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Instructors Solutions Manual for Additional Problems

Operations Management

E I G H T H E D I T I O N

Principles of Operations Management

S I X T H E D I T I O N

Upper Saddle River, New Jersey 07458

JAY HEIZER

Texas Lutheran University

BARRY RENDER

Rollins College

VP/Editorial Director: Jeff Shelstad Executive Editor: Mark Pfaltzgraff Senior Managing Editor: Alana Bradley Senior Editorial Assistant: Jane Avery Copyright 2006 by Pearson Education, Inc., Upper Saddle River, New Jersey, 07458. Pearson Prentice Hall. All rights reserved. Printed in the United States of America. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department. Pearson Prentice HallTM is a trademark of Pearson Education, Inc.

10 9 8 7 6 5 4 3 2 1

iii

Contents

Homework Problem Answers Chapter 1 Operations and Productivity ........................................................................... A-1 Chapter 3 Project Management ....................................................................................... A-3 Chapter 4 Forecasting ...................................................................................................... A-7 Chapter 5 Design of Goods and Services ...................................................................... A-11 Chapter 6 Managing Quality ......................................................................................... A-15 Supplement 6: Statistical Process Control ............................................................................ A-18 Chapter 7 Process Strategy ............................................................................................ A-20 Supplement 7: Capacity Planning ......................................................................................... A-23 Chapter 8 Location Strategies ....................................................................................... A-27 Chapter 9 Layout Strategy ............................................................................................. A-30 Supplement 10: Work Measurement ...................................................................................... A-34 Chapter 12 Inventory Management ................................................................................. A-36 Chapter 13 Aggregate Planning ...................................................................................... A-42 Chapter 14 Materials Requirements Planning (MRP) & ERP ........................................ A-46 Chapter 15 Short-Term Scheduling ................................................................................. A-51 Chapter 16 Just-In-Time and Lean Production Systems ................................................. A-55 Chapter 17 Maintenance and Reliability ......................................................................... A-57 Module A: Decision Making Tools ................................................................................ A-59 Module B: Linear Programming ..................................................................................... A-64 Module C: Transportation Modeling .............................................................................. A-70 Module D: Waiting Line Models .................................................................................... A-75 Module E: Learning Curves ........................................................................................... A-79 Module F: Simulation ..................................................................................................... A-80

A-1

1 CHAPTER

Operations and Productivity

1.1 a.

( )( ) ( )( )( )( ) ( )( )

Last yearsnumber of units of output

total factortotal dollar value of all inputs used

productivity

12,000 units

12,000 $2.00 14,000 $10.50

2,000 $8.00 4,000 $0.70 $30,000

12,000 units

$219,800

=

=

+ ++ +

= 0.0546 units dollar=

b.

( ) ( ) ( )( )( )( ) ( )( )

This yearsnumber of units of output

total factortotal dollar value of all inputs used

productivity

14,000 units

14,000 $2.05 16,000 $11.00

1,800 $7.50 3,800 $0.75 $26,000

14,000 units

$247,050

=

=

+ ++ +

= 0.0567 units dollar=

c.

This years Last years

total factor total factor

productivity productivity 0.0567 0.0546100% 100%

Last years 0.0546

total factor

productivity

3.84% 3.8%

=

= +

Answer : Total factor productivity increased by 3.798% this year as compared to last year.

A-2

1.2 ( ) ( )( )57,600

0.15160 12 L

= , where numberL = of laborers employed at the plant.

So ( )( )( )57,600

200160 12 0.15

L = =

Answer : 200 laborers

1.3 Output 28,000 customers= There are 4 approaches to solving the problem correctly:

1. Input 7 workers=

Then, 28,000

4,000 customers worker7

=

2. ( )Input 7 40 labor weeks= Then, ( )

28,000100 customers labor week

7 40=

3. ( )( )Input 7 40 50 labor hours= Then, ( )( )

28,0002 customers labor hour

7 40 50=

4. ( )( )Input 7 40 $250 dollars of worker wages= Then, ( )( )

28,0000.40 customers per dollar of labor

7 40 $250=

1.4 ( )( )

6,600 Cadillacs0.10

labor hours

66,000 labor hours

x

x

=

=

There are 300 laborers. So,

66,000 labor hours

220 labor hours laborer300 laborers

=

1.5 ( ) ( )

( )52 $90 198 $80$ output 20,520

$57.00labor hour 8 45 360

+= = = per labor hour

A-3

3 CHAPTER

Project Management

3.1

50 100 150 200

A

B

C

D

E

F

G

H

I

Hours

80

150

200

Gantt Chart

20

120

110

140

170

160

3.2 AON Network:

60

BPurchasing

30

DSawing

20

APlanning

100

CExcavation

20

EPlacement

10

FAssembly

20

GInfill

10

HOutfill

30

IDecoration

A-4

3.3 AOA Network:

Plan1

Purchase2

Saw3

Place4

Assemble5

Outfill6

Decorate8 9

7

Excavate

Infill DummyA B D E

C

F

G

H

I

3.4 Path Task Times (Hours) Total Hours 1 2 3 4 5 6 7 8 9

1 2 3 4 5 6 8 9 1 2 4 5 6 7 8 9 1 2 4 5 6 8 9

20 + 60 + 30 + 20 + 10 + 20 + 0 + 30 20 + 60 + 30 + 20 + 10 + 10 + 30 20 + 100 + 20 + 10 + 20 + 0 + 30 20 + 100 + 20 + 10 + 10 + 30

190 180 200 190

The longest path clearly is 1 2 4 5 6 7 8 9; hence, this is the critical path, and the

project will end after 200 hours.

Planning1

Excavate

Purchasing2

LF = 90LS = 30EF = 80ES = 20

Sawing3

LF = 120LS = 90EF = 110ES = 80

Placement4

LF = 140LS = 120EF = 140ES = 120

LF = 20LS = 0EF = 20ES = 0

Assembly5

LF = 150LS = 140EF = 150ES = 140

Outfill6 8

7

LF = 170LS = 160EF = 160ES = 150

Infill DummyLF = 170LS = 150EF = 170ES = 150

Decoration9

LF = 200LS = 170EF = 200ES = 170

LF = 120LS = 20EF = 120ES = 20

A

B

C

D E

F G H I

Answer : The critical path therefore is A C E F G I (200 hours). The activities that

can be delayed include ones with slack times > 0. Thus, B (10 hours), D (10 hours), and H (10 hours) can be delayed.

A-5

3.5

( )( )( )( )

2

2

2

2

2

4Mean: Variance Standard Deviation

6 6 6

20120 20A : 20 A : 11.11 A : 3.33

6 36 660360 60

B : 60 B: 100.00 B : 10.006 36 6

120600 120C : 100 C : 400.00 C : 20.00

6 36 610180 10

D : 30 D : 2.78 D : 1.676 36 6

a m b b a b a+ +

= = =

= = =

= = =

= = =

( )( )( )( )( )

2

2

2

2

2

10120 10E : 20 E : 2.78 E : 1.67

6 36 6060 0

F : 10 F : 0.00 F : 0.006 36 6

40120 40G : 20 G : 44.44 G : 6.67

6 36 6460 4

H : 10 H : 0.44 H : 0.676 36 6

40180 40I : 30 I : 44.44 I : 6.67

6 36 6

= = =

= = =

= = =

= = =

= = =

3.6 Since the critical path is A C E F G I, only those variances are along the critical

path are used. Therefore, the variances along critical path are 11.11, 400, 2.78, 0, 44.44, and 44.44 . So

the sum of these variances 502.77= .

Thus, the project completion standard deviation 502.77 22.4= .

= mean time of critical path 200 hrs= 22.4 hrs =

The z value 240 200 40

1.822.4 22.4

= = = . Using the cumulative normal distribution table in

Appendix I of the text, we observe that 96.4 percent of the distribution lies to the left of 1.8 standard deviations. Hence, there is a 100 96.4 3.6% = chance that it will take more than 240 hrs to build the garden/picnic area.

A-6

3.7 The critical path is A C E F G I. Hence, the project completion variance 11.11 400 2.78 0 44.44 44.44 502.77.= + + + + + =

So, the project completion standard deviation 502.77 22.4= . The cumulative normal distribution tells us that 90% of the area lies to the left of 1.29

standard deviations. Therefore, amount of time to build the garden/picnic area should be ( )200 22.4 1.29 200 29 229 hours+ = + = .

3.8 a. Activity on Nodes Diagram of the project.

A1

B1

C4

E2

F2

b. The critical path, listing all critical activities in chronological order:

( )A B E F 1 1 2 2 6 A C F 1 4 2 7. This is the CP.

not CP + + + = + + =

c. The project duration (in weeks): 7 (This is the length of CP.) d. The slack (in weeks) associated with any and all non-critical paths through the

project: Look at the paths that arent criticalonly 1 hereso from above: A B E F 7 6 1 = week slack.

3.9 We have only 1 activity with probabilistic duration.

( )8 1 4 2Due date 1

20.5 0.5

Z

+ +

= = = = (length of entire path is 7, not 4). For a 2z = ,

this means ( )Due date 8 97.72%P < = (table lookup) for the path so chance of being OVER 8 weeks is 2.28% (and we know non-CP path will be only 6 weeks)

3.10 Helps to modify the AON with the lowest costs to crash

1. CP is A C F ; C is cheapest to crash, so take it to 3 wks at $200. (and $200 < $250) 2. Now both paths through are critical. We would need to shorten A or F, or shorten C

and either B/E. This is not worth it, so we would not bother to crash any further.

A-7

4 CHAPTER

Forecasting

4.1 ( )Present period week 6.= So: ( ) ( ) ( ) ( )7 6 5 4 31 1 1 1 1 1 1 152 63 48 703 4 4 6 3 4 4 6F A A A A= + + + = + + + 56.75 patients= 4.2

1 120 2 136 3 114 1284 116 125

t tt A F

120 136 256128

2 2 Checking Data136 114 250

1252 2

+ = =

+

= =

5116 114 230

115 Answer2 2

F+

= = = =

4.3 Method 1: MAD : 0.20 0.05 0.05 0.20 0.5000 better+ + + = MSE : 0.04 0.0025 0.0025 0.04 0.0850+ + + = Method 2: MAD : 0.1 0.20 0.10 0.11 0.5100+ + + =

MSE : 0.01 0.04 0.01 0.0121 0.0721 better+ + + =

A-8

4.4 y a bx= +

4

1

2

1

58,538

75.75

191.5

23, 209

i ii

n

ii

x y

x

y

x

=

=

=

=

=

=

( ) ( )( )

( )2

58,538 4 75.75 191.5 513.502

256.7523, 209 4 75.75

191.5 2 75.75 40

40 2

85

210

b

a

y x

x

y

= = =

= =

+

=

4.5

t

Day Actual

Demand ForecastDemand

1 Monday 88 88 2 Tuesday 72 88 3 Wednesday 68 84 4 Thursday 48 80 5 Friday 72 Answer

( )1 1t t tF A F + = + . Let 14 = . Let Monday forecast demand = 88

( ) ( )( ) ( )( ) ( )( ) ( )

2

3

4

5

1 388 88 88

4 41 3

72 88 18 66 844 41 3

68 84 17 63 804 41 3

48 80 12 60 724 4

F

F

F

F

= + =

= + = + =

= + = + =

= + = + =

A-9

4.6 Winter Spring Summer Fall2001 1, 400 1,500 1,000 6002002 1, 200 1,400 2,100 7502003 1,000 1,600 2,000 6502004 900 1,500 1,900 500

4,500 6,000 7,000 2,500

Average over all seasons: 20,000

1,25016

=

Average over spring: 6,000

1,5004

=

Spring index: 1,500

1.21, 250

=

( )5,600Answer : 1.2 1,6804

=

sailboats

4.7 We need to find the smoothing constant . We know in general that ( )1 1t t tF A F + = + ,

1, 2, 3t = . Choose either 2t = or 3t = ( 1t = wont let us find because

( ) ( )2 50 50 1 50F = = + holds for any ). Lets pick, e.g., 2t = . Then ( ) ( )3 48 42 1 50F = = + . So

48 42 50 50

2 8

1.

