 lb
.60 .22
per month; it increases the break-even
volume from 65,789.47 lb per month
to 102,631.57 lb per month.
c
v p c
15.f v
f
v
Initial profit: (9,000)(.75)
4,000 (9,000)(.21) 6,750 4,000 1,890
$860 per month; increase in price:
(5,700)(.95) 4,000 (5,700)(.21) 5, 415
4,000 1,197 $218 per month; the dair
Z vp c vc
Z vp c
vc
= − − = −
− = − − =
= − −
= − − = −
− = y should not
raise its price.
16.−
f
v
35,000
= = = 1,750
30–10
c
v p c
The increase in fixed cost from $25,000 to
$35,000 will increase the break-even point from
1,250 to 1,750 or 500 dolls; thus, he should not
spend the extra $10,000 for advertising.
17. Original break-even point (from problem 7) = 1,250
New break-even point:= = =
− −
f
v
17,000 1,062.5
30 14
c
v p c
Reduces BE point by 187.5 dolls.
18. a)= = =
− −
f
v
$27,000 5,192.30 pizzas
8.95 3.75
c
v p c
b)=
5,192.3 259.6 days
20
c) Revenue for the first 30 days = 30(pv − vcv)
= 30[(8.95)(20) −
(20)(3.75)]
= $3,120
$27,000 − 3,120 = $23,880, portion of fixed cost
not recouped after 30 days.= = =
− −
f
v
$23,880
New 5,685.7 pizzas
7.95 3.75
c
v p c
Total break-even volume = 600 + 5,685.7 =
6,285.7 pizzas
1-35,685.7
Total time to break-even 30 20
314.3 days
= +
=
19. a) Cost of Regular plan = $55 + (.33)(260 minutes)
= $140.80
Cost of Executive plan = $100 + (.25)(60 minutes)
= $115
Select Executive plan.
b) 55 + (x − 1,000)(.33) = 100 + (x − 1,200)(.25)
− 275 + .33x = .25x − 200
x = 937.50 minutes per
month or 15.63 hrs.
20. a)= −
7,500
14,000 .35p
p = $0.89 to break even
b) If the team did not perform as well as expected
the crowds could be smaller; bad weather could
reduce crowds and/or affect what fans eat at the
game; the price she charges could affect demand.
c) This will be a subjective answer, but $1.25 seems
to be a reasonable price.
Z = vp − cf − vcv
Z = (14,000)(1.25) − 7,500 − (14,000)(0.35)
= 17,500 − 12,400
= $5,100
21. a) cf = $1,700
cv = $12 per pupil
p = $75= −
1,700
75 12
v
= 26.98 or 27 pupils
b) Z = vp − cf − vcv
$5,000 = v(75) − $1,700 − v(12)
63v = 6,700
v = 106.3 pupils
c) Z = vp − cf − vcv
$5,000 = 60p − $1,700 − 60(12)
60p = 7,420
p = $123.67
22. a) cf = $350,000
cv = $12,000
p = $18,000= −
f
v
c
v p c= −
350,000
18,000 12,000
= 58.33 or 59 students
b) Z = (75)(18,000) − 350,000 − (75)(12,000)
= $100,000
c) Z = (35)(25,000) − 350,000 − (35)(12,000)
= 105,000
This is approximately the same as the profit for
75 students and a lower tuition in part (b).
23. p = $400
cf = $8,000
cv = $75
Z = $60,000+
= −
f
v
Z c
v p c+
= −
60,000 8,000
400 75
v
v = 209.23 teams
24. Fixed cost (cf) = 875,000
Variable cost (cv) = $200
Price (p) = (225)(12) = $2,700
v = cf/(p – cv) = 875,000/(2,700 – 200)
= 350
With volume doubled to 700:
Profit (Z) = (2,700)(700) – 875,000 – (700)(200)
= $875,000
25. Fixed cost (cf) = 100,000
Variable cost (cv) = $(.50)(.35) + (.35)(.50) + (.15)(2.30)
= $0.695
Price (p) = $6
Profit (Z) = (6)(45,000) – 100,000 – (45,000)(0.695)
= $138,725
This is not the financial profit goal of $150,000.
The price to achieve the goal of $150,000 is,
p = (Z + cf + vcv)/v
= (150,000 + 100,000 + (45,000)(.695))/45,000
= $6.25
Total time to break-even 30 20
314.3 days
= +
=
19. a) Cost of Regular plan = $55 + (.33)(260 minutes)
= $140.80
Cost of Executive plan = $100 + (.25)(60 minutes)
= $115
Select Executive plan.
b) 55 + (x − 1,000)(.33) = 100 + (x − 1,200)(.25)
− 275 + .33x = .25x − 200
x = 937.50 minutes per
month or 15.63 hrs.
20. a)= −
7,500
14,000 .35p
p = $0.89 to break even
b) If the team did not perform as well as expected
the crowds could be smaller; bad weather could
reduce crowds and/or affect what fans eat at the
game; the price she charges could affect demand.
c) This will be a subjective answer, but $1.25 seems
to be a reasonable price.
Z = vp − cf − vcv
Z = (14,000)(1.25) − 7,500 − (14,000)(0.35)
= 17,500 − 12,400
= $5,100
21. a) cf = $1,700
cv = $12 per pupil
p = $75= −
1,700
75 12
v
= 26.98 or 27 pupils
b) Z = vp − cf − vcv
$5,000 = v(75) − $1,700 − v(12)
63v = 6,700
v = 106.3 pupils
c) Z = vp − cf − vcv
$5,000 = 60p − $1,700 − 60(12)
60p = 7,420
p = $123.67
22. a) cf = $350,000
cv = $12,000
p = $18,000= −
f
v
c
v p c= −
350,000
18,000 12,000
= 58.33 or 59 students
b) Z = (75)(18,000) − 350,000 − (75)(12,000)
= $100,000
c) Z = (35)(25,000) − 350,000 − (35)(12,000)
= 105,000
This is approximately the same as the profit for
75 students and a lower tuition in part (b).
23. p = $400
cf = $8,000
cv = $75
Z = $60,000+
= −
f
v
Z c
v p c+
= −
60,000 8,000
400 75
v
v = 209.23 teams
24. Fixed cost (cf) = 875,000
Variable cost (cv) = $200
Price (p) = (225)(12) = $2,700
v = cf/(p – cv) = 875,000/(2,700 – 200)
= 350
With volume doubled to 700:
Profit (Z) = (2,700)(700) – 875,000 – (700)(200)
= $875,000
25. Fixed cost (cf) = 100,000
Variable cost (cv) = $(.50)(.35) + (.35)(.50) + (.15)(2.30)
= $0.695
Price (p) = $6
Profit (Z) = (6)(45,000) – 100,000 – (45,000)(0.695)
= $138,725
This is not the financial profit goal of $150,000.
The price to achieve the goal of $150,000 is,
p = (Z + cf + vcv)/v
= (150,000 + 100,000 + (45,000)(.695))/45,000
= $6.25
Loading page 4...
1-4
The volume to achieve the goal of $150,000 is,
v = (Z + cf)/(p − cv)
= (150,000 + 100,000)/(6 − .695)
= 47,125
26. a) Monthly fixed cost (cf) = cost of van/60 months
+ labor (driver)/month
= (21,500/60) + (30.42
days/month)($8/hr)
(5 hr/day)
= 358.33 + 1,216.80
= $1,575.13
Variable cost (cv) = $1.35 + 15.00
= $16.35
Price (p) = $34
v = cf/(p − vc)
= (1,575.13)/(34 − 16.35)
v = 89.24 orders/month
b) 89.24/30.42 = 2.93 orders/day − Monday through
Thursday
Double for weekend = 5.86 orders/day − Friday
through Sunday
Orders per month = approximately (18 days)
(2.93 orders) + (12.4 days)(5.86 orders)
= 125.4 delivery orders per month
Profit = total revenue − total cost
= vp – (cf + vcv)
= (125.4)(34) − 1,575.13 – (125.4)(16.35)
= 638.18
27. a)= =
− −
f
v
500
30 14
c
v p c
v = 31.25 jobs
b) (8 weeks)(6 days/week)(3 lawns/day) = 144
lawns
Z = (144)(30) − 500 − (144)(14)
Z = $1,804
c) (8 weeks)(6 days/week)(4 lawns/day) = 192 lawns
Z = (192)(25) − 500 − (192)(14)
Z = $1,612
No, she would make less money than (b)
28. a)= =
− −
f
v
700
35 3
c
v p c
v = 21.88 jobs
b) (6 snows)(2 days/snow)(10 jobs/day) = 120 jobs
Z = (120)(35) − 700 − (120)(3)
Z = $3,140
c) (6 snows)(2 days/snow)(4 jobs/day) = 48 jobs
Z = (48)(150) − 1800 − (48)(28)
Z = $4,056
Yes, better than (b)
d) Z = (120)(35) − 700 − (120)(18)
Z = $1,340
Yes, still a profit with one more person
29. There are two possible answers, or solution points:
x = 25, y = 0 or x = 0, y = 50
Substituting these values in the objective function:
Z = 15(25) + 10(0) = 375
Z = 15(0) + 10(50) = 500
Thus, the solution is x = 0 and y = 50
This is a simple linear programming model, the
subject of the next several chapters. The student
should recognize that there are only two possible
solutions, which are the corner points of the
feasible solution space, only one of which is
optimal.
30. The solution is computed by solving
simultaneous equations,
x = 30, y = 10, Z = $1,400
It is the only, i.e., “optimal” solution because
there is only one set of values for x and y that
satisfy both constraints simultaneously.
The volume to achieve the goal of $150,000 is,
v = (Z + cf)/(p − cv)
= (150,000 + 100,000)/(6 − .695)
= 47,125
26. a) Monthly fixed cost (cf) = cost of van/60 months
+ labor (driver)/month
= (21,500/60) + (30.42
days/month)($8/hr)
(5 hr/day)
= 358.33 + 1,216.80
= $1,575.13
Variable cost (cv) = $1.35 + 15.00
= $16.35
Price (p) = $34
v = cf/(p − vc)
= (1,575.13)/(34 − 16.35)
v = 89.24 orders/month
b) 89.24/30.42 = 2.93 orders/day − Monday through
Thursday
Double for weekend = 5.86 orders/day − Friday
through Sunday
Orders per month = approximately (18 days)
(2.93 orders) + (12.4 days)(5.86 orders)
= 125.4 delivery orders per month
Profit = total revenue − total cost
= vp – (cf + vcv)
= (125.4)(34) − 1,575.13 – (125.4)(16.35)
= 638.18
27. a)= =
− −
f
v
500
30 14
c
v p c
v = 31.25 jobs
b) (8 weeks)(6 days/week)(3 lawns/day) = 144
lawns
Z = (144)(30) − 500 − (144)(14)
Z = $1,804
c) (8 weeks)(6 days/week)(4 lawns/day) = 192 lawns
Z = (192)(25) − 500 − (192)(14)
Z = $1,612
No, she would make less money than (b)
28. a)= =
− −
f
v
700
35 3
c
v p c
v = 21.88 jobs
b) (6 snows)(2 days/snow)(10 jobs/day) = 120 jobs
Z = (120)(35) − 700 − (120)(3)
Z = $3,140
c) (6 snows)(2 days/snow)(4 jobs/day) = 48 jobs
Z = (48)(150) − 1800 − (48)(28)
Z = $4,056
Yes, better than (b)
d) Z = (120)(35) − 700 − (120)(18)
Z = $1,340
Yes, still a profit with one more person
29. There are two possible answers, or solution points:
x = 25, y = 0 or x = 0, y = 50
Substituting these values in the objective function:
Z = 15(25) + 10(0) = 375
Z = 15(0) + 10(50) = 500
Thus, the solution is x = 0 and y = 50
This is a simple linear programming model, the
subject of the next several chapters. The student
should recognize that there are only two possible
solutions, which are the corner points of the
feasible solution space, only one of which is
optimal.
