Heizer Operation Management Solution Pdf 3n2i6m

Instructor’s Solutions Manual for Additional Problems

Operations Management

EIGHTH EDITION

Principles of

Operations Management

SIXTH EDITION

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 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

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: Module E: Module F:

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

iii

1 CHAPTER

1.1

a.

Operations and Productivity

 Last year’s  number of units of output    total factor  = total 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  =

b.

12, 000 units = 0.0546 units dollar $219,800

 This year’s  number of units of output   total factor =    productivity  total dollar value of all inputs used   =

14, 000 units (14, 000 )( $2.05 ) + (16, 000 )( $11.00 ) +

(1,800 )( $7.50 ) + ( 3,800 )( $0.75 ) + $26, 000 =

c.

14, 000 units = 0.0567 units dollar $247, 050

 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

57, 600 , where L = number of laborers employed at the plant. (160 )(12 )( L ) 57, 600 So L = = 200 (160 )(12 )( 0.15) 0.15 =

Answer : 200 laborers

1.3

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

Then, 3.

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

4.

28, 000 = 2 customers labor hour 7 ( 40 )( 50 )

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

1.4

28, 000 = 100 customers labor week 7 ( 40 )

28, 000 = 0.40 customers per dollar of labor 7 ( 40 )( $250 )

6, 600 Cadillacs = 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

1.5

52 ( $90 ) + 198 ( $80 ) 20,520 $ output = = = $57.00 per labor hour labor hour 8 ( 45 ) 360

A-2

3

Project Management

CHAPTER

Gantt Chart

3.1 A

20 80

B C

120

D

110

E

140

F

150

G

170

H

160

I

3.2

150

200 200

50

100 Hours

60

30

20

B Purchasing

D Sawing

G Infill

AON Network:

20

20

10

30

A Planning

E Placement

F Assembly

I Decoration

100 10 C Excavation

H Outfill

A-3

3.3

AOA Network: G Infill

1

A Plan

2

B Purchase

D Saw

3

E Place

4

5

F Assemble

6

7 Dummy Outfill H

8

I Decorate

9

C Excavate

3.4

Path 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

Task Times (Hours) 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

Total Hours 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.

1

A Planning ES = 0 EF = 20 LS = 0 LF = 20

2

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

3

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

4

ES = 120 EF = 140 LS = 120 LF = 140 Placement E

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

5

ES = 150 EF = 170 LS = 150 7 LF = 170 Infill Dummy F I G H Assembly Decoration Outfill 6 8 9 ES = 140 ES = 150 ES = 170 EF = 150 EF = 160 EF = 200 LS = 140 LS = 160 LS = 170 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

3.6

 a + 4m + b  Mean:   6   120 A: = 20 6 360 B: = 60 6 600 C: = 100 6 180 D: = 30 6 120 E: = 20 6 60 F: = 10 6 120 G: = 20 6 60 H: = 10 6 180 I: = 30 6

2

b−a Variance    6  2 20 ) ( A: = 11.11 36 2 60 ) ( B: = 100.00 36 2 120 ) ( C: = 400.00 36 2 10 ) ( D: = 2.78 36 2 10 ) ( E: = 2.78 36 2 0) ( F: = 0.00 36 2 40 ) ( G: = 44.44 36 2 4) ( H: = 0.44 36 2 40 ) ( I: = 44.44 36

b−a Standard Deviation    6  20 A: = 3.33 6 60 B: = 10.00 6 120 C: = 20.00 6 10 D: = 1.67 6 10 E: = 1.67 6 0 F : = 0.00 6 40 G: = 6.67 6 4 H : = 0.67 6 40 I: = 6.67 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 The z value = = = 1.8 . Using the cumulative normal distribution table in 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.

The critical path, listing all critical activities in chronological order: A → B → E → F 1 + 1 + 2 + 2 = 6 ( not ) A→C→F 1 + 4 + 2 = 7. This is the . The project duration (in weeks): 7 (This is the length of .) 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.

3.9

We have only 1 activity with probabilistic duration. Due date − µ 8 − (1 + 4 + 2 ) 1 = = = 2 (length of entire path is 7, not 4). For a z = 2 , Z= σ 0.5 0.5 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- path will be only 6 weeks)

3.10

Helps to modify the AON with the lowest costs to crash 1. 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 CHAPTER

4.1

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

4.2

Forecasting

t 1 2 3 4

1 1 1 1 1 1 1 1 A6 + A5 + A4 + A3 = ( 52 ) + ( 63) + ( 48 ) + ( 70 ) = 56.75 patients 3 4 4 6 3 4 4 6

At Ft 120 — 136 — 114 128 116 125

120 + 136 256  = = 128  2 2  Checking Data 136 + 114 250 = = 125  2 2 116 + 114 230 F5 = = = 115 = Answer 2 2 4.3

Method 1:

Method 2:

MAD : MSE : MAD : MSE :

0.20 + 0.05 + 0.05 + 0.20 = 0.5000 ← better 0.04 + 0.0025 + 0.0025 + 0.04 = 0.0850 0.1 + 0.20 + 0.10 + 0.11 = 0.5100 0.01 + 0.04 + 0.01 + 0.0121 = 0.0721 ← better

A-7

4.4

y = a + bx 4

∑x y i

i

= 58,538

i =1

x = 75.75 y = 191.5 n

∑x

2 i

i =1

b=

= 23, 209

58,538 − 4 ( 75.75)(191.5) 23, 209 − 4 ( 75.75 )

2

=

513.50 =2 256.75

a = 191.5 − 2 ( 75.75 ) = 40 y ≈ 40 + 2 x x = 85 ⇒ y ≈ 210 4.5

t 1 2 3 4 5

Day Monday Tuesday Wednesday Thursday Friday

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

Ft +1 = α At + (1 − α ) Ft . Let α =

1 . Let Monday forecast demand = 88 4

1 3 (88 ) + (88 ) = 88 4 4 1 3 F3 = ( 72 ) + ( 88 ) = 18 + 66 = 84 4 4 1 3 F4 = ( 68 ) + ( 84 ) = 17 + 63 = 80 4 4 1 3 F5 = ( 48 ) + ( 80 ) = 12 + 60 = 72 4 4 F2 =

A-8

4.6

4.7

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 = 1, 250 Average over all seasons: 16 6, 000 = 1,500 Average over spring: 4 1,500 = 1.2 Spring index: 1, 250  5,600  Answer :   (1.2 ) = 1, 680 sailboats  4  We need to find the smoothing constant α. We know in general that Ft +1 = α At + (1 − α ) Ft , t = 1, 2, 3 . Choose either t = 2 or t = 3 ( t = 1 won’t let us find α because

F2 = 50 = α ( 50 ) + (1 − α ) 50 holds for any α). Let’s pick, e.g., t = 2 . Then F3 = 48 = α ( 42 ) + (1 − α ) 50 . So

48 = 42α + 50 − 50α −2 = −8α 1 =α. 4 Now we can find F5 : F5 = α ( 46 ) + (1 − α ) 50 , with α = F5 =

1 3 ( 46 ) + ( 50 ) = 49 ← Answer 4 4

A-9

1 . So 4

4.8

Let X 1 , X 2 , … , X 6 be the prices; Y1 , Y2 , … , Y6 be the number sold. 6

∑X

X=

i

= Average price = 3.25833

i =1

6

(1)

6

∑Y

i

Y =

i =1

6

= Average number sold = 550.00

(2)

 All calculations to the    1  nearest  th   100, 000  

6

∑ X Y = 9, 783.00

(3)

∑X

(4)

i i

i =1 6

2 i

= 67.1925

i =1

Then y ≈ a + bx , where y = number sold , x = price , and 6

∑ X Y − n ( X )(Y ) ( 9, 783) − 6 ( 3.25833)( 550 ) −969.489 b= = = = −277.61395 3.49222 − 67.1925 6 3.25833 ( ) ∑ X − n( X ) i i

i =1

6

2

2

2 i

i =1

a = (Y ) − b ( X ) = 1, 454.5578

So at x = 1.80 , y = 1, 454.5578 − 277.61395 (1.80 ) = 954.85270 . Now round to the nearest integer: Answer : 955 dinners n

∑( A − F ) t

4.9

Tracking Signal = Month May June July August September October November December

t

t =1

At

100 80 110 115 105 110 125 120

MAD Ft 100 104 99 101 104 104 105 109

At − Ft

( At − Ft )

0 24 11 14 1 6 20 11 Sum: 87

0 –24 11 14 1 6 20 11 Sum: 39

87 = 10.875 8 39 1   Answer : th  = 3.586  to the nearest 10.875 1,000  

So: MAD :

A-10

5

Design of Goods and Services

CHAPTER

(0.80) 90 of 100 non-defect

5.1

$42,500

$27,500 (0.20) 70 of 100 non-defect

Use K1

$27,500 Use K2

–$32,500


$12,500


–$43,750


–$18,750


–$75,000

$4,062.50

Use K3

$24,375

Answer: $27,500—use K1 Outcome Calculations 90 10  ( 500 )( 300 )( $1.20 ) − ( 500 )( 300 )( $1.30 ) = −$100, 000 + 100 100  −$100, 000 + $162, 000 − $19, 500 = $42,500 70 30  (150, 000 )( $1.20 ) − (150, 000 )( $1.30 ) = −$100, 000 + 100 100  −$100, 000 + $126, 000 − $58,500 = −$32, 500

A-11

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

5.2

80.0

66.0 (0.7) U market

Use D0

(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 D2 − U : 750 ( 90, 000 ) − 800, 000

= $89, 200, 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) Demand falls

–$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) High demand (0.6)

Use B 180K Use C

–$50,000 $300,000

Low demand (0.4) $0 High demand (0.6)

$250,000

302K Low demand (0.4) D

$380,000

380K No D

$0

Note: K = $1,000’s

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

A-13

5.5

Bread & Rolls No Bread & Rolls Pies & Cakes No Full Service No

5.6

(p = 0.40)

$15,000

Bread & Rolls EMV = $12,000 No (p = 0.60) (p = 0.40) Bread & Rolls Pies & Cakes

$10,000 $25,000

EMV = $13,000 No (p = 0.60) (p = 0.40)

$5,000 $35,000

Full Service EMV = $7,500 No (p = 0.60)

–$10,000

Based upon this decision tree, Jeff should consider most seriously the medium-sized shop carrying bread, rolls, pies, and cakes.

