heizer and render 8thed. solutions manual
DESCRIPTION
Solution manual for Heizer and Render Operations Management, 8th ed.TRANSCRIPT
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
Upper Saddle River, New Jersey 07458
JAY HEIZER
Texas Lutheran University
BARRY RENDER
Rollins College
VP/Editorial Director: Jeff Shelstad Executive Editor: Mark Pfaltzgraff Senior Managing Editor: Alana Bradley Senior Editorial Assistant: Jane Avery Copyright 2006 by Pearson Education, Inc., Upper Saddle River, New Jersey, 07458. Pearson Prentice Hall. All rights reserved. Printed in the United States of America. This publication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department. Pearson Prentice HallTM is a trademark of Pearson Education, Inc.
10 9 8 7 6 5 4 3 2 1
iii
Contents
Homework Problem Answers Chapter 1 Operations and Productivity ........................................................................... A-1 Chapter 3 Project Management ....................................................................................... A-3 Chapter 4 Forecasting ...................................................................................................... A-7 Chapter 5 Design of Goods and Services ...................................................................... A-11 Chapter 6 Managing Quality ......................................................................................... A-15 Supplement 6: Statistical Process Control ............................................................................ A-18 Chapter 7 Process Strategy ............................................................................................ A-20 Supplement 7: Capacity Planning ......................................................................................... A-23 Chapter 8 Location Strategies ....................................................................................... A-27 Chapter 9 Layout Strategy ............................................................................................. A-30 Supplement 10: Work Measurement ...................................................................................... A-34 Chapter 12 Inventory Management ................................................................................. A-36 Chapter 13 Aggregate Planning ...................................................................................... A-42 Chapter 14 Materials Requirements Planning (MRP) & ERP ........................................ A-46 Chapter 15 Short-Term Scheduling ................................................................................. A-51 Chapter 16 Just-In-Time and Lean Production Systems ................................................. A-55 Chapter 17 Maintenance and Reliability ......................................................................... A-57 Module A: Decision Making Tools ................................................................................ A-59 Module B: Linear Programming ..................................................................................... A-64 Module C: Transportation Modeling .............................................................................. A-70 Module D: Waiting Line Models .................................................................................... A-75 Module E: Learning Curves ........................................................................................... A-79 Module F: Simulation ..................................................................................................... A-80
A-1
1 CHAPTER
Operations and Productivity
1.1 a.
( )( ) ( )( )
( )( ) ( )( )
Last year’snumber of units of output
total factortotal dollar value of all inputs used
productivity
12,000 units
12,000 $2.00 14,000 $10.50
2,000 $8.00 4,000 $0.70 $30,000
12,000 units
$219,800
=
= + +
+ +
= 0.0546 units dollar=
b.
( ) ( ) ( )( )
( )( ) ( )( )
This year’snumber of units of output
total factortotal dollar value of all inputs used
productivity
14,000 units
14,000 $2.05 16,000 $11.00
1,800 $7.50 3,800 $0.75 $26,000
14,000 units
$247,050
=
= + +
+ +
= 0.0567 units dollar=
c.
This year’s Last year’s
total factor total factor
productivity productivity 0.0567 0.0546100% 100%
Last year’s 0.0546
total factor
productivity
3.84% 3.8%
− − × = ×
= + ≈
Answer : Total factor productivity increased by 3.798% this year as compared to last year.
A-2
1.2 ( ) ( )( )57,600
0.15160 12 L
= , where numberL = of laborers employed at the plant.
So ( )( )( )57,600
200160 12 0.15
L = =
Answer : 200 laborers
1.3 Output 28,000 customers= There are 4 approaches to solving the problem correctly:
1. Input 7 workers=
Then, 28,000
4,000 customers worker7
=
2. ( )Input 7 40 labor weeks=
Then, ( )28,000
100 customers labor week7 40
=
3. ( )( )Input 7 40 50 labor hours=
Then, ( )( )28,000
2 customers labor hour7 40 50
=
4. ( )( )Input 7 40 $250 dollars of worker wages=
Then, ( )( )28,000
0.40 customers per dollar of labor7 40 $250
=
1.4 ( )( )
6,600 Cadillacs0.10
labor hours
66,000 labor hours
x
x
=
=
There are 300 laborers. So,
66,000 labor hours
220 labor hours laborer300 laborers
=
1.5 ( ) ( )
( )52 $90 198 $80$ output 20,520
$57.00labor hour 8 45 360
+= = = per labor hour
A-3
3 CHAPTER
Project Management
3.1
50 100 150 200
A
B
C
D
E
F
G
H
I
Hours
80
150
200
Gantt Chart
20
120
110
140
170
160
3.2 AON Network:
60
BPurchasing
30
DSawing
20
APlanning
100
CExcavation
20
EPlacement
10
FAssembly
20
GInfill
10
HOutfill
30
IDecoration
A-4
3.3 AOA Network:
Plan1
Purchase2
Saw3
Place4
Assemble5
Outfill6
Decorate8 9
7
Excavate
Infill DummyA B D E
C
F
G
H
I
3.4 Path Task Times (Hours) Total Hours 1 – 2 – 3 – 4 – 5 – 6 – 7 – 8 – 9
1 – 2 – 3 – 4 – 5 – 6 – 8 – 9 1 – 2 – 4 – 5 – 6 – 7 – 8 – 9 1 – 2 – 4 – 5 – 6 – 8 – 9
20 + 60 + 30 + 20 + 10 + 20 + 0 + 30 20 + 60 + 30 + 20 + 10 + 10 + 30 20 + 100 + 20 + 10 + 20 + 0 + 30 20 + 100 + 20 + 10 + 10 + 30
190 180 200 190
The longest path clearly is 1 – 2 – 4 – 5 – 6 – 7 – 8 – 9; hence, this is the critical path, and the
project will end after 200 hours.
Planning1
Excavate
Purchasing2
LF = 90LS = 30EF = 80ES = 20
Sawing3
LF = 120LS = 90EF = 110ES = 80
Placement4
LF = 140LS = 120EF = 140ES = 120
LF = 20LS = 0EF = 20ES = 0
Assembly5
LF = 150LS = 140EF = 150ES = 140
Outfill6 8
7
LF = 170LS = 160EF = 160ES = 150
Infill DummyLF = 170LS = 150EF = 170ES = 150
Decoration9
LF = 200LS = 170EF = 200ES = 170
LF = 120LS = 20EF = 120ES = 20
A
B
C
D E
F G H I
Answer : The critical path therefore is A – C – E – F – G – I (200 hours). The activities that
can be delayed include ones with slack times > 0. Thus, B (10 hours), D (10 hours), and H (10 hours) can be delayed.
A-5
3.5
( )
( )
( )
( )
2
2
2
2
2
4Mean: Variance Standard Deviation
6 6 6
20120 20A : 20 A : 11.11 A : 3.33
6 36 660360 60
B : 60 B: 100.00 B : 10.006 36 6
120600 120C : 100 C : 400.00 C : 20.00
6 36 610180 10
D : 30 D : 2.78 D : 1.676 36 6
a m b b a b a+ + − −
= = =
= = =
= = =
= = =
( )
( )
( )
( )
( )
2
2
2
2
2
10120 10E : 20 E : 2.78 E : 1.67
6 36 6060 0
F : 10 F : 0.00 F : 0.006 36 6
40120 40G : 20 G : 44.44 G : 6.67
6 36 6460 4
H : 10 H : 0.44 H : 0.676 36 6
40180 40I : 30 I : 44.44 I : 6.67
6 36 6
= = =
= = =
= = =
= = =
= = =
3.6 Since the critical path is A – C – E – F – G – I, only those variances are along the critical
path are used. Therefore, the variances along critical path are 11.11, 400, 2.78, 0, 44.44, and 44.44 . So
the sum of these variances 502.77= .
Thus, the project completion standard deviation 502.77 22.4= ≅ .
µ = mean time of critical path 200 hrs= 22.4 hrsσ =
The z value 240 200 40
1.822.4 22.4
−= = = . Using the cumulative normal distribution table in
Appendix I of the text, we observe that 96.4 percent of the distribution lies to the left of 1.8 standard deviations. Hence, there is a 100 96.4 3.6%− = chance that it will take more than 240 hrs to build the garden/picnic area.