30. The solution is computed by solving
simultaneous equations,
x = 30, y = 10, Z = $1,400
It is the only, i.e., “optimal” solution because
there is only one set of values for x and y that
satisfy both constraints simultaneously.
Loading page 5...
1-5
31.
Labor usage Clay usage Profit Possible
# bowls # mugs 12x + 15y < = 60 9x+5y < = 30 300x + 250y solution?
0 1 15 5 250 yes
1 0 12 9 300 yes
1 1 27 14 550 yes
0 2 30 10 500 yes
2 0 24 18 600 yes
1 2 42 19 800 yes
2 1 39 23 850 yes
2 2 54 28 1100 yes, best solution
0 3 45 15 750 yes
3 0 36 27 900 yes
1 3 57 24 1050 yes
3 1 51 32 1150 no
2 3 69 33 1350 no
3 2 66 37 1400 no
3 3 81 42 1650 no
4 0 48 36 1200 no
0 4 60 20 1000 yes
1 4 72 29 1300 no
4 1 63 41 1450 no
2 4 84 38 1600 no
4 2 78 46 1700 no
31.
Labor usage Clay usage Profit Possible
# bowls # mugs 12x + 15y < = 60 9x+5y < = 30 300x + 250y solution?
0 1 15 5 250 yes
1 0 12 9 300 yes
1 1 27 14 550 yes
0 2 30 10 500 yes
2 0 24 18 600 yes
1 2 42 19 800 yes
2 1 39 23 850 yes
2 2 54 28 1100 yes, best solution
0 3 45 15 750 yes
3 0 36 27 900 yes
1 3 57 24 1050 yes
3 1 51 32 1150 no
2 3 69 33 1350 no
3 2 66 37 1400 no
3 3 81 42 1650 no
4 0 48 36 1200 no
0 4 60 20 1000 yes
1 4 72 29 1300 no
4 1 63 41 1450 no
2 4 84 38 1600 no
4 2 78 46 1700 no
Loading page 6...
1-6
32. Maximize Z = $30xAN + 70xAJ + 40xBN + 60xBJ
subject to
xAN + xAJ = 400
xBN + xBJ = 400
xAN + xBN = 500
xAJ + xBJ = 300
The solution is xAN = 400, xBN = 100, xBJ = 300, and
Z = 34,000
This problem can be solved by allocating as much as
possible to the lowest cost variable, xAN = 400, then
repeating this step until all the demand has been met.
This is a similar logic to the minimum cell cost method.
33. This is virtually a straight linear relationship between
time and site visits; thus, a simple linear graph would
result in a forecast of approximately 34,500 site visits.
34. Determine logical solutions:
Cakes Bread Total Sales
1. 0 2 $12
2. 1 2 $22
3. 3 1 $36
4. 4 0 $40
Each solution must be checked to see if it violates the
constraints for baking time and flour. Some possible
solutions can be logically discarded because they are
obviously inferior. For example, 0 cakes and 1 loaf of
bread is clearly inferior to 0 cakes and 2 loaves of
bread. 0 cakes and 3 loaves of bread is not possible
because there is not enough flour for 3 loaves of bread.
Using this logic, there are four possible solutions as
shown. The best one, 4 cakes and 0 loaves of bread,
results in the highest total sales of $40.
36. The shortest route problem is one of the topics of
Chapter 7. At this point, the most logical “trial and
error” way that most students will probably
approach this problem is to identify all the feasible
routes and compute the total distance for each, as
follows:
1-2-6-9 = 228
1-2-5-9 = 213
1-3-5-9 = 211
1-3-8-9 = 276
1-4-7-8-9 = 275
Obviously inferior routes like 1-3-4-7-8-9 and
1-2-5-8-9 that include additional segments to the
routes listed above can be logically eliminated
from consideration. As a result, the route 1-3-5-9
is the shortest.
An additional aspect to this problem could be to
have the students look at these routes on a real
map and indicate which they think might
“practically” be the best route. In this case,
1-2-5-9 would likely be a better route, because
even though it’s two miles farther it is Interstate
highway the whole way, whereas 1-3-5-9
encompasses U.S. 4-lane highways and state
roads.
35. This problem demonstrates the cost trade-off
inherent in queuing analysis, the topic of Chapter 13.
In this problem the cost of service, i.e., the cost of
staffing registers, is added to the cost of customers
waiting, i.e., the cost of lost sales and ill will, as
shown in the following table.
Registers staffed 1 2 3 4 5 6 7 8
Waiting time (mins) 20 14 9 4 1.7 1 0.5 0.1
Cost of service ($) 60 120 180 240 300 360 420 480
Cost of waiting ($) 850 550 300 50 0 0 0 0
Total cost ($) 910 670 480 290 300 360 420 480
The total minimum cost of $290 occurs with 4 registers staffed
32. Maximize Z = $30xAN + 70xAJ + 40xBN + 60xBJ
subject to
xAN + xAJ = 400
xBN + xBJ = 400
xAN + xBN = 500
xAJ + xBJ = 300
The solution is xAN = 400, xBN = 100, xBJ = 300, and
Z = 34,000
This problem can be solved by allocating as much as
possible to the lowest cost variable, xAN = 400, then
repeating this step until all the demand has been met.
This is a similar logic to the minimum cell cost method.
33. This is virtually a straight linear relationship between
time and site visits; thus, a simple linear graph would
result in a forecast of approximately 34,500 site visits.
34. Determine logical solutions:
Cakes Bread Total Sales
1. 0 2 $12
2. 1 2 $22
3. 3 1 $36
4. 4 0 $40
Each solution must be checked to see if it violates the
constraints for baking time and flour. Some possible
solutions can be logically discarded because they are
obviously inferior. For example, 0 cakes and 1 loaf of
bread is clearly inferior to 0 cakes and 2 loaves of
bread. 0 cakes and 3 loaves of bread is not possible
because there is not enough flour for 3 loaves of bread.
Using this logic, there are four possible solutions as
shown. The best one, 4 cakes and 0 loaves of bread,
results in the highest total sales of $40.
36. The shortest route problem is one of the topics of
Chapter 7. At this point, the most logical “trial and
error” way that most students will probably
approach this problem is to identify all the feasible
routes and compute the total distance for each, as
follows:
1-2-6-9 = 228
1-2-5-9 = 213
1-3-5-9 = 211
1-3-8-9 = 276
1-4-7-8-9 = 275
Obviously inferior routes like 1-3-4-7-8-9 and
1-2-5-8-9 that include additional segments to the
routes listed above can be logically eliminated
from consideration. As a result, the route 1-3-5-9
is the shortest.
An additional aspect to this problem could be to
have the students look at these routes on a real
map and indicate which they think might
“practically” be the best route. In this case,
1-2-5-9 would likely be a better route, because
even though it’s two miles farther it is Interstate
highway the whole way, whereas 1-3-5-9
encompasses U.S. 4-lane highways and state
roads.
35. This problem demonstrates the cost trade-off
inherent in queuing analysis, the topic of Chapter 13.
In this problem the cost of service, i.e., the cost of
staffing registers, is added to the cost of customers
waiting, i.e., the cost of lost sales and ill will, as
shown in the following table.
Registers staffed 1 2 3 4 5 6 7 8
Waiting time (mins) 20 14 9 4 1.7 1 0.5 0.1
Cost of service ($) 60 120 180 240 300 360 420 480
Cost of waiting ($) 850 550 300 50 0 0 0 0
Total cost ($) 910 670 480 290 300 360 420 480
The total minimum cost of $290 occurs with 4 registers staffed
Loading page 7...
1-7
CASE SOLUTION: CLEAN CLOTHES
CORNER LAUNDRY
a)= = =
− −
f
v
1,700 2,000 items per month
1.10 .25
c
v p c
b) Solution depends on number of months; 36 used
here. $16,200 ÷ 36 = $450 per month, thus
monthly fixed cost is $2,150= = =
− −
f
v
2,150 2,529.4 items per month
1.10 .25
c
v p c
529.4 additional items per month
c) Z = vp − cf − vcv
= 4,300(1.10) − 2,150 − 4,300(.25)
= $1,505 per month
After 3 years, Z = $1,955 per month
d)= = =
− −
= − −
= − −
=
f
v
f v
1,700 2,297.3
.99 .25
3,800(.99) 1,700 3,800(.25)
$1,112 per month
c
v p c
Z vp c vc
e) With both options:
Z = vp − cf − vcv
= 4,700(.99) − 2,150 − 4,700(.25)
= $1,328
She should purchase the new equipment but not
decrease prices.
CASE SOLUTION: OCOBEE RIVER
RAFTING COMPANY=fAlternative 1: $3,000c= $20p=v $12c= = =
− −
f
1
v
3,000 375 rafts
20 12
c
v p c=fAlternative 2: $10,000c= $20p=v $8c= = =
− −
f
2
v
10,000 833.37
20 8
c
v p c
If demand is less than 375 rafts, the students should not
start the business.
If demand is less than 833 rafts, alternative 2 should not
be selected, and alternative 1 should be used if demand is
expected to be between 375 and 833.33 rafts.
If demand is greater than 833.33 rafts, which alternative
is best? To determine the answer, equate the two cost
functions.
3,000 + 12v = 10,000 + 8v
4v = 7,000
v = 1,750
This is referred to as the point of indifference between
the two alternatives. In general, for demand lower than this
point (1,750) the alternative should be selected with the
lowest fixed cost; for demand greater than this
point the alternative with the lowest variable cost should
be selected. (This general relationship can be observed by
graphing the two cost equations and seeing where they
intersect.)
Thus, for the Ocobee River Rafting Company, the
following guidelines should be used:
demand < 375, do not start business; 375 < demand
< 1,750, select alternative 1; demand > 1,750, select
alternative 2
Since Penny estimates demand will be approximately
1,000 rafts, alternative 1 should be selected.
Z = vp − cf − vcv
= (1,000)(20) − 3,000 − (1,000)(12)
Z = $5,000
CASE SOLUTION: CONSTRUCTING
A DOWNTOWN PARKING LOT
IN DRAPER
a) The annual capital recovery payment for a capital
expenditure of $4.5 million over 30 years at
8% is,
(4,500,000)[0.08(1 + .08)30] / (1 + .08)30 − 1
= $399,723.45
CASE SOLUTION: CLEAN CLOTHES
CORNER LAUNDRY
a)= = =
− −
f
v
1,700 2,000 items per month
1.10 .25
c
v p c
b) Solution depends on number of months; 36 used
here. $16,200 ÷ 36 = $450 per month, thus
monthly fixed cost is $2,150= = =
− −
f
v
2,150 2,529.4 items per month
1.10 .25
c
v p c
529.4 additional items per month
c) Z = vp − cf − vcv
= 4,300(1.10) − 2,150 − 4,300(.25)
= $1,505 per month
After 3 years, Z = $1,955 per month
d)= = =
− −
= − −
= − −
=
f
v
f v
1,700 2,297.3
.99 .25
3,800(.99) 1,700 3,800(.25)
$1,112 per month
c
v p c
Z vp c vc
e) With both options:
Z = vp − cf − vcv
= 4,700(.99) − 2,150 − 4,700(.25)
= $1,328
She should purchase the new equipment but not
decrease prices.
CASE SOLUTION: OCOBEE RIVER
RAFTING COMPANY=fAlternative 1: $3,000c= $20p=v $12c= = =
− −
f
1
v
3,000 375 rafts
20 12
c
v p c=fAlternative 2: $10,000c= $20p=v $8c= = =
− −
f
2
v
10,000 833.37
20 8
c
v p c
If demand is less than 375 rafts, the students should not
start the business.
If demand is less than 833 rafts, alternative 2 should not
be selected, and alternative 1 should be used if demand is
expected to be between 375 and 833.33 rafts.
If demand is greater than 833.33 rafts, which alternative
is best? To determine the answer, equate the two cost
functions.
3,000 + 12v = 10,000 + 8v
4v = 7,000
v = 1,750
This is referred to as the point of indifference between
the two alternatives. In general, for demand lower than this
point (1,750) the alternative should be selected with the
lowest fixed cost; for demand greater than this
point the alternative with the lowest variable cost should
be selected. (This general relationship can be observed by
graphing the two cost equations and seeing where they
intersect.)