A-14

6

Managing Quality

CHAPTER

6.1

1. 2. 3. 4. 5. 6. 7. 8. 9. Item Rated 1. 2. 3. 4. 5. 6. 7. 8. 9. Item Rated 1. 2. 3. 4. 5. 6. 7. 8. 9. a. b.

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

A 20 4 19 4 0 9 19 0 0

4 80 16 76 16 0 36 76 0 0

B 28 2 20 5 0 30 18 26 0

Overall Grade C 1 30 3 25 27 7 13 24 0

D 1 14 8 5 18 0 0 0 20

E 0 0 0 11 7 4 0 0 30

3 84 6 60 15 0 90 54 78 0

Weights 2 2 60 6 50 54 14 26 48 0

1 1 14 8 5 18 0 0 0 20

0 0 0 0 0 0 0 0 0 0

Highest rated is appearance of food; 2.61. Lowest rated is view of stage; 0.31. A-15

Total 167 96 150 86 72 140 156 126 20

Average 2.61 1.50 2.34 1.34 1.13 2.19 2.44 1.97 0.31

c.

A check sheet will help categorize the comment cards Check Sheet Positive 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

d.

6.2

! ! !

!! ! !!!!!! ! ! ! ! chilly

!

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.

a.

y minutes

14 13 12 11 10 9 8 0 b.

!!!! ! !

Negative

1

2

3

4

5

6

This is a scatter diagram.

A-16

x Trips

6.3

a.

b. 6.4

40 35 30 25 20 15 10 5 0

34

36

30 24

14 10 6 W

R

I

4 M

2 O

39% of complaints are W, demeaning towards women. Machines

Manpower

Incorrect measurement

Inadequate cleanup

Operator misreads display

Technician calculation off

Temperature controls off

Antiquated scales

Variability Equipment in disrepair

Inadequate flow controls Incorrect Formulation

Jars mislabeled Incorrect weights

Lack of clear instructions Damaged raw material Priority miscommunication

Incorrect maintenance Inadequate instructions

Methods

Materials

A-17

6

Statistical Process Control

SUPPLEMENT

S6.1

 σ   25  We are given a target of X = 420 . So LCL = X − Z   = 420 − 4   = 400 .  n  25   σ   25  UCL = X + Z   = 420 + 4   = 440 . Thus,  n  25  Answer : LCL = 400 calories

UCL = 440 calories S6.2

p=

7 250

5 9 + 250 + " + 250 7 + 5 +" + 9 300 = = = 0.040 30 7,500 7,500

UCL = p + Z LCL = p − Z

S6.3

p (1 − p ) n p (1 − p ) n

= 0.040 + 3 ( 0.01239 ) = 0.077 = 0.040 − 3 ( 0.01239 ) = 0.003

We want Z = 2 , since (1 − 0.0455 ) = 0.9545 which implies Z = 2 from the Normal Table. UCL = c + 2 c , where c = average number of breaks = 3 : 3 + 2 3 = 6.46 .

S6.4

Z = 3 for x -chart . Here, n = 4 so A2 = 0.729 (from Table S6.1). x = 2.0 , R = 0.1 ,

UCL x = x + A2 R = 2.0 + 0.729 ( 0.1) = 2.07 S6.5

C chart

0.0027 1.0000 − 0.0027 0.9973 = = 0.49865 ⇒ Z = 3 (see Normal Table) 2 2 UCL = c + 3 c = 1.5 + 3 1.5 = 5.17 A-18

S6.6

 σ  x±Z  = answers  n 384 x= = 16 lbs. 24  σ   0.12  Z  = 2  3  = 0.08    n 16.00 + 0.08 = 16.08 = UCL x 16.00 − 0.08 = 15.92 = LCL x

S6.7

x = 1.00 , R = 0.10 , A2 = 0.483 (from Table S6.1),

LCL = x − A2 R = 1 − ( 0.483)( 0.10 ) = 0.9517 weeks S6.8

R = 3.25 mph, Z = 3 , with n = 8 , from Table S6.1, UCL = 1.864 R = 6.058

LCL = 0.136 R = 0.442 30

S6.9

p=

∑

Number of defects 250

i =1

30

UCL p = p + 2 LCL p = p − 2

S6.10 a.

b.

=

300 = 0.04 , n = 250 7,500

( 0.04 )( 0.96 ) 250

( 0.04 )( 0.96 ) 250

= 0.04 + 2 ( 0.0124 ) = 0.0648 = 0.04 − 2 ( 0.0124 ) = 0.0152

We are counting attributes and we have no idea how many total observations there are (the proportion of drivers who weren’t offended enough to call!) This is a C-chart. 36 = 6 for c , as true c is unknown. Use mean of 6 weeks of observations 6 UCL = c + z c = 6 + 3 ( 2.45 ) = 13.3 LCL = c − z c = 6 − 3 ( 2.45 ) = −1.3 , or 0.

c.

It is in control because all weeks’ calls fall within interval of [ 0, 13] .

d.

Instead of using

36 = 6 , we now use c = 4 . UCL = 4 + 3 4 = 4 + 3 ( 2 ) = 10 . 6 LCL = 4 − 3 ( 2 ) = −2 , or 0. Week 4 (11 calls) exceeds UCL. Not in control.

A-19

7

Process Strategy

CHAPTER

7.1

a.

Find breakeven points, X p . Mass Customization: 1, 260, 000 + 60 X = 120 X → X p = 21, 000

b.

Intermittent:

1, 000, 000 + 70 X = 120 X → X p = 20, 000

Repetitive:

1, 625, 000 + 55 X = 120 X → X p = 25, 000

Continuous:

1,960, 000 + 50 X = 120 X → X p = 28, 000

Find least-cost process at X = 24, 000 units . Fixed cost VC Units Mass Customization: 1, 260, 000 + 60 ( 24, 000 ) = 2, 700, 000 Intermittent: Repetitive:

1, 000, 000 + 70 ( 24, 000 ) = 2, 680, 000 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.

24,000 #$% > 20,000 #$% ? yes! Anticipated Production Volume

Intermittent Process Breakeven Point

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-20

7.2

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

3 Cost R (Millions 2 MC of dollars) 1 I 0

V 0

10 P1 15P2 20

5

25

1,000’s of Ovens Finding value of P2: 1, 250, 000 + 50 ( P2 ) = 2, 000, 000 + 5 ( P2 ) . So P2 = 16, 666 (Note: P1 = 12, 500 ). Answer : For volumes of production V such that 16, 666 23 ≤ V ≤ 25,000 . 7.3

I R

14 12

C

C

10 Cost (millions)

R

8 6 4

I

2 0

Volume V 0

5,000 10,000 15,000 20,000 7,500

1, 000, 000 + 1, 650 x = 3, 000, 000 + 1, 250 x   400 x = 2, 000, 000 I&R   x = 5, 000 Intersect  1, 000, 000 + 1, 650 ( 5, 000 ) = $9, 250, 000  3, 000, 000 + 1, 250 x = 7,500, 000 + 650 x   600 x = 4,500, 000 R&C   x = 7,500 Intersect  3, 000, 000 + 1, 250 ( 7,500 ) = $12,375, 000  For all V between 5, 000 ≤ V ≤ 7,500

A-21

2 3

units .

7.4

Breakeven points R : 21, 000, 000 + 450 x = 750 x ⇒ x = 70, 000 a. C : 26, 250, 000 + 400 x = 750 x ⇒ x = 75, 000 M : 15, 000, 000 + 500 x = 750 x ⇒ x = 60, 000 b. Least cost process at x = 65, 000 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 Continuous : 2, 400, 000 + 20 x = 80 x ⇒ x = 40, 000 a. Repetitive : 1,950, 000 + 30 x = 80 x ⇒ x = 39, 000 Mass Customization : 1, 480, 000 + 40 x = 80 x ⇒ x = 37, 000 Intermittent : 1,800, 000 + 40 x = 80 x ⇒ x = 45, 000 b. Least cost process at x = 48, 000 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 (2,000; 300,000) (4,000; 860,000) (11,000; 1,350,000)

7.6

M $

R C I 2,000 4,000

11,000

Volume 15,000

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-22

7

Capacity Planning

SUPPLEMENT

S7.1

Problem is under risk and has two decisions, so use a decision tree: Medium demand 70 (0.4) Small expansion

No additional expansion 90

109 High demand (0.6)

148

135

Additional minor expansion 135 Large expansion

Medium demand 40 (0.4) 148 High demand (0.6)

220

(Payoffs and Expected Payoffs are in $1,000’s) Answer : Ralph should undertake a large expansion. Then the annual expected profit will equal $148,000.

A-23

No additional minor expansion

S7.2

$90,000

High demand 140,000 (0.3) Small expansion

Additional minor expansion

70,000

Medium demand (0.7)

$40,000

$70,000 High demand (0.3) Large expansion

$105,000

14,000

Medium demand (0.7)

–$25,000

Maximum value = $70,000 S7.3

No expand

$0 Demand up small (0.4)

$18,000

Small expand

16,000 Demand up medium (0.6)

Large expand

$10,000

Demand up medium (0.3)

$20,000

–$10,000

18,000 Demand up large (0.7)

Answer : $18,000

A-24

$34,000

$140,000

S7.4

(3) (1)

300,000

(2) 200,000

(2) (1)

(2) (1)

100,000 50,000 (3)

x

250 400

1,000

2,000

Cap level (2) is lowest for all x so 1, 000 ≤ x ≤ 2, 000 S7.5

Effective Capacity (text Equation S7-3) Efficiency 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 =

Actual (or expected) Output =

  200 cars    38.4 cars per bay  S7.6

a.

b.

F 450 = = 878.05 ∑ 1 − (Vi Pi ) Wi 0.5125 Breakeven ( $ ) = $878.05 Number of pizzas required at breakeven: Whole pizzas = ( 878.05 × 0.30 ) 5.00 = 52.7 → 53 BEP ( $ ) =

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.

that Yr 0 has no discounting. Initial coat $1,000,0000 yearly maint Salvage cost $50,000 yearly dues Discount rate 0.100 Year 0 1 2 3 4 5

Cost $1,075,000 75,000 75,000 75,000 75,000 75,000

Revenues $300,000 300,000 300,000 300,000 300,000 350,000 undisc. Profit A-25

75,000 $300,000 500

Profit $–775,000 225,000 225,000 225,000 225,000 275,000 400,000

dues/member $600

PV Mult PV Profit 1 –$775,000 0.9 $202,500 0.81 $182,250 0.729 $164,025 0.6561 $147,623 0.59049 $162,385 PV Profit $83,782

b.