4

= +

=

=

Now we can find 5F : ( ) ( )5 46 1 50F = + , with 14 = . So ( ) ( )5 1 346 50 49 Answer4 4F = + =

A-10

4.8 Let 1 2 6, , , X X X be the prices; 1 2 6, , , Y Y Y be the number sold.

6

1 Average price 3.258336

ii

XX == = =

(1)

6

1 Average number sold 550.006

ii

YY == = =

(2)

All calculations to the

1nearest th

100,000

6

1

9,783.00i ii

X Y=

= (3)

6

2

1

67.1925ii

X=

= (4) Then y a bx + , where number soldy = , pricex = , and

( ) ( )( )

( ) ( )( )( )

( ) ( )

6

16 2

22

1

9,783 6 3.25833 550 969.489277.61395

3.4922267.1925 6 3.25833

1,454.5578

i ii

ii

X Y n X Yb

X n X

a Y b X

=

=

= = = =

= =

So at 1.80x = , ( )1,454.5578 277.61395 1.80 954.85270y = = . Now round to the nearest

integer: Answer : 955 dinners

4.9 Tracking Signal ( )

1

MAD

n

t tt

A F=

=

Month tA tF t tA F ( )t tA F May 100 100 0 0 June 80 104 24 24 July 110 99 11 11 August 115 101 14 14 September 105 104 1 1 October 110 104 6 6 November 125 105 20 20 December 120 109 11 11 Sum: 87 Sum: 39

So: 87

MAD : 10.8758=

39 1

Answer : 3.586 to the nearest th10.875 1,000

=

A-11

5 CHAPTER

Design of Goods and Services

5.1

$27,500

$27,500

Use K1

(0.80)

(0.20)

90 of 100non-defect

70 of 100non-defect

$42,500

$32,500

$4,062.50

(0.85)

(0.15)

90 of 100non-defect

75 of 100non-defect

$12,500

$43,750

Use K2

$24,375

(0.90)

(0.10)

95 of 100non-defect

80 of 100non-defect

$18,750

$75,000

Use K3

Answer: $27,500use K1

Outcome Calculations

( )( )( ) ( )( )( )90 10$100,000 500 300 $1.20 500 300 $1.30

100 100$100,000 $162,000 $19,500 $42,500

+ =

+ =

( )( ) ( )( )70 30$100,000 150,000 $1.20 150,000 $1.30

100 100$100,000 $126,000 $58,500 $32,500

+ =

+ =

A-12

( )( ) ( )( )90 10$130,000 150,000 $1.20 150,000 $1.30

100 100$130,000 $162,000 $19,500 $12,500

+ =

+ =

( )( ) ( )( )75 25$130,000 150,000 $1.20 150,000 $1.30

100 100$130,000 $35,000 $48,750 $43,750

+ =

+ =

( )( ) ( )( )95 5$180,000 150,000 $1.20 150,000 $1.30

100 100$180,000 $171,000 $9,750 $18,750

+ =

+ =

( )( ) ( )( )80 20$180,000 150,000 $1.20 150,000 $1.30

100 100$180,000 $144,000 $39,000 $75,000

+ =

+ =

5.2

84.0 84.0Use D1

(0.4)F market 99.0

(0.6)U market

74.0

66.0

(0.3)F market 80.0

(0.7)U market

60.0

80.2

(0.6)F market 89.2

(0.4)U market

66.7

Use D0

Use D2

(All $ figures in millions in tree)

( )( )( )( )( )( )

$ Profits : D0 F : 1,000 80,000 $80,000,000

D0 U : 750 80,000 $60,000,000

D1 F : 1,000 100,000 1,000,000 $99,000,000D1 U : 750 100,000 1,000,000 $74,000,000

D2 F : 1,000 90,000 800,000 $89, 200,000D2 U : 750 90,000 800

=

=

=

=

=

,000 $66,700,000=

Answer : Answer: Design D1 has an expected profit of $84,000,000.

A-13

5.3

$14,000

$10,000

(0.3)Demand rises $30,000

$20,000Purchase

overhead hoist

(0.5)Demand stays

same

(0.2)Demand falls

$10,000

$14,000

(0.4)Demand rises $20,000

(0.6)Demand stays

same $10,000

Purchaseforklift

$0

Donothing

Answer : Maximum expected payoff $14,000=

5.4

Low demand (0.4)

$380,000Upgrade to D

160K $50,000Use A Low demand (0.4)

$300,000High demand (0.6)

180K $0

$300,000High demand (0.6)Use B

302K

$250,000High demand (0.6)Use C

380K

$0No upgrade to D

Low demand (0.4)

$0Do nothing

Note: K = $1,000s Answer : Use Design C. If demands turns out to be low, upgrade to Design D.

A-14

5.5

Bread & RollsPies & Cakes

Support

Support

No support

No support

Bread & Rolls

Support

No support

Full Service

5.6

Bread & RollsPies & Cakes

$15,000

$10,000

Bread & Rolls

Full Service

Support ( = 0.40)

No support ( = 0.60)EMV = $12,000

$25,000

$5,000

Support ( = 0.40)

No support ( = 0.60)EMV = $13,000

$35,000

$10,000

Support ( = 0.40)

No support ( = 0.60)EMV = $7,500

p

p

p

p

p

p

Based upon this decision tree, Jeff should consider most seriously the medium-sized shop

carrying bread, rolls, pies, and cakes.

A-15

6 CHAPTER

Managing Quality

6.1 1. Appearance of food

2. Portion size 3. Lighting 4. Speed of service 5. Knowledge of server 6. Quality of service 7. Appearance of room 8. Appropriate amount of space 9. View of stage and audio

Item Overall Grade Rated A B C D E 1. 20 28 1 1 0 2. 4 2 30 14 0 3. 19 20 3 8 0 4. 4 5 25 5 11 5. 0 0 27 18 7 6. 9 30 7 0 4 7. 19 18 13 0 0 8. 0 26 24 0 0 9. 0 0 0 20 30 Item Weights Rated 4 3 2 1 0 Total Average 1. 80 84 2 1 0 167 2.61 2. 16 6 60 14 0 96 1.50 3. 76 60 6 8 0 150 2.34 4. 16 15 50 5 0 86 1.34 5. 0 0 54 18 0 72 1.13 6. 36 90 14 0 0 140 2.19 7. 76 54 26 0 0 156 2.44 8. 0 78 48 0 0 126 1.97 9. 0 0 0 20 0 20 0.31

a. Highest rated is appearance of food; 2.61. b. Lowest rated is view of stage; 0.31.

A-16

c. A check sheet will help categorize the comment cards

Check Sheet Positive Negative Appearance of food Portion size ! ! ! ! ! Lighting ! Speed of service ! ! Knowledge of server ! Quality of service ! ! Appearance of room Appropriate amount of space ! View of stage and audio ! ! ! ! ! ! Other ! ! ! ! ! chilly

d. The written comments are not always consistent: Portion size is highly rated in

comments, but 5th in overall grade. View/audio is lowest rated in both. 6.2 a.

8

9

10

11

12

13

14

1 2 3 4 5 6

x

yminutes

Trips0

b. This is a scatter diagram.

A-17

6.3 a.

5

10

15

20

25

30

35

W R I M O

40

02

461014

24

3034

36

b. 39% of complaints are W, demeaning towards women. 6.4

mislabeled

Manpower

Incorrectmeasurement

Operatormisreads display

Inadequatecleanup

Techniciancalculation off

Machines

Temperaturecontrols off

Variability

Antiquatedscales

Inadequateflow controls

Equipmentin disrepair

IncorrectFormulation

Materials

Jars

Incorrectweights

Damagedraw material

instructions

Methods

Lack of clear

Prioritymiscommunication

Incorrectmaintenance

Inadequate instructions

A-18

6 SUPPLEMENT

Statistical Process Control

S6.1 We are given a target of 420X = . So 25

LCL 420 4 40025

X Zn

= = =

.

25UCL 420 4 440

25X Z

n

= + = + =

. Thus,

Answer : LCL 400 calories

UCL 440 calories

=

=

S6.2 7 5 9

250 250 250 7 5 9 300 0.04030 7,500 7,500

p+ + + + + +

= = = =

" "

( ) ( )( ) ( )

1UCL 0.040 3 0.01239 0.077

1LCL 0.040 3 0.01239 0.003

p pp Z

n

p pp Z

n

= + = + =

= = =

S6.3 We want 2Z = , since ( )1 0.0455 0.9545 = which implies 2Z = from the Normal Table.

UCL 2c c= + , where average number of breaks 3: 3 2 3 6.46c = = + = . S6.4 3Z = for -chartx . Here, 4n = so 2 0.729A = (from Table S6.1). 2.0x = , 0.1R = ,

( )2UCL 2.0 0.729 0.1 2.07x x A R= + = + = S6.5 C chart

0.0027

1.0000 0.0027 0.9973

0.49865 32 2

Z

= = = (see Normal Table)

UCL 3 1.5 3 1.5 5.17c c= + = + =

A-19

S6.6 answersx Zn

=

38416 lbs.

24

0.122 0.08

3

16.00 0.08 16.08 UCL

16.00 0.08 15.92 LCLx

x

x

Zn

= =

= =

+ = =

= =

S6.7 1.00x = , 0.10R = , 2 0.483A = (from Table S6.1),

( )( )2LCL 1 0.483 0.10 0.9517x A R= = = weeks S6.8 3.25R = mph, 3Z = , with 8n = , from Table S6.1,

UCL 1.864 6.058

LCL 0.136 0.442

R

R

= =

= =

S6.9

30Number of defects

2501 300 0.04

30 7,500ip == = =

, 250n =

( )( ) ( )( ) ( ) ( )

0.04 0.96UCL 2 0.04 2 0.0124 0.0648

250

0.04 0.96LCL 2 0.04 2 0.0124 0.0152

250

p

p

p

p

= + = + =

= = =

S6.10 a. We are counting attributes and we have no idea how many total observations there are

(the proportion of drivers who werent offended enough to call!) This is a C-chart.

b. Use mean of 6 weeks of observations 36

66= for c , as true c is unknown.