A-6
3.7 The critical path is A – C – E – F – G – I. Hence, the project completion variance 11.11 400 2.78 0 44.44 44.44 502.77.= + + + + + =
So, the project completion standard deviation 502.77 22.4= ≅ . The cumulative normal distribution tells us that 90% of the area lies to the left of 1.29
standard deviations. Therefore, amount of time to build the garden/picnic area should be ( )200 22.4 1.29 200 29 229 hours+ = + = .
3.8 a. Activity on Nodes Diagram of the project.
A1
B1
C4
E2
F2
b. The critical path, listing all critical activities in chronological order:
( )A B E F 1 1 2 2 6 A C F 1 4 2 7. This is the CP.
not CP→ → → + + + =→ → + + =
c. The project duration (in weeks): 7 (This is the length of CP.) d. The slack (in weeks) associated with any and all non-critical paths through the
project: Look at the paths that 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.
( )8 1 4 2Due date 1
20.5 0.5
Zµ
σ− + +−= = = = (length of entire path is 7, not 4). For a 2z = ,
this means ( )Due date 8 97.72%P < = (table lookup) for the path so chance of being OVER
8 weeks is 2.28% (and we know non-CP path will be only 6 weeks) 3.10 Helps to modify the AON with the lowest costs to crash
1. CP is A C F→ → ; C is cheapest to crash, so take it to 3 wks at $200. (and $200 < $250) 2. Now both paths through are critical. We would need to shorten A or F, or shorten C
and either B/E. This is not worth it, so we would not bother to crash any further.
A-7
4 CHAPTER
Forecasting
4.1 ( )Present period week 6.=
So: ( ) ( ) ( ) ( )7 6 5 4 3
1 1 1 1 1 1 1 152 63 48 70
3 4 4 6 3 4 4 6F A A A A= + + + = + + + 56.75 patients=
4.2
1 120 —2 136 —3 114 1284 116 125
t tt A F
120 136 256128
2 2 Checking Data136 114 250
1252 2
+ = = + = =
5
116 114 230115 Answer
2 2F
+= = = =
4.3 Method 1: MAD : 0.20 0.05 0.05 0.20 0.5000 better+ + + = ←
MSE : 0.04 0.0025 0.0025 0.04 0.0850+ + + = Method 2: MAD : 0.1 0.20 0.10 0.11 0.5100+ + + =
MSE : 0.01 0.04 0.01 0.0121 0.0721 better+ + + = ←
A-8
4.4 y a bx= +
4
1
2
1
58,538
75.75
191.5
23, 209
i ii
n
ii
x y
x
y
x
=
=
=
==
=
∑
∑
( ) ( )( )
( )
2
58,538 4 75.75 191.5 513.502
256.7523, 209 4 75.75
191.5 2 75.75 40
40 2
85
210
b
a
y x
x
y
−= = =
−
= − =≈ +=
⇒ ≈
4.5
t
Day Actual
Demand ForecastDemand
1 Monday 88 88 2 Tuesday 72 88 3 Wednesday 68 84 4 Thursday 48 80 5 Friday 72 Answer←
( )1 1t t tF A Fα α+ = + − . Let 1
4α = . Let Monday forecast demand = 88
( ) ( )
( ) ( )
( ) ( )
( ) ( )
2
3
4
5
1 388 88 88
4 41 3
72 88 18 66 844 41 3
68 84 17 63 804 41 3
48 80 12 60 724 4
F
F
F
F
= + =
= + = + =
= + = + =
= + = + =
A-9
4.6 Winter Spring Summer Fall2001 1, 400 1,500 1,000 6002002 1, 200 1,400 2,100 7502003 1,000 1,600 2,000 6502004 900 1,500 1,900 500
4,500 6,000 7,000 2,500
Average over all seasons: 20,000
1,25016
=
Average over spring: 6,000
1,5004
=
Spring index: 1,500
1.21, 250
=
( )5,600Answer : 1.2 1,680
4 =
sailboats
4.7 We need to find the smoothing constant α. We know in general that ( )1 1t t tF A Fα α+ = + − ,
1, 2, 3t = . Choose either 2t = or 3t = ( 1t = won’t let us find α because
( ) ( )2 50 50 1 50F α α= = + − holds for any α). Let’s pick, e.g., 2t = . Then
( ) ( )3 48 42 1 50F α α= = + − . So
48 42 50 50
2 8
1.
4
α αα
α
= + −− = −
=
Now we can find 5F : ( ) ( )5 46 1 50F α α= + − , with 1
4α = . So
( ) ( )5
1 346 50 49 Answer
4 4F = + = ←
A-10
4.8 Let 1 2 6, , , X X X… be the prices; 1 2 6, , , Y Y Y… be the number sold.
6
1 Average price 3.258336
ii
XX == = =
∑ (1)
6
1 Average number sold 550.006
ii
YY == = =
∑ (2)
All calculations to the
1nearest th
100,000
6
1
9,783.00i ii
X Y=
=∑ (3)
6
2
1
67.1925ii
X=
=∑ (4)
Then y a bx≈ + , where number soldy = , pricex = , and
( ) ( )
( )( ) ( )( )
( )
( ) ( )
6
16 2
22
1
9,783 6 3.25833 550 969.489277.61395
3.4922267.1925 6 3.25833
1,454.5578
i ii
ii
X Y n X Yb
X n X
a Y b X
=
=
− − −= = = = −−−
= − =
∑
∑
So at 1.80x = , ( )1,454.5578 277.61395 1.80 954.85270y = − = . Now round to the nearest
integer: Answer : 955 dinners
4.9 Tracking Signal ( )
1
MAD
n
t tt
A F=
−=∑
Month tA tF t tA F− ( )t tA F−
May 100 100 0 0 June 80 104 24 –24 July 110 99 11 11 August 115 101 14 14 September 105 104 1 1 October 110 104 6 6 November 125 105 20 20 December 120 109 11 11 Sum: 87 Sum: 39
So: 87
MAD : 10.8758
=
39 1
Answer : 3.586 to the nearest th10.875 1,000
=
A-11
5 CHAPTER
Design of Goods and Services
5.1
$27,500
$27,500
Use K1
(0.80)
(0.20)
90 of 100non-defect
70 of 100non-defect
$42,500
–$32,500
$4,062.50
(0.85)
(0.15)
90 of 100non-defect
75 of 100non-defect
$12,500
–$43,750
Use K2
$24,375
(0.90)
(0.10)
95 of 100non-defect
80 of 100non-defect
–$18,750
–$75,000
Use K3
Answer: $27,500—use K1
Outcome Calculations
( )( )( ) ( )( )( )90 10
$100,000 500 300 $1.20 500 300 $1.30100 100
$100,000 $162,000 $19,500 $42,500
− + − =− + − =
( )( ) ( )( )70 30
$100,000 150,000 $1.20 150,000 $1.30100 100
$100,000 $126,000 $58,500 $32,500
− + − =− + − = −
A-12
( )( ) ( )( )90 10
$130,000 150,000 $1.20 150,000 $1.30100 100
$130,000 $162,000 $19,500 $12,500
− + − =− + − =
( )( ) ( )( )75 25
$130,000 150,000 $1.20 150,000 $1.30100 100
$130,000 $35,000 $48,750 $43,750
− + − =− + − = −
( )( ) ( )( )95 5
$180,000 150,000 $1.20 150,000 $1.30100 100
$180,000 $171,000 $9,750 $18,750
− + − =− + − = −
( )( ) ( )( )80 20
$180,000 150,000 $1.20 150,000 $1.30100 100
$180,000 $144,000 $39,000 $75,000
− + − =− + − = −
5.2
84.0 84.0Use D1
(0.4)F market 99.0
(0.6)U market
74.0
66.0
(0.3)F market 80.0
(0.7)U market
60.0
80.2
(0.6)F market 89.2
(0.4)U market
66.7
Use D0
Use D2
(All $ figures in millions in tree)
( )( )
( )( )
( )( )
$ Profits : D0 F : 1,000 80,000 $80,000,000
D0 U : 750 80,000 $60,000,000
D1 F : 1,000 100,000 1,000,000 $99,000,000D1 U : 750 100,000 1,000,000 $74,000,000
D2 F : 1,000 90,000 800,000 $89, 200,000D2 U : 750 90,000 800
− =− =
− − =− − =
− − =− − ,000 $66,700,000=
Answer : Answer: Design D1 has an expected profit of $84,000,000.