Thus, for the Ocobee River Rafting Company, the
following guidelines should be used:
demand < 375, do not start business; 375 < demand
< 1,750, select alternative 1; demand > 1,750, select
alternative 2
Since Penny estimates demand will be approximately
1,000 rafts, alternative 1 should be selected.
Z = vp − cf − vcv
= (1,000)(20) − 3,000 − (1,000)(12)
Z = $5,000
CASE SOLUTION: CONSTRUCTING
A DOWNTOWN PARKING LOT
IN DRAPER
a) The annual capital recovery payment for a capital
expenditure of $4.5 million over 30 years at
8% is,
(4,500,000)[0.08(1 + .08)30] / (1 + .08)30 − 1
= $399,723.45
Loading page 8...
2-1
Chapter Two: Linear Programming: Model Formulation and Graphical Solution
PROBLEM SUMMARY
1. Maximization (1–28 continuation), graphical
solution
2. Maximization, graphical solution
3. Minimization, graphical solution
4. Sensitivity analysis (2–3)
5. Minimization, graphical solution
6. Maximization, graphical solution
7. Slack analysis (2–6)
8. Sensitivity analysis (2–6)
9. Maximization, graphical solution
10. Slack analysis (2–9)
11. Maximization, graphical solution
12. Minimization, graphical solution
13. Maximization, graphical solution
14. Sensitivity analysis (2–13)
15. Sensitivity analysis (2–13)
16. Maximization, graphical solution
17. Sensitivity analysis (2–16)
18. Maximization, graphical solution
19. Sensitivity analysis (2–18)
20. Maximization, graphical solution
21. Standard form (2–20)
22. Maximization, graphical solution
23. Standard form (2–22)
24. Maximization, graphical solution
25. Constraint analysis (2–24)
26. Minimization, graphical solution
27. Sensitivity analysis (2–26)
28. Sensitivity analysis (2–26)
29. Sensitivity analysis (2–22)
30. Minimization, graphical solution
31. Minimization, graphical solution
32. Sensitivity analysis (2–31)
33. Minimization, graphical solution
34. Maximization, graphical solution
35. Minimization, graphical solution
36. Maximization, graphical solution
37. Sensitivity analysis (2–34)
38. Minimization, graphical solution
39. Maximization, graphical solution
40. Maximization, graphical solution
41. Sensitivity analysis (2–38)
42. Maximization, graphical solution
43. Sensitivity analysis (2–40)
44. Maximization, graphical solution
45. Sensitivity analysis (2–42)
46. Minimization, graphical solution
47. Sensitivity analysis (2–44)
48. Maximization, graphical solution
49. Sensitivity analysis (2–46)
50. Maximization, graphical solution
51. Sensitivity analysis (2–48)
52. Maximization, graphical solution
53. Minimization, graphical solution
54. Sensitivity analysis (2–53)
55. Minimization, graphical solution
56. Sensitivity analysis (2–55)
57. Maximization, graphical solution
58. Minimization, graphical solution
59. Sensitivity analysis (2–52)
60. Maximization, graphical solution
61. Sensitivity analysis (2–54)
62. Multiple optimal solutions
63. Infeasible problem
64. Unbounded problem
PROBLEM SOLUTIONS
1. a) x1 = # cakes
x2 = # loaves of bread
maximize Z = $10x1 + 6x2
subject to
3x1 + 8x2 ≤ 20 cups of flour
45x1 + 30x2 ≤ 180 minutes
x1,x2 ≥ 0
Chapter Two: Linear Programming: Model Formulation and Graphical Solution
PROBLEM SUMMARY
1. Maximization (1–28 continuation), graphical
solution
2. Maximization, graphical solution
3. Minimization, graphical solution
4. Sensitivity analysis (2–3)
5. Minimization, graphical solution
6. Maximization, graphical solution
7. Slack analysis (2–6)
8. Sensitivity analysis (2–6)
9. Maximization, graphical solution
10. Slack analysis (2–9)
11. Maximization, graphical solution
12. Minimization, graphical solution
13. Maximization, graphical solution
14. Sensitivity analysis (2–13)
15. Sensitivity analysis (2–13)
16. Maximization, graphical solution
17. Sensitivity analysis (2–16)
18. Maximization, graphical solution
19. Sensitivity analysis (2–18)
20. Maximization, graphical solution
21. Standard form (2–20)
22. Maximization, graphical solution
23. Standard form (2–22)
24. Maximization, graphical solution
25. Constraint analysis (2–24)
26. Minimization, graphical solution
27. Sensitivity analysis (2–26)
28. Sensitivity analysis (2–26)
29. Sensitivity analysis (2–22)
30. Minimization, graphical solution
31. Minimization, graphical solution
32. Sensitivity analysis (2–31)
33. Minimization, graphical solution
34. Maximization, graphical solution
35. Minimization, graphical solution
36. Maximization, graphical solution
37. Sensitivity analysis (2–34)
38. Minimization, graphical solution
39. Maximization, graphical solution
40. Maximization, graphical solution
41. Sensitivity analysis (2–38)
42. Maximization, graphical solution
43. Sensitivity analysis (2–40)
44. Maximization, graphical solution
45. Sensitivity analysis (2–42)
46. Minimization, graphical solution
47. Sensitivity analysis (2–44)
48. Maximization, graphical solution
49. Sensitivity analysis (2–46)
50. Maximization, graphical solution
51. Sensitivity analysis (2–48)
52. Maximization, graphical solution
53. Minimization, graphical solution
54. Sensitivity analysis (2–53)
55. Minimization, graphical solution
56. Sensitivity analysis (2–55)
57. Maximization, graphical solution
58. Minimization, graphical solution
59. Sensitivity analysis (2–52)
60. Maximization, graphical solution
61. Sensitivity analysis (2–54)
62. Multiple optimal solutions
63. Infeasible problem
64. Unbounded problem
PROBLEM SOLUTIONS
1. a) x1 = # cakes
x2 = # loaves of bread
maximize Z = $10x1 + 6x2
subject to
3x1 + 8x2 ≤ 20 cups of flour
45x1 + 30x2 ≤ 180 minutes
x1,x2 ≥ 0
Loading page 9...
2-2
b)
2. a) Maximize Z = 6x1 + 4x2 (profit, $)
subject to
10x1 + 10x2 ≤ 100 (line 1, hr)
7x1 + 3x2 ≤ 42 (line 2, hr)
x1,x2 ≥ 0
b)
3. a) Minimize Z = .05x1 + .03x2 (cost, $)
subject to
8x1 + 6x2 ≥ 48 (vitamin A, mg)
x1 + 2x2 ≥ 12 (vitamin B, mg)
x1,x2 ≥ 0
b)
4. 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
5. a) Minimize Z = 3x1 + 5x2 (cost, $)
subject to
10x1 + 2x2 ≥ 20 (nitrogen, oz)
6x1 + 6x2 ≥ 36 (phosphate, oz)
x2 ≥ 2 (potassium, oz)
x1,x2 ≥ 0
b)
6. a) Maximize Z = 400x1 + 100x2 (profit, $)
subject to
8x1 + 10x2 ≤ 80 (labor, hr)
2x1 + 6x2 ≤ 36 (wood)
x1 ≤ 6 (demand, chairs)
x1,x2 ≥ 0
b)
2. a) Maximize Z = 6x1 + 4x2 (profit, $)
subject to
10x1 + 10x2 ≤ 100 (line 1, hr)
7x1 + 3x2 ≤ 42 (line 2, hr)
x1,x2 ≥ 0
b)
3. a) Minimize Z = .05x1 + .03x2 (cost, $)
subject to
8x1 + 6x2 ≥ 48 (vitamin A, mg)
x1 + 2x2 ≥ 12 (vitamin B, mg)
x1,x2 ≥ 0
b)
4. 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
5. a) Minimize Z = 3x1 + 5x2 (cost, $)
subject to
10x1 + 2x2 ≥ 20 (nitrogen, oz)
6x1 + 6x2 ≥ 36 (phosphate, oz)
x2 ≥ 2 (potassium, oz)
x1,x2 ≥ 0
b)
6. a) Maximize Z = 400x1 + 100x2 (profit, $)
subject to
8x1 + 10x2 ≤ 80 (labor, hr)
2x1 + 6x2 ≤ 36 (wood)
x1 ≤ 6 (demand, chairs)
x1,x2 ≥ 0
Loading page 10...
2-3
b)
7. 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
8x1 + 10x2 ≤ 80 hr
8(6) + 10(3.2) ≤ 80
48 + 32 ≤ 80
80 ≤ 80
There is no labor left unused.
Wood
2x1 + 6x2 ≤ 36
2(6) + 6(3.2) ≤ 36
12 + 19.2 ≤ 36
31.2 ≤ 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,
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.
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
resource constraints for flour and sugar and
determine how much of each resource is left
over.
Flour
5x1 + 5x2 ≤ 25 lb
5(0) + 5(4) ≤ 25
20 ≤ 25
25 − 20 = 5
There are 5 lb of flour left unused.
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
3x1 + x2 ≥ 6 (antibiotic 1, units)
x1 + x2 ≥ 4 (antibiotic 2, units)
2x1 + 6x2 ≥ 12 (antibiotic 3, units)
x1,x2 ≥ 0
b)
b)
7. 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
8x1 + 10x2 ≤ 80 hr
8(6) + 10(3.2) ≤ 80
48 + 32 ≤ 80
80 ≤ 80
There is no labor left unused.
Wood
2x1 + 6x2 ≤ 36
2(6) + 6(3.2) ≤ 36
12 + 19.2 ≤ 36
31.2 ≤ 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,
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.
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
resource constraints for flour and sugar and
determine how much of each resource is left
over.
Flour
5x1 + 5x2 ≤ 25 lb
5(0) + 5(4) ≤ 25
20 ≤ 25
25 − 20 = 5
There are 5 lb of flour left unused.
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
3x1 + x2 ≥ 6 (antibiotic 1, units)
x1 + x2 ≥ 4 (antibiotic 2, units)
2x1 + 6x2 ≥ 12 (antibiotic 3, units)
x1,x2 ≥ 0
b)
Loading page 11...
2-4
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
b)
14. The new objective function, Z = 300x1 +
600x2, 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.
The feasible solution space will change. The
new constraint line, 3x1 + 4x2 = 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.
15. a) Optimal solution: x1 = 4 necklaces, x2 = 3
bracelets. The maximum demand is not
achieved by the amount of one bracelet.
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 “slope”
must change such that it is equal to or greater
than 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 − 3x1
x2 = 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
objective function is −3/4:
400x2 = Z − 300x1
x2 = 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 − 600x1
x2 = Z/400 − 3/2x1
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.
16. a) Maximize Z = 50x1 + 40x2 (profit, $) subject
to
3x1 + 5x2 ≤ 150 (wool, yd2)
10x1 + 4x2 ≤ 200 (labor, hr)
x1,x2 ≥ 0
b)
17. The feasible solution space changes from the
area 0ABC to 0AB'C', as shown on the
following graph.
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
b)
14. The new objective function, Z = 300x1 +
600x2, 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.
The feasible solution space will change. The
new constraint line, 3x1 + 4x2 = 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.
15. a) Optimal solution: x1 = 4 necklaces, x2 = 3
bracelets. The maximum demand is not
achieved by the amount of one bracelet.
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 “slope”
must change such that it is equal to or greater
than 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 − 3x1
x2 = 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
objective function is −3/4:
400x2 = Z − 300x1
x2 = 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 − 600x1
x2 = Z/400 − 3/2x1
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.
16. a) Maximize Z = 50x1 + 40x2 (profit, $) subject
to
3x1 + 5x2 ≤ 150 (wool, yd2)
10x1 + 4x2 ≤ 200 (labor, hr)
x1,x2 ≥ 0
b)
17. The feasible solution space changes from the
area 0ABC to 0AB'C', as shown on the
following graph.
