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Instructor’s 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



JAY HEIZER



Texas Lutheran University

BARRY RENDER Upper Saddle River, New Jersey 07458



Rollins College



VP/Editorial Director:Jeff Shelstad Executive Editor:Mark Pfaltzgraff Senior Editor:Alana Bradley Senior Managing 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. TM

Pearson Prentice Hall

is a trademark of Pearson Education, Inc.

10 9 8 7 6 5 4 3 2 1 

Contents Homework Problem Answers Chapter 1 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Supplement 6: Chapter 7 Supplement 7: Chapter 8 Chapter 9 Supplement 10: Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Module A: Module B: Module C: Module D:

Operations and Productivity ........................................................................... A-1 Project Management ....................................................................................... A-3 Forecasting ...................................................................................................... A-7 Design of Goods and Services ...................................................................... A-11 Managing Quality ......................................................................................... A-15 Statistical Process Control ............................................................................ A-18 Process Strategy ............................................................................................ A-20 Capacity Planning ......................................................................................... A-23 Location Strategies ....................................................................................... A-27 Layout Strategy ............................................................................................. A-30 Work Measurement ...................................................................................... A-34 Inventory Management ................................................................................. A-36 Aggregate Planning ...................................................................................... A-42 Materials Requirements Planning (MRP) & ERP ........................................ A-46 Short-Term Scheduling ................................................................................. A-51 Just-In-Time and Lean Production Systems ................................................. A-55 Maintenance and Reliability ......................................................................... A-57 Decision Making Tools ................................................................................ A-59 Linear Programming ..................................................................................... A-64 Transportation Modeling .............................................................................. A-70 Waiting Line Models .................................................................................... A-75

Module E: Module F:

Learning Curves ........................................................................................... A-79 Simulation ..................................................................................................... A-80

iii



1

Operations and Productivity

CHAPTER

1.1

! Last year’s " number of units of output # total factor $ # $ = total dollar value of all inputs used #productivity$ % &

a.



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

! This year’s " number of units of output # total factor $ = # $ #productivity $ total dollar value of all inputs used % &

b.



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

c.



= 0.0567 units dollar

! This year’s " ! Last year’s " # total factor $ − # total factor  $ # $ # $ #productivity$ # productivity$ % & % & × 100% = 0.0567− 0.0546× 100%  0.0546 ! Last year’s " # total factor $ #productivity$ % & = + 3.84% ≈ 3.8%

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

A-1

1.2

0.15 = So L =

57,600

(160)(12)(L )

, where L = numberof laborers employed at the plant.

57,600

(160 )(12)( 0.15)

= 200 

Answer : 200 laborers 1.3

Output = 28,000 customers  There are 4 approaches to solving the problem correctly: 1. Input = 7 workers  28,000 Then, = 4,000 customers worker 7 Input = 7 ( 40) labor weeks  2. Then, 3.



1.5

= 100 customers labor week

28,000 7 (40 )( 50 )

= 2 customers labor hour

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

1.4

7 ( 40 )

Input = 7 ( 40)( 50) labor hours  Then,

4.

28,000

28,000 7 (40 )($250 )

6, 600 Cadillacs

= 0.40 customers per dollar of labor

= 0.10

(x )( labor hours)

 x = 66, 000 labor hours There are 300 laborers. So, 66,000 labor hours = 220 labor hours laborer 300 laborers $ output labor hour

=

52 ( $90) + 198( $80) 8 ( 45 )

=

20, 520 360

= $57.00per labor hour

A-2

Project Management

3

CHAPTER

Gantt Chart

3.1

A

20

B

80 120

C

110

D

140

E F

150 170

G

160

H I 50

3.2

100 Hours

150

200 200 

AON Network: 60

30

20

B Purchasing

D Sawing

G Infill

20

20

10

30

A Planning

E Placement

F Assembly

I Decoration

100 10

C Excavation

H Outfill



A-3

3.3

AOA Network:

Excavate



3.4

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.