A-13
5.3
$14,000
$10,000
(0.3)Demand rises $30,000
–$20,000Purchase
overhead hoist
(0.5)Demand stays
same
(0.2)Demand falls
$10,000
$14,000
(0.4)Demand rises $20,000
(0.6)Demand stays
same $10,000
Purchaseforklift
$0
Donothing
Answer : Maximum expected payoff $14,000=
5.4
Low demand (0.4)
$380,000Upgrade to D
160K –$50,000Use A Low demand (0.4)
$300,000High demand (0.6)
180K $0
$300,000High demand (0.6)Use B
302K
$250,000High demand (0.6)Use C
380K
$0No upgrade to D
Low demand (0.4)
$0Do nothing
Note: K = $1,000’s Answer : Use Design C. If demands turns out to be low, upgrade to Design D.
A-14
5.5
Bread & RollsPies & Cakes
Support
Support
No support
No support
Bread & Rolls
Support
No support
Full Service
5.6
Bread & RollsPies & Cakes
$15,000
$10,000
Bread & Rolls
Full Service
Support ( = 0.40)
No support ( = 0.60)EMV = $12,000
$25,000
$5,000
Support ( = 0.40)
No support ( = 0.60)EMV = $13,000
$35,000
–$10,000
Support ( = 0.40)
No support ( = 0.60)EMV = $7,500
p
p
p
p
p
p
Based upon this decision tree, Jeff should consider most seriously the medium-sized shop
carrying bread, rolls, pies, and cakes.
A-15
6 CHAPTER
Managing Quality
6.1 1. Appearance of food
2. Portion size 3. Lighting 4. Speed of service 5. Knowledge of server 6. Quality of service 7. Appearance of room 8. Appropriate amount of space 9. View of stage and audio
Item Overall Grade Rated A B C D E 1. 20 28 1 1 0 2. 4 2 30 14 0 3. 19 20 3 8 0 4. 4 5 25 5 11 5. 0 0 27 18 7 6. 9 30 7 0 4 7. 19 18 13 0 0 8. 0 26 24 0 0 9. 0 0 0 20 30 Item Weights Rated 4 3 2 1 0 Total Average 1. 80 84 2 1 0 167 2.61 2. 16 6 60 14 0 96 1.50 3. 76 60 6 8 0 150 2.34 4. 16 15 50 5 0 86 1.34 5. 0 0 54 18 0 72 1.13 6. 36 90 14 0 0 140 2.19 7. 76 54 26 0 0 156 2.44 8. 0 78 48 0 0 126 1.97 9. 0 0 0 20 0 20 0.31
a. Highest rated is appearance of food; 2.61. b. Lowest rated is view of stage; 0.31.
A-16
c. A check sheet will help categorize the comment cards
Check Sheet Positive Negative Appearance of food Portion size ! ! ! ! ! Lighting ! Speed of service ! ! Knowledge of server ! Quality of service ! ! Appearance of room Appropriate amount of space ! View of stage and audio ! ! ! ! ! ! Other ! ! ! ! ! chilly
d. The written comments are not always consistent: Portion size is highly rated in
comments, but 5th in overall grade. View/audio is lowest rated in both. 6.2 a.
8
9
10
11
12
13
14
1 2 3 4 5 6
x
yminutes
Trips0
b. This is a scatter diagram.
A-17
6.3 a.
5
10
15
20
25
30
35
W R I M O
40
02
461014
24
3034
36
b. 39% of complaints are W, demeaning towards women. 6.4
mislabeled
Manpower
Incorrectmeasurement
Operatormisreads display
Inadequatecleanup
Techniciancalculation off
Machines
Temperaturecontrols off
Variability
Antiquatedscales
Inadequateflow controls
Equipmentin disrepair
IncorrectFormulation
Materials
Jars
Incorrectweights
Damagedraw material
instructions
Methods
Lack of clear
Prioritymiscommunication
Incorrectmaintenance
Inadequate instructions
A-18
6 SUPPLEMENT
Statistical Process Control
S6.1 We are given a target of 420X = . So 25
LCL 420 4 40025
X Zn
σ = − = − =
.
25UCL 420 4 440
25X Z
n
σ = + = + =
. Thus,
Answer : LCL 400 calories
UCL 440 calories
=
=
S6.2 7 5 9
250 250 250 7 5 9 3000.040
30 7,500 7,500p
+ + + + + += = = =" "
( ) ( )
( ) ( )
1UCL 0.040 3 0.01239 0.077
1LCL 0.040 3 0.01239 0.003
p pp Z
n
p pp Z
n
−= + = + =
−= − = − =
S6.3 We want 2Z = , since ( )1 0.0455 0.9545− = which implies 2Z = from the Normal Table.
UCL 2c c= + , where average number of breaks 3: 3 2 3 6.46c = = + = . S6.4 3Z = for -chartx . Here, 4n = so 2 0.729A = (from Table S6.1). 2.0x = , 0.1R = ,
( )2UCL 2.0 0.729 0.1 2.07x x A R= + = + =
S6.5 C chart
0.0027
1.0000 0.0027 0.9973
0.49865 32 2
Z− = = ⇒ = (see Normal Table)
UCL 3 1.5 3 1.5 5.17c c= + = + =
A-19
S6.6 answersx Zn
σ ± =
38416 lbs.
24
0.122 0.08
3
16.00 0.08 16.08 UCL
16.00 0.08 15.92 LCLx
x
x
Zn
σ
= =
= = + = =− = =
S6.7 1.00x = , 0.10R = , 2 0.483A = (from Table S6.1),
( )( )2LCL 1 0.483 0.10 0.9517x A R= − = − = weeks
S6.8 3.25R = mph, 3Z = , with 8n = , from Table S6.1,
UCL 1.864 6.058
LCL 0.136 0.442
R
R
= == =
S6.9
30Number of defects
2501 300
0.0430 7,500
ip == = =∑
, 250n =
( )( ) ( )
( ) ( ) ( )
0.04 0.96UCL 2 0.04 2 0.0124 0.0648
250
0.04 0.96LCL 2 0.04 2 0.0124 0.0152
250
p
p
p
p
= + = + =
= − = − =
S6.10 a. We are counting attributes and we have no idea how many total observations there are
(the proportion of drivers who weren’t offended enough to call!) This is a C-chart.
b. Use mean of 6 weeks of observations 36
66
= for c , as true c is unknown.
( )UCL 6 3 2.45 13.3c z c= + = + =
( )LCL 6 3 2.45 1.3c z c= − = − = − , or 0.
c. It is in control because all weeks’ calls fall within interval of [ ]0, 13 .
d. Instead of using 36
66
= , we now use 4c = . ( )UCL 4 3 4 4 3 2 10= + = + = .
( )LCL 4 3 2 2= − = − , or 0. Week 4 (11 calls) exceeds UCL. Not in control.
A-20
7 CHAPTER
Process Strategy
7.1 a. Find breakeven points, pX .
Mass Customization: 1,260,000 60 120 21,000pX X X+ = → =
Intermittent: 1,000,000 70 120 20,000pX X X+ = → =
Repetitive: 1,625,000 55 120 25,000pX X X+ = → =
Continuous: 1,960,000 50 120 28,000pX X X+ = → =
b. Find least-cost process at 24,000 unitsX = . Fixed cost VC Units Mass Customization: ( )1,260,000 60 24,000 2,700,000+ =
Intermittent: ( )1,000,000 70 24,000 2,680,000+ =
Repetitive: ( )1,625,000 55 24,000 2,945,000+ =
Continuous: ( )1,960,000 50 24,000 3,160,000+ =
The least-cost process: Intermittent Process. c.
Anticipated IntermittentProduction Process
Volume Breakeven Point
24,000 20,000 ? yes!>#$% #$%
Annual Profit Using Intermittent Process: ( )$ 120 24,000 2,680,000 $200,000 − =
Answer : The intermittent process will maximize annual profit.
Annual Profit: $200,000
A-21
7.2 Use a crossover chart. First graph. Then solve for breakpoint(s).
5
1
2
3
10 15 20 25
V
RMC
00
1,000’s of Ovens
I
P2P1
Cost(Millionsof dollars)
Finding value of P2: ( ) ( )1,250,000 50 P2 2,000,000 5 P2+ = + . So 2
3P2 16,666 units= .
(Note: P1 12,500= ).
Answer : For volumes of production V such that 2316,666 25,000V≤ ≤ .