( )UCL 6 3 2.45 13.3c z c= + = + = ( )LCL 6 3 2.45 1.3c z c= = = , or 0. c. It is in control because all weeks calls fall within interval of [ ]0, 13 . d. Instead of using

366

6= , we now use 4c = . ( )UCL 4 3 4 4 3 2 10= + = + = .

( )LCL 4 3 2 2= = , or 0. Week 4 (11 calls) exceeds UCL. Not in control.

A-20

7 CHAPTER

Process Strategy

7.1 a. Find breakeven points, pX .

Mass Customization: 1,260,000 60 120 21,000pX X X+ = =

Intermittent: 1,000,000 70 120 20,000pX X X+ = =

Repetitive: 1,625,000 55 120 25,000pX X X+ = =

Continuous: 1,960,000 50 120 28,000pX X X+ = =

b. Find least-cost process at 24,000 unitsX = . Fixed cost VC Units Mass Customization: ( )1,260,000 60 24,000 2,700,000+ = Intermittent: ( )1,000,000 70 24,000 2,680,000+ = Repetitive: ( )1,625,000 55 24,000 2,945,000+ = Continuous: ( )1,960,000 50 24,000 3,160,000+ = The least-cost process: Intermittent Process. c.

Anticipated IntermittentProduction Process

Volume Breakeven Point

24,000 20,000 ? yes!>#$% #$%

Annual Profit Using Intermittent Process: ( )$ 120 24,000 2,680,000 $200,000 = Answer : The intermittent process will maximize annual profit.

Annual Profit: $200,000

A-21

7.2 Use a crossover chart. First graph. Then solve for breakpoint(s).

5

1

2

3

10 15 20 25

V

RMC

00

1,000s of Ovens

I

P2P1

Cost(Millionsof dollars)

Finding value of P2: ( ) ( )1,250,000 50 P2 2,000,000 5 P2+ = + . So 23P2 16,666 units= .

(Note: P1 12,500= ).

Answer : For volumes of production V such that 2316,666 25,000V . 7.3

2

4

6

8

10

12

14

5,000 15,000

VVolume

10,000 20,000

I RC

I

R

C

Cost(millions)

7,500

00

( )

1,000,000 1,650 3,000,000 1,250

400 2,000,000I&R5,000Intersect

1,000,000 1,650 5,000 $9,250,000

x x

x

x

+ = +

=

= + =

( )

3,000,000 1, 250 7,500,000 650

600 4,500,000R&C7,500Intersect

3,000,000 1, 250 7,500 $12,375,000

x x

x

x

+ = +

=

= + =

For all V between 5,000 7,500V

A-22

7.4 Breakeven points a. : 21,000,000 450 750 70,000R x x x+ = = : 26, 250,000 400 750 75,000C x x x+ = = : 15,000,000 500 750 60,000M x x x+ = = b. Least cost process at 65,000x = Cost R: $50,250,000 C: $52,250,000 M: $47,500,000 lowest cost with Mass Customization c. 65,000 demand > 60,000 breakeven for M

7.5 Breakeven points

a. Continuous : 2,400,000 20 80 40,000x x x+ = = Repetitive : 1,950,000 30 80 39,000x x x+ = = Mass Customization : 1,480,000 40 80 37,000x x x+ = = Intermittent : 1,800,000 40 80 45,000x x x+ = = b. Least cost process at 48,000x = Continuous: $3,360,000 least cost Repetitive: $3,390,000 Mass Customization: $3,400,000 Intermittent: $3,720,000 c. Is 48,000 > 40,000? Yes, so we use continuous process. Annual profit $480,000=

7.6

4,000 11,000Volume

15,000I

R

C

$

2,000

M

(11,000; 1,350,000)(4,000; 860,000)

(2,000; 300,000)

widest Repetitive has the widest production volume range over which it is a least-cost process. 7.7 Total profit now: Profit 40,000 2.00 20,000 40,000 0.75 80,000 20,000 30,000 30,000= = = Total profit with new machine: Profit 50,000 2.00 2,000 50,000 1.25 100,000 25,000 62,500 12,500= = = Since profit decreases with the new piece of equipment added to the line, purchase of the

machine probably would not be a good investment.

A-23

7 SUPPLEMENT

Capacity Planning

S7.1 Problem is under risk and has two decisions, so use a decision tree:

109

High demand

70

135(0.6)

Medium demand(0.4)

No additional expansion 90

135Additional minor expansion

148

High demand(0.6)

Medium demand(0.4) 40

220

148

Smallexpansion

Largeexpansion

(Payoffs and Expected Payoffs are in $1,000s) Answer : Ralph should undertake a large expansion. Then the annual expected profit will

equal $148,000.

A-24

S7.2

High demand 140,000

(0.3)

No additional

$90,000

$140,000

70,000

$70,000

Smallexpansion

Largeexpansion

minor expansion

Additionalminor expansion

Medium demand(0.7) $40,000

High demand(0.3)

14,000

Medium demand(0.7)

$105,000

$25,000

Maximum value = $70,000 S7.3

$18,000 Smallexpand

Demand upsmall (0.4)

16,000

Demand upmedium (0.6)

$10,000

$20,000

$0Noexpand

Demand upmedium (0.3)

18,000

Demand uplarge (0.7)

$10,000

$34,000

Largeexpand

Answer : $18,000

A-25

S7.4

(1)

50,000

100,000

200,000

300,000

400

x

1,000 2,000

(1)

250

(2)

(1)

(2)

(3)

(3)

(2)

Cap level (2) is lowest for all x so 1,000 2,000x

S7.5 Actual (or expected)Effective Capacity

Output = Efficiency

(text Equation S7-3)

4.8 cars = 5.5 cars 0.880. Therefore in one 8 hour day one bay accommodates 38.4 cars = ( )8 hrs 4.8 cars per hr and to do 200 cars per day requires 5.25 bays or 6 bays =

200 cars38.4 cars per bay

S7.6 a. ( ) ( )450

BEP $ 878.050.51251 i i i

F

V P W= = =

Breakeven ( )$ $878.05= b. Number of pizzas required at breakeven: Whole pizzas ( )878.05 0.30 5.00 52.7 53= = Slices ( )878.05 0.05 0.75 58.5 59= = Whole pizzas to make slices 59 6 9.8 10= = Therefore, he needs a total of 63 pizzas. He does not have sufficient capacity. S7.7 a. Remember that Yr 0 has no discounting.

Initial coat $1,000,0000 yearly maint 75,000 members dues/memberSalvage cost $50,000 yearly dues $300,000 500 $600 Discount rate 0.100

Year Cost Revenues Profit PV Mult PV Profit 0 $1,075,000 $300,000 $775,000 1 $775,000 1 75,000 300,000 225,000 0.9 $202,500 2 75,000 300,000 225,000 0.81 $182,250 3 75,000 300,000 225,000 0.729 $164,025 4 75,000 300,000 225,000 0.6561 $147,623 5 75,000 350,000 275,000 0.59049 $162,385 undisc. Profit 400,000 PV Profit $83,782

A-26

Assume dues are collected at the beginning of each year. This is a simplificationin reality, people are likely to join throughout the year. (Technically, if equipment is sold at the end of year 5, it should probably appear as a final revenue stream in year 6 but the difference is only $2,952.45.

b. Special deal comparison: $3,000 for all 6 years. Compare the PV cash stream of yearly dues from one member to that of the deal. Since we specified the club will always be full, we can make the assumption that the member (or her replacement) will always be paying the annual fee.

Initial cost $0 yearly maint $0 Salvage cost $0 yearly dues $600 Discount rate 0.100

(Membership fee) Year Cost Revenues Profit PV Mult PV Profit 0 $0 $600 $600 1 $600 1 0 600 600 0.9 $540 2 0 600 600 0.81 $486 3 0 600 600 0.729 $437 4 0 600 600 0.6561 $394 5 0 600 600 0.59049 $354 undisc. Profit 3,600 PV Profit $2,811

Since this is less than $3K, the special deal is worth more to the Health Club. Note also: If Health Club member is using same discount rates, its better for her to pay yearly.

S7.8 Breakeven: Costs = Revenues 500 0.50 0.75b b+ = where b = number of units at breakeven or ( )0.75 0.50 500b = ,

and 500

2,000 units0.25

b = =

a. breakeven in units = 2,000 units b. breakeven in dollars = $0.75 2,000 $1,500 =

A-27

8 CHAPTER

Location Strategies

8.1

( ) ( ) ( )( ) ( )( ) ( )( )( ) ( ) ( )( ) ( ) ( )

2,000 2.5 5,000 2.5 10,000 5.5 7,000 5.0

10,000 8.0 20,000 7.0 14,000 9.06.67

2,000 5,000 10,000 7,000 10,000 20,000 14,000xC

+ + +

+ + += =

+ + + + + +

( )( ) ( )( ) ( )( ) ( )( )( )( ) ( )( ) ( )( )

2,000 4.5 5,000 2.5 10,000 4.5 7,000 2.0

10,000 5.0 20,000 2.0 14,000 2.53.02

68,000yC

+ + +

+ + += =

8.2 Site Total Weighted Score A 174 B 185 C 187 D 165

8.3 Population weights 5,000=

( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( )( )0 2,050 1 550 2 1,025 3 775 2 250 2 350

0.5255,000x

C+ + + + +

= =

( )( ) ( )( ) ( )( ) ( ) ( ) ( ) ( ) ( )( )0 2,050 1.5 550 1 1,025 3 775 3 250 1 350

0.2055,000y

C+ + + + +

= =

Coordinates: ( )0.525, 0.205 8.4

25

50

75

100

125

150

1 2 3 4 5 6

V

$ cost(millions)

(2, 60)

(1, 35)

(1/2, 30)(0, 25)

(0, 20)(0, 10)

10,000s of Autos = V0

A

B

C

For all V such that 10,000 60,000V .

A-28

8.5 Site Score A 5 320w+ B 4 330w+ C 3 370w+ D 5 255w+ Find all w from 130 so that:

3 370 5 320 50 2 25

3 370 4 330 40 40

3 370 5 255 115 2 57.5

w w w w

w w w w

w w w w

+ + + + + +

Answer : For all w such that 1.0 25.0w 8.6

2

4

6

8

10

12

5 10 15 20 25 30

V

TC(millions $)

(1, 35)

(30, 9.5)(20, 7)

35 (thousands)

L.A.K.C.Charlotte

Char

K.C.

