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Solutions Manual For Introduction to Management Science 13th Edition By Bernard W. Taylor (All Chapters, 100% Original Verified, A+ Grade) $28.49   Add to cart

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Solutions Manual For Introduction to Management Science 13th Edition By Bernard W. Taylor (All Chapters, 100% Original Verified, A+ Grade)

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Solutions Manual For Introduction to Management Science 13th Edition By Bernard W. Taylor (All Chapters, 100% Original Verified, A+ Grade) Solutions Manual For Introduction to Management Science 13e By Bernard W. Taylor (All Chapters, 100% Original Verified, A+ Grade)

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  • February 7, 2024
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  • Introduction to Management Science 13e Bernard W.
  • Introduction to Management Science 13e Bernard W.
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Copyright © 201 9 Pearson Education, Inc. 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 chang e (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 ; profit analysis
21.Break -even analysis ; indifference (1−20)
22.Break -even analysis
23.Break -even analysis; volume and
price analysis
24.Break -even analysis
25.Break -even analysis
26.Break -even analysis; prof it analysis
27.Break -even analysis; price and volume analysis
28.Break -even analysis; profit analysis
29.Break -even analysis; profit analysis
30.Break -even analysis; profit analysis
31.Break -even analysis
32.Multiproduct b reak-even analysis33.Decision analysis
34.Expected value
35.Linear programming
36.Linear programming
37.Linear programming
38.Linear programming
39.Forecasting/statistics
40.Linear programming
41.Waiting lines
42.Shortest route
PROBLEM SOLUTIONS 1.a)
==
==
= + = + =
= = =
= − =f
v
fv300, $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 monthvc
cp
c vc
vp
Z
b) f
v8,00069.56 tables per month180 65cvpc= = =−−
2.a)
fv
fv12,000, $18,000, $0.90,
$3.20;
TC
18,000 (12,000)(0.90)
$28,800;
TR (12,000)($3.20) $38, 400;
$38, 400 28,800 $9,600 per yearv c c
p
c vc
vp
Z= = =
=
=+
=+
=
= = =
= − =
b) f
v18,0007,8263.20 0.90= = =−−cvpc
3.a)
= = =
=
= + = + =
= = =
= − =−fv
fv18,000, $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.
= = =
= = =−−fv
f
v$25,000, $.40, $.15,
25,000100,000 lb per month.40 .15c p c
cvpc
f
v21,00024,705.88 yd per month1.30 .45cvpc= = =−−Introduction to Management Science 13e Bernard W. Taylor (Solutions Manual All Chapters, 100% Original Verified, A+ Grade) All Chapters/
Supplement files download link at the end of this file. 1-2 Copyright © 201 9 Pearson Education, Inc. 5. 6. 7. −−f
v$25,000= = = 1,250 dolls30 10cvpc 8. Break-even volume as percentage of capac ity
7,826.652 65.2%12,000= = = =v
k 9. = = = =Break-even volume as percentage of capac ity
24,750.88.988 98.8%25,000v
k 10. = = = =Break-even volume as percentage of
100,000capacity .833 83.3%120,000v
k 11. f
v18,0009,729.7cupcakes2.75 0.90
It increases the break-even
volumefrom 7,826 to 9,729.7
per year.cvpc= = =−− 12. = = =−−f
v25,00055,555.55 lb.60 .15
per month; it reduces the break-even
volume from 100,000 lb per month
to 55,555.55 lb.cvpc 13. f
v25,00065,789.47 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.cvpc= = =−− 14. = = =−−f
v39,000102,613.57 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.cvpc 15. fv
f
vInitial 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 dairZ vp c vc
Z vp c
vc= − − = −
− = − − =
= − −
= − − = −
−= y should not
raise its price. 16. −f
v35,000= = = 1,75030–10cvpc 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
v17,0001,062.530 14cvpc Reduces BE point by 187.5 dolls. 18. a) = = =−−f
v$27,0005,192.30 pizzas8.95 3.75cvpc b) =5,192.3259.6 days20 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$23,880New 5,685.7 pizzas7.95 3.75cvpc 1-3 Copyright © 201 9 Pearson Education, Inc. Total break -even volume = 600 + 5,685.7 = 6,285.7 pizzas 5,685.7Total time to break-even 3020
314.3 days=+
= 19. a) Cost of Regular pla n = $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 + .33 x = .25 x − 200 x = 937.50 minutes per month or 15.63 hrs. 20. cf = $26,000 cv = $0.67 ($5.36/8 = 0.67) p = $3.75 =−26,000
3.75 0.67v = 8,442 slices Forecasted annual demand = (540)(52) = 28,080 Z = $ 91,260 – 44,813.6 = 46,446.4 21. OLD New 26,000 + (.67) v = 30,000 + (.48) v .19v = 4,000 v = 21,053 slices Z = New profit – old profit Z = $47,781.60 – 46,446.40 = $1,335.20 Purchase equipment 22. a) =−7,50014,000.35p 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 an d/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 23. a) cf = $1,700 cv = $12 per pupil p = $75 =−1,700
75 12v = 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 = 60 p − $1,700 − 60(12) 60p = 7,420 p = $123.67 24. a) cf = $350,000 cv = $12,000 p = $18,000 =−f
vcvpc =−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) = 105,00 0 This is approximately the same as the profit for 75 students and a lower tuition in part (b). 25. p = $400 cf = $8,000 cv = $75 Z = $60,000 +=−f
vZcvpc +=−60,000 8,000
400 75v v = 209.23 teams 26. 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 1-4 Copyright © 201 9 Pearson Education, Inc. With volume doubled to 700: Profit ( Z) = (2 ,700)(700) – 875,000 – (700)(200) = $875,000 27. 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 The volume to achieve the goal of $150,000 is, v = (Z + cf)/(p − cv) = (150 ,000 + 100 ,000)/(6 − .695) = 47,125 28. a) Monthly fixed cost ( cf) = cost of van/60 months + labor (driver)/month = (21,500/60) + (3 0.42 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) + (1 2.4 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 29. a) ==−−f
v500
30 14cvpc v = 31.25 jobs b) (8 weeks)(6 d ays/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) 30. a) ==−−f
v700
35 3cvpc v = 21.88 job s 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 31. cf = $7,500 Monthly cf = ($2,300)(12) = $27,600 Total cf = $35,100 cv = 0 p = $0.24 fc
vp35,100= = = 146,250 hits per year.24 v = 12,188 hits per month $45,000 = v(.24) – (12)(3,500) – (0)v .24v = 87,000 v = 362,500 v = 30,208 hits per month 32. This is a “multiproduct” break -even problem. The formula for the break -even volume is, =
   −       vTotal fixed cost
weighted average weighted average
selling price variable cost =+ − +18,000
[(3.20)(.70) (2.50)(.30)] [(.90)(.70) (.45) (.30)]v v = 8,089.89 units

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