7.3
2
4
6
8
10
12
14
5,000 15,000
VVolume
10,000 20,000
I R C
I
R
C
Cost(millions)
7,500
00
( )
1,000,000 1,650 3,000,000 1,250
400 2,000,000I&R5,000Intersect
1,000,000 1,650 5,000 $9,250,000
x x
x
x
+ = + = = + =
( )
3,000,000 1, 250 7,500,000 650
600 4,500,000R&C7,500Intersect
3,000,000 1, 250 7,500 $12,375,000
x x
x
x
+ = + = = + =
For all V between 5,000 7,500V≤ ≤
A-22
7.4 Breakeven points a. : 21,000,000 450 750 70,000R x x x+ = ⇒ = : 26, 250,000 400 750 75,000C x x x+ = ⇒ = : 15,000,000 500 750 60,000M x x x+ = ⇒ = b. Least cost process at 65,000x = Cost R: $50,250,000 C: $52,250,000 M: $47,500,000 ← lowest cost with Mass Customization c. 65,000 demand > 60,000 breakeven for M
7.5 Breakeven points
a. Continuous : 2,400,000 20 80 40,000x x x+ = ⇒ = Repetitive : 1,950,000 30 80 39,000x x x+ = ⇒ = Mass Customization : 1,480,000 40 80 37,000x x x+ = ⇒ = Intermittent : 1,800,000 40 80 45,000x x x+ = ⇒ = b. Least cost process at 48,000x = Continuous: $3,360,000 ← least cost Repetitive: $3,390,000 Mass Customization: $3,400,000 Intermittent: $3,720,000 c. Is 48,000 > 40,000? Yes, so we use continuous process. Annual profit $480,000=
7.6
4,000 11,000Volume
15,000I
R
C
$
2,000
M
(11,000; 1,350,000)(4,000; 860,000)
(2,000; 300,000)
widest Repetitive has the widest production volume range over which it is a least-cost process. 7.7 Total profit now: Profit 40,000 2.00 20,000 40,000 0.75 80,000 20,000 30,000 30,000= × − − × = − − = Total profit with new machine: Profit 50,000 2.00 2,000 50,000 1.25 100,000 25,000 62,500 12,500= × − − × = − − = Since profit decreases with the new piece of equipment added to the line, purchase of the
machine probably would not be a good investment.
A-23
7 SUPPLEMENT
Capacity Planning
S7.1 Problem is under risk and has two decisions, so use a decision tree:
109
High demand
70
135(0.6)
Medium demand(0.4)
No additional expansion 90
135Additional minor expansion
148
High demand(0.6)
Medium demand(0.4) 40
220
148
Smallexpansion
Largeexpansion
(Payoffs and Expected Payoffs are in $1,000’s) Answer : Ralph should undertake a large expansion. Then the annual expected profit will
equal $148,000.
A-24
S7.2
High demand 140,000
(0.3)
No additional
$90,000
$140,000
70,000
$70,000
Smallexpansion
Largeexpansion
minor expansion
Additionalminor expansion
Medium demand(0.7) $40,000
High demand(0.3)
14,000
Medium demand(0.7)
$105,000
–$25,000
Maximum value = $70,000 S7.3
$18,000 Smallexpand
Demand upsmall (0.4)
16,000
Demand upmedium (0.6)
$10,000
$20,000
$0Noexpand
Demand upmedium (0.3)
18,000
Demand uplarge (0.7)
–$10,000
$34,000
Largeexpand
Answer : $18,000
A-25
S7.4
(1)
50,000
100,000
200,000
300,000
400
x
1,000 2,000
(1)
250
(2)
(1)
(2)
(3)
(3)
(2)
Cap level (2) is lowest for all x so 1,000 2,000x≤ ≤
S7.5 Actual (or expected)Effective Capacity
Output = Efficiency
(text Equation S7-3)
4.8 cars = 5.5 cars × 0.880. Therefore in one 8 hour day one bay accommodates 38.4 cars = ( )8 hrs 4.8 cars per hr× and to do 200 cars per day requires 5.25 bays or 6 bays =
200 cars38.4 cars per bay
S7.6 a. ( ) ( )450
BEP $ 878.050.51251 i i i
F
V P W= = =
− ∑
Breakeven ( )$ $878.05=
b. Number of pizzas required at breakeven: Whole pizzas ( )878.05 0.30 5.00 52.7 53= × = →
Slices ( )878.05 0.05 0.75 58.5 59= × = →
Whole pizzas to make slices 59 6 9.8 10= = → Therefore, he needs a total of 63 pizzas. He does not have sufficient capacity. S7.7 a. Remember that Yr 0 has no discounting.
Initial coat $1,000,0000 yearly maint 75,000 members dues/memberSalvage cost $50,000 yearly dues $300,000 500 $600 Discount rate 0.100
Year Cost Revenues Profit PV Mult PV Profit 0 $1,075,000 $300,000 $–775,000 1 –$775,000 1 75,000 300,000 225,000 0.9 $202,500 2 75,000 300,000 225,000 0.81 $182,250 3 75,000 300,000 225,000 0.729 $164,025 4 75,000 300,000 225,000 0.6561 $147,623 5 75,000 350,000 275,000 0.59049 $162,385 undisc. Profit 400,000 PV Profit $83,782
A-26
Assume dues are collected at the beginning of each year. This is a simplification—in reality, people are likely to join throughout the year. (Technically, if equipment is sold at the end of year 5, it should probably appear as a final revenue stream in year 6 but the difference is only $2,952.45.
b. Special deal comparison: $3,000 for all 6 years. Compare the PV cash stream of yearly dues from one member to that of the deal. Since we specified the club will always be full, we can make the assumption that the member (or her replacement) will always be paying the annual fee.
Initial cost $0 yearly maint $0 Salvage cost $0 yearly dues $600 Discount rate 0.100
(Membership fee) Year Cost Revenues Profit PV Mult PV Profit 0 $0 $600 $600 1 $600 1 0 600 600 0.9 $540 2 0 600 600 0.81 $486 3 0 600 600 0.729 $437 4 0 600 600 0.6561 $394 5 0 600 600 0.59049 $354 undisc. Profit 3,600 PV Profit $2,811
Since this is less than $3K, the special deal is worth more to the Health Club. Note also: If Health Club member is using same discount rates, it’s better for her to pay yearly.
S7.8 Breakeven: Costs = Revenues 500 0.50 0.75b b+ × = × where b = number of units at breakeven or ( )0.75 0.50 500b − = ,
and 500
2,000 units0.25
b = =
a. breakeven in units = 2,000 units b. breakeven in dollars = $0.75 2,000 $1,500× =
A-27
8 CHAPTER
Location Strategies
8.1
( ) ( ) ( )( ) ( )( ) ( )( )( ) ( ) ( )( ) ( ) ( )
2,000 2.5 5,000 2.5 10,000 5.5 7,000 5.0
10,000 8.0 20,000 7.0 14,000 9.06.67
2,000 5,000 10,000 7,000 10,000 20,000 14,000xC
+ + +
+ + += =
+ + + + + +
( )( ) ( )( ) ( )( ) ( )( )( )( ) ( )( ) ( )( )
2,000 4.5 5,000 2.5 10,000 4.5 7,000 2.0
10,000 5.0 20,000 2.0 14,000 2.53.02
68,000yC
+ + +
+ + += =
8.2 Site Total Weighted Score A 174 B 185 C 187 D 165
8.3 Population weights 5,000=∑
( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( )( )0 2,050 1 550 2 1,025 3 775 2 250 2 350
0.5255,000xC
+ − + + + − + −= =
( )( ) ( )( ) ( )( ) ( ) ( ) ( ) ( ) ( )( )0 2,050 1.5 550 1 1,025 3 775 3 250 1 350
0.2055,000yC
+ + − + + − + −= =
Coordinates: ( )0.525, 0.205 8.4
25
50
75
100
125
150
1 2 3 4 5 6
V
$ cost(millions)
(2, 60)
(1, 35)
(1/2, 30)(0, 25)
(0, 20)(0, 10)
10,000’s of Autos = V0
A
B
C
For all V such that 10,000 60,000V≤ ≤ .