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2-5
The extreme points to evaluate are now A,
B', and C'.
A: x1 = 0
x2 = 30
Z = 1,200
*B': x1 = 15.8
x2 = 20.5
Z = 1,610
C': x1 = 24
x2 = 0
Z = 1,200
Point B' is optimal
18. a) Maximize Z = 23x1 + 73x2
subject to
x1 ≤ 40
x2 ≤ 25
x1 + 4x2 ≤ 120
x1,x2 ≥ 0
b)
19. 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
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
Profit will go down from (c)
20.
21. Maximize Z = 1.5x1 + x2 + 0s1 + 0s2 + 0s3
subject 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
22.
23. Maximize Z = 5x1 + 8x2 + 0s1 + 0s3 + 0s4
subject to
3x1 + 5x2 + s1 = 50
The extreme points to evaluate are now A,
B', and C'.
A: x1 = 0
x2 = 30
Z = 1,200
*B': x1 = 15.8
x2 = 20.5
Z = 1,610
C': x1 = 24
x2 = 0
Z = 1,200
Point B' is optimal
18. a) Maximize Z = 23x1 + 73x2
subject to
x1 ≤ 40
x2 ≤ 25
x1 + 4x2 ≤ 120
x1,x2 ≥ 0
b)
19. 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
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
Profit will go down from (c)
20.
21. Maximize Z = 1.5x1 + x2 + 0s1 + 0s2 + 0s3
subject 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
22.
23. Maximize Z = 5x1 + 8x2 + 0s1 + 0s3 + 0s4
subject to
3x1 + 5x2 + s1 = 50
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2-6
2x1 + 4x2 + s2 = 40
x1 + s3 = 8
x2 + s4 = 10
x1,x2 ≥ 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
24.
25. 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 solution
space boundary.
26. 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)
27. Changing the pay for a full-time claims
processor from $64 to $54 will change the
solution to point A in the graphical solution
where x1 = 28.125 and x2 = 0, i.e., there will
be no part-time operators. 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.
28. 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.
29. The solution becomes infeasible; there are
not enough workstations to handle the
increase in the volume of claims.
30.
31.
32. The problem becomes infeasible.
2x1 + 4x2 + s2 = 40
x1 + s3 = 8
x2 + s4 = 10
x1,x2 ≥ 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
24.
25. 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 solution
space boundary.
26. 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)
27. Changing the pay for a full-time claims
processor from $64 to $54 will change the
solution to point A in the graphical solution
where x1 = 28.125 and x2 = 0, i.e., there will
be no part-time operators. 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.
28. 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.
29. The solution becomes infeasible; there are
not enough workstations to handle the
increase in the volume of claims.
30.
31.
32. The problem becomes infeasible.
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2-7
33.
34.
35.
36. a) Maximize Z = $4.15x1 + 3.60x2 (profit, $)
subject to1 2
1 2
1
1 2
2
1 2
115 (freezer space, gals.)
0.93 0.75 90 (budget, $)
2 or 2 0 (demand)
1
, 0
x x
x x
x x x
x
x x
+
+
−
b)
37. No additional profit, freezer space is not a
binding constraint.
38. a) Minimize Z = 200x1 + 160x2 (cost, $)
subject to
6x1 + 2x2 ≥ 12 (high-grade ore, tons)
2x1 + 2x2 ≥ 8 (medium-grade ore, tons)
4x1 + 12x2 ≥ 24 (low-grade ore, tons)
x1,x2 ≥ 0
b)
33.
34.
35.
36. a) Maximize Z = $4.15x1 + 3.60x2 (profit, $)
subject to1 2
1 2
1
1 2
2
1 2
115 (freezer space, gals.)
0.93 0.75 90 (budget, $)
2 or 2 0 (demand)
1
, 0
x x
x x
x x x
x
x x
+
+
−
b)
37. No additional profit, freezer space is not a
binding constraint.
38. a) Minimize Z = 200x1 + 160x2 (cost, $)
subject to
6x1 + 2x2 ≥ 12 (high-grade ore, tons)
2x1 + 2x2 ≥ 8 (medium-grade ore, tons)
4x1 + 12x2 ≥ 24 (low-grade ore, tons)
x1,x2 ≥ 0
b)
Loading page 15...
2-8
39. a) Maximize Z = 800x1 + 900x2 (profit, $)
subject to
2x1 + 4x2 ≤ 30 (stamping, days)
4x1 + 2x2 ≤ 30 (coating, days)
x1 + x2 ≥ 9 (lots)
x1,x2 ≥ 0
b)
40. a) Maximize Z = 30x1 + 70x2 (profit, $) subject
to
4x1 + 10x2 ≤ 80 (assembly, hr)
14x1 + 8x2 ≤ 112 (finishing, hr)
x1 + x2 ≤ 10 (inventory, units)
x1,x2 ≥ 0
b)
41. The slope of the original objective function
is computed as follows:
Z = 30x1 + 70x2
70x2 = Z − 30x1
x2 = Z/70 − 3/7x1
slope = −3/7
The slope of the new objective function is
computed as follows:
Z = 90x1 + 70x2
70x2 = Z − 90x1
x2 = Z/70 − 9/7x1
slope = −9/7
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 C: x1 = 5.3
x2 = 8 x2 = 4.7
Z = 560 Z = 806
B: x1 = 3.3 D: x1 = 8
x2 = 6.7 x2 = 0
Z = 766 Z = 720
42. a) Maximize Z = 9x1 + 12x2 (profit, $1,000s)
subject to
4x1 + 8x2 ≤ 64 (grapes, tons)
5x1 + 5x2 ≤ 50 (storage space, yd3)
15x1 + 8x2 ≤ 120 (processing time, hr)
x1 ≤ 7 (demand, Nectar)
x2 ≤ 7 (demand, Red)
x1,x2 ≥ 0
b)
39. a) Maximize Z = 800x1 + 900x2 (profit, $)
subject to
2x1 + 4x2 ≤ 30 (stamping, days)
4x1 + 2x2 ≤ 30 (coating, days)
x1 + x2 ≥ 9 (lots)
x1,x2 ≥ 0
b)
40. a) Maximize Z = 30x1 + 70x2 (profit, $) subject
to
4x1 + 10x2 ≤ 80 (assembly, hr)
14x1 + 8x2 ≤ 112 (finishing, hr)
x1 + x2 ≤ 10 (inventory, units)
x1,x2 ≥ 0
b)
41. The slope of the original objective function
is computed as follows:
Z = 30x1 + 70x2
70x2 = Z − 30x1
x2 = Z/70 − 3/7x1
slope = −3/7
The slope of the new objective function is
computed as follows:
Z = 90x1 + 70x2
70x2 = Z − 90x1
x2 = Z/70 − 9/7x1
slope = −9/7
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 C: x1 = 5.3
x2 = 8 x2 = 4.7
Z = 560 Z = 806
B: x1 = 3.3 D: x1 = 8
x2 = 6.7 x2 = 0
Z = 766 Z = 720
42. a) Maximize Z = 9x1 + 12x2 (profit, $1,000s)
subject to
4x1 + 8x2 ≤ 64 (grapes, tons)
5x1 + 5x2 ≤ 50 (storage space, yd3)
15x1 + 8x2 ≤ 120 (processing time, hr)
x1 ≤ 7 (demand, Nectar)
x2 ≤ 7 (demand, Red)
x1,x2 ≥ 0
b)
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2-9
43. a) 15(4) + 8(6) ≤ 120 hr
60 + 48 ≤ 120
108 ≤ 120
120 − 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.
44. a) Maximize Z = .28x1 + .19x21 2
2
1
1 2
96 cans
2
, 0
x x
x
x
x x
+
b)
45. The model formulation would become,
maximize Z = $0.23x1 + 0.19x2
subject to
x1 + x2 ≤ 96
–1.5x1 + x2 ≥ 0
x1,x2 ≥ 0
The solution is x1 = 38.4, x2 = 57.6, and
Z = $19.78
The discount would reduce profit.
46. a) Minimize Z = $0.46x1 + 0.35x2
subject to
.91x1 + .82x2 = 3,500
x1 ≥ 1,000
x2 ≥ 1,000
.03x1 − .06x2 ≥ 0
x1,x2 ≥ 0
b)
47. a) Minimize Z = .09x1 + .18x2
subject to
.46x1 + .35x2 ≤ 2,000
x1 ≥ 1,000
x2 ≥ 1,000
.91x1 − .82x2 = 3,500
x1,x2 ≥ 0
b) 477 − 445 = 32 fewer defective items
48. a) Maximize Z = $2.25x1 + 1.95x2
subject to
8x1 + 6x2 ≤ 1,920
3x1 + 6x2 ≤ 1,440
3x1 + 2x2 ≤ 720
x1 + x2 ≤ 288
x1,x2 ≥ 0
43. a) 15(4) + 8(6) ≤ 120 hr
60 + 48 ≤ 120
108 ≤ 120
120 − 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.
44. a) Maximize Z = .28x1 + .19x21 2
2
1
1 2
96 cans
2
, 0
x x
x
x
x x
+
b)
45. The model formulation would become,
maximize Z = $0.23x1 + 0.19x2
subject to
x1 + x2 ≤ 96
–1.5x1 + x2 ≥ 0
x1,x2 ≥ 0
The solution is x1 = 38.4, x2 = 57.6, and
Z = $19.78
The discount would reduce profit.
46. a) Minimize Z = $0.46x1 + 0.35x2
subject to
.91x1 + .82x2 = 3,500
x1 ≥ 1,000
x2 ≥ 1,000
.03x1 − .06x2 ≥ 0
x1,x2 ≥ 0
b)
47. a) Minimize Z = .09x1 + .18x2
subject to
.46x1 + .35x2 ≤ 2,000
x1 ≥ 1,000
x2 ≥ 1,000
.91x1 − .82x2 = 3,500
x1,x2 ≥ 0
b) 477 − 445 = 32 fewer defective items
48. a) Maximize Z = $2.25x1 + 1.95x2
subject to
8x1 + 6x2 ≤ 1,920
3x1 + 6x2 ≤ 1,440
3x1 + 2x2 ≤ 720
x1 + x2 ≤ 288
x1,x2 ≥ 0
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2-10
b)
49. A new constraint is added to the model in1
2
1.5
x
x
The solution is x1 = 160, x2 = 106.67,
Z = $568
50. 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)
51. 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 = 25.71
x2 = 37 x2 = 14.29
Z = 11,100 Z = 14,571
B': x1 = 3 D': x1 = 26
x2 = 37 x2 = 0
Z = 12,300 Z = 10,400
52. a) Maximize Z = $7,600x1 + 22,500x2
subject to
x1 + x2 ≤ 3,500
x2/(x1 + x2) ≤ .40
.12x1 + .24x2 ≤ 600
x1,x2 ≥ 0
b)
49. A new constraint is added to the model in1
2
1.5
x
x
The solution is x1 = 160, x2 = 106.67,
Z = $568
50. 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)
51. 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 = 25.71
x2 = 37 x2 = 14.29
Z = 11,100 Z = 14,571
B': x1 = 3 D': x1 = 26
x2 = 37 x2 = 0
Z = 12,300 Z = 10,400
52. a) Maximize Z = $7,600x1 + 22,500x2
subject to
x1 + x2 ≤ 3,500
x2/(x1 + x2) ≤ .40
.12x1 + .24x2 ≤ 600
x1,x2 ≥ 0
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2-11
b)
53. a) Minimize Z = $(.05)(8)x1 + (.10)(.75)x2
subject to
5x1 + x2 ≥ 8001
2
5 1.5
x
x =
8x1 + .75x2 ≤ 1,200
x1, x2 ≥ 0
x1 = 96
x2 = 320
Z = $62.40
b)
54. The new solution is
x1 = 106.67
x2 = 266.67
Z = $62.67
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
wine, then there would logically be no waste
for wine but only for beer. This amount
“logically” 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.