L.A.

cutoff0

4,100,000 180 1,000,000 300

For all so: 4,100,000 180 2,000,000 250

0 35,0003,100,000 120

2,100,000 70

0 35,0001

25,833 Answer : 30,000 35,0003

30,000

V V

V V V

V

V

V

V

VV

V

+ +

+ +

A-29

8.7 weights 14,000= (or 14 for calculations below)

( )( ) ( ) ( ) ( ) ( )2 3.5 8 7 6 3.5

6.014x

C+ +

= =

( )( ) ( )( ) ( )( )6 3.5 1 7 2 3.5

2.514y

C+ +

= =

A-30

9 CHAPTER

Layout Strategy

9.1 a. ( )( )60 60 sec 3,600

Cycle time 20 sec per PLA180 PLAs 180

= = =

b. Theoretical number of work stations task time 60

3cycle time 20

= = =

c. Yes, it is feasible. 9.2 1 2

3 4

D AC B

Department pair

( )( ) ( )( ) ( )( )( )( )( )( )( )( ) ( )

121212

12

1212

Weekly$

Cost: 8 800 3,200: 6 700 2,100: 4 400 800: 10 300 1,500: 7 200 700: 9 600 2,700

$11,000

AB

AC

AD

BC

BD

CD

=

=

=

=

=

=

A-31

9.3 a. ( )

( )274 seconds

Cycle time Cycle time secondsitn = =

( ) ( )60 60 secondsC.T. 60 seconds per truck60 trucks

= = so 274

4.5667 560

n n= = = .

Answer : 5

b. Steps 1 and 2: Sample Answer

60 seconds

From (a)number of stations is at least 5

c =

Precedence diagram:

A

40

B

30 D

40

E

6 H

20

C

50 F

25

G

15 I

18

J

30

Step 3

Task

Number of Successors

Task

Number of Successors

A 9 F 2 B 4 G 2 C 4 H 1 D 2 I 1 E 2 J 0

A-32

Step 4

Available

Available and Fit

Assigned

Station 1 A A A B, C Station 2 B, C B, C C (Broke a tie) B, F, G Station 3 B, F, G B, F, G B D, E, F, G E, F, G F (Broke ties) D, E, G Station 4 D, E, G D, E, G D (Broke ties) E, G E, G G (Broke a tie) E, I Station 5 E, I E, I E I, H I, H H (Broke a tie) I I I J Station 6 J J J

Answer : Station Tasks(Other answers possible, 1 Adepending upon how ties 2 C

are broken in above 3 B, Fprocedure) 4 D, G

5 E, H, I6 J

c. 6 work stations are in our answer.n =

( ) ( )274

Efficiency 0.7611C.T. 6 60

it

n= = =

9.4 a. First assignment costs ( ) 18,000 7,200 1,600 4,800 8,000 8002

$15,200

= + + + + +

=

b. New layout costs ( ) 18,000 9,600 6,400 3,600 2,000 800 $15, 2002

= + + + + + =

No improvementboth yield the same cost.

A-33

9.5 Cost of 3 attempts: Work Area Attempt 1 2 3 Cost a. 1 S D M $1,088 b. 2 D S M $1,142

c. 3 M D S $1,100

( )( ) ( ) ( ) ( )( )2 23 10 32 5 20 8 $1,100 + + =

9.6 a. Theoretical minimum number of stations task times

cycle time=

Cycle time 60

12 minutes5

= = . So minimum number of stations 48

4 stations12

= =

b.

A

10 min

WS #1

B

12 min

WS #2

C

8 min

WS #3

D

6 min

WS #4E

F

12 min

WS #5

This requires 5 stationsit cannot be done with 4.

c. Efficiency 48 48

80%5 12 60

= = =

for 5 stations.

9.7 There are three alternatives: Station Alternative 1 Tasks Alternative 2 Tasks Alternative 3 Tasks 1 A, B, F A, B A, F, G 2 C, D C, D H, B 3 E G, H C, D 4 G, H E E 5 I I I Each alternative has an efficiency of 86.67%.

A-34

10 SUPPLEMENT

Work Measurement

S10.1 Required sample size 2

Zsn

hx

= =

where 0.15s = , 0.4x = , 1.96z = (for 95% confidence),

10%h = accuracy level ( )( )( )( )

21.96 0.15

54.0225 540.10 0.4

n

= =

S10.2 a. 2

Zsn

hx

=

Thus, ( )( ) ( )0.10 0.40 12

0.9240.15

hx nZ

s= = =

Referring to Appendix I (Standard Normal Table), Area 0.64 64%= = . The confidence level when 12n = is 64%, as opposed to 95% when 54n = (in Problem S10.1)

b. Average observed time 0.331 0.243 0.484

0.4484 minutes12

+ + += =

"

Normal time = Average time perf. rating = 0.4484 0.90 0.4036 minutes =

Standard time Normal time 0.4036

0.429 minutes1 allowance factor 1 0.06

= = =

S10.3 Average observed time 100 hours 60 minutes 0.75

22.5 minutes200 units

= =

Normal time 22.5 minutes 1.1 24.75 minutes= =

Standard time for job Normal time for process 24.75

29.12 minute unit1 Allowance fraction 1 0.15

= = =

S10.4 a. sum of times 1.74

Observed time 0.10875 minutes 6.525 secondsnumber of cycles 16

= = = =

b.

( ) ( )Normal time Observed time Performance rating factor 6.525 95%6.2 seconds

= =

=

c. normal time 6.2 6.2

Standard time 6.739 seconds1 allowance factor 1 8% 92%

= = = =

A-35

S10.5 Normal time 10 minutes 0.90 9 minutes= =

Allowance fraction Personal Fatigue Delay 5 3 1 9

0.1560 minutes 60 60

+ + + += = = =

Normal time 9

Standard time 10.59 minutes1 Allowance fraction 1 0.15

= = =

S10.6 Observation (Minutes Per Cycle)

Element

Rating 1

2

3

4

5

Average Time

Normal Time

1 100% 1.5 1.6 1.4 1.5 1.5 1.5 1.50 2 90% 2.3 2.5 2.1 2.2 2.4 2.3 2.07 3 120% 1.7 1.9 1.9 1.4 1.6 1.7 2.04 4 100% 3.5 3.6 3.6 3.6 3.2 3.5 3.50 Normal time for lab test = 9.11

Standard time for lab test Normal time for process 9.11

11.1 minutes1 Allowance fraction 1 0.18

= = =

A-36

12 CHAPTER

Inventory Management

12.1 An ABC system classifies the top 70% of dollar volume items as A, the next 20% as B, and

the remaining 10% as C items. Similarly, A items constitute 20% of total number of items, B items are 30%; and C items are 50%.

Item Code Number Average Dollar Volume Percent of Total $ Volume1289 400 3.75

2347 300 4.00

2349 120 2.50

2363 75 1.50

2394 60 1.75

2395 30 2.00

6782 20 1.15

7844 12 2.05

8210 8 1.80

8310 7 2.00

9111 6

=

=

=

=

=

=

=

=

=

=

1,500.00 44.0%

1, 200.00 36.0%

300.00 9.0%

112.50 3.3%

105.00 3.1%

60.00 1.8%

23.00 0.7%

24.60 0.7%

14.40 0.4%

14.00 0.4%

3.00 18.00 0.5%

$3,371.50 100%

=

Answer : The company can make the following classification:

A: 1289, 2347 B: 2349, 2363, 2394, 2395 C: 6782, 7844, 8210, 8310, 9111 12.2 D (Annual Demand) = 4,800 units, P (Purchase Price/Unit) = $27, H (Holding Cost) = $2

S (Ordering Cost) = $30. So, ( ) ( )( )2 4,800 302Order Quantity 2402

DSQ

H

= = = .

( ) ( ) 2 240 30 4,800Thus, Total Annual Cost 4,800 27

2 2 2 240

129,600 240 600 $130,440

HQ SDTC PD

= + + = + +

= + + =

A-37

12.3 D (Annual Demand) = 14,558, P (Purchase Price/Unit) = $5, H (Holding Cost/Unit) = $4,

S (Ordering Cost/Order) = $22, 2 2 14,558 22

4004

DSQ

H

= = = tons per order

( ) 4 400 22 14,5585 14,558 72,790 800 800.69

2 2 2 400

$74,390.69

HQ SDTC PD

= + + = + + = + +

=

Answer : The optimal order quantity ( ) 400 tons orderQ = ; total annual inventory cost ( ) $74,391TC = .

12.4 D (Annual Demand) = 400 12 = 4,800, P (Purchase Price/Unit) $350 unit= , H (Holding Cost/Unit) $35 unit year= , S (Ordering Cost/Order) $120 order= . So,

2 2 4,800 120181.42 181

35

DSQ

H

= = = = units (rounded off).

( ) ( ) 35 181 120 4,800Thus, Total Cost 4,800 325

2 2 181

1,560,000 3,168 3,182 $1,566,350

HQ SDTC PD

Q

= + + = + +

= + + =

However, if Bell Computers orders 200 units,

( ) 35 200 120 4,8004,800 325 1,440,000 3,500 2,880 $1, 446,3802 200

TC

= + + = + + =

Answer : Bell Computers should order 200 units for a minimum total cost of $1,446,380.

12.5 12 2 4,800 120

181 units35

DSQ

H

= = =

22 2 4,800 120

188 units32.5

DSQ

H

= = =

32 2 4,800 120

196 units30

DSQ

H

= = =

181 units cannot be bought at $350, hence that isnt feasible. 196 units cannot be bought at $300, hence that isnt possible either. So, 188 unitsEOQ = .

( ) ( ) 32.5 188 120 4,800Thus, 188 units 325 4,8002 2 188

1,560,000 3,055 3,064 $1,566,119

HQ SDTC PD

Q

= + + = + +

= + + =

( ) ( ) 30 200 120 4,800200 units 300 4,8002 2 200

1,440,000 3,000 2,880 $1, 445,880

HQ SDTC PD

Q

= + + = + +

= + + =

Answer : The minimum order quantity is 200 units yet again, since the overall cost of

$1,445,880 is less than ordering 188 units which has an overall cost of $1,566,119.

A-38

12.6 12,500 yearD = , so 50 dayd = , 300 dayp = , $30 orderS = , $2 unit yearH =

a. 2 2 12,500 30 300

612.37 1.095 6712 300 50

DS pQ

H p d

= = = =

b. Number of production runs ( ) 12,500 18.63671

DN

Q= = =

c. Maximum inventory level ( )max 50 11 671 1 671 1 559300 6d

I Qp

= = = =

d. Days of demand satisfied by each production run 250

13.4218.63

= = days in demand

only mode

Time in production for each order 671

2.24300

Q

p= = = days in production for each

order. Total time 13.42= days per cycle.

Thus, percent of time in production 2.24

16.7%13.42

= = .

12.7 $2 unit yearH = , $10 orderS = , 4,000

4,000 16250

D d

= = =

. So,

2 2 4,000 10200

2

DSQ

H

= = = . ROP (reorder point) l d= , l (lead time = 5 days).

Thus, ROP 5 16 80 units= = . Answer : Saveola, Inc. should place an order for 200 frames every time the inventory of

frames falls to 80 units. This will be their inventory policy.

12.8 ( )2 5, 400 34

300H

= ; Square both sides 367,200

90,000H

= , 367,200

$4.0890,000

H = =

12.9 a. ( )( )2 6,000 302

189.74 units10

DSEOQ

H= = =

b. Average inventory = 94.87 c. Optimal number of orders/year = 31.62

d. Optimal days between orders 250

7.9131.62

= =

e. Total annual inventory cost 601,897.37= (including the $600,000 cost of goods)

12.10 a. Holding cost = $530.33 b. Set up cost = $530.33 c. Unit costs = $56,250.00 d. Total costs = $57,310.66 e. Order quantity = 16,970.56 units Thus, order 10,001 units for a total cost of $57,310.66.