A-28
8.5 Site Score A 5 320w + B 4 330w + C 3 370w + D 5 255w + Find all w from 1–30 so that:
3 370 5 320 50 2 25
3 370 4 330 40 40
3 370 5 255 115 2 57.5
w w w w
w w w w
w w w w
+ ≥ + ⇔ ≥ ⇔ ≤+ ≥ + ⇔ ≥ ⇔ ≤+ ≥ + ⇔ ≥ ⇔ ≤
Answer : For all w such that 1.0 25.0w≤ ≤
8.6
2
4
6
8
10
12
5 10 15 20 25 30
V
TC(millions $)
(1, 35)
(30, 9.5)(20, 7)
35 (thousands)
L.A.K.C.Charlotte
Char
K.C.
L.A.
cutoff0
4,100,000 180 1,000,000 300
For all so: 4,100,000 180 2,000,000 250
0 35,0003,100,000 120
2,100,000 70
0 35,0001
25,833 Answer : 30,000 35,0003
30,000
V V
V V V
V
V
V
V
VV
V
+ ≤ + + ≤ + ≤ ≤
≤⇔ ≤ ≤ ≤ ≤⇔ ⇒ ≤ ≤ ≤
A-29
8.7 weights 14,000=∑ (or 14 for calculations below)
( )( ) ( ) ( ) ( ) ( )2 3.5 8 7 6 3.5
6.014xC
+ += =
( )( ) ( )( ) ( )( )6 3.5 1 7 2 3.5
2.514yC
+ += =
A-30
9 CHAPTER
Layout Strategy
9.1 a. ( )( )60 60 sec 3,600
Cycle time 20 sec per PLA180 PLAs 180
= = =
b. Theoretical number of work stations task time 60
3cycle time 20
= = =∑
c. Yes, it is feasible. 9.2 1 2
3 4
D AC B
Department pair
( )( ) ( )( ) ( )( )( )
( )( )( )( )( ) ( )
121212
12
1212
Weekly$
Cost: 8 800 3,200: 6 700 2,100: 4 400 800: 10 300 1,500: 7 200 700: 9 600 2,700
$11,000
AB
AC
AD
BC
BD
CD
======
A-31
9.3 a. ( )
( )274 seconds
Cycle time Cycle time secondsitn = =∑
( ) ( )60 60 seconds
C.T. 60 seconds per truck60 trucks
= = so 274
4.5667 560
n n= = → = .
Answer : 5
b. Steps 1 and 2: Sample Answer
60 seconds
From (a)number of stations is at least 5
c =
Precedence diagram:
A
40
B
30 D
40
E
6 H
20
C
50 F
25
G
15 I
18
J
30
Step 3
Task
Number of Successors
Task
Number of Successors
A 9 F 2 B 4 G 2 C 4 H 1 D 2 I 1 E 2 J 0
A-32
Step 4
Available
Available and Fit
Assigned
Station 1 A A A B, C — — Station 2 B, C B, C C (Broke a tie) B, F, G — — Station 3 B, F, G B, F, G B D, E, F, G E, F, G F (Broke ties) D, E, G — — Station 4 D, E, G D, E, G D (Broke ties) E, G E, G G (Broke a tie) E, I — — Station 5 E, I E, I E I, H I, H H (Broke a tie) I I I J — — Station 6 J J J
Answer : Station Tasks(Other answers possible, 1 Adepending upon how ties 2 C
are broken in above 3 B, Fprocedure) 4 D, G
5 E, H, I6 J
c. 6 work stations are in our answer.n =
( ) ( )274
Efficiency 0.7611C.T. 6 60
it
n= = =∑
9.4 a. First assignment costs ( ) 18,000 7,200 1,600 4,800 8,000 800
2$15,200
= + + + + + ×
=
b. New layout costs ( ) 18,000 9,600 6,400 3,600 2,000 800 $15, 200
2= + + + + + × =
No improvement—both yield the same cost.
A-33
9.5 Cost of 3 attempts: Work Area Attempt 1 2 3 Cost a. 1 S D M $1,088 b. 2 D S M $1,142
c. 3 M D S $1,100
↑
( )( ) ( ) ( ) ( )( )2 23 10 32 5 20 8 $1,100 + + =
9.6 a. Theoretical minimum number of stations task times
cycle time= ∑
Cycle time 60
12 minutes5
= = . So minimum number of stations 48
4 stations12
= =
b.
A
10 min
WS #1
B
12 min
WS #2
C
8 min
WS #3
D
6 min
WS #4E
F
12 min
WS #5
This requires 5 stations—it cannot be done with 4.
c. Efficiency 48 48
80%5 12 60
= = =×
for 5 stations.
9.7 There are three alternatives: Station Alternative 1 Tasks Alternative 2 Tasks Alternative 3 Tasks 1 A, B, F A, B A, F, G 2 C, D C, D H, B 3 E G, H C, D 4 G, H E E 5 I I I Each alternative has an efficiency of 86.67%.
A-34
10 SUPPLEMENT
Work Measurement
S10.1 Required sample size 2
Zsn
hx = =
where 0.15s = , 0.4x = , 1.96z = (for 95% confidence),
10%h = accuracy level ( )( )( )( )
21.96 0.15
54.0225 540.10 0.4
n
= = ≈
S10.2 a. 2
Zsn
hx =
Thus, ( )( ) ( )0.10 0.40 12
0.9240.15
hx nZ
s= = =
Referring to Appendix I (Standard Normal Table), Area 0.64 64%= = . The confidence level when 12n = is 64%, as opposed to 95% when 54n = (in Problem S10.1)
b. Average observed time 0.331 0.243 0.484
0.4484 minutes12
+ + += ="
Normal time = Average time × perf. rating = 0.4484 0.90 0.4036 minutes× =
Standard time Normal time 0.4036
0.429 minutes1 allowance factor 1 0.06
= = =− −
S10.3 Average observed time 100 hours 60 minutes 0.75
22.5 minutes200 units
× ×= =
Normal time 22.5 minutes 1.1 24.75 minutes= × =
Standard time for job Normal time for process 24.75
29.12 minute unit1 Allowance fraction 1 0.15
= = =− −
S10.4 a. sum of times 1.74
Observed time 0.10875 minutes 6.525 secondsnumber of cycles 16
= = = =
b.
( ) ( )Normal time Observed time Performance rating factor 6.525 95%
6.2 seconds
= × = ×
=
c. normal time 6.2 6.2
Standard time 6.739 seconds1 allowance factor 1 8% 92%
= = = =− −
A-35
S10.5 Normal time 10 minutes 0.90 9 minutes= × =
Allowance fraction Personal Fatigue Delay 5 3 1 9
0.1560 minutes 60 60
+ + + += = = =
Normal time 9
Standard time 10.59 minutes1 Allowance fraction 1 0.15
= = =− −
S10.6 Observation (Minutes Per Cycle)
Element
Rating 1
2
3
4
5
Average Time
Normal Time
1 100% 1.5 1.6 1.4 1.5 1.5 1.5 1.50 2 90% 2.3 2.5 2.1 2.2 2.4 2.3 2.07 3 120% 1.7 1.9 1.9 1.4 1.6 1.7 2.04 4 100% 3.5 3.6 3.6 3.6 3.2 3.5 3.50 Normal time for lab test = 9.11
Standard time for lab test Normal time for process 9.11
11.1 minutes1 Allowance fraction 1 0.18
= = =− −
A-36
12 CHAPTER
Inventory Management
12.1 An ABC system classifies the top 70% of dollar volume items as A, the next 20% as B, and
the remaining 10% as C items. Similarly, A items constitute 20% of total number of items, B items are 30%; and C items are 50%.
Item Code Number Average Dollar Volume Percent of Total $ Volume1289 400 3.75
2347 300 4.00
2349 120 2.50
2363 75 1.50
2394 60 1.75
2395 30 2.00
6782 20 1.15
7844 12 2.05
8210 8 1.80
8310 7 2.00
9111 6
→ × =→ × =→ × =→ × =→ × =→ × =→ × =→ × =→ × =→ × =→ ×
1,500.00 44.0%
1, 200.00 36.0%
300.00 9.0%
112.50 3.3%
105.00 3.1%
60.00 1.8%
23.00 0.7%
24.60 0.7%
14.40 0.4%
14.00 0.4%
3.00 18.00 0.5%
$3,371.50 100%
=
Answer : The company can make the following classification:
A: 1289, 2347 B: 2349, 2363, 2394, 2395 C: 6782, 7844, 8210, 8310, 9111 12.2 D (Annual Demand) = 4,800 units, P (Purchase Price/Unit) = $27, H (Holding Cost) = $2
S (Ordering Cost) = $30. So, ( ) ( )( )2 4,800 302Order Quantity 240
2
DSQ
H∗ = = = .