55. a) Minimize Z = 3700x1 + 5100x2
subject to
x1 + x2 = 45
(32x1 + 14x2) / (x1 + x2) ≤ 21
.10x1 + .04x2 ≤ 61
1 2
.25
( )
x
x x
+2
1 2
.25
( )
x
x x
+
x1, x2 ≥ 0
b)
b)
53. a) Minimize Z = $(.05)(8)x1 + (.10)(.75)x2
subject to
5x1 + x2 ≥ 8001
2
5 1.5
x
x =
8x1 + .75x2 ≤ 1,200
x1, x2 ≥ 0
x1 = 96
x2 = 320
Z = $62.40
b)
54. The new solution is
x1 = 106.67
x2 = 266.67
Z = $62.67
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
wine, then there would logically be no waste
for wine but only for beer. This amount
“logically” 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.
55. a) Minimize Z = 3700x1 + 5100x2
subject to
x1 + x2 = 45
(32x1 + 14x2) / (x1 + x2) ≤ 21
.10x1 + .04x2 ≤ 61
1 2
.25
( )
x
x x
+2
1 2
.25
( )
x
x x
+
x1, x2 ≥ 0
b)
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2-12
56. a) No, the solution would not change
b) No, the solution would not change
c) Yes, the solution would change to China (x1)
= 22.5, Brazil (x2) = 22.5, and
Z = $198,000.
57. a) x1 = $ invested in stocks
x2 = $ invested in bonds
maximize Z = $0.18x1 + 0.06x2 (average
annual return)
subject to
x1 + x2 ≤ $720,000 (available funds)
x1/(x1 + x2) ≤ .65 (% of stocks)
.22x1 + .05x2 ≤ 100,000 (total possible loss)
x1,x2 ≥ 0
b)
58. x1 = exams assigned to Brad
x2 = exams assigned to Sarah
minimize Z = .10x1 + .06x2
subject to
x1 + x2 = 120
x1 ≤ (720/7.2) or 100
x2 ≤ 50(600/12)
x1,x2 ≥ 0
59. If the constraint for Sarah’s time became x2
≤ 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 Brad’s time became x1 ≤ 108.33 with an
additional hour then the solution point (A)
would not change. All of Brad’s time is not
being used anyway so assigning him more
time would not have an effect.
One more hour of Sarah’s 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 Brad’s is
zero.
60. a) x1 = # cups of Pomona
x2 = # 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 = 3/2
x1,x2 ≥ 0
56. a) No, the solution would not change
b) No, the solution would not change
c) Yes, the solution would change to China (x1)
= 22.5, Brazil (x2) = 22.5, and
Z = $198,000.
57. a) x1 = $ invested in stocks
x2 = $ invested in bonds
maximize Z = $0.18x1 + 0.06x2 (average
annual return)
subject to
x1 + x2 ≤ $720,000 (available funds)
x1/(x1 + x2) ≤ .65 (% of stocks)
.22x1 + .05x2 ≤ 100,000 (total possible loss)
x1,x2 ≥ 0
b)
58. x1 = exams assigned to Brad
x2 = exams assigned to Sarah
minimize Z = .10x1 + .06x2
subject to
x1 + x2 = 120
x1 ≤ (720/7.2) or 100
x2 ≤ 50(600/12)
x1,x2 ≥ 0
59. If the constraint for Sarah’s time became x2
≤ 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 Brad’s time became x1 ≤ 108.33 with an
additional hour then the solution point (A)
would not change. All of Brad’s time is not
being used anyway so assigning him more
time would not have an effect.
One more hour of Sarah’s 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 Brad’s is
zero.
60. a) x1 = # cups of Pomona
x2 = # 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 = 3/2
x1,x2 ≥ 0
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2-13
b) Solution:
x1 = 87.3 cups
x2 = 130.9 cups
Z = $421.09
61. 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.
One more pound of Colombian would
increase sales from $421.09 to $463.20.
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.
62.
Multiple optimal solutions; A and B alternate
optimal
63.
64.
b) Solution:
x1 = 87.3 cups
x2 = 130.9 cups
Z = $421.09
61. 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.
One more pound of Colombian would
increase sales from $421.09 to $463.20.
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.
62.
Multiple optimal solutions; A and B alternate
optimal
63.
64.
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2-14
CASE SOLUTION: METROPOLITAN
POLICE PATROL
The linear programming model for this case
problem is
Minimize Z = x/60 + y/45
subject to
2x + 2y ≥ 5
2x + 2y ≤ 12
y ≥ 1.5x
x, 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 1.5
miles by 1 mile and the response time is
0.05 hr, 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 ≥ 0
x2/(x1 + x2) ≥ .10 or .90x2 − .10x1 ≥ 0
x1x2 ≥ 0
The graphical solution is shown as follows.
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 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
= 29.75x1 + 58.24x2
subject to1 2
1
2
2
1
1 2
175 208 $120, 000
.33
2
, 0
+ =
x x
x
x
x
x
x x
x1 = 203
x2 = 406
Z = $29,691.372
1
Eliminating the constraint .33
x
x
will have no effect on the solution.1
2
Eliminating the constraint 2
x
x
will change the solution to x1 = 149,
x2 = 451.55, Z = $30,731.52.
CASE SOLUTION: METROPOLITAN
POLICE PATROL
The linear programming model for this case
problem is
Minimize Z = x/60 + y/45
subject to
2x + 2y ≥ 5
2x + 2y ≤ 12
y ≥ 1.5x
x, 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 1.5
miles by 1 mile and the response time is
0.05 hr, 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 ≥ 0
x2/(x1 + x2) ≥ .10 or .90x2 − .10x1 ≥ 0
x1x2 ≥ 0
The graphical solution is shown as follows.
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 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
= 29.75x1 + 58.24x2
subject to1 2
1
2
2
1
1 2
175 208 $120, 000
.33
2
, 0
+ =
x x
x
x
x
x
x x
x1 = 203
x2 = 406
Z = $29,691.372
1
Eliminating the constraint .33
x
x
will have no effect on the solution.1
2
Eliminating the constraint 2
x
x
will change the solution to x1 = 149,
x2 = 451.55, Z = $30,731.52.
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2-15
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 “shadow” or “dual” price as described
in Chapter 3.
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 “shadow” or “dual” price as described
in Chapter 3.
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3-1
Chapter Three: Linear Programming: Computer Solution and Sensitivity Analysis
PROBLEM SUMMARY
1. QM for Windows
2. QM for Windows and Excel
3. Excel
4. Excel solution
5. Excel
6. Graphical solution; sensitivity analysis
7. Model formulation
8. Graphical solution; sensitivity analysis (3–7)
9. Sensitivity analysis (3–7)
10. Model formulation
11. Graphical solution; sensitivity analysis (3–10)
12. Sensitivity analysis (3–10)
13. Model formulation
14. Graphical solution; sensitivity analysis (3–13)
15. Computer solution; sensitivity analysis (3–13)
16. Model formulation
17. Graphical solution; sensitivity analysis (3–16)
18. Computer solution; sensitivity analysis (3–16)
19. Model formulation
20. Graphical solution; sensitivity analysis (3–19)
21. Computer solution; sensitivity analysis (3–19)
22. Model formulation
23. Graphical solution; sensitivity analysis (3–22)
24. Computer solution; sensitivity analysis (3–22)
25. Model formulation
26. Graphical solution; sensitivity analysis (3–25)
27. Computer solution; sensitivity analysis (3–25)
28. Model formulation
29. Graphical solution; sensitivity analysis (3–28)
30. Computer solution; sensitivity analysis (3–28)
31. Model formulation; graphical solution
32. Computer solution; sensitivity analysis (3–31)
33. Model formulation
34. Graphical solution; computer solution; sensitivity
analysis (3–33).
35. Model formulation
36. Graphical solution; sensitivity analysis (3–35)
37. Computer solution; sensitivity analysis (3–35)
38. Model formulation; computer solution
39. Model formulation; computer solution
40. Computer solution; sensitivity analysis
41. Model formulation
42. Graphical solution; sensitivity analysis (3–41)
43. Computer solution; sensitivity analysis (3–41)
44. Model formulation; computer solution
45. Sensitivity analysis (3–44)
46. Model formulation; computer solution
47. Sensitivity analysis (3–46)
48. Model formulation
49. Computer solution; sensitivity analysis (3–48)
50. Model formulation
51. Computer solution; sensitivity analysis (3–50)
52. Model formulation
53. Computer solution; sensitivity analysis (3–52)
54. Model formulation
55. Computer solution; sensitivity analysis (3–54)
56. Computer solution
57. Model formulation
58. Sensitivity analysis (3–57)
59. Model formulation
60. Computer solution; sensitivity analysis (3–59)
61. Model formulation; graphical solution
62. Computer solution; sensitivity analysis (3–61)
PROBLEM SOLUTIONS
1.
2. QM for Windows establishes a “template” for
the linear programming model based on the
user’s specification of the type of objective
function, the number of constraints, and
number of variables. Then the model
parameters are input and the problem is solved.
In Excel the model “template” must be
developed by the user.
Chapter Three: Linear Programming: Computer Solution and Sensitivity Analysis
PROBLEM SUMMARY
1. QM for Windows
2. QM for Windows and Excel
3. Excel
4. Excel solution
5. Excel
6. Graphical solution; sensitivity analysis
7. Model formulation
8. Graphical solution; sensitivity analysis (3–7)
9. Sensitivity analysis (3–7)
10. Model formulation
11. Graphical solution; sensitivity analysis (3–10)
12. Sensitivity analysis (3–10)
13. Model formulation
14. Graphical solution; sensitivity analysis (3–13)
15. Computer solution; sensitivity analysis (3–13)
16. Model formulation
17. Graphical solution; sensitivity analysis (3–16)
18. Computer solution; sensitivity analysis (3–16)
19. Model formulation
20. Graphical solution; sensitivity analysis (3–19)
21. Computer solution; sensitivity analysis (3–19)
22. Model formulation
23. Graphical solution; sensitivity analysis (3–22)
24. Computer solution; sensitivity analysis (3–22)
25. Model formulation
26. Graphical solution; sensitivity analysis (3–25)
27. Computer solution; sensitivity analysis (3–25)
28. Model formulation
29. Graphical solution; sensitivity analysis (3–28)
30. Computer solution; sensitivity analysis (3–28)
31. Model formulation; graphical solution
32. Computer solution; sensitivity analysis (3–31)
33. Model formulation
34. Graphical solution; computer solution; sensitivity
analysis (3–33).
35. Model formulation
36. Graphical solution; sensitivity analysis (3–35)
37. Computer solution; sensitivity analysis (3–35)
38. Model formulation; computer solution
39. Model formulation; computer solution
40. Computer solution; sensitivity analysis
41. Model formulation
42. Graphical solution; sensitivity analysis (3–41)
43. Computer solution; sensitivity analysis (3–41)
44. Model formulation; computer solution
45. Sensitivity analysis (3–44)
46. Model formulation; computer solution
47. Sensitivity analysis (3–46)
48. Model formulation
49. Computer solution; sensitivity analysis (3–48)
50. Model formulation
51. Computer solution; sensitivity analysis (3–50)
52. Model formulation
53. Computer solution; sensitivity analysis (3–52)
54. Model formulation
55. Computer solution; sensitivity analysis (3–54)
56. Computer solution
57. Model formulation
58. Sensitivity analysis (3–57)
59. Model formulation
60. Computer solution; sensitivity analysis (3–59)
61. Model formulation; graphical solution
62. Computer solution; sensitivity analysis (3–61)
PROBLEM SOLUTIONS
1.
2. QM for Windows establishes a “template” for
the linear programming model based on the
user’s specification of the type of objective
function, the number of constraints, and
number of variables. Then the model
parameters are input and the problem is solved.
In Excel the model “template” must be
developed by the user.
Loading page 24...