Assume dues are collected at the beginning of each year. This is a simplification—in reality, people are likely to 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. 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 Year 0 1 2 3 4 5

Cost $0 0 0 0 0 0

(hip fee) Revenues Profit $600 $600 600 600 600 600 600 600 600 600 600 600 undisc. Profit 3,600

PV Mult 1 0.9 0.81 0.729 0.6561 0.59049 PV Profit

PV Profit $600 $540 $486 $437 $394 $354 $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, it’s better for her to pay yearly. S7.8

Breakeven: Costs = Revenues 500 + 0.50 × b = b × 0.75 where b = number of units at breakeven or b ( 0.75 − 0.50 ) = 500 , 500 = 2,000 units 0.25 breakeven in units = 2,000 units breakeven in dollars = $0.75 × 2, 000 = $1, 500

and b = a. b.

A-26

8 CHAPTER

8.1

Cx

( 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 )( 9.0 ) = = 6.67 2, 000 + 5, 000 + 10, 000 + 7, 000 + 10, 000 + 20, 000 + 14, 000 ( 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 )( 2.5 )

Cy = 8.2

8.3

Location Strategies

68, 000

Site A B C D

Total Weighted Score 174 185 187 165

∑ Population weights = 5, 000 Cx = Cy =

( 0 )( 2, 050 ) + ( −1)( 550 ) + ( 2 )(1, 025 ) + ( 3)( 775 ) + ( −2 )( 250 ) + ( −2 )( 350 ) = 0.525 5, 000 ( 0 )( 2, 050 ) + (1.5)( 550 ) + ( −1)(1, 025) + ( 3)( 775) + ( −3)( 250 ) + ( −1)( 350 ) 5, 000

Coordinates: ( 0.525, 0.205 ) 8.4

= 3.02

A 150 (0, 25) B 125 (1/2, 30) $ cost 100 (millions) 75 (2, 60) C 50 25 (1, 35) (0, 20) (0, 10) 0 1 2 3 4 5 6 10,000’s of Autos = V For all V such that 10, 000 ≤ V ≤ 60, 000 . A-27

V

= 0.205

8.5

Site A B C D

Score 5w + 320 4 w + 330 3w + 370 5w + 255

Find all w from 1–30 so that: 3w + 370 ≥ 5w + 320 ⇔ 50 ≥ 2w ⇔ w ≤ 25 3w + 370 ≥ 4 w + 330 ⇔ 40 ≥ w ⇔ w ≤ 40 3w + 370 ≥ 5w + 255 ⇔ 115 ≥ 2w ⇔ w ≤ 57.5 Answer : For all w such that 1.0 ≤ w ≤ 25.0 (1, 35)

8.6 12 10 TC (millions $)

Char

8

L.A. K.C. Charlotte

K.C. (20, 7) (30, 9.5)

6 4

L.A.

2

V 0

5

10

15

20

25

30

35 (thousands) cutoff

4,100, 000 + 180V ≤ 1, 000, 000 + 300V  For all V so: 4,100, 000 + 180V ≤ 2, 000, 000 + 250V  0 ≤ V ≤ 35, 000   3,100, 000 ≤ 120V  ⇔ 2,100, 000 ≤ 70V  0 ≤ V ≤ 35, 000  1  25,833 ≤ V ⇔ ⇒ Answer : 30, 000 ≤ V ≤ 35,000 3   30, 000 ≤ V

A-28

8.7

∑ weights = 14, 000 (or 14 for calculations below) Cx = Cy =

( 2 )( 3.5 ) + ( 8)( 7 ) + ( 6 )( 3.5) = 6.0 14 ( 6 )( 3.5) + (1)( 7 ) + ( 2 )( 3.5) 14

= 2.5

A-29

9 CHAPTER

9.1

9.2

Layout Strategy ( 60 )( 60 sec ) = 3,600 = 20 sec per PLA

a.

Cycle time =

b.

Theoretical number of work stations =

c.

Yes, it is feasible.

180 PLAs

180

1

D A2 3 C B4

Department pair Weekly Cost AB : 12 ( 8)( 800 ) = AC : 12 ( 6 )( 700 ) = AD : 12 ( 4 )( 400 ) = BC : 12 (10 )( 300 ) = BD : 12 ( 7 )( 200 ) = CD : 12 ( 9 )( 600 ) =

($) 3, 200 2,100 800 1,500 700 2, 700 $11, 000

A-30

∑ task time = 60 = 3 cycle time

20

9.3

a.

n=

∑t

Cycle time

C.T. =

b.

i

=

274 ( seconds )

Cycle time ( seconds )

60 ( 60 ) seconds

= 60 seconds ( per truck ) so n =

60 trucks Answer : 5 Steps 1 and 2: Sample Answer c = 60 seconds   From (a) number of stations is at least 5 Precedence diagram:

274 = 4.5667 → n = 5 . 60

40 30 B

D

20 H

6 E

40 A 50 C

30

25 F 15

J 18 I

G

Step 3

Task A B C D E

Number of Successors 9 4 4 2 2

Task F G H I J

A-31

Number of Successors 2 2 1 1 0

Step 4

Station 1 Station 2 Station 3

Station 4

Station 5

Station 6

c.

9.4

a.

Available A B, C B, C B, F, G B, F, G D, E, F, G D, E, G D, E, G E, G E, I E, I I, H I J J

Available and Fit A — B, C — B, F, G E, F, G — D, E, G E, G — E, I I, H I — J

Assigned A — C — B F — D G — E H I — J

(Broke a tie)

(Broke ties) (Broke ties) (Broke a tie)

(Broke a tie)

Answer : Station Tasks (Other answers possible, 1 A depending upon how ties 2 C are broken in above 3 B, F procedure) 4 D, G 5 E, H, I 6 J n = 6 work stations are in our answer. ∑ ti = 274 = 0.7611 Efficiency = n ( C.T.) 6 ( 60 )

First assignment costs = ( 8, 000 + 7, 200 + 1, 600 + 4,800 + 8, 000 + 800 ) ×

1 2

= $15, 200 b.

New layout costs = ( 8, 000 + 9, 600 + 6, 400 + 3, 600 + 2, 000 + 800 ) × No improvement—both yield the same cost.

A-32

1 = $15, 200 2

9.5

Cost of 3 attempts:

Attempt 1 2 3

a. b. c.

1 S D M

Work Area 2 D S D

3 M M S

Cost $1,088 $1,142 $1,100 ↑

2 ( 23)(10 ) + ( 32 )( 5) + ( 20 )( 8)  = $1,100

9.6

a.

Theoretical minimum number of stations = Cycle time =

b.

∑ task times cycle time

60 48 = 12 minutes . So minimum number of stations = = 4 stations 5 12

WS #1

WS #2

10 min

12 min

A

B

WS #5 D

8 min

F

6 min WS #4 E

C

12 min WS #3

c. 9.7

This requires 5 stations—it cannot be done with 4. 48 48 Efficiency = = = 80% for 5 stations. 5 × 12 60

There are three alternatives: Station 1 2 3 4 5

Alternative 1 Tasks A, B, F C, D E G, H I

Alternative 2 Tasks A, B C, D G, H E I

Each alternative has an efficiency of 86.67%.

A-33

Alternative 3 Tasks A, F, G H, B C, D E I

10 SUPPLEMENT

Work Measurement

2

 Zs  S10.1 Required sample size = n =   where s = 0.15 , x = 0.4 , z = 1.96 (for 95% confidence),  hx 

 (1.96 )( 0.15 )  n=  = 54.0225 ≈ 54  ( 0.10 )( 0.4 )  2

h = 10% accuracy level

S10.2 a.

b.

( )

2 hx n ( 0.10 )( 0.40 ) 12  Zs  = = 0.924 n =   Thus, Z = s 0.15  hx  Referring to Appendix I (Standard Normal Table), Area = 0.64 = 64% . The confidence level when n = 12 is 64%, as opposed to 95% when n = 54 (in Problem S10.1) 0.331 + 0.243 + " + 0.484 = 0.4484 minutes Average observed time = 12 Normal time = Average time × perf. rating = 0.4484 × 0.90 = 0.4036 minutes Normal time 0.4036 Standard time = = = 0.429 minutes 1 − allowance factor 1 − 0.06

100 hours × 60 minutes × 0.75 = 22.5 minutes 200 units Normal time = 22.5 minutes × 1.1 = 24.75 minutes Normal time for process 24.75 Standard time for job = = = 29.12 minute unit 1 − Allowance fraction 1 − 0.15

S10.3 Average observed time =

S10.4 a.

b.

c.

sum of times 1.74 = = 0.10875 minutes = 6.525 seconds number of cycles 16 Normal time = ( Observed time ) × ( Performance rating factor ) = 6.525 × 95%

Observed time =

= 6.2 seconds normal time 6.2 6.2 Standard time = = = = 6.739 seconds 1 − allowance factor 1 − 8% 92%

A-34

S10.5 Normal time = 10 minutes × 0.90 = 9 minutes Personal + Fatigue + Delay 5 + 3 + 1 9 = = = 0.15 Allowance fraction = 60 minutes 60 60 Normal time 9 Standard time = = = 10.59 minutes 1 − Allowance fraction 1 − 0.15

Observation (Minutes Per Cycle)

S10.6

Element 1 2 3 4

Rating 100% 90% 120% 100%

1 1.5 2.3 1.7 3.5

2 1.6 2.5 1.9 3.6

3 1.4 2.1 1.9 3.6

4 1.5 2.2 1.4 3.6

5 1.5 2.4 1.6 3.2

Average Normal Time Time 1.5 1.50 2.3 2.07 1.7 2.04 3.5 3.50

Normal time for lab test = 9.11 Normal time for process 9.11 Standard time for lab test = = = 11.1 minutes 1 − Allowance fraction 1 − 0.18

A-35

12 CHAPTER

12.1

Inventory Management

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 $ Volume 1289 → 400 × 3.75 = 1, 500.00 44.0% 2347

→ 300 × 4.00 =

1, 200.00

36.0%

2349

→ 120 × 2.50 =

300.00

9.0%

2363

→

75 × 1.50 =

112.50

3.3%

2394

→

60 ×1.75 =

105.00

3.1%

2395

→

30 × 2.00 =

60.00

1.8%

6782

→

20 ×1.15 =

23.00

0.7%

7844

→

12 × 2.05 =

24.60

0.7%

8210

→

8 × 1.80 =

14.40

0.4%

8310

→

7 × 2.00 =

14.00

0.4%

9111

→

6 × 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

2 ( 4,800 )( 30 ) 2 DS = . H 2 HQ SD  2 × 240   30 × 4,800  Thus, TC ( Total Annual Cost ) = PD + + = ( 4,800 × 27 ) +  +  2 2 240   2   = 129, 600 + 240 + 600 = $130, 440

S (Ordering Cost) = $30. So, Q∗ ( Order Quantity ) = 240 =

A-36

12.3

D (Annual Demand) = 14,558, P (Purchase Price/Unit) = $5, H (Holding Cost/Unit) = $4, 2 DS 2 × 14,558 × 22 = = 400 tons per order S (Ordering Cost/Order) = $22, Q = H 4 HQ SD  4 × 400   22 × 14,558  + = ( 5 × 14,558 ) +  TC = PD + +  = 72, 790 + 800 + 800.69 2 2 400  2    = $74,390.69 Answer : The optimal order quantity ( Q ) = 400 tons order ; total annual inventory cost

(TC ) = $74,391 .