( ) ( ) 2 240 30 4,800
Thus, Total Annual Cost 4,800 272 2 2 240
129,600 240 600 $130,440
HQ SDTC PD
× × = + + = × + +
= + + =
A-37
12.3 D (Annual Demand) = 14,558, P (Purchase Price/Unit) = $5, H (Holding Cost/Unit) = $4,
S (Ordering Cost/Order) = $22, 2 2 14,558 22
4004
DSQ
H
× ×= = = tons per order
( ) 4 400 22 14,5585 14,558 72,790 800 800.69
2 2 2 400
$74,390.69
HQ SDTC PD
× × = + + = × + + = + +
=
Answer : The optimal order quantity ( ) 400 tons orderQ = ; total annual inventory cost
( ) $74,391TC = .
12.4 D (Annual Demand) = 400 × 12 = 4,800, P (Purchase Price/Unit) $350 unit= ,
H (Holding Cost/Unit) $35 unit year= , S (Ordering Cost/Order) $120 order= . So,
2 2 4,800 120181.42 181
35
DSQ
H
× ×= = = = units (rounded off).
( ) ( ) 35 181 120 4,800
Thus, Total Cost 4,800 3252 2 181
1,560,000 3,168 3,182 $1,566,350
HQ SDTC PD
Q
× × = + + = × + +
= + + =
However, if Bell Computers orders 200 units,
( ) 35 200 120 4,8004,800 325 1,440,000 3,500 2,880 $1, 446,380
2 200TC
× × = × + + = + + =
Answer : Bell Computers should order 200 units for a minimum total cost of $1,446,380.
12.5 1
2 2 4,800 120181 units
35
DSQ
H
× ×= = =
2
2 2 4,800 120188 units
32.5
DSQ
H
× ×= = =
3
2 2 4,800 120196 units
30
DSQ
H
× ×= = =
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, 188 unitsEOQ = .
( ) ( ) 32.5 188 120 4,800Thus, 188 units 325 4,800
2 2 188
1,560,000 3,055 3,064 $1,566,119
HQ SDTC PD
Q
× × = + + = × + +
= + + =
( ) ( ) 30 200 120 4,800200 units 300 4,800
2 2 200
1,440,000 3,000 2,880 $1, 445,880
HQ SDTC PD
Q
× × = + + = × + +
= + + =
Answer : The minimum order quantity is 200 units yet again, since the overall cost of
$1,445,880 is less than ordering 188 units which has an overall cost of $1,566,119.
A-38
12.6 12,500 yearD = , so 50 dayd = , 300 dayp = , $30 orderS = , $2 unit yearH =
a. 2 2 12,500 30 300
612.37 1.095 6712 300 50
DS pQ
H p d
× ×= × = × = × =− −
b. Number of production runs ( ) 12,50018.63
671D
NQ
= = =
c. Maximum inventory level ( )max
50 11 671 1 671 1 559
300 6d
I Qp
= − = − = − =
d. Days of demand satisfied by each production run 250
13.4218.63
= = days in demand
only mode
Time in production for each order 671
2.24300
Q
p= = = days in production for each
order. Total time 13.42= days per cycle.
Thus, percent of time in production 2.24
16.7%13.42
= = .
12.7 $2 unit yearH = , $10 orderS = , 4,000
4,000 16250
D d = = =
. So,
2 2 4,000 10200
2
DSQ
H
× ×= = = . ROP (reorder point) l d= × , l (lead time = 5 days).
Thus, ROP 5 16 80 units= × = . Answer : Saveola, Inc. should place an order for 200 frames every time the inventory of
frames falls to 80 units. This will be their inventory policy.
12.8 ( )2 5, 400 34
300H
= ; Square both sides 367,200
90,000H
= , 367,200
$4.0890,000
H = =
12.9 a. ( )( )2 6,000 302
189.74 units10
DSEOQ
H= = =
b. Average inventory = 94.87 c. Optimal number of orders/year = 31.62
d. Optimal days between orders 250
7.9131.62
= =
e. Total annual inventory cost 601,897.37= (including the $600,000 cost of goods)
12.10 a. Holding cost = $530.33 b. Set up cost = $530.33 c. Unit costs = $56,250.00 d. Total costs = $57,310.66 e. Order quantity = 16,970.56 units Thus, order 10,001 units for a total cost of $57,310.66.
A-39
12.11 Inventory Item
$Value
per Case
#Ordered
per Week
Total $
Value/Week
(52 Weeks) Total
($*Weeks)
Rank
Percent of Inventory
CumulativePercent ofInventory
Fish Fillets 143 10 $1,430.00 $74,360.00 1 17.54% 34.43% French Fries 43 32 $1,376.00 $71,552.00 2 16.88% 47.31% Chickens 75 14 $1,050.00 $54,600.00 3 12.88% 59.53% Prime Rib 166 6 $996.00 $51,792.00 4 12.22% 69.83% Lettuce (case) 35 24 $840.00 $43,680.00 5 10.31% 78.85% Lobster Tail 245 3 $735.00 $38,220.00 6 9.02% 83.82% Rib Eye Steak 135 3 $405.00 $21,060.00 7 4.97% 87.25% Bacon 56 5 $280.00 $14,560.00 8 3.44% 90.64% Pasta 23 12 $276.00 $14,352.00 9 3.39% 93.74% Tomato Sauce 23 11 $253.00 $13,156.00 10 3.10% 95.71% Table Cloths 32 5 $160.00 $8,320.00 11 1.96% 97.60% Eggs (case) 22 7 $154.00 $8,008.00 12 1.89% 98.28% Oil 28 2 $56.00 $2,912.00 13 0.69% 98.72% Trash Can Liners 12 3 $36.00 $1,872.00 14 0.44% 99.13% Garlic Powder 11 3 $33.00 $1,716.00 15 0.40% 99.42% Napkins 12 2 $24.00 $1,248.00 16 0.29% 99.72% Order Pads 12 2 $24.00 $1,248.00 17 0.29% 99.83% Pepper 3 3 $9.00 $468.00 18 0.11% 99.93% Sugar 4 2 $8.00 $416.00 19 0.10% 99.93% Salt 3 2 $6.00 $312.00 20 0.07% 100.0% $8,151.00 $423,852.00 100.00%
a. Fish filets total $74,360 b. C items are items 10 through 20 in the above list (although this can be one or two
items more or less) c. Total annual $ volume = $423,852
12.12 Incremental Costs Safety Stock Carrying Cost Stockout Cost Total Cost 0 0 70(100 × 0.4 + 200 × 0.2) = 5,600 5,600 100 100 × 15 = 1,500 (100 × 0.2) × (70) = 1,400 2,900 200 200 × 15 = 3,000 0 3,000 The safety stock which minimizes total incremental cost is 100 kilos. The re-order point then
becomes 200 kilos + 100 kilos or, 300 kilos.
A-40
12.13 S $10 order= , 4 daysLT = , 80 dayd = , 20σ = , H 10%= of $250,
( )for term 80 100 8,000D = × =
a. 2
800 calzonesDS
QH
= = . Order every days 10 daysQ
d=
b. ROP for calzone, ( )demand during lead time 40LTσ σ= × = , 1Z = (from Table)
for 0.1587, ROP 360dLTd LT Z σ= × + × =
c. On hand = 85, days left = 1. Since ROP daltd LT Z σ= × + × , ROP d LT
ZLTσ
− ×= .
Thus, Z of 0.25 gives 0.5987. So there is a 40.13% chance of stockout.
d. 400 average inventory2
Q = = , 10 orders termD
Q= .