3-2
3. Set Target cell: B13
Changing cells: B10:B12
Profit: = B10 * C4 + B11 * D4 + B12 * E4
Constraints: B10:B12 ≥ 0
G6 ≤ F6
G7 ≤ F7
4. x1 = 0
x2 = 9
Z = $54
5. F6: = C6 * B12 + D6 * B13
F7: = C7 * B12 + D7 * B13
F8: = C8 * B12 + D8 * B13
F9: = C9 * B12 + D9 * B13
G6: = E6 − F6
G7: = E7 − F7
G8: = E8 − F8
G9: = E9 − F9
B14: = C4 * B12 + D4 * B13
x1 = 0
x2 = 5.2
Z = 81.6
6. The slope of the constraint line is −70/60. The
optimal solution is at point A where x1 = 0 and
x2 = 70. To change the solution to B, c1 must
increase such that the slope of the objective
function is at least as great as the slope of the
constraint line,
−c1/50 = −70/60
c1 = 58.33
Alternatively, c2 must decrease such that the
slope of the objective function is at least as
great as the slope of the constraint line,
–30/c2 = –70/60
c2 = 25.71
Thus, if c1 increases to greater than 58.33 or c2
decreases to less than 25.71, B will become optimal.
7. a) x1 = no. of basketballs
x2 = no. of footballs
maximize Z = 12x1 + 16x2
subject to
3x1 + 2x2 ≤ 500
4x1 + 5x2 ≤ 800
x1, x2 ≥ 0
b) maximize Z = 12x1 + 16x2 + 0s1 + 0s2
subject to
3x1 + 2x2 + s1 = 500
4x1 + 5x2 + s2 = 800
x1, x2, s1, s2 ≥ 0
8.
a) A: 3(0) + 2(160) + s1 = 500
s1 = 180
4(0) + 5(160) + s2 = 800
s2 = 0
B: 3(128.5) + 2(57.2) + s1 = 500
s1 = 0
4(128.5) + 2(57.2) + s2 = 800
s2 = 0
C: 2(167) + 2(0) + s1 = 500
s1 = 0
4(167) + 5(0) + s2 = 800
s2 = 132
b) Z = 12x1 + 16x2
and,
x2 = Z/16 − 12x1/16
The slope of the objective function, −12/16,
would have to become steeper (i.e., greater)
than the slope of the constraint line 4x1 + 5x2 =
800, for the solution to change.
The profit, c1, for a basketball that would
change the solution point is,
−4/5 = −c1/16
5c1 = 64
c1 = 12.8
Since $13 > 12.8 the solution point would
change to B where x1 = 128.5, x2 = 57.2.
The new Z value is $2,585.70.
For a football,
−4/5 = −12/c2
4c2 = 60
c2 = 15
Thus, if the profit for a football decreased to
$15 or less, point B will also be optimal (i.e.,
multiple optimal solutions). The solution at B
is x1 = 128.5, x2 = 57.2, and Z = $2,400.
3. Set Target cell: B13
Changing cells: B10:B12
Profit: = B10 * C4 + B11 * D4 + B12 * E4
Constraints: B10:B12 ≥ 0
G6 ≤ F6
G7 ≤ F7
4. x1 = 0
x2 = 9
Z = $54
5. F6: = C6 * B12 + D6 * B13
F7: = C7 * B12 + D7 * B13
F8: = C8 * B12 + D8 * B13
F9: = C9 * B12 + D9 * B13
G6: = E6 − F6
G7: = E7 − F7
G8: = E8 − F8
G9: = E9 − F9
B14: = C4 * B12 + D4 * B13
x1 = 0
x2 = 5.2
Z = 81.6
6. The slope of the constraint line is −70/60. The
optimal solution is at point A where x1 = 0 and
x2 = 70. To change the solution to B, c1 must
increase such that the slope of the objective
function is at least as great as the slope of the
constraint line,
−c1/50 = −70/60
c1 = 58.33
Alternatively, c2 must decrease such that the
slope of the objective function is at least as
great as the slope of the constraint line,
–30/c2 = –70/60
c2 = 25.71
Thus, if c1 increases to greater than 58.33 or c2
decreases to less than 25.71, B will become optimal.
7. a) x1 = no. of basketballs
x2 = no. of footballs
maximize Z = 12x1 + 16x2
subject to
3x1 + 2x2 ≤ 500
4x1 + 5x2 ≤ 800
x1, x2 ≥ 0
b) maximize Z = 12x1 + 16x2 + 0s1 + 0s2
subject to
3x1 + 2x2 + s1 = 500
4x1 + 5x2 + s2 = 800
x1, x2, s1, s2 ≥ 0
8.
a) A: 3(0) + 2(160) + s1 = 500
s1 = 180
4(0) + 5(160) + s2 = 800
s2 = 0
B: 3(128.5) + 2(57.2) + s1 = 500
s1 = 0
4(128.5) + 2(57.2) + s2 = 800
s2 = 0
C: 2(167) + 2(0) + s1 = 500
s1 = 0
4(167) + 5(0) + s2 = 800
s2 = 132
b) Z = 12x1 + 16x2
and,
x2 = Z/16 − 12x1/16
The slope of the objective function, −12/16,
would have to become steeper (i.e., greater)
than the slope of the constraint line 4x1 + 5x2 =
800, for the solution to change.
The profit, c1, for a basketball that would
change the solution point is,
−4/5 = −c1/16
5c1 = 64
c1 = 12.8
Since $13 > 12.8 the solution point would
change to B where x1 = 128.5, x2 = 57.2.
The new Z value is $2,585.70.
For a football,
−4/5 = −12/c2
4c2 = 60
c2 = 15
Thus, if the profit for a football decreased to
$15 or less, point B will also be optimal (i.e.,
multiple optimal solutions). The solution at B
is x1 = 128.5, x2 = 57.2, and Z = $2,400.
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3-3
c) If the constraint line for rubber changes to 3x1
+ 2x2 = 1,000, it moves outward, eliminating
points B and C. However, since A is the optimal
point, it will not change and the optimal solution
remains the same, x1 = 0, x2 = 160, and Z = 2,560.
There will be an increase in slack, s1, to 680 lbs.
If the constraint line for leather changes to 4x1 +
5x2 = 1,300, point A will move to a new
location, x1 = 0, x2 = 250, Z = $4,000.
9. a) For c1 the upper limit is computed as
−4/5 = −c1/16
5c1 = 64
c1 = 12.8
and the lower limit is unlimited.
For c2 the lower limit is,
−4/5 = −12/c2
4c2 = 60
c2 = 15
and the upper limit is unlimited.
Summarizing,
∞ ≤ c1 ≤ 12.8
15 ≤ c2 ≤ ∞
For q1 the upper limit is ∞ since no matter how
much q1 increases the optimal solution point A
will not change.
The lower limit for q1 is at the point where the
constraint line (3x1 + 2x2 = q1) intersects with
point A where x1 = 0, x2 = 160,
3x1 + 2x2= q1
3(0) + 2(160) = q1
q1 = 320
For q2 the upper limit is at the point where the
rubber constraint line (3x1 + 2x2 = 500)
intersects with the leather constraint line
(4x1 + 5x2 = 800) along the x2 axis, i.e.,
x1 = 0, x2 = 250,
4x1 + 5x2 = q2
4(0) + 5(250) = q2
q2 = 1,250
The lower limit is 0 since that is the lowest
point on the x2 axis the constraint line can
decrease to.
Summarizing,
320 ≤ q1 ≤ ∞
0 ≤ q2 ≤ 1,250
c) Z = 2,560.000
Variable Value
Reduced
Cost
x1 0.00 0.800
x2 160.000 0.000
Constraint Slack/Surplus
Shadow
Price
c1 180.00 0.00
c2 0.00 3.20
The shadow price for rubber is $0. Since there
is slack rubber left over at the optimal point,
extra rubber would have no marginal value.
The shadow price for leather is $3.20. For each
additional ft.2 of leather that the company can
obtain profit would increase by $3.20, up to
the upper limit of the sensitivity range for
leather (i.e., 1,250 ft.2).
10. a) x1 = no. of units of A
x2 = no. of units of B
maximize Z = 9x1 + 7x2
subject to
12x1 + 4x2 ≤ 60
4x1 + 8x2 ≤ 40
x1, x2 ≥ 0
b) maximize Z = 9x1 + 7x2 + 0s1 + 0s2
subject to
12x1 + 4x2 + s1 = 60
4x1 + 8x2 + s2 = 40
x1, x2, s1, s2 ≥ 0
b)
Objective Coefficient Ranges
Variables Lower Limit Current Values Upper Limit Allowable Increase Allowable Decrease
x1 No limit 12.000 12.800 0.800 No limit
x2 15.000 16.000 No limit No limit 1.000
Right Hand Side Ranges
Constraints Lower Limit Current Values Upper Limit Allowable Increase Allowable Decrease
c1 320.000 500.000 No limit No limit 180.000
c2 0.000 800.000 1,250.000 450.000 800.000
c) If the constraint line for rubber changes to 3x1
+ 2x2 = 1,000, it moves outward, eliminating
points B and C. However, since A is the optimal
point, it will not change and the optimal solution
remains the same, x1 = 0, x2 = 160, and Z = 2,560.
There will be an increase in slack, s1, to 680 lbs.
If the constraint line for leather changes to 4x1 +
5x2 = 1,300, point A will move to a new
location, x1 = 0, x2 = 250, Z = $4,000.
9. a) For c1 the upper limit is computed as
−4/5 = −c1/16
5c1 = 64
c1 = 12.8
and the lower limit is unlimited.
For c2 the lower limit is,
−4/5 = −12/c2
4c2 = 60
c2 = 15
and the upper limit is unlimited.
Summarizing,
∞ ≤ c1 ≤ 12.8
15 ≤ c2 ≤ ∞
For q1 the upper limit is ∞ since no matter how
much q1 increases the optimal solution point A
will not change.
The lower limit for q1 is at the point where the
constraint line (3x1 + 2x2 = q1) intersects with
point A where x1 = 0, x2 = 160,
3x1 + 2x2= q1
3(0) + 2(160) = q1
q1 = 320
For q2 the upper limit is at the point where the
rubber constraint line (3x1 + 2x2 = 500)
intersects with the leather constraint line
(4x1 + 5x2 = 800) along the x2 axis, i.e.,
x1 = 0, x2 = 250,
4x1 + 5x2 = q2
4(0) + 5(250) = q2
q2 = 1,250
The lower limit is 0 since that is the lowest
point on the x2 axis the constraint line can
decrease to.
Summarizing,
320 ≤ q1 ≤ ∞
0 ≤ q2 ≤ 1,250
c) Z = 2,560.000
Variable Value
Reduced
Cost
x1 0.00 0.800
x2 160.000 0.000
Constraint Slack/Surplus
Shadow
Price
c1 180.00 0.00
c2 0.00 3.20
The shadow price for rubber is $0. Since there
is slack rubber left over at the optimal point,
extra rubber would have no marginal value.
The shadow price for leather is $3.20. For each
additional ft.2 of leather that the company can
obtain profit would increase by $3.20, up to
the upper limit of the sensitivity range for
leather (i.e., 1,250 ft.2).
10. a) x1 = no. of units of A
x2 = no. of units of B
maximize Z = 9x1 + 7x2
subject to
12x1 + 4x2 ≤ 60
4x1 + 8x2 ≤ 40
x1, x2 ≥ 0
b) maximize Z = 9x1 + 7x2 + 0s1 + 0s2
subject to
12x1 + 4x2 + s1 = 60
4x1 + 8x2 + s2 = 40
x1, x2, s1, s2 ≥ 0
b)
Objective Coefficient Ranges
Variables Lower Limit Current Values Upper Limit Allowable Increase Allowable Decrease
x1 No limit 12.000 12.800 0.800 No limit
x2 15.000 16.000 No limit No limit 1.000
Right Hand Side Ranges
Constraints Lower Limit Current Values Upper Limit Allowable Increase Allowable Decrease
c1 320.000 500.000 No limit No limit 180.000
c2 0.000 800.000 1,250.000 450.000 800.000
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3-4
11.
a) A: 12(0) + 4(5) + s1 = 60
s1 = 40
4(0) + 8(5) + s2 = 40
s2 = 0
B: 12(4) + 4(3) = 60
s1 = 0
4(4) + 8(3) + s2 = 40
s2 = 0
C: 12(5) + 4(0) + s1 = 60
s1 = 0
4(5) + 8(0) + s2 = 40
s2 = 20
b) The constraint line 12x1 + 4x2 = 60 would move
inward resulting in a new location for point B
at x1 = 2, x2 = 4, which would still be optimal.
c) In order for the optimal solution point to change
from B to A the slope of the objective function
must be at least as flat as the slope of the
constraint line, 4x1 + 8x2 = 40, which is −1/2.