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 DS 2 × 4,800 × 120 = = 181.42 = 181 units (rounded off). H 35 HQ SD  35 × 181   120 × 4,800  Thus, TC ( Total Cost ) = PD + + = ( 4,800 × 325 ) +  +  Q 2 181  2    = 1,560, 000 + 3,168 + 3,182 = $1,566,350 However, if Bell Computers orders 200 units,  35 × 200   120 × 4,800  TC = ( 4,800 × 325 ) +  +  = 1, 440, 000 + 3,500 + 2,880 = $1, 446,380 2 200     Answer : Bell Computers should order 200 units for a minimum total cost of $1,446,380. Q=

12.5

2 DS 2 × 4,800 × 120 = = 181 units H 35 2 DS 2 × 4,800 ×120 Q2 = = = 188 units H 32.5 2 DS 2 × 4,800 × 120 Q3 = = = 196 units H 30 181 units cannot be bought at $350, hence that isn’t feasible. 196 units cannot be bought at $300, hence that isn’t possible either. So, EOQ = 188 units . HQ SD  32.5 × 188   120 × 4,800  Thus, TC (188 units ) = PD + + = ( 325 × 4,800 ) +  +  Q 2 2 188     = 1,560, 000 + 3, 055 + 3, 064 = $1,566,119 HQ SD  30 × 200   120 × 4,800  + = ( 300 × 4,800 ) +  TC ( 200 units ) = PD + +  Q 2 2 200     = 1, 440, 000 + 3, 000 + 2,880 = $1, 445,880 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. Q1 =

A-37

12.6

D = 12,500 year , so d = 50 day , p = 300 day , S = $30 order , H = $2 unit year

a. b. c. d.

12.7

p 2 × 12, 500 × 30 300 = × = 612.37 × 1.095 = 671 p−d 2 300 − 50 D 12,500 Number of production runs ( N ) = = = 18.63 671 Q  d 50    1 Maximum inventory level ( I max ) = Q 1 −  = 6711 −  = 6711 −  = 559 p  300   6  250 Days of demand satisfied by each production run = = 13.42 days in demand 18.63 only mode Q 671 Time in production for each order = = = 2.24 days in production for each p 300 order. Total time = 13.42 days per cycle. 2.24 = 16.7% . Thus, percent of time in production = 13.42 Q=

2 DS × H

4, 000   H = $2 unit year , S = $10 order , D = 4, 000  d = = 16  . So, 250   2 DS 2 × 4, 000 ×10 Q= = = 200 . ROP (reorder point) = l × d , l (lead time = 5 days). H 2 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

300 =

12.9

a. b. c. d. e.

12.10 a. b. c. d. e.

2 ( 5, 400 ) 34 H

90, 000 =

; Square both sides

367, 200 , H

H=

367, 200 = $4.08 90,000

2 ( 6, 000 )( 30 ) 2 DS = = 189.74 units H 10 Average inventory = 94.87 Optimal number of orders/year = 31.62 250 Optimal days between orders = = 7.91 31.62 Total annual inventory cost = 601,897.37 (including the $600,000 cost of goods) EOQ =

Holding cost = $530.33 Set up cost = $530.33 Unit costs = $56,250.00 Total costs = $57,310.66 Order quantity = 16,970.56 units Thus, order 10,001 units for a total cost of $57,310.66. A-38

12.11

(52 Weeks) Cumulative Inventory $Value #Ordered Total $ Total Percent of Percent of Item per per Value/Week ($*Weeks) Rank Inventory Inventory Case Week 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% $312.00 20 0.07% 100.0% Salt 3 2 $6.00 $8,151.00 $423,852.00 100.00% a. b. c.

Fish filets total $74,360 C items are items 10 through 20 in the above list (although this can be one or two items more or less) Total annual $ volume = $423,852

12.12

Safety Stock 0 100 200

Carrying Cost 0 100 × 15 = 1,500 200 × 15 = 3,000

Incremental Costs Stockout Cost 70(100 × 0.4 + 200 × 0.2) = 5,600 (100 × 0.2) × (70) = 1,400 0

Total Cost 5,600 2,900 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-39

12.13 S = $10 order , LT = 4 days , d = 80 day , σ = 20 , H = 10% of $250,

D ( for term ) = 80 × 100 = 8, 000

a. b.

2 DS Q days = 10 days = 800 calzones . Order every H d ROP for calzone, σ ( demand during lead time ) = LT × σ = 40 , Z = 1 (from Table) Q=

for 0.1587, ROP = d × LT + Z × σ dLT = 360 c.

d.

ROP − d × LT . σ LT Thus, Z of 0.25 gives 0.5987. So there is a 40.13% chance of stockout. Q D = 400 = average inventory , = 10 orders term . 2 Q Q Holding cost term = H = ( 400 )( 0.25 ) = $100 2 D Order cost term = S = (10 )(10 ) = $100 Q On hand = 85, days left = 1. Since ROP = d × LT + Z × σ dalt , Z =

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

a. b. c.

Q=

2 DS = 1,131.37 H (1 − d p )

Q 1,131.37 = = 14.14 d 80 Q 1,131 = 7.07 days days = Run production for 160 p

Cycle =

A-40

12.15 Under present price of $6.40 per box: 2 DS 2 × 5000 × 25 Economic Order Quantity: Q∗ = = = 395.3 or 395 boxes where H 0.25 × 6.40 D = annual demand, S = set-up or order cost, H = holding cost DS QH + + CD Total cost = order cost + holding cost + purchase cost = Q 2 5, 000 × 25 395 × 0.25 × 6.40 = + + 6.4 × 5, 000 = 316.46 + 316.00 + 32, 000 395 2 = $32, 632.46 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: DS QH Total cost = order cost + holding cost + purchase cost = + Q 2 5, 000 × 25 5, 000 × 0.25 × 6.00 = + + 5, 000 × 6.00 = 41.67 + 3, 750.00 + 30, 000 3, 000 2 = $33, 791.67 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 2 DS 2 × 10, 000 × 200 Q∗ = = = 2,309.4 or 2,309 units 50 d 1.00 (1 − 200 H 1− p )

(

)

A-41

13 CHAPTER

13.1

Aggregate Planning

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. Beginning Month Inventory January 0 February 200 March 300 April 400 May 400 June 400 July 300 August 100 September 0 October 0 November 0 December 0

Production 700 700 700 700 700 700 700 700 700 700 700 700 8,400

Production Cost $49,000 49,000 49,000 49,000 49,000 49,000 49,000 49,000 49,000 49,000 49,000 49,000 $588,000

Demand 500 600 600 700 700 800 900 900 800 700 600 600 8,400

Ending Inventory 200 300 400 400 400 300 100 0 0 0 0 0 2,100

Inventory Shortage Cost 0 $600 0 900 0 1,200 0 1,200 0 1,200 0 900 0 300 100 0 200 0 200 0 100 0 0 $ 0 600 $ 6,300

The total cost of this plan is the sum of the three costs, or $600,300.

A-42

Shortage Cost $0 0 0 0 0 0 0 1,000 2,000 2,000 1,000 0 $6,000

13.2

a.

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

Quarter Winter Spring Summer Fall Totals

Hotel (Room) Demand 800,000 2,200,000 3,300,000 700,000 7,000,000

Personnel Required 33 33 33 33 132

Restaurant Demand Req. 160,000 12 800,000 12 960,000 12 160,000 12 2,080,000 48

Personal cost = 180 quarters of labor × 5,000 = Hiring cost = 45 hires at $1,000 = Termination cost = none = b.

Total 45 45 45 45 180

Hire Terminate 45 – – – – – – – 45

$900,000 45,000 0 $945,000

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 Hotel (Room) Quarter Demand Winter 800,000 Spring 2,200,000 Summer 3,300,000 Fall 700,000 Totals 7,000,000

Personnel

TermiTermi- Restaurant Demand Req. Hire nate Req. Hire nate 8 8 160,000 2 2 22 14 800,000 10 8 33 11 960,000 12 2 7 0 26 160,000 2 0 10 70 2,080,000 26

Personnel cost = 96 quarters of labor × 5,000 = Hiring cost = 45 hires @ $1,000 = Termination cost = 36 terminations @ $2,000 = c.

Totals Quarter Quarter Terminations Hires 10 22 13 36 45 36

$480,000 45,000 72,000 $597,000 The chase plan developed in part b is most economical ($945,000 vs. $597,000)

A-43

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 Fall Spring Summer Fall Totals

Hotel (Room) Req. Hire Demand 800,000 8 7 2,200,000 22 0 3,300,000 33 0 700,000 7 0 7,000,000 28*

Personnel from Agency 1 15 26 0 42

Restaurant Demand 160,000 800,000 960,000 160,000 2,080,000

Req. 2 10 12 2 26

Quarter Personnel Quarter from from Hires Agency Hire Agency 2 0 2 1 0 8 0 23 0 10 0 36 0 0 0 0 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 = Hiring cost = 9 ( = 7 + 2 ) hires @ $1,000 = Termination cost = 0 terminations @ $2,000 = Staffing agency costs = 60 @ 6,500 =

13.4

Plan A: Month Demand Production Mar 1,000 900 Apr 1,200 1,200 May 1,400 1,400 June 1,200 1,200 July 1,500 1,500 Aug 1,300 1,300 Total extra cost: $112,000

Hire

$180,000 9,000 0 390,000 $579,000

Fire 700

Extra Cost 56,000 12,000 8,000 8,000 12,000 16,000

300 200 200 300 200

Plan B: Month Demand Production Inventory Mar 1,000 1,100 200 Apr 1,200 1,100 100 May 1,400 1,100 June 1,200 1,100 July 1,500 1,100 Aug 1,300 1,100 Total extra cost: $39,000 Therefore, Plan B would be preferred.