A-39

12.11 Inventory Item

$Value

per Case

#Ordered

per Week

Total $

Value/Week

(52 Weeks) Total

($*Weeks)

Rank

Percent of Inventory

CumulativePercent ofInventory

Fish Fillets 143 10 $1,430.00 $74,360.00 1 17.54% 34.43% French Fries 43 32 $1,376.00 $71,552.00 2 16.88% 47.31% Chickens 75 14 $1,050.00 $54,600.00 3 12.88% 59.53% Prime Rib 166 6 $996.00 $51,792.00 4 12.22% 69.83% Lettuce (case) 35 24 $840.00 $43,680.00 5 10.31% 78.85% Lobster Tail 245 3 $735.00 $38,220.00 6 9.02% 83.82% Rib Eye Steak 135 3 $405.00 $21,060.00 7 4.97% 87.25% Bacon 56 5 $280.00 $14,560.00 8 3.44% 90.64% Pasta 23 12 $276.00 $14,352.00 9 3.39% 93.74% Tomato Sauce 23 11 $253.00 $13,156.00 10 3.10% 95.71% Table Cloths 32 5 $160.00 $8,320.00 11 1.96% 97.60% Eggs (case) 22 7 $154.00 $8,008.00 12 1.89% 98.28% Oil 28 2 $56.00 $2,912.00 13 0.69% 98.72% Trash Can Liners 12 3 $36.00 $1,872.00 14 0.44% 99.13% Garlic Powder 11 3 $33.00 $1,716.00 15 0.40% 99.42% Napkins 12 2 $24.00 $1,248.00 16 0.29% 99.72% Order Pads 12 2 $24.00 $1,248.00 17 0.29% 99.83% Pepper 3 3 $9.00 $468.00 18 0.11% 99.93% Sugar 4 2 $8.00 $416.00 19 0.10% 99.93% Salt 3 2 $6.00 $312.00 20 0.07% 100.0% $8,151.00 $423,852.00 100.00%

a. Fish filets total $74,360 b. C items are items 10 through 20 in the above list (although this can be one or two

items more or less) c. Total annual $ volume = $423,852

12.12 Incremental Costs Safety Stock Carrying Cost Stockout Cost Total Cost 0 0 70(100 0.4 + 200 0.2) = 5,600 5,600 100 100 15 = 1,500 (100 0.2) (70) = 1,400 2,900 200 200 15 = 3,000 0 3,000 The safety stock which minimizes total incremental cost is 100 kilos. The re-order point then

becomes 200 kilos + 100 kilos or, 300 kilos.

A-40

12.13 S $10 order= , 4 daysLT = , 80 dayd = , 20 = , H 10%= of $250,

( )for term 80 100 8,000D = = a.

2800 calzones

DSQ

H= = . Order every days 10 days

Q

d=

b. ROP for calzone, ( )demand during lead time 40LT = = , 1Z = (from Table) for 0.1587, ROP 360dLTd LT Z = + =

c. On hand = 85, days left = 1. Since ROP daltd LT Z = + , ROP d LT

ZLT

= .

Thus, Z of 0.25 gives 0.5987. So there is a 40.13% chance of stockout.

d. 400 average inventory2

Q= = , 10 orders term

D

Q= .

( )( )Holding cost term 400 0.25 $1002

QH= = =

( )( )Order cost term 10 10 $100DSQ

= = =

12.14 $16S = , $0.40 calzone termH = , 160p = , 80d = , 8,000D = for 100 days of the term

a. ( )2

1,131.371

DSQ

H d p= =

b. 1,131.37

Cycle 14.1480

Q

d= = =

c. Run production for 1,131

days 7.07 days160

Q

p= =

A-41

12.15 Under present price of $6.40 per box:

Economic Order Quantity: 2 2 5000 25

395.3 or 395 boxes0.25 6.40

DSQ

H = = =

where

D = annual demand, S = set-up or order cost, H = holding cost

Total cost order cost holding cost purchase cost2

5,000 25 395 0.25 6.406.4 5,000 316.46 316.00 32,000

395 2$32,632.46

DS QHCD

Q= + + = + +

= + + = + +

=

Note: Order and carrying costs are not exactly equal due to rounding of the EOQ to a whole number. Under the quantity discount price of $6.00 per box:

Total cost order cost holding cost purchase cost2

5,000 25 5,000 0.25 6.005,000 6.00 41.67 3,750.00 30,000

3,000 2

$33,791.67

DS QH

Q= + + = +

= + + = + +

=

Therefore, the old supplier with whom they would incur a total cost of $32,632.46, is preferable.

12.16 Economic Order Quantity, non-instantaneous delivery: where: D = period demand, S = set-up or order cost, H = holding cost, d = daily demand rate, p = daily production rate

( ) ( )502002 2 10,000 200

2,309.41.00 11 dp

DSQ

H = = =

or 2,309 units

A-42

13 CHAPTER

Aggregate Planning

13.1 The total production required over the year is 8,400 units, of 700 per month. Thus, the

schedule is to produce 700 per month and have no costs associated with work force variation. The only costs incurred will be the monthly production cost, the inventory cost, and the shortage cost. The costs are calculated as follows.

Month Beginning Inventory

Produc- tion

Production Cost

Demand

Ending Inventory

Shortage

Inventory Cost

Shortage Cost

January 0 700 $49,000 500 200 0 $600 $0 February 200 700 49,000 600 300 0 900 0 March 300 700 49,000 600 400 0 1,200 0 April 400 700 49,000 700 400 0 1,200 0 May 400 700 49,000 700 400 0 1,200 0 June 400 700 49,000 800 300 0 900 0 July 300 700 49,000 900 100 0 300 0 August 100 700 49,000 900 0 100 0 1,000 September 0 700 49,000 800 0 200 0 2,000 October 0 700 49,000 700 0 200 0 2,000 November 0 700 49,000 600 0 100 0 1,000 December 0 700 49,000 600 0 0 $ 0 0 8,400 $588,000 8,400 2,100 600 $ 6,300 $6,000 The total cost of this plan is the sum of the three costs, or $600,300.

A-43

13.2 a. Total hotel demand for the year = 7,000,000 Total restaurant demand for the year = 2,080,000

Level staffing

Quarter

Hotel (Room) Demand

Personnel Required

RestaurantDemand

Req.

Total

Hire

Terminate Winter 800,000 33 160,000 12 45 45 Spring 2,200,000 33 800,000 12 45 Summer 3,300,000 33 960,000 12 45 Fall 700,000 33 160,000 12 45 Totals 7,000,000 132 2,080,000 48 180 45

Personal cost = 180 quarters of labor 5,000 = $900,000 Hiring cost = 45 hires at $1,000 = 45,000 Termination cost = none = 0 $945,000 b. Total hotel demand for the year = 7,000,000 Total restaurant demand for the year = 2,080,000

Chase staffing (staffing to meet the forecasted demand)

Personnel Personnel Totals

Quarter

Hotel (Room) Demand

Req.

Hire

Termi-

nate

RestaurantDemand

Req.

Hire

Termi-

nate

Quarter Hires

Quarter Termi- nations

Winter 800,000 8 8 160,000 2 2 10 Spring 2,200,000 22 14 800,000 10 8 22 Summer 3,300,000 33 11 960,000 12 2 13 Fall 700,000 7 0 26 160,000 2 0 10 36 Totals 7,000,000 70 2,080,000 26 45 36

Personnel cost = 96 quarters of labor 5,000 = $480,000 Hiring cost = 45 hires @ $1,000 = 45,000 Termination cost = 36 terminations @ $2,000 = 72,000 $597,000 c. The chase plan developed in part b is most economical ($945,000 vs. $597,000)

A-44

13.3 Total hotel demand for the year = 7,000,000 Total restaurant demand for the year = 2,080,000

Hiring from local staffing agency all personnel above base requirements.

Quarter

Hotel (Room) Demand

Req.

Hire

Personnel from

Agency

Restaurant Demand

Req.

Hire

Personnel from

Agency

Quarter Hires

Quarterfrom

Agency Fall 800,000 8 7 1 160,000 2 2 0 2 1 Spring 2,200,000 22 0 15 800,000 10 0 8 0 23 Summer 3,300,000 33 0 26 960,000 12 0 10 0 36 Fall 700,000 7 0 0 160,000 2 0 0 0 0 Totals 7,000,000 28* 42 2,080,000 26 8** 18 35 60 *On Hotel Grand payroll (7 each quarter 4 quarters = 28) ** On Hotel Grand payroll (2 each quarter 4 quarters = 8) Total on Grand Hotel payroll = 28+ 8 = 36

Personnel cost = 36 quarters of labor 5,000 = $180,000 Hiring cost = 9 ( )7 2= + hires @ $1,000 = 9,000 Termination cost = 0 terminations @ $2,000 = 0 Staffing agency costs = 60 @ 6,500 = 390,000 $579,000 13.4 Plan A: Month Demand Production Hire Fire Extra Cost Mar 1,000 900 700 56,000 Apr 1,200 1,200 300 12,000 May 1,400 1,400 200 8,000 June 1,200 1,200 200 8,000 July 1,500 1,500 300 12,000 Aug 1,300 1,300 200 16,000 Total extra cost: $112,000

Plan B: Month Demand Production Inventory Sub-Contracting Extra Cost Mar 1,000 1,100 200 2,000 Apr 1,200 1,100 100 1,000 May 1,400 1,100 200 8,000 June 1,200 1,100 100 4,000 July 1,500 1,100 400 16,000 Aug 1,300 1,100 200 8,000

Total extra cost: $39,000 Therefore, Plan B would be preferred.

A-45

13.5 Plan Number 1: Month Demand Production Inventory Sub-Contracting Extra Cost 1 1,000 1,200 300 3,000 2 1,200 1,200 300 3,000 3 1,400 1,200 100 1,000 4 1,200 1,200 100 1,000 5 1,500 1,200 200 8,000 6 1,300 1,200 100 4,000 Total extra cost: $112,000

Plan Number 2: Month Demand Production Inventory Overtime Extra Cost 1 1,000 1,200 300 3,000 2 1,200 1,200 300 3,000 3 1,400 1,200 100 1,000 4 1,200 1,200 100 1,000 5 1,500 1,200 200 2,000 6 1,300 1,200 100 1,000 Total extra cost: $39,000 Therefore, Plan Number 2 would be preferred. 13.6 Plan Y: Month Demand Production Hire Fire Extra Cost 1 1,100 1,100 400 32,000 2 1,600 1,600 500 20,000 3 2,200 2,200 600 24,000 4 2,100 2,100 100 8,000 5 1,800 1,800 300 24,000 6 1,900 1,900 100 4,000 Total extra cost: $112,000

Plan Z: Month Demand Production Inventory Sub-Contracting Extra Cost 1 1,100 1,600 600 6,000 2 1,600 1,600 600 6,000 3 2,200 1,600 4 2,100 1,600 500 20,000 5 1,800 1,600 200 8,000 6 1,900 1,600 300 12,000 Total extra cost: $52,000 Therefore, Plan Z would be preferred.