Thus, the profit for product B would have to be,
−9/c2 = −1/2
c2 = 18
If the profit for product B is increased to $15
the optimal solution point will not change,
although Z would change from $57 to $81.
If the profit for product B is increased to $20
the solution point will change from B to A,
x1 = 0, x2 = 5, Z = $100.
12. a) For c1 the upper limit is computed as,
−c1/7 = −3
c1 = 21
and the lower limit is,
−c1/7 = −1/2
c1 = 3.50
For c2 the upper limit is,
−9/c2 = −1/2
c2 = 18
and the lower limit is,
−9/c2 = −3
c2 = 3
Summarizing,
3.50 ≤ c1 ≤ 21
3 ≤ c2 ≤ 18
b)
Z = 57.000
Variable Value
Reduced
Cost
x1 4.000 0.000
x2 3.000 0.000
Constraint Slack/Surplus
Shadow
Price
c1 0.000 0.550
c 2 0.000 0.600
Objective Coefficient Ranges
Variables
Lower
Limit
Current
Values
Upper
Limit
Allowable
Increase
Allowable
Decrease
x1 3.500 9.000 21.000 12.000 5.500
x2 3.000 7.000 18.000 11.000 4.000
Right Hand Side Ranges
Constraints
Lower
Limit
Current
Values
Upper
Limit
Allowable
Increase
Allowable
Decrease
c1 20.000 60.000 120.000 60.000 40.000
c 2 20.000 40.000 120.000 80.000 20.000
c) The shadow price for line 1 time is $0.55 per
11.
a) A: 12(0) + 4(5) + s1 = 60
s1 = 40
4(0) + 8(5) + s2 = 40
s2 = 0
B: 12(4) + 4(3) = 60
s1 = 0
4(4) + 8(3) + s2 = 40
s2 = 0
C: 12(5) + 4(0) + s1 = 60
s1 = 0
4(5) + 8(0) + s2 = 40
s2 = 20
b) The constraint line 12x1 + 4x2 = 60 would move
inward resulting in a new location for point B
at x1 = 2, x2 = 4, which would still be optimal.
c) In order for the optimal solution point to change
from B to A the slope of the objective function
must be at least as flat as the slope of the
constraint line, 4x1 + 8x2 = 40, which is −1/2.
Thus, the profit for product B would have to be,
−9/c2 = −1/2
c2 = 18
If the profit for product B is increased to $15
the optimal solution point will not change,
although Z would change from $57 to $81.
If the profit for product B is increased to $20
the solution point will change from B to A,
x1 = 0, x2 = 5, Z = $100.
12. a) For c1 the upper limit is computed as,
−c1/7 = −3
c1 = 21
and the lower limit is,
−c1/7 = −1/2
c1 = 3.50
For c2 the upper limit is,
−9/c2 = −1/2
c2 = 18
and the lower limit is,
−9/c2 = −3
c2 = 3
Summarizing,
3.50 ≤ c1 ≤ 21
3 ≤ c2 ≤ 18
b)
Z = 57.000
Variable Value
Reduced
Cost
x1 4.000 0.000
x2 3.000 0.000
Constraint Slack/Surplus
Shadow
Price
c1 0.000 0.550
c 2 0.000 0.600
Objective Coefficient Ranges
Variables
Lower
Limit
Current
Values
Upper
Limit
Allowable
Increase
Allowable
Decrease
x1 3.500 9.000 21.000 12.000 5.500
x2 3.000 7.000 18.000 11.000 4.000
Right Hand Side Ranges
Constraints
Lower
Limit
Current
Values
Upper
Limit
Allowable
Increase
Allowable
Decrease
c1 20.000 60.000 120.000 60.000 40.000
c 2 20.000 40.000 120.000 80.000 20.000
c) The shadow price for line 1 time is $0.55 per
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3-5
hour, while the shadow price for line 2 time is
$0.60 per hour. The company would prefer to
obtain more line 2 time since it would result in
the greater increase in profit.
13. a) x1 = no. of yards of denim
x2 = no. of yards of corduroy
maximize Z = $2.25x1 + 3.10x2
subject to
5.0x1 + 7.5x2 ≤ 6,500
3.0x1 + 3.2x2 ≤ 3,000
x2 ≤ 510
x1, x2 ≥ 0
b) maximize Z = $2.25x1 + 3.10x2 + 0s1 + 0s2 + 0s3
subject to
5.0x1 + 7.5x2 + s1 = 6,500
3.0x1 + 3.2x2 + s2 = 3,000
x2 + s3 = 510
x1, x2, s1, s2, s3 ≥ 0
14.
a) 5.0(456) + 7.5(510) + s1 = 6,500
s1 = 6,500 − 6,105
s1 = 395 lbs.
3.0(456) + 3.2(510) + s2 = 3,000
s2 = 0 hrs.
510 + s3 = 510
s3 = 0
Therefore demand for corduroy is met.
b) In order for the optimal solution point to
change from B to C the slope of the objective
function must be at least as great as the slope
of the constraint line, 3.0x1 + 3.2x2 = 3,000,
which is −3/3.2. Thus, the profit for denim
would have to be,
−c1/3.0 = −3/3.2
c1 = 2.91
If the profit for denim is increased from $2.25 to
$3.00 the optimal solution would change to point
C, where x1 = 1,000, x2 = 0, Z = 3,000.
Profit for corduroy has no upper limit that
would change the optimal solution point.
c) The constraint line for cotton would move
inward as shown in the following graph where
point C is optimal.
15. Z = 2,607.000
Variable Value
Reduced
Cost
x1 456.000 0.000
x2 510.000 0.000
Constraint Slack/Surplus
Shadow
Price
c1 395.000 0.000
c2 0.000 0.750
c3 0.000 0.700
hour, while the shadow price for line 2 time is
$0.60 per hour. The company would prefer to
obtain more line 2 time since it would result in
the greater increase in profit.
13. a) x1 = no. of yards of denim
x2 = no. of yards of corduroy
maximize Z = $2.25x1 + 3.10x2
subject to
5.0x1 + 7.5x2 ≤ 6,500
3.0x1 + 3.2x2 ≤ 3,000
x2 ≤ 510
x1, x2 ≥ 0
b) maximize Z = $2.25x1 + 3.10x2 + 0s1 + 0s2 + 0s3
subject to
5.0x1 + 7.5x2 + s1 = 6,500
3.0x1 + 3.2x2 + s2 = 3,000
x2 + s3 = 510
x1, x2, s1, s2, s3 ≥ 0
14.
a) 5.0(456) + 7.5(510) + s1 = 6,500
s1 = 6,500 − 6,105
s1 = 395 lbs.
3.0(456) + 3.2(510) + s2 = 3,000
s2 = 0 hrs.
510 + s3 = 510
s3 = 0
Therefore demand for corduroy is met.
b) In order for the optimal solution point to
change from B to C the slope of the objective
function must be at least as great as the slope
of the constraint line, 3.0x1 + 3.2x2 = 3,000,
which is −3/3.2. Thus, the profit for denim
would have to be,
−c1/3.0 = −3/3.2
c1 = 2.91
If the profit for denim is increased from $2.25 to
$3.00 the optimal solution would change to point
C, where x1 = 1,000, x2 = 0, Z = 3,000.
Profit for corduroy has no upper limit that
would change the optimal solution point.
c) The constraint line for cotton would move
inward as shown in the following graph where
point C is optimal.
15. Z = 2,607.000
Variable Value
Reduced
Cost
x1 456.000 0.000
x2 510.000 0.000
Constraint Slack/Surplus
Shadow
Price
c1 395.000 0.000
c2 0.000 0.750
c3 0.000 0.700
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3-6
Objective Coefficient Ranges
Variables Lower Limit Current Values Upper Limit Allowable Increase Allowable Decrease
x1 0.000 2.250 2.906 0.656 2.250
x2 2.400 3.100 No limit No limit 0.700
Right Hand Side Ranges
Constraints Lower Limit Current Values Upper Limit Allowable Increase Allowable Decrease
c1 6,015.000 6,500.000 No limit No limit 395.000
c2 1,632.000 3,000.000 3,237.000 237.000 1,368.000
c3 0.000 510.000 692.308 182.308 510.000
a) The company should select 237 additional
hours of processing time, with a shadow price
of $0.75 per hour. Cotton has a shadow price
of $0 because there is already extra (slack)
cotton available and not being used, so any
more would have no marginal value.
b) 0 ≤ c1 ≤ 2.906 6,105 ≤ q1 ≤ ∞
2.4 ≤ c2 ≤ ∞ 1,632 ≤ q2 ≤ 3,237
0 ≤ q3 ≤ 692.308
The demand for corduroy can decrease to zero
or increase to 692.308 yds. without changing
the current solution mix of denim and
corduroy. If the demand increases beyond
692.308 yds., then denim would no longer be
produced and only corduroy would be
produced.
16. x1 = no. of days to operate Mill 1
x2 = no. of days to operate Mill 2
minimize Z = 6,000x1 + 7,000x2
subject to
6x1 + 2x2 ≥ 12
2x1 + 2x2 ≥ 8
4x1 + 10x2 ≥ 5
x1,x2 ≥ 0
17.
a) 6(4) + 2(0) − s1 = 12
s1 = 12
2(4) + 2(0) − s2 = 8
s2 = 0
4(4) + 10(0) − s3 = 5
s3 = 11
b) The slope of the objective
function, −6,000/7,000, must become flatter
(i.e., less) than the slope of the constraint line,
2x1 + 2x2 = 8, for the solution to
change. The cost of operating Mill 1, c1, that
would change the solution point is,
−c1/7,000 = −1
c1 = 7,000
Since $7,500 > $7,000, the solution point will
change to B, where x1 = 1, x2 = 3, Z = $28,500.
c) If the constraint line for high-grade aluminum
changes to 6x1 + 2x2 = 10, it moves inward but
does not change the optimal variable mix. B
remains optimal but moves to a new location, x1
= 0.5, x2 = 3.5, Z = $27,500.
18. Z = 24,000
Variable Value
x1 4.000
x2 0.000
Constraint Slack/Surplus
Shadow
Price
c1 12.000 0.000
c2 0.000 −3,000.000
c3 11.000 0.000
Objective Coefficient Ranges
Variables Lower Limit Current Values Upper Limit Allowable Increase Allowable Decrease
x1 0.000 2.250 2.906 0.656 2.250
x2 2.400 3.100 No limit No limit 0.700
Right Hand Side Ranges
Constraints Lower Limit Current Values Upper Limit Allowable Increase Allowable Decrease
c1 6,015.000 6,500.000 No limit No limit 395.000
c2 1,632.000 3,000.000 3,237.000 237.000 1,368.000
c3 0.000 510.000 692.308 182.308 510.000
a) The company should select 237 additional
hours of processing time, with a shadow price
of $0.75 per hour. Cotton has a shadow price
of $0 because there is already extra (slack)
cotton available and not being used, so any
more would have no marginal value.
b) 0 ≤ c1 ≤ 2.906 6,105 ≤ q1 ≤ ∞
2.4 ≤ c2 ≤ ∞ 1,632 ≤ q2 ≤ 3,237
0 ≤ q3 ≤ 692.308
The demand for corduroy can decrease to zero
or increase to 692.308 yds. without changing
the current solution mix of denim and
corduroy. If the demand increases beyond
692.308 yds., then denim would no longer be
produced and only corduroy would be
produced.