ES = 20

LF = 20

ES = 80

ES = 120

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

ES = 150 EF = 170 LS = 150

LF = 150

LF = 170

LF = 200



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-4

3.5

! a + 4m + b " !b −a" Variance # $ $ 6 % & % 6 &

A:

120 6

D: E:

6 120 6 60

F: G: H: I:

180

= 20

6 60 6

180 6

= 30 = 20

= 10

6 120

! b − a" $ % 6 &

Standard Deviation#

2

B : 360 = 60 6 600 C: = 100 6

3.6

2

Mean: #

= 20

= 10 = 30

A:

( 20 ) 36

= 11.11

A:

2

20 6

= 3.33

B : ( 60 ) = 100.00 B : 60 = 10.00 36 6 2 (120 ) 120 C: = 400.00 C : = 20.00 36 6 2 (10 ) 10 D: D: = 2.78 = 1.67 36 6 2 (10) 10 = 2.78 =1.67 E: E: 36 6 2 (0) 0 = 0.00 F: F : = 0.00 36 2 6 ( 40 ) 40 = 44.44 = 6.67 G: G: 36 6 2 (4) 4 = 0.44 H: H : = 0.67 36 6 2 ( 40 ) 40 = 44.44 = 6.67 I: I: 36 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 . µ= mean

time of critical path = 200 hrs σ = 22.4 hrs  240 − 200 40 = = 1.8. Using the cumulative normal distribution table in Thezvalue = 22.4 22.4 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-5

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. B 1

E 2

A 1

F 2 C 4

b.

c. d.

3.9



The critical path, listing all critical activities in chronological order: A → B → E → F 1 + 1 + 2 + 2 = 6 ( not CP)  A→ C→ F 1+ 4 + 2 = 7. This is the CP. The project duration (in weeks): 7 (This is the length of CP.) The slack (in weeks) associated with any and all non-critical paths through the project:Look at the paths that aren’t critical—only 1 here—so from above: A → B → E → F  7 − 6 = 1 week slack.

We have only 1 activity with probabilistic duration. Due date − µ 8 − (1 + 4 + 2 ) 1 Z = = = = 2 (length of entire path is 7, not 4). For a z= 2 , 0.5 0.5 σ this means P(Due date < 8 ) = 97.72% (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-6

4

Forecasting

CHAPTER

4.1

Present = period ( week ) 6.  So: F7 =

4.2

t 1 2 3 4

1 3

A6 +

1 4

A5 +

1 4

A4 +

1 6

A3 =

1 3

1

1

1

4

4

6

(52 ) + ( 63) + ( 48 ) + ( 70 ) = 56.75 patients 

At Ft 120 —  136 —  114 128 116 125

120 + 136

+ , 2 2 - Checking Data  136 + 114 250 , = = 125 ,. 2 2 =

256 

116  +114 F = 5

4.3

2

Method 1:

= 128,

230 = 2 = 115 = Answer  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-7

4.4

y = a + bx  4

/x y i

i =1

i

= 58,538

x = 75.75



y = 191.5 n

/x

2 i

= 23,209

i =1

58, 538 − 4 (75.75 )(191.5 )

b=

23, 209 − 4( 75.75)

2

=

513.50 =2 256.75

a =191.5 − 2 ( 75.75) = 40 y ≈ 40 + 2 x



x = 85

0  ≈ y

210 Actual Forecast  Demand Demand 88 88 72 88 68 84 48 80 ← Answer 72

4.5

Day Monday Tuesday Wednesday Thursday Friday

t 1 2 3 4 5

1 F

t  = αA

t +1

F2 = F3 = F4 = F5 =

1 4 1 4 1 4 1 4

) F. Let = 4 . Let Monday forecast demand = 88

1 − α

+(

t

α

3

( 88) + ( 88) = 88 4 3

( 72) + ( 88) = 18+ 66 = 84 4 3



( 68) + ( 84) = 17 + 63 = 80 4 3

( 48) + ( 80) = 12 + 60 = 72 4

A-8

4.6

Winter Spring Summer Fall 2001 1, 400 1,500 1, 000 600 2002 1, 200 1, 400 2,100 750  2003 1, 000 1,600 2, 000 650 2004 900 1,500 1,900 500 4,500 6,000 7,000 2,500 20,000 Average over all seasons: = 1,250 16 6,000 = 1,500 Average over spring: 4 1,500 = 1.2 Spring index: 1, 250

! 5,600 " $ (1.2 ) = 1, 680 sailboats % 4 &

Answer : #

4.7

We need to find the smoothing constant α. We know in general thatFt +1 = α At + (1 −α )Ft , t= 1, 2, 3 . Choose either t= 2 or t = 3 ( t= 1 won’t let us find αbecause F2 = 50  = α ( 50 ) + (1 − α) 50 holds for anyα). Let’s pick, e.g., t = 2 . Then F  = α( 42) +( 1 − α) 50 . So 3 = 48

48 = 42α + 50 − 50α −2 = − 8α



1 4 =α .  α (46 ) + (1 − α ) 50 , with Now we can findF5 : F5 =

F 5=

1 4

3

(46 ) + (50 ) =49 ←Answer  4

A-9

α =

1 4

. So

4.8

Let X 1, X 2,

, X6 be the prices; Y1 , Y2 ,

!