A-44

Sub-Contracting

200 100 400 200

Extra Cost 2,000 1,000 8,000 4,000 16,000 8,000

13.5

Plan Number 1: Month Demand Production Inventory 1 1,000 1,200 300 2 1,200 1,200 300 3 1,400 1,200 100 4 1,200 1,200 100 5 1,500 1,200 6 1,300 1,200 Total extra cost: $112,000 Plan Number 2: Month Demand Production Inventory 1 1,000 1,200 300 2 1,200 1,200 300 3 1,400 1,200 100 4 1,200 1,200 100 5 1,500 1,200 6 1,300 1,200 Total extra cost: $39,000

Sub-Contracting

200 100

Overtime

200 100

Extra Cost 3,000 3,000 1,000 1,000 8,000 4,000

Extra Cost 3,000 3,000 1,000 1,000 2,000 1,000

Therefore, Plan Number 2 would be preferred. 13.6

Plan Y: Month Demand Production 1 1,100 1,100 2 1,600 1,600 3 2,200 2,200 4 2,100 2,100 5 1,800 1,800 6 1,900 1,900 Total extra cost: $112,000

Hire

Fire 400

Extra Cost 32,000 20,000 24,000 8,000 24,000 4,000

500 600 100 300 100

Plan Z: Month Demand Production Inventory 1 1,100 1,600 600 2 1,600 1,600 600 3 2,200 1,600 4 2,100 1,600 5 1,800 1,600 6 1,900 1,600 Total extra cost: $52,000 Therefore, Plan Z would be preferred.

A-45

Sub-Contracting

Extra Cost 6,000 6,000

500 200 300

20,000 8,000 12,000

14

Materials Requirements Planning (MRP) & ERP

CHAPTER

14.1

14.2

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 baby’s breath 7,000 pink roses Lot Size

Lot for Lot

Lot for Lot

Lot for Lot

Lot for Lot

Lot for Lot

Lead Time

1

1

1

1

2

On Safety AlloHand Stock cated

1,000

—

—

—

—

—

—

—

—

—

—

—

—

—

—

LowLevel Code

0

1

1

1

2

Item ID

Period (week)

CD Case 1

2

5

6

7

Gross Requirements 650 300 Scheduled Receipts CD Projected On Hand 1,000 1,000 1,000 1,000 350 Case Net Requirements Planned Order Receipts Planned Order Releases 500

550

400

500

50 500 500 400

400 400 500

500 500

Gross Requirements Scheduled Receipts CD Projected On Hand Top Net Requirements Planned Order Receipts Planned Order Releases

500

400

500

500 500 400

400 400 500

500 500

500

400

500

500 500 400

400 400 500

500 500

500

400

500

400 400 500

500 500

500

500 500 400

500

400

500

500 500 500

400 500

100 500 500

500

Gross Requirements Scheduled Receipts CD Projected On Hand Bottom Net Requirements Planned Order Receipts Planned Order Releases

500

Gross Requirements Scheduled Receipts CD Projected On Hand Insert Net Requirements Planned Order Receipts Planned Order Releases Gross Requirements Scheduled Receipts Projected On Hand Black Net Requirements Dye Planned Order Receipts Planned Order Releases

A-46

3

500

500

4

100

100

8

100

14.3

LowLot Lead On Safety AlloLevel Size Time Hand Stock cated Code

Lot for Lot

Lot for Lot

Lot for Lot

Lot for Lot

Lot for Lot

Lot for Lot

Lot for Lot

Lot for Lot

1

1

3

1

1

1

1

1

—

—

—

—

—

3,000

3,000

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

Item ID

1

1

2

3

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

0

1

Period (day)

Ball Point Pens

10,000

10,000 10,000 2,000 — — 2,000 2,000 2,000

Gross Requirements Scheduled Receipts Projected On Hand Body Net Requirements Planned Order Receipts Planned Order Releases Gross Requirements Scheduled Receipts Clip Projected On Hand Fine Net Requirements Pt Planned Order Receipts Planned Order Releases Gross Requirements Scheduled Receipts Std. Projected On Hand 3,000 Clip Net Requirements Planned Order Receipts Planned Order Releases Gross Requirements Scheduled Receipts Std. Projected On Hand 3,000 Ball Point Net Requirements Planned Order Receipts Planned Order Releases

Fine Point Ball Point

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

A-47

5

10,000 — — 10,000 10,000 10,000 10,000

Gross Requirements Scheduled Receipts Projected On Hand Cap Net Requirements Planned Order Receipts Planned Order Releases Gross Requirements Scheduled Receipts Projected On Hand Ink CC Net Requirements Planned Order Receipts Planned Order Releases

4

10,000

10,000 10,000 10,000 5,000

5,000 5,000 5,000 5,000 3,000 2,000 2,000 2,000 5,000 3,000 2,000 2,000 2,000 5,000

5,000 5,000 5,000

6

7

8

14.4

LowLot Lead On Safety AlloLevel Size Time Hand Stock cated Code

Lot for Lot

Lot for Lot

Lot for Lot

Lot for Lot

Lot for Lot

Lot for Lot

Lot for Lot

Lot for Lot

Lot for Lot

1

1

1

1

1

1

1

1

1

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

—

0

1

1

1

1

2

2

2

3

Item ID

Period (week, day)

Ball Point Pens 1

2

3

Gross Requirements Scheduled Receipts Coffee Projected On Hand Table Net Requirements Planned Order Receipts Planned Order Releases Gross Requirements Scheduled Receipts Projected On Hand Top Net Requirements Planned Order Receipts Planned Order Releases

640

Gross Requirements Scheduled Receipts Stain Projected On Hand gal Net Requirements Planned Order Receipts Planned Order Releases Gross Requirements Scheduled Receipts Glue Projected On Hand 100 gal Net Requirements Planned Order Receipts Planned Order Releases

80

100

100

Gross Requirements Scheduled Receipts Projected On Hand Base Net Requirements Planned Order Receipts Planned Order Releases Gross Requirements Scheduled Receipts Long Projected On Hand Braces Net Requirements Planned Order Receipts Planned Order Releases Gross Requirements Scheduled Receipts Short Projected On Hand Braces Net Requirements Planned Order Receipts Planned Order Releases Gross Requirements Scheduled Receipts Projected On Hand Leg Net Requirements Planned Order Receipts Planned Order Releases Gross Requirements Scheduled Receipts Brass Projected On Hand Caps Net Requirements Planned Order Receipts Planned Order Releases

A-48

100

640

4

5

6

7

8

640

640

128

128

640 640 128

128 128 128

128 128

640

640 640 640

640

640

128

128

640 640 640

640 640 128

128 128 128

128 128

80

80

16

16

80 80 80

80 80 16

16 16 16

16 16

40

40

8

8

100

60

20

12

640

640

128

128

640 640 640

640 640 128

128 128 128

128 128

880

880

880

880

4

1,280

1,280 1,280 1,280 1,280

1,280 1,280 1,280 5,120

5,120 5,120 5,120 5,120 880 5,120 6,000 6,000

880

14.5

Coffee Table Master Schedule

Hours Required

Table Assembly 2 Top Preparation 2 Assemble Base 1 Long Braces (2) 0.25 Short Braces (2) 0.25 Legs (4) 0.25 Total Hours Employees needed @ 8 hrs. each

14.6

Lead Time

Day 1

Day 2

Day 3

Day 4

320 320 640 1280 160

1,280 640 320 320 640 3,200 400

1,280 1,280 640 64 64 128 3,456 432

1 1 1 1 1 1 0 0

Day 5 640 1,280 256 128 64 64 128 1,920 240

Day 6 640 256 256 128

Day 7 128 256

640 80

256 32

Day 8 128

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 PG-A SA-B SA-C SA-D E F

On-Hand 0 0 0 0 10 5

LT (weeks) 1 1 2 2 1 3

Components SA-B, SA-C(2), SA-D(2) SA-D(2) E, F(2) E (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-C(2)

[LT =2] SA-D(2) E(3) [LT = 1]

a.

[LT =2] SA-D(2) E(3) [LT = 1]

E [LT = 1]

F(2) [LT = 3]

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 = 2 ×1 + 2 4 SA-D 14 E = 2 × 3 + 2 ×1 + 2 × 3 = 6 + 2 + 6 = 2× 2 4 F

A-49

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.

c.

Longest branch is 6 weeks = 1wk FGA + 2wks SA-C + 3wks F 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. b.

17 × 2 (level 2) = 34 [17 × 3 (B’s)] – 15 = 51 – 15 = 36 36 × 5 (D’s) = 180

14.8

a. b.

17 × 2 (level 2) + 17 × 2 × 3 (level 3) = 136 170 – 12 × 2 (level 2 on hand) = 170 – 24 = 146

A-50

15 CHAPTER

Short-Term Scheduling

15.1

A dummy task is added to balance the problem. The assignment is May—Task 1 Gray—Task 2 Ray—Task 3 Total time = 1 + 1 + 1 = 3 hours

15.2

Convert the minutes into $,

Chris

Marketing $80

Finance $120

Operations Human Resources $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 Steve Juana Rebecca

Finance Marketing Human Resources Operation

$120 $ 20 $ 45 $ 25 $210

15.3

The best pairs are assigned as follows: Ajay—Jackie Jack—Barbara Gray—Stella Raul—Dana Total compatibility score (overall) = 90 + 70 + 50 + 20 = 230

A-51

Gantt Chart

15.4 F

6

D

10 13

C B

20 28

E A

37 10

Project A B C D E F a. b.

15.5

20 Hour

30

Time 9 7 3 4 8 6

40

Due Date 22 17 16 13 16 9

Late Days 15 3 0 0 12 0

6 + 10 + 13 + 20 + 28 + 37 114 = = 19 days 6 6 total late days 15 + 3 + 12 Average lateness = = = 5 days number of jobs 6 Average flow time =

c.

Maximum lateness = 15 days (for job A—Gantt Table)

a.

The shortest processing time ( SPT ) =DBACE D Flow Time Due Hours Late Hours

0 0 0

B 1 12 —

A 3 4 —

41 = 8.2 hours 5 Number of deliveries late = C and E = 2 5 + 13 18 Average hours late = = = 9 hours 2 2

Average flow time =

A-52

C 6 8 —

E 11 6 5

20 7 13

= 41 = 18

b.