A-46

14 CHAPTER

Materials Requirements Planning (MRP) & ERP

14.1 Requirement for 3,500 Get Well bud vases. 3,500 Vases 3,500 8" white ribbons (2,333 ft.) 3,500 8" red ribbons (2,333 ft.) 3,500 signature cards 7,000 sprigs of babys breath 7,000 pink roses 14.2 Period (week)

Lot Size

Lead Time

On Hand

Safety Stock

Allo-cated

Low-Level Code

Item ID

CD Case 1 2 3 4 5 6 7 8

Gross Requirements 650 300 550 400 500 Scheduled Receipts Projected On Hand 1,000 1,000 1,000 1,000 350 50 Net Requirements 500 400 500 Planned Order Receipts 500 400 500

Lot for Lot

1

1,000

0

CD Case

Planned Order Releases 500 400 500 Gross Requirements 500 400 500 Scheduled Receipts Projected On Hand Net Requirements 500 400 500 Planned Order Receipts 500 400 500

Lot for Lot

1

1

CD Top


Lot for Lot

1

1

CD Bottom


Lot for Lot

1

1

CD Insert

Planned Order Releases 500 400 500 Gross Requirements 500 400 500 Scheduled Receipts Projected On Hand 100 100 100 100 Net Requirements 500 400 500 Planned Order Receipts 500 500 500

Lot for Lot

2

2

BlackDye

Planned Order Releases 500 500 500

A-47

14.3 Period (day)

Lot Size

Lead Time

On Hand

Safety Stock

Allo-cated

Low-Level Code

Item ID

Ball Point Pens 1 2 3 4 5 6 7 8

Gross Requirements 10,000 Scheduled Receipts Projected On Hand Net Requirements 10,000 Planned Order Receipts 10,000

Lot for Lot

1

0

Planned Order Releases 10,000 10,000 Gross Requirements 10,000 Scheduled Receipts Projected On Hand Net Requirements Planned Order Receipts 10,000

Lot for Lot

1

1

Cap

Planned Order Releases 10,000 Gross Requirements 2,000 Scheduled Receipts Projected On Hand Net Requirements 2,000 Planned Order Receipts 2,000

Lot for Lot

3

1

Ink CC


Lot for Lot

1

Body


Lot for Lot

1

Clip Fine Pt

Planned Order Releases 5,000 Gross Requirements 5,000 Scheduled Receipts Projected On Hand 3,000 3,000 Net Requirements 2,000 Planned Order Receipts 2,000

Lot for Lot

1

3,000

Std. Clip

Planned Order Releases 2,000 Gross Requirements 5,000 Scheduled Receipts Projected On Hand 3,000 3,000 Net Requirements 2,000 Planned Order Receipts 2,000

Lot for Lot

1

3,000

Std. Ball Point


Lot for Lot

1

Fine Point Ball Point

Planned Order Releases 5,000

A-48

14.4 Period (week, day)

Lot Size

Lead Time

On Hand

Safety Stock

Allo-cated

Low-Level Code

Item ID

Ball Point Pens 1 2 3 4 5 6 7 8

Gross Requirements 640 640 128 128 Scheduled Receipts Projected On Hand Net Requirements 640 640 128 128 Planned Order Receipts 640 640 128 128

Lot for Lot

1

0

Coffee Table

Planned Order Releases 640 640 128 128 Gross Requirements 640 640 128 128 Scheduled Receipts Projected On Hand Net Requirements 640 640 128 128 Planned Order Receipts 640 640 128 128

Lot for Lot

1

1

Top

Planned Order Releases 640 640 128 128 Gross Requirements 80 80 16 16 Scheduled Receipts Projected On Hand Net Requirements 80 80 16 16 Planned Order Receipts 80 80 16 16

Lot for Lot

1

1

Stain gal

Planned Order Releases 80 80 16 16 Gross Requirements 40 40 8 8 Scheduled Receipts Projected On Hand 100 100 100 100 100 60 20 12 4 Net Requirements Planned Order Receipts

Lot for Lot

1

1

Glue gal

Planned Order Releases Gross Requirements 640 640 128 128 Scheduled Receipts Projected On Hand Net Requirements 640 640 128 128 Planned Order Receipts 640 640 128 128

Lot for Lot

1

1

Base

Planned Order Releases 640 640 128 128 Gross Requirements 1,280 Scheduled Receipts Projected On Hand Net Requirements 1,280 Planned Order Receipts 1,280

Lot for Lot

1

2

Long Braces


Lot for Lot

1

2

Short Braces


Lot for Lot

1

2

Leg

Planned Order Releases 5,120 Gross Requirements 5,120 Scheduled Receipts Projected On Hand 880 880 880 880 880 880 Net Requirements 5,120 Planned Order Receipts 6,000

Lot for Lot

1

3

Brass Caps

Planned Order Releases 6,000

A-49

14.5 Coffee Table Hours Lead Master Schedule Required Time Day 1 Day 2 Day 3 Day 4 Day 5 Day 6 Day 7 Day 8 640 640 128 128 Table Assembly 2 1 1,280 1,280 256 256 Top Preparation 2 1 1,280 1,280 256 256 Assemble Base 1 1 640 640 128 128 Long Braces (2) 0.25 1 320 320 64 64 Short Braces (2) 0.25 1 320 320 64 64 Legs (4) 0.25 1 640 640 128 128 Total Hours 0 1280 3,200 3,456 1,920 640 256 Employees needed @ 8 hrs. each 0 160 400 432 240 80 32

14.6 The following table lists the components used in assembling FG-A. Also included for each

component are the following information: the on-hand supply, lead time, and direct components.

Item On-Hand LT (weeks) Components PG-A 0 1 SA-B, SA-C(2), SA-D(2) SA-B 0 1 SA-D(2) SA-C 0 2 E, F(2) SA-D 0 2 E (3) E 10 1 F 5 3 While not required as part of the question, we recommend you make a product structure

tree to help you answer the questions on the Bill-of-Materials and lead time.

1 FGA [LT =1]

[LT =1]SA-B

[LT =2]SA-D(2)

E(3) [LT = 1]

[LT =2]SA-C(2)

E [LT = 1] F(2) [LT = 3]

[LT =2]SA-D(2)

E(3) [LT = 1]

a. Bill-of-Material associated with 1 unit of FG-A 1 SA-B Note: this is just the master recipe. Not subtracting off on on-hand quantities YET 2 SA-C 4 SA-D 2 1 2= + 14 E 2 3 2 1 2 3 6 2 6= + + = + + 4 F 2 2=

A-50

b. Total lead time (in weeks) associated with making an item of FG-A, assuming we had no starting on-hand for any part? 6 weeks

Look at all branches of the product structure tree (or an assembly time chart, if we

had one). The most common mistake people tend to make is to forget the LT associated with final assembly.

Longest branch is 6 weeks 1wk FGA 2wks SA-C 3wks F= + + c. Yes, if we wanted to make one FG-A, we need to order more of either E or F.

We have enough F on-hand, but need 4 more E. Now we look at the on-hand quantities and see we have 10 E and 5 F. Compare that

to the BOM calculated above. No subassemblies like SA-B, SA-C or SA-D, so we are going to need all the component parts. Our on-hand records show we have 14 E and 4 F, so we are short 4 E.

14.7 a. 17 2 (level 2) = 34

b. [17 3 (Bs)] 15 = 51 15 = 36 36 5 (Ds) = 180

14.8 a. 17 2 (level 2) + 17 2 3 (level 3) = 136

b. 170 12 2 (level 2 on hand) = 170 24 = 146

A-51

15 CHAPTER

Short-Term Scheduling

15.1 A dummy task is added to balance the problem. The assignment is MayTask 1 GrayTask 2 RayTask 3 Total time = 1 + 1 + 1 = 3 hours 15.2 Convert the minutes into $, Marketing Finance Operations Human Resources Chris $80 $120 $125 $140 Steve $20 $115 $145 $160 Juana $40 $100 $35 $45 Rebecca $65 $35 $25 $75 (Now subtract smallest number in each column from every number.) The Minimum Cost Solution =

Chris Finance $120

Steve Marketing $ 20

Juana Human Resources $ 45

Rebecca Operation $ 25

$210

15.3 The best pairs are assigned as follows: AjayJackie JackBarbara GrayStella RaulDana Total compatibility score (overall) = 90 + 70 + 50 + 20 = 230

A-52

15.4

10 20 30 40

F

D

C

B

E

A

Hour

6

10

20

37

Gantt Chart

13

28

Project Time Due Date Late Days A 9 22 15 B 7 17 3 C 3 16 0 D 4 13 0 E 8 16 12 F 6 9 0

a. 6 10 13 20 28 37 114

Average flow time 19 days6 6

+ + + + += = =

b. total late days 15 3 12

Average lateness 5 daysnumber of jobs 6

+ += = =

c. Maximum lateness 15 days= (for job AGantt Table) 15.5 a. The shortest processing time ( )SPT =DBACE

D B A C E Flow Time 0 1 3 6 11 20 = 41 Due Hours 0 12 4 8 6 7 Late Hours 0 5 13 = 18

Average flow time 41

8.2 hours5

= =

Number of deliveries late C and E 2= =

Average hours late 5 13 18

9 hours2 2

+= = =

A-53

b. The EDD schedule = BCEAD

B C E A D Flow Time 0 2 7 16 19 20 Due Hours 0 4 6 7 8 12 Late Hours 0 1 9 11 8

Average flow time 64

12.8 hours5

= =

Number of deliveries late C, E, A, and D 4= =

Average hours late 29

7.25 days4

= =

15.6 5

Cut & sew

Deliver

10 15 20 25

5 10 15 20 25

1

1

2

2

3

3

17114

6 18 23

[123 schedule]

5

Cut & sew

Deliver

10 15 20 25

5 10 15 20 25

3 2 1

17136

6 11 22

[321 schedule]

Answer : The 321 schedule finishes in 22 days, 1 day faster than the 123 schedule, which

finishes in 23 days. 15.7 a. We begin by taking 5 empty slots

b. Next, we find the shortest time in the table. It is product 2678 on Machine B. Since this is on the second machine, this product can be done as late as possible.

2678

c. Then we find the shortest again. It is 2800 on B 2800 2678 d. Then it is 2731 on A. Since it is on the first machine, it is the earliest job.

2731 2800 2678

A-54

e. Finally, we get 2731 2134 2387 2800 2678

5

A

12 19 22

B

2731 2134 2387 2800 2678

7 14 21 25

2731 2134

33 35

2800 26782387 Total time 33 hours=

15.8 a. The jobs should be processed in the sequence: 3, 6, 2, 7, 5, 1, 4

b. Time = 61 hours 15.9 a. Jobs should be processed in the sequence A, B, C, D, E if scheduled by the FCFS

scheduling rule. b. Jobs should be processed in the sequence A, B, C, D, E if scheduled by the EDD

scheduling rule. c. Jobs should be scheduled in the sequence B, E, A, C, D if scheduled by the SPT

scheduling rule. d. Jobs should be processed in the sequence D, C, A, B, E if scheduled by the LPT

scheduling rule. 15.10 Job Due Date Duration (Days) Critical Ratio 103 214 10 1.4 205 223 7 3.3 309 217 11 1.5 410 219 5 3.8 517 217 15 1.1 Jobs should be scheduled in the sequence 517, 103, 309, 205, 412 if scheduled by the

critical ratio scheduling rule.