16. x1 = no. of days to operate Mill 1
x2 = no. of days to operate Mill 2
minimize Z = 6,000x1 + 7,000x2
subject to
6x1 + 2x2 ≥ 12
2x1 + 2x2 ≥ 8
4x1 + 10x2 ≥ 5
x1,x2 ≥ 0
17.
a) 6(4) + 2(0) − s1 = 12
s1 = 12
2(4) + 2(0) − s2 = 8
s2 = 0
4(4) + 10(0) − s3 = 5
s3 = 11
b) The slope of the objective
function, −6,000/7,000, must become flatter
(i.e., less) than the slope of the constraint line,
2x1 + 2x2 = 8, for the solution to
change. The cost of operating Mill 1, c1, that
would change the solution point is,
−c1/7,000 = −1
c1 = 7,000
Since $7,500 > $7,000, the solution point will
change to B, where x1 = 1, x2 = 3, Z = $28,500.
c) If the constraint line for high-grade aluminum
changes to 6x1 + 2x2 = 10, it moves inward but
does not change the optimal variable mix. B
remains optimal but moves to a new location, x1
= 0.5, x2 = 3.5, Z = $27,500.
18. Z = 24,000
Variable Value
x1 4.000
x2 0.000
Constraint Slack/Surplus
Shadow
Price
c1 12.000 0.000
c2 0.000 −3,000.000
c3 11.000 0.000
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3-7
Objective Coefficient Ranges
Variables Lower Limit Current Values Upper Limit Allowable Increase Allowable Decrease
x1 0.000 6,000.000 7,000.000 1,000.000 6,000.000
x2 6,000.000 7,000.000 No limit No limit 1,000.000
Right Hand Side Ranges
Constraints Lower Limit Current Values Upper Limit Allowable Increase Allowable Decrease
c1 No limit 12.000 24.000 12.000 No limit
c2 4.000 8.000 No limit No limit 4.000
c3 No limit 5.000 16.000 11.000 No limit
a) There is surplus high-grade and low-grade
aluminum so the shadow price is $0 for both.
The shadow price for medium-grade
aluminum is $3,000, indicating that for every
ton that this constraint could be reduced, cost
will decrease by $3,000.
b) 0 ≤ c1 ≤ 7,000 ∞ ≤ q1 ≤ 24
6,000 ≤ c2 ≤ ∞ 4 ≤ q2 ≤ ∞
∞ ≤ q3 ≤ 16
c) There will be no change.
19. x1 = no. of acres of corn
x2 = no. of acres of tobacco
maximize Z = 300x1 + 520x2
subject to
x1 + x2 ≤ 410
105x1 + 210x2 ≤ 52,500
x2 ≤ 100
x1,x2 ≥ 0
20.
a) x1 = 320, x2 = 90
320 + 90 + s1 = 410
s1 = 0 acres uncultivated
90 + s3 = 100
s3 = 10 acres of tobacco
allotment unused
b) At point D only corn is planted. In order for
point D to be optimal the slope of the
objective function will have to be at least as
great (i.e., steep) as the slope of the constraint
line, x1 + x2 = 410, which is −1. Thus, the
profit for corn is computed as,
−c1/520 = −1
c1 = 520
The profit for corn must be greater than $520 for
the Bradleys to plant only corn.
c) If the constraint line changes from
x1 + x2 = 410 to x1 + x2 = 510, it will move
outward to a location which changes the
solution to the point where 105x1 +
210x2 = 52,500 intersects with the axis. This
new point is x1 = 500, x2 = 0, Z = $150,000.
d) If the constraint line changes from x1 +
x2 = 410 to x1 + x2 = 360, it moves inward to a
location which changes the solution point to
the intersection of x1 + x2 = 360 and 105x1 +
210x2 = 52,500. At this point x1 = 260,
x2 = 100, and Z = $130,000.
21. Z = 142,800.000
Variable Value
x1 320.000
x2 90.000
Constraint Slack/Surplus
Shadow
Price
c1 0.000 80.000
c2 0.000 2.095
c3 10.000 0.000
Objective Coefficient Ranges
Variables Lower Limit Current Values Upper Limit Allowable Increase Allowable Decrease
x1 0.000 6,000.000 7,000.000 1,000.000 6,000.000
x2 6,000.000 7,000.000 No limit No limit 1,000.000
Right Hand Side Ranges
Constraints Lower Limit Current Values Upper Limit Allowable Increase Allowable Decrease
c1 No limit 12.000 24.000 12.000 No limit
c2 4.000 8.000 No limit No limit 4.000
c3 No limit 5.000 16.000 11.000 No limit
a) There is surplus high-grade and low-grade
aluminum so the shadow price is $0 for both.
The shadow price for medium-grade
aluminum is $3,000, indicating that for every
ton that this constraint could be reduced, cost
will decrease by $3,000.
b) 0 ≤ c1 ≤ 7,000 ∞ ≤ q1 ≤ 24
6,000 ≤ c2 ≤ ∞ 4 ≤ q2 ≤ ∞
∞ ≤ q3 ≤ 16
c) There will be no change.
19. x1 = no. of acres of corn
x2 = no. of acres of tobacco
maximize Z = 300x1 + 520x2
subject to
x1 + x2 ≤ 410
105x1 + 210x2 ≤ 52,500
x2 ≤ 100
x1,x2 ≥ 0
20.
a) x1 = 320, x2 = 90
320 + 90 + s1 = 410
s1 = 0 acres uncultivated
90 + s3 = 100
s3 = 10 acres of tobacco
allotment unused
b) At point D only corn is planted. In order for
point D to be optimal the slope of the
objective function will have to be at least as
great (i.e., steep) as the slope of the constraint
line, x1 + x2 = 410, which is −1. Thus, the
profit for corn is computed as,
−c1/520 = −1
c1 = 520
The profit for corn must be greater than $520 for
the Bradleys to plant only corn.
c) If the constraint line changes from
x1 + x2 = 410 to x1 + x2 = 510, it will move
outward to a location which changes the
solution to the point where 105x1 +
210x2 = 52,500 intersects with the axis. This
new point is x1 = 500, x2 = 0, Z = $150,000.
d) If the constraint line changes from x1 +
x2 = 410 to x1 + x2 = 360, it moves inward to a
location which changes the solution point to
the intersection of x1 + x2 = 360 and 105x1 +
210x2 = 52,500. At this point x1 = 260,
x2 = 100, and Z = $130,000.
21. Z = 142,800.000
Variable Value
x1 320.000
x2 90.000
Constraint Slack/Surplus
Shadow
Price
c1 0.000 80.000
c2 0.000 2.095
c3 10.000 0.000
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3-8
Objective Coefficient Ranges
Variables Lower Limit Current Values Upper Limit Allowable Increase Allowable Decrease
x1 260.000 300.000 520.000 220.000 40.000
x2 300.000 520.000 600.000 80.000 220.000
Right Hand Side Ranges
Constraints Lower Limit Current Values Upper Limit Allowable Increase Allowable Decrease
c1 400.000 410.000 500.000 90.000 10.000
c2 43,050.000 52,500.000 53,550.000 1,050.000 9,450.000
c3 90.000 100.000 No limit No limit 10.000
a) No, the shadow price for land is $80 per acre
indicating that profit will increase by no more
than $80 for each additional acre obtained. The
maximum price the Bradleys should pay is $80
and the most they should obtain is at the upper
limit of the sensitivity range for land. This limit
is 500 acres, or 90 additional acres. Beyond 90
acres the shadow price would change.
b) The shadow price for the budget is $2.095.
Thus, for every $1 borrowed they could
expect a profit increase of $2.095. If they
borrowed $1,000 it would change the amount
of corn and tobacco they plant; x1 = 310.5
acres of corn and x2 = 99.5 acres of tobacco.
22. x1 = no. of sausage biscuits
x2 = no. of ham biscuits
maximize Z = .60x1 + .50x2
subject to
.10x1 ≤ 30
.15x2 ≤ 30
.04x1 + .04x2 ≤ 16
.01x1 + .024x2 ≤ 6
x1, x2 ≥ 0
23.
a) x1 = 300, x2 = 100, Z = $230
.10(300) + s1 = 30
s1 = 0 leftover sausage
.15(100) + s2 = 30
s2 = 15 lbs. leftover ham
.01(300) + .024(100) + s4 = 6
s4 = 0.6 hr.
b) The slope of the objective function, −6/5,
must become flatter (i.e., less) than the slope
of the constraint line, .04x1 + .04x2 = 16, for
the solution to change. The profit for ham, c2,
that would change the solution point is,
−0.6/c2 = −1
c2 = .60
Thus, an increase in profit for ham of 0.60
will create a second optimal solution point at
C where x1 = 257, x2 = 143, and Z = $240. (Point
D would also continue to be optimal, i.e.,
multiple optimal solutions.)
c) A change in the constraint line from .04x1 +
.04x2 = 16 to .04x1 + .04x2 = 18 would move the
line outward, eliminating both points C and D.
The new solution point occurs at the intersection
of 0.01x1 + .024x2 = 6 and .10x = 30. This point
is x1 = 300, x2 = 125, and Z = $242.50.
24. Z = 230.000
Variable Value
x1 300.000
x2 100.000
Constraint Slack/Surplus Shadow
Price
c1 0.000 1.000
c2 15.000 0.000
c3 0.000 12.500
c4 0.600 0.000
Objective Coefficient Ranges
Variables Lower Limit Current Values Upper Limit Allowable Increase Allowable Decrease
x1 260.000 300.000 520.000 220.000 40.000
x2 300.000 520.000 600.000 80.000 220.000
Right Hand Side Ranges
Constraints Lower Limit Current Values Upper Limit Allowable Increase Allowable Decrease
c1 400.000 410.000 500.000 90.000 10.000
c2 43,050.000 52,500.000 53,550.000 1,050.000 9,450.000
c3 90.000 100.000 No limit No limit 10.000
a) No, the shadow price for land is $80 per acre
indicating that profit will increase by no more
than $80 for each additional acre obtained. The
maximum price the Bradleys should pay is $80
and the most they should obtain is at the upper
limit of the sensitivity range for land. This limit
is 500 acres, or 90 additional acres. Beyond 90
acres the shadow price would change.
b) The shadow price for the budget is $2.095.
Thus, for every $1 borrowed they could
expect a profit increase of $2.095. If they
borrowed $1,000 it would change the amount
of corn and tobacco they plant; x1 = 310.5
acres of corn and x2 = 99.5 acres of tobacco.
22. x1 = no. of sausage biscuits
x2 = no. of ham biscuits
maximize Z = .60x1 + .50x2
subject to
.10x1 ≤ 30
.15x2 ≤ 30
.04x1 + .04x2 ≤ 16
.01x1 + .024x2 ≤ 6
x1, x2 ≥ 0
23.
a) x1 = 300, x2 = 100, Z = $230
.10(300) + s1 = 30
s1 = 0 leftover sausage
.15(100) + s2 = 30
s2 = 15 lbs. leftover ham
.01(300) + .024(100) + s4 = 6
s4 = 0.6 hr.
b) The slope of the objective function, −6/5,
must become flatter (i.e., less) than the slope
of the constraint line, .04x1 + .04x2 = 16, for
the solution to change. The profit for ham, c2,
that would change the solution point is,
−0.6/c2 = −1
c2 = .60
Thus, an increase in profit for ham of 0.60
will create a second optimal solution point at
C where x1 = 257, x2 = 143, and Z = $240. (Point
D would also continue to be optimal, i.e.,
multiple optimal solutions.)
c) A change in the constraint line from .04x1 +
.04x2 = 16 to .04x1 + .04x2 = 18 would move the
line outward, eliminating both points C and D.
The new solution point occurs at the intersection
of 0.01x1 + .024x2 = 6 and .10x = 30. This point
is x1 = 300, x2 = 125, and Z = $242.50.
24. Z = 230.000
Variable Value
x1 300.000
x2 100.000
Constraint Slack/Surplus Shadow
Price
c1 0.000 1.000
c2 15.000 0.000
c3 0.000 12.500
c4 0.600 0.000
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Subject
Mathematics