The EDD schedule = BCEAD B Flow Time Due Hours Late Hours

C

0 0 0

E

2 4 —

7 6 1

A 16 7 9

D 19 8 11

20 12 8

64 = 12.8 hours 5 Number of deliveries late = C, E, A, and D = 4 29 = 7.25 days Average hours late = 4

Average flow time =

15.6

1011 1517 20

45 Cut & sew 1

2

25

3 [123 schedule]

Deliver

Cut & sew

1

2

3

15 18 20 2325

56

10

56

10 13 1517 20

3

2

25

1 [321 schedule]

Deliver 56

1011 15

2022 25

Answer : The 321 schedule finishes in 22 days, 1 day faster than the 123 schedule, which finishes in 23 days.

15.7

a. b.

We begin by taking 5 empty slots 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. d.

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

A-53

e.

Finally, we get 2731 2134 2387 2800 2678 5

12

19

22

A 2731 2134 2387 2800 2678 B

2731 2134 7

14

21 25 33 35 2387 2800 2678

Total time = 33 hours

15.8

a. b.

The jobs should be processed in the sequence: 3, 6, 2, 7, 5, 1, 4 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. Jobs should be processed in the sequence A, B, C, D, E if scheduled by the EDD scheduling rule. Jobs should be scheduled in the sequence B, E, A, C, D if scheduled by the SPT scheduling rule. Jobs should be processed in the sequence D, C, A, B, E if scheduled by the LPT scheduling rule.

b. c. d.

15.10

Job 103 205 309 410 517

Due Date 214 223 217 219 217

Duration (Days) 10 7 11 5 15

Critical Ratio 1.4 3.3 1.5 3.8 1.1

Jobs should be scheduled in the sequence 517, 103, 309, 205, 412 if scheduled by the critical ratio scheduling rule.

A-54

16

Just-In-Time and Lean Production Systems

CHAPTER

16.1

a.

Q=

2× D× S H 1 − dp

Q2 =

2× D× S H 1 − dp

S=

(

(

)

(

)

Q 2 ( H ) 1 − dp 2D

) = (1, 000 ) (10 ) (1 − ) = (1, 000, 000 )(10 )(.2 ) = 2, 000, 000 = $10 2

400 500

2 × 100, 000

200, 000

200, 000

10 1 ⇒ hr ⇒ 10 min 60 6 Note that the lead time was not needed in this problem.

b.

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

Q=

2 × (1, 250, 000 ) ×15 2× D× S 37,500, 000 = = = 3, 750, 000 = 1,936 units 5,000 d 10 20 (1 − 10,000 ) H 1− p

(

)

1, 936 per container = 19.36 containers 100

A-55

16.5

Note: S = 6 minutes (or Q=

2× D× S

(

H 1 − dp

)

=

1 10

hour) ×$100 shop labor cost = .1× $100 = $10

( 2 )(12, 500 )(10 ) = 50 50 (1 − 500 )

250, 000 = 5, 555 = 74 45

1 = EOQ + Safety stock + Lead time = 74 + 10 +   ( 500 ) 2 = 334 oil pumps 334 No. 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): S=

(

Q 2 × H × 1 − dp

) = 150

2

150 × 10 × (1 − 1,000 )

=

22,500 × 10 × (1 − 0.15 )

2D 2 × 40, 000 80, 000 191, 250 = = $2.39 80, 000 $2.39 set-up × 60 minute hour Set-up time = = 2.69 minute set-up $50 hour

A-56

17 CHAPTER

17.1

17.2

Maintenance and Reliability

Failure Rate Analysis Station No. 1: N = 1,000 F = 22 22 FR = = 0.022 1, 000 R = 1 − 0.022 = 0.978 = 97.8% Station No. 2: N = 1,000 F = 51 51 FR = = 0.051 1, 000 R = 1 − 0.051 = 0.949 = 94.5% Answer : Unit #1 is clearly better, with a lower failure rate and better reliability. Reliability ( Rs ) = R1 × R2 × R3 ×! × Rn . So, R5 = ( 0.98 )( 0.99 )( 0.96 ) = 0.9304 Answer : The reliability is 93.04%

A-57

17.3

Task 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Total

Errors 1 6 4 2 3 1 0 2 1 2 1 1 0 2 3 29

FR 0.001 0.006 0.004 0.002 0.003 0.001 0.000 0.002 0.001 0.002 0.001 0.001 0.000 0.002 0.003

Reliability 0.999 0.994 0.996 0.998 0.997 0.999 1.000 0.998 0.999 0.998 0.999 0.999 1.000 0.998 0.997

Overall Reliability ( Rs ) = Product of all reliabilities above = 0.971 = 97.1% 17.4

Reliability (stick) = 1 − (1 − 0.99 ) = 0.9999 2

Reliability (cloth) = 1 − (1 − 0.98 ) = 0.999998 2

The reliability of the snake-charmer’s work = 0.9999 × 0.999998 = 0.9998 = 99.98% 17.5

.90 .96

.93 .99

.90

.96

.93 .99

.90

.96

.93 .90

Mountain test’s reliability = 1 − (1 − 0.96 ) = 0.999936 3

Bump test’s reliability = 1 − (1 − 0.99 ) = 0.999900 2

Speed test’s reliability = 1 − (1 − 0.93) = 0.999657 3

Sudden brake test’s reliability = 1 − (1 − 0.90 ) = 0.999900 4

So, overall reliability ( Rs ) = 0.999936 × 0.999900 × 0.999657 × 0.999900 = 0.9994 = 99.94%

A-58

A

Decision Making Tools

MODULE

States of Nature

A.1

Prob.

0.35 0.65 Competes Doesn’t Compete

Decision Alternatives

Expected Value

Assembly Line

$20,000

$30,000

$26,500

Plant Addition

–$10,000

$50,000

$29,000

No New Product

$0

$0

$0

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

States of Nature Slight Major Increase Increase

A.2

Fixed N 0 M –4,000 Decision Alternatives L –10,000 O –50,000 Use “maximin: criterion. No floor space (N). A.3

Increasing capacity Using overtime Buying equipment

2,000 8,000 6,000 4,000

Row Average $700,000 $700,000 $733,333 ←

Using equally likely, “Buying equipment” is the best option.

A-59

3,000 9,000 20,000 40,000

Minimum 0 –4,000 –10,000 –50,000

3 2 35 15 ( 50 ) + ( 20 ) + (100 ) + (100 ) 10 10 100 100 = 15 + 4 + 35 + 15 = 69

A.4

Expected value under certainty =

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.

1 1 3 1 ( 50 ) + ( 92 ) + ( 40 ) + ( 64 ) = 10 + 23 + 12 + 16 = 61 5 4 10 4

A.6

Expected value under certainty =

A.7

Solution Approach: Decision tree, since problem is under risk and has more than one decision. Buy now 33

$30,000

Unavailable $0 (.40)

Wait 1 day 33

Buy now

Avail. (.60) 55

$55,000

Unavailable $0 (.70)

Wait 1 day 21

Buy now

$70,000

Avail. (.30) 70 Wait 1 day

$0

(Numbers in Nodes are in $1,000’s) 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-60

A.8

We use a decision table, since expected marginal value of perfect information is asked for. Profits (= Payoffs)

D1: D2: D3:

Good Market: Bad Market: Good Market: Bad Market: Good Market: Bad Market:

80(400) – 25,000 = $7,000 70(375) – 25,000 = $1,250 85(450) – 30,000 = $8,250 80(425) – 30,000 = $4,000 90(475) – 33,000 = $9,750 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

States of Nature Good Market Bad Market (0.6) (0.4)

Decision Alternatives

D1 D2 D3 N

7,000 8,250 9,750 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-61

1,250 4,000 1,000 0

EV 4,700 6,550 6,250 0

A.9

More than one decision is involved, and problem is under risk, so use a decision tree approach. Low (0.4) demand

200 Expand

242 Small Facility

High (0.6) demand

270

270

Don’t expand Modest (0.3) response

544

Low (0.4) demand

Large Facility

223

20

160

160

Sizable (0.7) response

544 Don’t High (0.6) demand

220 40 800

Note: Payoffs and expected payoffs are in $1000’s. a. Build the large facility. If demand proves to be low, then to stimulate demand. If demand proves to be high, no advertising option is available (so don’t ). Expected Payoff: $544,000. b. A.10

Approach: Set up and solve via a decision table. $16.50 hr Cost per dozen bagels: Labor: = $1.65 dozen 10 dozen hr + $0.85 hr Ingredients: $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 B = dozens baked in A.M. Suppose: D = dozens sold during the day (= dozens demanded) $3D − $1( B − D ) if B ≥ D Then, total profit =  if B ≤ D $3B Also, since demand is between 40–43 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-62

Dozens Demanded D (= States of Nature) 40 41 42 43 (0.20) (0.30) (0.35) (0.15) 40 41 42 43

Dozens Baked B = Decision Alternatives

120 119 118 117

120 123 122 121

120 123 126 125

120 123 126 129

EV 120 122.2 123.2 122.8

Answer : 42 dozen bagels

A.11

a.

Payoff Rev – Cost 10 – 6 =4

EMV = 0.6(0.4) + 0.4(–1) = 2 P(high) = 0.6 EMV = 0.5(0) + 0.5(–3) = –1.5 P(success) = 0.5

8 – (6 + 2) = 0

Build 2

Build?

P(low) = 0.4

Prob (not success) = 0.5

5 – (6 + 2) = –3

Ad? Don’t

5 – 6 = –1

Don’t Build

b. c. A.12

$0

Build studio, but don’t (even if demand is low) Expected value = $2 million Build

a. Build P(high) = 0.6

10 – 6 = 4

Don’t Build

0

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

P(low) = 0.4

Ad?

We don’t really care about this part of the tree (which has a –1 EMV anyways)

Build? Don’t Build

b.

EVPI = EV with PI – EMV = 24 – 2 = 0.4 = $400,000

A-63

0

B MODULE

B.1

a.

b.

Linear Programming

Let T = number of trucks to produce per day Let C = number of cars to produce per day max z = 300T + 220C 1 1 S.T. T + C ≤1 40 60 1 1 T + C ≤1 50 50 T, C ≥ 0 Graph feasible region: C 60

A

50 40

B

30 20 10 O

C

T

10 20 30 40 50 60 70 80 90 (1) (2)

c.

Point Coordinates O ( 0, 0 ) A ( 0, 50 ) C ( 40, 0 ) B Solve 2 equations in 2 unknowns derived from (1) and ( 2 ) to obtain ( 20, 30 )

d.