A-55

16 CHAPTER

Just-In-Time and Lean Production Systems

16.1 a. ( )2

1 dp

D SQ

H

=

( )2 2

1 dp

D SQ

H

=

( )( ) ( ) ( )( ) ( )( )( )22 4005001 1,000 10 1 1,000,000 10 .2 2,000,000 $10

2 2 100,000 200,000 200,000

dpQ H

SD

= = = = =

b. 10 1

hr 10 min60 6

Note that the lead time was not needed in this problem. 16.2 10 min. 2 min. = 8 min. improvement required 16.3 How to improve setups (from Figure 16.4 in Heizer/Render text).

Step 1: Separate setup into preparation and actual setup, doing as much as possible while the machine/process is operating.

Step 2: Move material closer and improve material handling Step 3: Standardize and improve tooling Step 4: Use one-touch system to eliminate adjustments Step 5: Train operators and standardize work procedures

16.4 ( )( )

( )5,00010,0002 1, 250,000 152 37,500,000

3,750,000 1,936 units1020 11 dp

D SQ

H

= = = = =

1,936

per container 19.36 containers100

=

A-56

16.5 Note: 6S = minutes (or 110 hour) $100 shop labor cost .1 $100 $10= =

( )( )( )( )

( )( )

50500

2 12,500 102 250,0005,555 74

50 1 451

1EOQ Safety stock Lead time 74 10 500

2

334 oil pumps

dp

D SQ

H

= = = = =

= + + = + +

=

334No. of Kanban containers 9.27 9 or 10 containers required

36= = =

16.6 Where: D = annual demand, S = set-up or order cost, H = holding cost, d = daily demand rate, p = daily production rate. Solving for S (set-up cost):

( ) ( ) ( )2 2 1501,0001 150 10 1 22,500 10 1 0.152 2 40,000 80,000

191,250$2.39

80,000

$2.39 set-up 60 minute hourSet-up time 2.69 minute set-up

$50 hour

dpQ H

SD

= = =

= =

= =

A-57

17 CHAPTER

Maintenance and Reliability

17.1 Failure Rate Analysis Station No. 1: N = 1,000 F = 22

22

0.0221,000

1 0.022 0.978 97.8%

FR

R

= =

= = =

Station No. 2: N = 1,000 F = 51

51

0.0511,000

1 0.051 0.949 94.5%

FR

R

= =

= = =

Answer : Unit #1 is clearly better, with a lower failure rate and better reliability.

17.2 Reliability ( ) 1 2 3s nR R R R R= ! . So, ( )( )( )5 0.98 0.99 0.96 0.9304R = = Answer : The reliability is 93.04%

A-58

17.3 Task Errors FR Reliability 1 1 0.001 0.999 2 6 0.006 0.994 3 4 0.004 0.996 4 2 0.002 0.998 5 3 0.003 0.997 6 1 0.001 0.999 7 0 0.000 1.000 8 2 0.002 0.998 9 1 0.001 0.999 10 2 0.002 0.998 11 1 0.001 0.999 12 1 0.001 0.999 13 0 0.000 1.000 14 2 0.002 0.998 15 3 0.003 0.997 Total 29

Overall Reliability ( ) Product of all reliabilities above 0.971 97.1%sR = = =

17.4 Reliability (stick) ( )21 1 0.99 0.9999= = Reliability (cloth) ( )21 1 0.98 0.999998= = The reliability of the snake-charmers work 0.9999 0.999998 0.9998 99.98%= = = 17.5

.96

.96

.96

.99

.99

.93

.93

.93

.90

.90

.90

.90

Mountain tests reliability ( )31 1 0.96 0.999936= = Bump tests reliability ( )21 1 0.99 0.999900= = Speed tests reliability ( )31 1 0.93 0.999657= = Sudden brake tests reliability ( )41 1 0.90 0.999900= = So, overall reliability ( ) 0.999936 0.999900 0.999657 0.999900 0.9994 99.94%sR = = =

A-59

A MODULE

Decision Making Tools

A.1

$20,000 $30,000 $26,500AssemblyLine

$10,000 $50,000 $29,000PlantAddition

$0 $0 $0No NewProduct

0.35Competes

0.65Doesnt Compete

ExpectedValue

Prob.States of Nature

DecisionAlternatives

Answer : ( ) ( )0.35 $20,000 0.65 $50,000 $29,000 $10,500+ = A.2

0 2,000 3,000N

4,000 8,000 9,000M

10,000 6,000 20,000L

FixedSlight

IncreaseMajor

Increase

States of Nature


Minimum

0

4,000

10,000

50,000 4,000 40,000O 50,000 Use maximin: criterion. No floor space (N). A.3 Row Average Increasing capacity $700,000 Using overtime $700,000 Buying equipment $733,333 Using equally likely, Buying equipment is the best option.

A-60

A.4 ( ) ( ) ( ) ( )3 2 35 15Expected value under certainty 50 20 100 10010 10 100 10015 4 35 15 69

= + + +

= + + + =

A.5 ( ) ( ) ( ) ( )E A 0.4 40 0.2 100 0.4 60 60= + + = ( ) ( ) ( ) ( )E B 0.4 85 0.2 60 0.4 70 74= + + = ( ) ( ) ( ) ( )E C 0.4 60 0.2 70 0.4 70 66= + + = ( ) ( ) ( ) ( )E D 0.4 65 0.2 75 0.4 70 69= + + = ( ) ( ) ( ) ( )E E 0.4 70 0.2 65 0.4 80 73= + + = Choose Alternative B.

A.6 ( ) ( ) ( ) ( )1 1 3 1Expected value under certainty 50 92 40 64 10 23 12 16 615 4 10 4

= + + + = + + + =

A.7 Solution Approach: Decision tree, since problem is under risk and has more than one

decision.

Wait 1 day

Buy now $70,000

Buy now $55,000

$0

Avail. (.30)

21Wait 1 day

Unavailable(.70)55

$0

Avail. (.60)

33

33

Buy now $30,000

Wait 1 day

Unavailable(.40)

70

$0

(Numbers in Nodes are in $1,000s) Answer : Maximum expected profit: $33,000. Manny should wait 1 day. Then, if an XPO2

is available, he should buy it. Otherwise, he should stop pursuing an XPO2 on the wholesale market.

A-61

A.8 We use a decision table, since expected marginal value of perfect information is asked for.

Profits (= Payoffs)

D1: Good Market: 80(400) 25,000 = $7,000 Bad Market: 70(375) 25,000 = $1,250 D2: Good Market: 85(450) 30,000 = $8,250 Bad Market: 80(425) 30,000 = $4,000 D3: Good Market: 90(475) 33,000 = $9,750 Bad Market: 80(425) 33,000 = $1,000

Decision Alternatives : D1, D2, D3, N (N = Do Nothing)

States of Nature : Good Market, Bad Market

Decision Table and Solution

7,000 1,250D1

8,250 4,000D2

9,750 1,000D3

Good Market(0.6)

Bad Market(0.4)

States of Nature


EV

4,700

6,550

6,250

0 0N 0 The car enthusiast should use design D2. Exp. value under certainty: ( )( ) ( )( )0.6 9,750 0.4 4,000+ = $7,450 Minus Max Expected Value: $6,550 Expected (Marginal) Value $900

A-62

A.9 More than one decision is involved, and problem is under risk, so use a decision tree approach.

20Modest (0.3)

response

160

160

220Sizable (0.7)response

Advertise

40Dont Advertise544

800

Low (0.4)demand

High (0.6)demand

544

LargeFacility

270

242

223

270

Dont expand

Expand

Low (0.4)demand

High (0.6)demand

SmallFacility

200

Note: Payoffs and expected payoffs are in $1000s.

a. Build the large facility. If demand proves to be low, then advertise to stimulate demand. If demand proves to be high, no advertising option is available (so dont advertise).

b. Expected Payoff: $544,000. A.10 Approach: Set up and solve via a decision table.

Cost per dozen bagels: Labor: $16.50 hr

$1.65 dozen10 dozen hr

=

Ingredients: $0.85 hr

$2.50 dozen

+

$3.00 if sold during the day

Profit per dozen bagels$1.00 if not

=

$3.00 $5.50 $2.50

$1.00 $1.50 $2.50

=

=

Suppose: B = dozens baked in A.M. D = dozens sold during the day (= dozens demanded)

Then, ( )$3 $1 iftotal profit$3 if

D B D B DB B D

=

Also, since demand is between 4043 dozen bagels, inclusive, always more profitable to bake 40 dozen than fewer than 40 dozen, AND always more profitable to bake 43 dozen

than more than 43 dozen. From work above, decision table, expected values, and answer

are as follows.

A-63

120 12040

119 12341

118 12242

40(0.20)

41(0.30)

(= States of Nature)

DozensBaked B

EV

120

122.2

123.2

117 12143 122.8

Dozens Demanded D

120 120

123 123

126 126

42(0.35)

43(0.15)

125 129

= DecisionAlternatives

Answer : 42 dozen bagels

A.11 a.

8 (6 + 2) = 0

Ad?

5 (6 + 2) = 3

P(success) = 0.5

Prob (not success) = 0.5

advertise

5 6 = 1Advertise

Dont

EMV = 0.5(0) + 0.5(3) = 1.5

10 6 = 4Rev Cost

PayoffEMV = 0.6(0.4) + 0.4(1) = 2

P(high) = 0.6

P(low) = 0.4Build?

$0Dont Build

Build

2

b. Build studio, but dont advertise (even if demand is low) c. Expected value = $2 million

A.12 a.

0

10 6 = 4

P(high) = 0.6

P(low) = 0.4

Demand

Dont Build

Build

Build

Build?

Ad?Build

EMV = 0.6(4) +0.4(0) = 2.4

0

We dont really careabout this part of the tree(which has a 1 EMV anyways)

Dont Build b. EVPI = EV with PI EMV = 24 2 = 0.4 = $400,000

A-64

B MODULE

Linear Programming

B.1 a. Let T = number of trucks to produce per day

Let C = number of cars to produce per day max z = 300T + 220C

S.T. 1 1

140 60

T C+

1 1

150 50

T C+

, 0T C b. Graph feasible region:

10

20

30

40

50

60

10 20 30 40 50 60

T

C

70 80 90

A

B

C

(1) (2)O

c.

( )( )( )

( ) ( )( )

Point Coordinates valueO 0, 0 0A 0, 50 11,000C 40, 0 12,000B Solve 2 equations in 2 unknowns 12,600

derived from 1 and 2to obtain 20, 30

z

d. Produce 20 trucks and 30 cars daily for a profit of $12,600 per day.