( )( )Holding cost term 400 0.25 $1002
QH= = =
( )( )Order cost term 10 10 $100D
SQ
= = =
12.14 $16S = , $0.40 calzone termH = , 160p = , 80d = , 8,000D = for 100 days of the term
a. ( )2
1,131.371
DSQ
H d p= =
−
b. 1,131.37
Cycle 14.1480
Q
d= = =
c. Run production for 1,131
days 7.07 days160
Q
p= =
A-41
12.15 Under present price of $6.40 per box:
Economic Order Quantity: 2 2 5000 25
395.3 or 395 boxes0.25 6.40
DSQ
H∗ × ×= = =
× where
D = annual demand, S = set-up or order cost, H = holding cost
Total cost order cost holding cost purchase cost2
5,000 25 395 0.25 6.406.4 5,000 316.46 316.00 32,000
395 2$32,632.46
DS QHCD
Q= + + = + +
× × ×= + + × = + +
=
Note: Order and carrying costs are not exactly equal due to rounding of the EOQ to a whole number. Under the quantity discount price of $6.00 per box:
Total cost order cost holding cost purchase cost2
5,000 25 5,000 0.25 6.005,000 6.00 41.67 3,750.00 30,000
3,000 2
$33,791.67
DS QH
Q= + + = +
× × ×= + + × = + +
=
Therefore, the old supplier with whom they would incur a total cost of $32,632.46, is preferable.
12.16 Economic Order Quantity, non-instantaneous delivery: where: D = period demand, S = set-up or order cost, H = holding cost, d = daily demand rate, p = daily production rate
( ) ( )50200
2 2 10,000 2002,309.4
1.00 11 dp
DSQ
H∗ × ×= = =
−− or 2,309 units
A-42
13 CHAPTER
Aggregate Planning
13.1 The total production required over the year is 8,400 units, of 700 per month. Thus, the
schedule is to produce 700 per month and have no costs associated with work force variation. The only costs incurred will be the monthly production cost, the inventory cost, and the shortage cost. The costs are calculated as follows.
Month Beginning Inventory
Produc- tion
Production Cost
Demand
Ending Inventory
Shortage
Inventory Cost
Shortage Cost
January 0 700 $49,000 500 200 0 $600 $0 February 200 700 49,000 600 300 0 900 0 March 300 700 49,000 600 400 0 1,200 0 April 400 700 49,000 700 400 0 1,200 0 May 400 700 49,000 700 400 0 1,200 0 June 400 700 49,000 800 300 0 900 0 July 300 700 49,000 900 100 0 300 0 August 100 700 49,000 900 0 100 0 1,000 September 0 700 49,000 800 0 200 0 2,000 October 0 700 49,000 700 0 200 0 2,000 November 0 700 49,000 600 0 100 0 1,000 December 0 700 49,000 600 0 0 $ 0 0 8,400 $588,000 8,400 2,100 600 $ 6,300 $6,000 The total cost of this plan is the sum of the three costs, or $600,300.
A-43
13.2 a. Total hotel demand for the year = 7,000,000 Total restaurant demand for the year = 2,080,000
Level staffing
Quarter
Hotel (Room) Demand
Personnel Required
RestaurantDemand
Req.
Total
Hire
Terminate Winter 800,000 33 160,000 12 45 45 – Spring 2,200,000 33 800,000 12 45 – – Summer 3,300,000 33 960,000 12 45 – – Fall 700,000 33 160,000 12 45 – – Totals 7,000,000 132 2,080,000 48 180 45
Personal cost = 180 quarters of labor × 5,000 = $900,000 Hiring cost = 45 hires at $1,000 = 45,000 Termination cost = none = 0 $945,000 b. Total hotel demand for the year = 7,000,000 Total restaurant demand for the year = 2,080,000
Chase staffing (staffing to meet the forecasted demand)
Personnel Personnel Totals
Quarter
Hotel (Room) Demand
Req.
Hire
Termi-
nate
RestaurantDemand
Req.
Hire
Termi-
nate
Quarter Hires
Quarter Termi- nations
Winter 800,000 8 8 160,000 2 2 10 Spring 2,200,000 22 14 800,000 10 8 22 Summer 3,300,000 33 11 960,000 12 2 13 Fall 700,000 7 0 26 160,000 2 0 10 36 Totals 7,000,000 70 2,080,000 26 45 36
Personnel cost = 96 quarters of labor × 5,000 = $480,000 Hiring cost = 45 hires @ $1,000 = 45,000 Termination cost = 36 terminations @ $2,000 = 72,000 $597,000 c. The chase plan developed in part b is most economical ($945,000 vs. $597,000)
A-44
13.3 Total hotel demand for the year = 7,000,000 Total restaurant demand for the year = 2,080,000
Hiring from local staffing agency all personnel above base requirements.
Quarter
Hotel (Room) Demand
Req.
Hire
Personnel from
Agency
Restaurant Demand
Req.
Hire
Personnel from
Agency
Quarter Hires
Quarterfrom
Agency Fall 800,000 8 7 1 160,000 2 2 0 2 1 Spring 2,200,000 22 0 15 800,000 10 0 8 0 23 Summer 3,300,000 33 0 26 960,000 12 0 10 0 36 Fall 700,000 7 0 0 160,000 2 0 0 0 0 Totals 7,000,000 28* 42 2,080,000 26 8** 18 35 60 *On Hotel Grand payroll (7 each quarter × 4 quarters = 28) ** On Hotel Grand payroll (2 each quarter × 4 quarters = 8) Total on Grand Hotel payroll = 28+ 8 = 36
Personnel cost = 36 quarters of labor × 5,000 = $180,000 Hiring cost = 9 ( )7 2= + hires @ $1,000 = 9,000
Termination cost = 0 terminations @ $2,000 = 0 Staffing agency costs = 60 @ 6,500 = 390,000 $579,000 13.4 Plan A: Month Demand Production Hire Fire Extra Cost Mar 1,000 900 700 56,000 Apr 1,200 1,200 300 12,000 May 1,400 1,400 200 8,000 June 1,200 1,200 200 8,000 July 1,500 1,500 300 12,000 Aug 1,300 1,300 200 16,000 Total extra cost: $112,000
Plan B: Month Demand Production Inventory Sub-Contracting Extra Cost Mar 1,000 1,100 200 2,000 Apr 1,200 1,100 100 1,000 May 1,400 1,100 200 8,000 June 1,200 1,100 100 4,000 July 1,500 1,100 400 16,000 Aug 1,300 1,100 200 8,000
Total extra cost: $39,000 Therefore, Plan B would be preferred.
A-45
13.5 Plan Number 1: Month Demand Production Inventory Sub-Contracting Extra Cost 1 1,000 1,200 300 3,000 2 1,200 1,200 300 3,000 3 1,400 1,200 100 1,000 4 1,200 1,200 100 1,000 5 1,500 1,200 200 8,000 6 1,300 1,200 100 4,000 Total extra cost: $112,000
Plan Number 2: Month Demand Production Inventory Overtime Extra Cost 1 1,000 1,200 300 3,000 2 1,200 1,200 300 3,000 3 1,400 1,200 100 1,000 4 1,200 1,200 100 1,000 5 1,500 1,200 200 2,000 6 1,300 1,200 100 1,000 Total extra cost: $39,000 Therefore, Plan Number 2 would be preferred. 13.6 Plan Y: Month Demand Production Hire Fire Extra Cost 1 1,100 1,100 400 32,000 2 1,600 1,600 500 20,000 3 2,200 2,200 600 24,000 4 2,100 2,100 100 8,000 5 1,800 1,800 300 24,000 6 1,900 1,900 100 4,000 Total extra cost: $112,000
Plan Z: Month Demand Production Inventory Sub-Contracting Extra Cost 1 1,100 1,600 600 6,000 2 1,600 1,600 600 6,000 3 2,200 1,600 4 2,100 1,600 500 20,000 5 1,800 1,600 200 8,000 6 1,900 1,600 300 12,000 Total extra cost: $52,000 Therefore, Plan Z would be preferred.