, Y6 be the number sold.

!

6

/X i

 i =1

X =

= Average price = 3.25833 

6

(1)

6

Y =

/Y = Average number sold = 550.00 i =1

i

6

(2)

!# All calculations to the"$  1 ## nearest th $$ 100,000 % &

6

/X Y = 9,783.00  i =1

6

/X i =1

(3)

i i

2 i

= 67.1925 

(4)

Then y ≈ a + bx , where y = number sold, x= price , and

/X Y − n ( X )( Y) 6

b=

i i

i =1

6

/X i =1

2 i

=

2

− n ( X)

( 9, 783) − 6 ( 3.25833 )( 550) − 969.489 = = − 277.61395 2 3.49222 67.1925 − 6( 3.25833) 

a = ( Y ) − b ( X ) = 1,454.5578

So at x = 1.80 , y =1, 4 54.5578 −277.61395( 1.80) = 954.85270 . Now round to the nearest integer: Answer : 955 dinners

/ (A − F ) n

t

4.9

Tracking Signal =

t

t= 1



Month

MAD At  Ft 

May June July August September October

100 80 110 115 105 110

100 104 99 101 104 104

0 24 11 14 1 6

0 –24 11 14 1 6

November December

125 120

105 109

20 11 Sum: 87

20 11 Sum: 39

87 = 10.875  8 39 Answer : = 3.586  10.875

At − Ft 

So: MAD :

1 ! " # to the nearest 1,000 th $  % & A-10

(At − Ft ) 

5.1

(0.80) 90 of 100

$42,500

–$32,500

$12,500

–$43,750

–$18,750

80 of 100 non-defect

–$75,000



Answer: $27,500—use K1 Outcome Calculations

90 10 1 ( 500)( 300)( $1.20) − ( 500)( 300)( $1.30) = ,−$100, 000 + 100  2,−$100, 000 + 100 $162, 000− $19, 500 = $42, 500 3 70 30 1 ( 150, 000) ( $1.20) − ( 150, 000)( $1.30) = ,−$100, 000 +  100 100 2 ,3−$100, 000 + $126, 000− $58, 500= − $32, 500

A-11 

90 10 1 (150, 000)( $1.20) − ( 150, 000)( $1.30) = ,−$130, 000 +  100 100 2 ,3−$130, 000+ $162, 000− $19, 500= $12, 500 1 $130, 000 75 150, 000 $1.20 25 150, 000 $1.30 + ( )( )− ( )( )= ,−  100 ,23−$130, 000 + 100 $35, 000− $48, 750 = − $43, 750 95 5 1 (150, 000)( $1.20) − ( 150, 000)( $1.30) = ,−$180, 000 +  100 100 2 ,3−$180, 000 + $171, 000− $9, 750= − $18, 750 80 20 1 (150, 000)( $1.20) − ( 150, 000)( $1.30) = ,−$180, 000 +  100 100 2 ,−$180, 000 + $144, 000− $39, 000 = − $75, 000 3 (0.3) F market

5.2

80.0

66.0 Use D0

(0.7) U market (0.4) F market

84.0

Use D1

60.0

99.0

84.0 (0.6) U market (0.6) F market

Use D2

74.0

89.2

80.2 (0.4) U market

66.7

(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, 000 D1− U : 750( 100, 000) − 1, 000, 000 = $74, 000, 000 D2 − F : 1, 000( 90, 000) − 800, 000

= $89, 200, 000

D2 − U : 750( 90, 000) − 800, 000

= $66, 700,000

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

A-12

(0.3) Demand rises

5.3

$10,000 Purchase overhead hoist

$14,000

(0.5) Demand stays $10,000 same

(0.2)falls Demand

–$20,000

(0.4) Demand rises

$20,000

Purchase forklift $14,000

Do nothing

$30,000

(0.6) Demand stays $10,000 same

$0



Answer : Maximum expected payoff = $14,000  5.4

$0 Do nothing High demand (0.6) $300,000 Use A

160K

Low demand (0.4)

–$50,000

High demand (0.6) $300,000

Use B 180K Use C

Low demand (0.4) $0 High demand (0.6) $250,000

302K Low demand (0.4) Upgrade to D

$380,000

380K $0 No upgrade to D Note: K = $1,000’s



Answer : Use Design C. If demands turns out to be low, upgrade to Design D.

A-13 