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

A-64

z value 0 11, 000 12, 000 12, 600 ←

B.2

We solve this problem by the isocost line method: x2

18 16 14 12 10 8

(6, 6)

6 4 (0, 4) 2

x1 2

–2

4

6

z=4

8 10 12 14 16 18 z = 11 z = 12

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. (5) x2 8

B

(3) (1)

6 4

(4)

2 A –2

(2) x1 2

4

6

–4 –6 –8 z=0

z=5

z = 20

Answer : Unique optimal solution is ( x1 , x2 ) = ( 3, 5) , with objective function value 20.

A-65

B.4

Using the isoprofit line method.

B 9 8 (5) Opt. sol. 7 6 (2) 5 4 3 2 (3) (4) 1

z=9

(1)

z = –4 z = –6

1 2 3 4 5 6 7 8 9

–1 –2 –3 –4

(6) A

Answer : Unique optimal solution is ( A, B ) = (1, 5 ) . It has objective function 9.0.

B.5

Feasible region is same as in Problem B.4. Use Isoprofit line method. B

–1 –2 –3 –4

z=4

z=1

9 8 7 6 5 4 3 2 1

A 1

2

3

4

5

6

7

Answer : Problem is feasible and unbounded.

A-66

8

9

B.6

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

maximum: z = 10 S + 8 D 1 S + D ≤ 300 2 2 S + D ≤ 360 3 S, D ≥ 0

( A) ( B)

D z = $3,840

(0, 540)

400 z = 2,000 z = 1,000

300

(A) (240, 180)

200 100 100

200

(B)

(360, 0)

300

400

Extreme (Corner) Points Point Profit (0, 0) $0 (0, 300) $2,400 (360, 0) $3,600 $3,840 ← optimal solution and answer (240, 180)

A-67

S 500

600

y

B.7 (–1, 3 1/2) 5 4 (–2, 4) (–2, 3) 3 (2)

(4) (1)

(3)

2

(4, 1)

1 –2 –1

(5)1

x 2

3

4

5

6

x≤4 −x ≤ 2 x + 2y ≤ 6 −x + 2 y ≤ 8 y≥0

( line 1) ( line 2 ) ( line 3) ( line 4 ) ( line 5 )

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

Foods Apple Sauce Canned Corn Fried Chicken French Fries Mac & Cheese Turkey Breast Garden Salad

Cost Servings

Cost/ Serving $0.30 $0.40 $0.90 $0.20 $0.50 $1.50 $0.90

Calories/ Serving 100 150 250 400 430 300 100

Percent Protein 0% 20% 55% 5% 20% 67% 15%

Percent Carbs 100% 80% 5% 35% 30% 0% 40%

Percent Fat 0% 0% 40% 60% 50% 33% 45%

AS CC FC FF MC TB GS 0.3 0.4 0.9 0.2 0.5 1.5 0.9 0 1.333333 0.457143 0 1.129568 0 0

Constraints AS Cals min 100 Cals max 100 Protein min 0 Carb min 100 Fat max 0 Fruit/Veg Min 100

CC 150 150 30 120 0 150

FC 250 250 137.5 12.5 100 0

FF 400 400 20 140 240 0 A-68

MC 430 430 86 129 215 0

TB 300 300 201 0 99 0

GS 100 100 15 40 45 100

Fruit/ Vegetable 1 1 0 0 0 0 1

$1.51 LHS RHS 800 500 800 800 200 200 311.4286 200 288.5714 400 200 200

Target Cell (Min)

Answer Report (Relevant Section) Original Value Final Value $2.91 $1.51

Cell $I$14

Name servings

Cell $B$14 $C$14 $D$14 $E$14 $F$14 $G$14 $H$14

Name serving A serving C serving FC serving FF serving M serving T serving G

Cell $I$17 $I$18 $I$19 $I$20 $I$21 $I$22

Name Cals min LI Cals max L Protein min Carb min L Fat max LI Fruit + Veg I

Cell $B$14 $C$14 $D$14 $E$14 $F$14 $G$14 $H$14

Sensitivity Report (Relevant Section) Final Reduced Objective Allowable Name Value Cost Coefficient Increase serving A 0 0.172602 0.3 1E + 30 serving C 1.333333 0 0.4 0.258904 serving FC 0.457143 0 0.9 0.105072 serving FF 0 0.152691 0.2 1E + 30 serving M 1.129568 0 0.5 0.062909 serving T 0 0.169316 1.5 1E + 30 serving G 0 0.666151 0.9 1E + 30

Allowable Decrease 0.172602 0.225581 0.100581 0.152891 0.707833 0.169318 0.688151

Cell $I$17 $I$18 $I$19 $I$20 $I$21 $I$22

Final Shadow Constraint Name Value Price R.H. Side Cals min LI 800 0 500 Cals max L 800 –0.00023 800 Protein min 200 0.008983 200 Carb min L 311.4286 0 200 Fat max LI 288.5714 0 400 Fruit + Veg I 200 0.001504 200

Allowable Decrease 1E + 30 251.6129 40 1E + 30 111.4286 200

Adjustable Cells Original Value 1.50 0.00 1.33 0.00 0.00 0.00 1.40

Final Value 0.00 1.33 0.46 0.00 1.13 0.00 0.00

Cell Value 800 800 200 311.4286 288.5714 200

Formula $I$17 ≥ $J$1 $I$18 ≤ $J$1 $I$19 ≥ $J$1 $I$20 ≥ $J$2 $I$21 ≤ $J$2 $I$22 ≥ $J$2

Constraints

Adjustable Cells

Status Not Binding Binding Binding Not Binding Not Binding Binding

Slack 300 0 0 111.4286 111.4286 0

Constraints

A-69

Allowable Increase 300 200 155 111.4285 1E + 30 485.7143

C

Transportation Modeling

MODULE

C.1

Destination 2 3 10 9 8 5 3 1

A Source

4

B Demand

5 X

3

4

Supply 7 8 5 0

2

1

1

6

5

4 3 X

6 X

5 X

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

Change in Cost if “Opened Up” 0 Since these numbers 0 are all nonnegative, 0 the solution is optimal.

Answer : Yes

A-70

C.2

First, apply the optimality test:

Cell C-MI C-J H-L B-MI B-D

Change in Cost if Route is “Opened Up” +3 +4 +2 –1 +4

Resulting new basic feasible solution: Destination San Miami Denver Lincoln Jose

7 CHI.

2 55

3 Source

HOU.

40

BUF.

30

Demand

70

4

5

45 1

100 5

2

35 6

75 9

7

4 50

90

Supply

45

80

50

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

A-71

C.3

Total supply = 450 < 500 = Total demand.

W

Destination A B C Supply 4 9 6 200 200 10

X Source

0 12

5 100 7

Y

Demand

175 6

75 0

Dummy = Z

8 75

0

75 0

50 250

50 100

150

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. A

C.4

B 4

1

21 5

8

6

9

2

1

4

3

3

17 7 1

9 1

9 4

D

2

2 3

C

25 8

9 1

2 13

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

A-72

C.5

A 12 1

4

2

2

Source:

Destination D B C 8 5 10 4 8

6

11

3

7

E 4 2 9

12

Cost = $48 + 64 + 40 + 8 + 12 + 36 = $208

C.6

The only cell with a negative cost improvement index is Houston–Miami. It achieves a “–1”. Allocate 10 to that cell. The result is: Denver 0 20 0

Houston St. Louis Chicago

Yuma 0 0 20

Miami 10 0 10

Total cost = $170 2

C.7

4

5

9

4 6

7

8 1

1 3

3 4

10 1

11

12 3

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

A-73

C.8

Cell B-2 C-1

Improvement Index –9 +6

Result:

A

Destinations 1 2 2 8 6 6

9 Sources: B

1

3 7

C

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

A-74

4 7

2 7

6

5

2

D MODULE

D.1

Waiting Line Models

CURRENT MACHINE: λ = 40 , µ = 60 40 1. Utilization ( ρ ) = = 67% 60 2.

( 40 ) 1 Average number of customers waiting ( Lq ) = =1 60 ( 60 − 40 ) 3

3.

Average number of customers in system ( Ls ) =

4.

Average time waiting (Wq ) =

2

5.

customers

40 = 2 customers 60 − 40

40 = 0.033 hours = 2 minutes 60 ( 60 − 40 ) 1 1 = hours = 3 minutes Average time in system (Ws ) = 60 − 40 20

PROPOSED MACHINE: λ = 40 , µ = 90 40 1. Utilization ( ρ ) = = 44% 90 2.

( 40 ) = 0.356 Average number of customers waiting ( Lq ) = 90 ( 90 − 40 )

3.

Average number of customers in system ( Ls ) =

2

Average time waiting (Wq ) =

customers

40 = 0.8 customers 90 − 40

40 = 0.0089 hours = 0.533 minutes 90 ( 90 − 40 ) 1 1 = hours = 1.2 minutes 5. Average time in system (Ws ) = 90 − 40 50 Answer : Thus, we observe that the proposed machine will give better results with a decrease in queries as well as time in system.

4.

A-75

D.2

Two machine system λ 40 1 ρ= = = . Each machine is busy, on average, 33 percent of the time. a. M µ 2 × 60 3 b. Lq = average number of customers waiting in line = 0.081 c. d.

e.

D.3

Ls = Lq + Lq

λ = 0.081 + 0.667 = 0.748 customers in the system µ

0.081 = 0.002 hours = 0.1215 minutes λ 40 = 7.3 seconds average waiting time in the queue 1 1 Ws = Wq + = 0.002 + = 0.0187 hours = 1.12 minutes 60 µ = 67.3 seconds average time in the system The 2-machine system seems to be the best overall, with even further reduction in queues and waiting times. Wq =

=

This model is, again, the M/M/1 model. The service time of 30 seconds means that the service rate, µ, is 120 per hour. λ 40 1 ρ= = = . In this example and in Problem D.2, the system utilization is the a. µ 120 3 same. λ2 402 1 Lq = = = = 0.167 customers. The average number of b. µ ( µ − λ ) 120 (120 − 40 ) 6 customers waiting in line is twice as large as the average number (0.081) found in Problem D.2. λ 40 1 Ls = = = customer. On the average, the number of customers in c. µ − λ 120 − 40 2 1 3 the system is , rather than the (0.748) found in Problem D.2. 2 4 λ 40 1 1 Wq = = = d. hour = minutes = 15 seconds , which 4 µ ( µ − λ ) 120 (120 − 40 ) 240 exceeds Wq in Problem D.2. e.