A-65

B.2 We solve this problem by the isocost line method:

2

4

6

8

10

12

2 4 6 8 10 12

x

14 16 18

(6, 6)

1

x2

14

16

18

(0, 4)

z = 12z = 4 z = 112

Answer : Unique optimal solution is (0, 4) with z = 4

B.3 Feasible region is a line segment AB, where A = (0,0), B = (3, 5). Solution via isoprofit line

method is shown.

2

4

6

8

2 4 6

x

(1)

1

x2

8

6

4

z = 20

2

z = 0z = 5

(2)

(3)

A

B

(4)

(5)

Answer : Unique optimal solution is ( ) ( )1 2, 3, 5x x = , with objective function value 20.

A-66

B.4 Using the isoprofit line method.

123456

1 2 3 4 5 6 7 8 9

Opt. sol.789

(2)

1234

(5)

(6)

(4)(3)

(1)z = 9

z = 4 = 6z

A

B

Answer : Unique optimal solution is ( ) ( ), 1, 5A B = . It has objective function 9.0. B.5 Feasible region is same as in Problem B.4. Use Isoprofit line method.

123456

1 2 3 4 5 6 7 8 9

789

1234

z = 1

A

z = 4B

Answer : Problem is feasible and unbounded.

A-67

B.6 Let S = number of standard bags to produce per week Let D = number of deluxe bags to produce per week

( )( )

maximum: 10 8

1300

22

3603, 0

z S D

S D A

S D B

S D

= +

+

+

100

200

300

100 200 300 400 500 600

400

S

D

(A)

(B)

(0, 540)

(240, 180)

(360, 0)

z = $3,840

z = 2,000

z = 1,000

Extreme (Corner) Points Point Profit (0, 0) $0 (0, 300) $2,400 (360, 0) $3,600

(240, 180) $3,840 optimal solution and answer

A-68

B.7

1

2

3

1 2 3 4 5 6

4

x

y

(2, 4)

12

(2, 3)

(2)(3)

(4)

(1)

(4, 1)

(1, 3 1/2)

(5)

5

( )( )( )( )( )

4 line 1

2 line 22 6 line 3

2 8 line 40 line 5

x

x

x y

x y

y

+ +

There are 5 corner (extreme) points. B.8

Foods Cost/

Serving Calories/ Serving

Percent Protein

Percent Carbs

Percent Fat

Fruit/ Vegetable

Apple Sauce $0.30 100 0% 100% 0% 1 Canned Corn $0.40 150 20% 80% 0% 1 Fried Chicken $0.90 250 55% 5% 40% 0 French Fries $0.20 400 5% 35% 60% 0 Mac & Cheese $0.50 430 20% 30% 50% 0 Turkey Breast $1.50 300 67% 0% 33% 0 Garden Salad $0.90 100 15% 40% 45% 1 AS CC FC FF MC TB GS Cost 0.3 0.4 0.9 0.2 0.5 1.5 0.9 Servings 0 1.333333 0.457143 0 1.129568 0 0 $1.51 Constraints AS CC FC FF MC TB GS LHS RHS Cals min 100 150 250 400 430 300 100 800 500 Cals max 100 150 250 400 430 300 100 800 800 Protein min 0 30 137.5 20 86 201 15 200 200 Carb min 100 120 12.5 140 129 0 40 311.4286 200 Fat max 0 0 100 240 215 99 45 288.5714 400 Fruit/Veg Min 100 150 0 0 0 0 100 200 200

A-69

Target Cell (Min) Answer Report (Relevant Section) Cell Name Original Value Final Value $I$14 servings $2.91 $1.51 Adjustable Cells Cell Name Original Value Final Value $B$14 serving A 1.50 0.00 $C$14 serving C 0.00 1.33 $D$14 serving FC 1.33 0.46 $E$14 serving FF 0.00 0.00 $F$14 serving M 0.00 1.13 $G$14 serving T 0.00 0.00 $H$14 serving G 1.40 0.00 Constraints Cell Name Cell Value Formula Status Slack $I$17 Cals min LI 800 $I$17 $J$1 Not Binding 300 $I$18 Cals max L 800 $I$18 $J$1 Binding 0 $I$19 Protein min 200 $I$19 $J$1 Binding 0 $I$20 Carb min L 311.4286 $I$20 $J$2 Not Binding 111.4286 $I$21 Fat max LI 288.5714 $I$21 $J$2 Not Binding 111.4286 $I$22 Fruit + Veg I 200 $I$22 $J$2 Binding 0 Adjustable Cells Sensitivity Report (Relevant Section)

Cell

Name Final Value

Reduced Cost

Objective Coefficient

Allowable Increase

Allowable Decrease

$B$14 serving A 0 0.172602 0.3 1E + 30 0.172602 $C$14 serving C 1.333333 0 0.4 0.258904 0.225581 $D$14 serving FC 0.457143 0 0.9 0.105072 0.100581 $E$14 serving FF 0 0.152691 0.2 1E + 30 0.152891 $F$14 serving M 1.129568 0 0.5 0.062909 0.707833 $G$14 serving T 0 0.169316 1.5 1E + 30 0.169318 $H$14 serving G 0 0.666151 0.9 1E + 30 0.688151 Constraints

Cell

Name Final Value

Shadow Price

Constraint R.H. Side

Allowable Increase

Allowable Decrease

$I$17 Cals min LI 800 0 500 300 1E + 30 $I$18 Cals max L 800 0.00023 800 200 251.6129 $I$19 Protein min 200 0.008983 200 155 40 $I$20 Carb min L 311.4286 0 200 111.4285 1E + 30 $I$21 Fat max LI 288.5714 0 400 1E + 30 111.4286 $I$22 Fruit + Veg I 200 0.001504 200 485.7143 200

A-70

C MODULE

Transportation Modeling

C.1

105

93

8 7

1 2 3 4 SupplyDestination

A

Source 46

31

2 1B

Demand

5

5 4 6 53XX

X X

8 5 0

12 7 1 X

Answer :

a. Final Solution: Shown in the final tableau above.

b. ( ) ( ) ( ) ( ) ( )Total Cost : 5 10 3 9 1 3 6 2 5 1 $97+ + + + = c. Perform optimality test: Result :

Cell

A-3

A-4

A-1

if Opened Up

0

0

0

Change in Cost

Since these numbers

are all nonnegative,

the solution is optimal. Answer : Yes

A-71

C.2 First, apply the optimality test:

Cell Change in Cost if Route

is Opened Up C-MI +3 C-J +4 H-L +2 B-MI 1 B-D +4 Resulting new basic feasible solution:

745

255

4 5

Miami Denver LincolnSan

Supply

Destination

CHI.

Source3 1

355 2

HOU.

Demand

40

Jose

6 950

7 4BUF. 30

70 90 45 50

100

75

80

TOTAL COST: ( ) ( ) ( ) ( ) ( ) ( )30 6 40 3 35 1 55 2 45 4 50 4 $825 Answer+ + + + + = =

A-72

C.3 Total supply = 450 < 500 = Total demand.

6

75

4200

9

A B C Supply

Destination

W

Source

10 5100

8X

Demand

0

12 7 6Y 75

250 100 150

200

175

75

0 0 0Z 50 50Dummy =

Optimal solution and meaning :

Ship 200 tons of grain from W to A Ship 100 tons of grain from X to B Ship 75 tons of grain from X to C Ship 75 tons of grain from Y to C.

Results in demands being met at destinations B and C, but in a shortfall of 50 tons at destination A. Total cost of optimal shipping plan: $2,750.

C.4

4

1

821

6

A B C

1

5 2

2

12

7 9 33 25

9 8 94

17

9

D

4

3

2

1

131

Cost $84 16 34 9 75 9 26 $253= + + + + + + = . This is also optimal.

A-73

C.5

128

84

5 10

A B C

Destination

1

Source: 6 1112

3 72 2

D

4

9

E

4 2

Cost $48 64 40 8 12 36 $208= + + + + + = C.6 The only cell with a negative cost improvement index is HoustonMiami. It achieves a 1. Allocate 10 to that cell. The result is: Denver Yuma Miami Houston 0 0 10 St. Louis 20 0 0 Chicago 0 20 10 Total cost = $170 C.7 2

44 5 9

6 71

8 101

14

33

11 123

Cost $3 1 4 3 4 5 1 8 1 10 3 12 $89= + + + + + =

A-74

C.8 Cell Improvement Index B-2 9 C-1 +6 Result:

2 86

1 2

Destinations

A

Sources:9 6

3B 1

7 7C

6

4

22

57 Cost $12 9 18 14 $53= + + + =

A-75

D MODULE

Waiting Line Models

D.1 CURRENT MACHINE: 40 = , 60 =

1. Utilization ( ) 40 67%60

= =

2. Average number of customers waiting ( ) ( )( )2

40 11

60 60 40 3qL = =

customers

3. Average number of customers in system ( ) 40 260 40s

L = =

customers

4. Average time waiting ( ) ( )40

0.033 hours 2 minutes60 60 40q

W = = =

5. Average time in system ( ) 1 1 hours 3 minutes60 40 20s

W = = =

PROPOSED MACHINE: 40 = , 90 =

1. Utilization ( ) 40 44%90

= =

2. Average number of customers waiting ( ) ( )( )2

400.356

90 90 40qL = =

customers

3. Average number of customers in system ( ) 40 0.890 40s

L = =

customers

4. Average time waiting ( ) ( )40

0.0089 hours 0.533 minutes90 90 40q

W = = =

5. Average time in system ( ) 1 1 hours 1.2 minutes90 40 50s

W = = =

Answer : Thus, we observe that the proposed machine will give better results with a

decrease in queries as well as time in system.

A-76

D.2 Two machine system

a. 40 1

2 60 3M

= = =

. Each machine is busy, on average, 33 percent of the time.

b. qL = average number of customers waiting in line = 0.081

c. 0.081 0.667 0.748s qL L

= + = + = customers in the system

d.

0.0810.002 hours 0.1215 minutes

407.3 seconds average waiting time in the queue

qq

LW = = = =

=

e.

1 10.002 0.0187 hours 1.12 minutes

60

67.3 seconds average time in the system

s qW W = + = + = =

=

The 2-machine system seems to be the best overall, with even further reduction in queues and waiting times.

D.3 This model is, again, the M/M/1 model. The service time of 30 seconds means that the

service rate, , is 120 per hour. a.

40 1120 3

= = = . In this example and in Problem D.2, the system utilization is the

same.

b. ( ) ( )2 240 1

0.167120 120 40 6q

L

= = = = customers. The average number of

customers waiting in line is twice as large as the average number (0.081) found in Problem D.2.

c. 40 1

120 40 2sL

= = = customer. On the average, the number of customers in

the system is 1

2, rather than the

3

4 (0.748) found in Problem D.2.

d. ( ) ( )40 1 1

hour minutes 15 seconds120 120 40 240 4q

W

= = = = = , which

exceeds qW in Problem D.2.

e. 1 1 1 3

hour minutes 45 seconds120 40 80 4s

W = = = = = . Thus, average time

spent in the system is 15

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