1 1 1 3 = = hour = minutes = 45 seconds . Thus, average time 4 µ − λ 120 − 40 80 spent in the system is 15 seconds waiting plus 30 seconds service. The 45 seconds calculated here is less than the 67.3 seconds found in Problem D.2.

Ws =

A-76

D.4

EMPLOYEE SYSTEM: The current system is actually an M/D/1 system, because service times are constant. λ 10 ρ = = = 0.83 . The toll collector is busy 83 percent of the time. a. µ 12 b.

Lq =

λ2 100 100 = = = 2.08 drivers. The average number of 2 µ ( µ − λ ) 2 (12 )(12 − 10 ) 48

drivers waiting to pay the toll is 2.08. c.

Ls = Lq +

λ = 2.08 + 0.83 = 2.91 drivers. The average number of drivers at any one µ

toll booth is 2.91.

λ 10 10 = = = 0.208 minutes = 12.5 seconds . The 2µ ( µ − λ ) 2 (12 )(12 − 10 ) 48

d.

Wq =

e.

average time drivers spend waiting is 12.5 seconds. Ws = waiting time + service time = 12.5 seconds + 5 seconds = 17.5 seconds

AUTOMATED SYSTEM: The proposed system, although automated, is an M/M/1 system. λ 10 ρ = = = 0.83 . The utilization of the new system is the same as that of the old a. µ 12 system. λ2 100 100 Lq = = = = 4.16 drivers. Twice as many drivers are in b. µ ( µ − λ ) 12 (12 − 10 ) 24 line under the new, automated system. c.

d.

e.

Ls = Lq +

λ = 5 drivers. On average, 5 customers are in the system, which is 67 µ

percent more customers than the current system allows. λ 10 10 Wq = = = = 0.417 minutes = 25 seconds . The waiting time µ ( µ − λ ) 12 (12 − 10 ) 24 will double with the new system. 1 Ws = Wq + = 30 seconds. The time in the system will rise by 67 percent.

µ

The employee system seems to be better overall.

A-77

D.5

a. b., c.

D.6

Each server handles 60 registrants hour , so it takes 4 servers to handle 200 arrivals hour λ = 200 , µ = 60 , M = 4 servers yields: Lq = 3.29 , Wq = 0.0164 hours Server cost = 4 × $15 = $60 Wait cost = ( 3.29 people )( 0.0164 hour )( $100 ) = $5.38 ( rounded ) Total cost = $65.30 With 5 servers, Lq = 0.65 , Wq = 0.0033

d.

Wait cost = ( 0.65 )( 0.0033)( $100 ) = $0.21 , Server cost = 5 × 15 = $75 Total cost = $75.21 The system is optimal with 4 servers. Server utilization rate = 83.33% .

a. b.

The optimal number of servers is again 4. The wait cost is now $50 person hour × Lq × Wq (which are the same as in Problem D.5). Entertainment cost = $15 hour , Wait cost = ( 3.29 )( 0.0164 )( $50 ) = $2.70 ( rounded ) Total cost = $15 + $60 + 2.70 = $77.70

D.7

D.8

λ = 15 hour , µ = 20 hour a.

Wq = 0.075 hours = 4.5 minutes

b.

Lq = 1.125 people

λ = 10 hour , µ = 30 hour a.

Wq = 0.0083 hours =

b.

Lq = 0.083 people

1 minute 2

A-78

E MODULE

E.1

Learning Curves

In order to estimate the learning curve rate, we take the ratios of the units that have doubled. report report report report report report report report

2 56 = = 0.848 1 66 4 49 = = 0.860 2 56 6 45 = = 0.849 3 53 8 42 = = 0.857 4 49

The learning curve rates are not identical for each paired comparison, but since they are in the range of 84 percent to 86 percent, we can safely use 85 percent. E.2

Since the 6th report took 45 minutes, the 12th unit should take 85 percent of 45 minutes, or 38.25 minutes. Multiplying again by 85 percent yields a time of 32.5 minutes for the 24th report. One more multiplication yields a time of 27.6 minutes for the 48th report. An even more exact answer can be found using Excel OM or POM for Windows software. Or a third approach is the formula TN = T1C where C ≈ 0.402 (from Table E.3);

T48 = ( 66 )( 0.402 ) = 26.5 minutes. E.3

TN = T1C where C = 0.400 (from Table E.3), T50 = ( 66 )( 0.400 ) = 26.4 minutes

E.4

TN = T1C where C = 25.513 and TN is now cumulative time. T50 = ( 66 )( 25.513) = 1, 683.85 minutes = 28.06 hours

A-79

F

Simulation

MODULE

F.1

We will use the following random number intervals when simulating demand and lead time. We will select column 1 of text Table F.4 to get the random numbers for demand, while we will use column 2 of the same table to find the lead time whenever an order is placed. Probability 0.20 0.40 0.20 0.15 0.05

Cumulative Probability 0.20 0.60 0.80 0.95 1.00

RN Interval 01–20 21–60 61–80 81–95 96–00

Demand 0 1 2 3 4

Probability 0.15 0.35 0.50

Cumulative Probability 0.15 0.50 1.00

RN Interval 01–15 16–50 51–00

Lead Time 1 2 3

The results are: Units Received

10

10

Begin Inv. 5 4 3 0 10 6 2 1 0 10

RN 52 37 82 69 98 96 33 50 88 90

Demand 1 1 3 2 4 4 1 1 3 3 Total

End Inv. 4 3 0 0 6 2 1 0 0 7 23

Lost Sales 0 0 0 2 0 0 0 0 3 0 5

Order?

RN

Lead time

Yes

06

1

Yes

63

3

The total stock out cost = 5($40) = $200. The total holding cost = 23($1) = $23. A-80

F.2

If the reorder point Problem F.1 is changed to 4 units, we have: Units Received

10

10

Begin Inv. 5 4 13 10 8 4 0 0 10 7

RN 52 37 82 69 98 96 33 50 88 90

End Inv. 4 3 10 8 4 0 0 0 7 4 40

Demand 1 1 3 2 4 4 1 1 3 3 Total

Lost Sales 0 0 0 0 0 0 1 1 0 0 2

Order? Yes

RN 6

Lead time 1

Yes

63

3

Yes

57

3

The total stock out cost = 2($40) = $80. The total holding cost = 40($1) = $40. The total cost is $120 with a reorder point of 4 and $223 with a reorder point of 2. (Same random numbers were used as in Problem F.1). F.3

Since average waiting time is a variable of concern, a next event time increment model should be used. (1) Customer Number 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

(2) Random Number 50 28 68 36 90 62 27 50 18 36 61 21 46 01 14

(3) Interval to Arrival 2 1 2 1 4 2 1 2 1 1 2 1 2 0 1

(4) Time of Arrival 9:02 9:03 9:05 9:06 9:10 9:12 9:13 9:15 9:16 9:17 9:19 9:20 9:22 9:22 9:23

(5) Random Number 52 37 82 69 98 96 33 50 88 90 50 27 45 81 66

(6) Service Time 3 2 3 3 4 4 2 3 4 4 3 2 2 3 3

(7) Start Service 9:02 9:05 9:07 9:10 9:13 9:17 9:21 9:23 9:26 9:30 9:34 9:37 9:39 9:41 9:44

(8) End Service 9:05 9:07 9:10 9:13 9:17 9:21 9:23 9:26 9:30 9:34 9:37 9:39 9:41 9:44 9:47

(9) Wait Time 0 2 2 4 3 5 8 8 10 13 15 17 17 19 21

(10) Idle Time 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Read the data as in the following example for the first row: Column 1: Number of customer. Column 2: From third column of random number Table F.4. Column 3: Time interval corresponding to random number (random number of 50 implies a 2-minute interval). Column 4: Starting at 9 A.M. the first arrival is at 9:02. Column 5. From the first column of the random number Table F.4. Column 6: Teller time corresponding to random number 52 is 3 minutes. Column 7: Teller is available and can start at 9:02. Column 8: Teller completes work at 9:05 (9:02 + 0:03). Column 9: Wait time for customer is 0 as the teller was available. Column 10: Idle time for the teller was 2 minutes (9:00 to 9:02).

The drive-in window clearly does not meet the manager’s criteria for an average wait time of 2 minutes. As a matter of fact, we can observe an increasing queue buildup after only a few customer simulations. This observation can be confirmed by expected value calculations on both arrival and service rates. A-81

F.4

Time Until Next Arrival (Minutes) 0 1 2 3 Service Time (Minutes) 1 3 5 10

F.5

Probability 0.1 0.2 0.3 0.4

Probability 0.1 0.5 0.2 0.2

Cumulative Probability 0.1 0.3 0.6 1.00

Random Number Range 00–09 10–29 30–59 60–99

Cumulative Probability 0.1 0.6 0.8 1.00

Random Number Range 00–09 10–59 60–79 80–99

Simulating employment of one detailer: *

Random Number 52 37 82 69 98 96 33 50 27 45 81

**

IAT 2 3 2 0 3 3 2 2 1 2 3

Random Number 06 63 57 02 94 52 69 33 32 30 48

Service Time 1 3 2 1 4 2 3 2 2 2 2

Arrival Time 2 5 7 7 10 13 15 17 18 20 23

Detailer 1 Time Time In Out 2 3 5 8 8 10 10 11 11 15 15 17 17 20 20 22 22 24 24 26 26 28

Wait Time

1 3 1 2 2 3 4 4 3

Maximum waiting time: 4 minutes. Total waiting time: 23 minutes. 4 minutes will not meet the 2.5 minute criteria. *

From Column 1 of Table F.4 in text. From Column 2 of Table F.4 in text.

**

A-82

F.6

Simulating employment of two detailers:

Random Number 52 37 82 69 98 96 33 50 27 45 81

IAT 2 3 2 0 3 3 2 2 1 2 3

Random Service Arrival Number Time Time 06 1 2 63 3 5 57 2 7 02 1 7 94 4 10 52 2 13 69 3 15 33 2 17 32 2 18 30 2 20 48 2 23

Detailer 1 Time Time In Out 2 3 5 8

Detailer 2 Time Time In Out

7 8 10

9

9 14

15

18

18 20 23

20 22 25

1 13

15

17

19

Maximum waiting time: 1 minute. Total waiting time: 1 minute. Two detailers with a waiting time of one (1) minute will meet the 2.5 minute criteria.

A-83

Wait Time

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