A-46
14 CHAPTER
Materials Requirements Planning (MRP) & ERP
14.1 Requirement for 3,500 “Get Well” bud vases. 3,500 Vases 3,500 × 8" white ribbons (2,333 ft.) 3,500 × 8" red ribbons (2,333 ft.) 3,500 signature cards 7,000 sprigs of baby’s breath 7,000 pink roses 14.2 Period (week)
Lot Size
Lead Time
On Hand
Safety Stock
Allo-cated
Low-Level Code
Item ID
CD Case 1 2 3 4 5 6 7 8
Gross Requirements 650 300 550 400 500 Scheduled Receipts Projected On Hand 1,000 1,000 1,000 1,000 350 50 Net Requirements 500 400 500 Planned Order Receipts 500 400 500
Lot for Lot
1
1,000
—
—
0
CD Case
Planned Order Releases 500 400 500 Gross Requirements 500 400 500 Scheduled Receipts Projected On Hand Net Requirements 500 400 500 Planned Order Receipts 500 400 500
Lot for Lot
1
—
—
—
1
CD Top
Planned Order Releases 500 400 500 Gross Requirements 500 400 500 Scheduled Receipts Projected On Hand Net Requirements 500 400 500 Planned Order Receipts 500 400 500
Lot for Lot
1
—
—
—
1
CD Bottom
Planned Order Releases 500 400 500 Gross Requirements 500 400 500 Scheduled Receipts Projected On Hand Net Requirements 500 400 500 Planned Order Receipts 500 400 500
Lot for Lot
1
—
—
—
1
CD Insert
Planned Order Releases 500 400 500 Gross Requirements 500 400 500 Scheduled Receipts Projected On Hand 100 100 100 100 Net Requirements 500 400 500 Planned Order Receipts 500 500 500
Lot for Lot
2
—
—
—
2
BlackDye
Planned Order Releases 500 500 500
A-47
14.3 Period (day)
Lot Size
Lead Time
On Hand
Safety Stock
Allo-cated
Low-Level Code
Item ID
Ball Point Pens 1 2 3 4 5 6 7 8
Gross Requirements 10,000 Scheduled Receipts — Projected On Hand — Net Requirements 10,000 Planned Order Receipts 10,000
Lot for Lot
1
—
—
—
0
Planned Order Releases 10,000 10,000 Gross Requirements 10,000 Scheduled Receipts Projected On Hand Net Requirements Planned Order Receipts 10,000
Lot for Lot
1
—
—
—
1
Cap
Planned Order Releases 10,000 Gross Requirements 2,000 Scheduled Receipts — Projected On Hand — Net Requirements 2,000 Planned Order Receipts 2,000
Lot for Lot
3
—
—
—
1
Ink CC
Planned Order Releases 2,000 Gross Requirements 10,000 Scheduled Receipts Projected On Hand Net Requirements 10,000 Planned Order Receipts 10,000
Lot for Lot
1
—
—
—
Body
Planned Order Releases 10,000 Gross Requirements 5,000 Scheduled Receipts Projected On Hand Net Requirements 5,000 Planned Order Receipts 5,000
Lot for Lot
1
—
—
—
Clip Fine Pt
Planned Order Releases 5,000 Gross Requirements 5,000 Scheduled Receipts Projected On Hand 3,000 3,000 Net Requirements 2,000 Planned Order Receipts 2,000
Lot for Lot
1
3,000
—
—
Std. Clip
Planned Order Releases 2,000 Gross Requirements 5,000 Scheduled Receipts Projected On Hand 3,000 3,000 Net Requirements 2,000 Planned Order Receipts 2,000
Lot for Lot
1
3,000
—
—
Std. Ball Point
Planned Order Releases 2,000 Gross Requirements 5,000 Scheduled Receipts Projected On Hand Net Requirements 5,000 Planned Order Receipts 5,000
Lot for Lot
1
—
—
—
Fine Point Ball Point
Planned Order Releases 5,000
A-48
14.4 Period (week, day)
Lot Size
Lead Time
On Hand
Safety Stock
Allo-cated
Low-Level Code
Item ID
Ball Point Pens 1 2 3 4 5 6 7 8
Gross Requirements 640 640 128 128 Scheduled Receipts Projected On Hand Net Requirements 640 640 128 128 Planned Order Receipts 640 640 128 128
Lot for Lot
1
—
—
—
0
Coffee Table
Planned Order Releases 640 640 128 128 Gross Requirements 640 640 128 128 Scheduled Receipts Projected On Hand Net Requirements 640 640 128 128 Planned Order Receipts 640 640 128 128
Lot for Lot
1
—
—
—
1
Top
Planned Order Releases 640 640 128 128 Gross Requirements 80 80 16 16 Scheduled Receipts Projected On Hand Net Requirements 80 80 16 16 Planned Order Receipts 80 80 16 16
Lot for Lot
1
—
—
—
1
Stain gal
Planned Order Releases 80 80 16 16 Gross Requirements 40 40 8 8 Scheduled Receipts Projected On Hand 100 100 100 100 100 60 20 12 4 Net Requirements Planned Order Receipts
Lot for Lot
1
—
—
—
1
Glue gal
Planned Order Releases Gross Requirements 640 640 128 128 Scheduled Receipts Projected On Hand Net Requirements 640 640 128 128 Planned Order Receipts 640 640 128 128
Lot for Lot
1
—
—
—
1
Base
Planned Order Releases 640 640 128 128 Gross Requirements 1,280 Scheduled Receipts Projected On Hand Net Requirements 1,280 Planned Order Receipts 1,280
Lot for Lot
1
—
—
—
2
Long Braces
Planned Order Releases 1,280 Gross Requirements 1,280 Scheduled Receipts Projected On Hand Net Requirements 1,280 Planned Order Receipts 1,280
Lot for Lot
1
—
—
—
2
Short Braces
Planned Order Releases 1,280 Gross Requirements 5,120 Scheduled Receipts Projected On Hand Net Requirements 5,120 Planned Order Receipts 5,120
Lot for Lot
1
—
—
—
2
Leg
Planned Order Releases 5,120 Gross Requirements 5,120 Scheduled Receipts Projected On Hand 880 880 880 880 880 880 Net Requirements 5,120 Planned Order Receipts 6,000
Lot for Lot
1
—
—
—
3
Brass Caps
Planned Order Releases 6,000
A-49
14.5 Coffee Table Hours Lead
Master Schedule Required Time Day 1 Day 2 Day 3 Day 4 Day 5 Day 6 Day 7 Day 8 640 640 128 128 Table Assembly 2 1 1,280 1,280 256 256 Top Preparation 2 1 1,280 1,280 256 256 Assemble Base 1 1 640 640 128 128 Long Braces (2) 0.25 1 320 320 64 64 Short Braces (2) 0.25 1 320 320 64 64 Legs (4) 0.25 1 640 640 128 128 Total Hours 0 1280 3,200 3,456 1,920 640 256 Employees needed @ 8 hrs. each 0 160 400 432 240 80 32
14.6 The following table lists the components used in assembling FG-A. Also included for each
component are the following information: the on-hand supply, lead time, and direct components.
Item On-Hand LT (weeks) Components PG-A 0 1 SA-B, SA-C(2), SA-D(2) SA-B 0 1 SA-D(2) SA-C 0 2 E, F(2) SA-D 0 2 E (3) E 10 1 — F 5 3 — While not required as part of the question, we recommend you make a product structure
tree to help you answer the questions on the Bill-of-Materials and lead time.
1 FGA [LT =1]
[LT =1]SA-B
[LT =2]SA-D(2)
E(3) [LT = 1]
[LT =2]SA-C(2)
E [LT = 1] F(2) [LT = 3]
[LT =2]SA-D(2)
E(3) [LT = 1]
a. Bill-of-Material associated with 1 unit of FG-A 1 SA-B Note: this is just the master recipe. Not subtracting off on on-hand quantities YET 2 SA-C 4 SA-D 2 1 2= × + 14 E 2 3 2 1 2 3 6 2 6= × + × + × = + + 4 F 2 2= ×
A-50
b. Total lead time (in weeks) associated with making an item of FG-A, assuming we had no starting on-hand for any part? 6 weeks
Look at all branches of the product structure tree (or an assembly time chart, if we
had one). The most common mistake people tend to make is to forget the LT associated with final assembly.
Longest branch is 6 weeks 1wk FGA 2wks SA-C 3wks F= + + c. Yes, if we wanted to make one FG-A, we need to order more of either E or F.
We have enough F on-hand, but need 4 more E. Now we look at the on-hand quantities and see we have 10 E and 5 F. Compare that
to the BOM calculated above. No subassemblies like SA-B, SA-C or SA-D, so we are going to need all the component parts. Our on-hand records show we have 14 E and 4 F, so we are short 4 E.
14.7 a. 17 × 2 (level 2) = 34
b. [17 × 3 (B’s)] – 15 = 51 – 15 = 36 36 × 5 (D’s) = 180
14.8 a. 17 × 2 (level 2) + 17 × 2 × 3 (level 3) = 136
b. 170 – 12 × 2 (level 2 on hand) = 170 – 24 = 146
A-51
15 CHAPTER
Short-Term Scheduling
15.1 A dummy task is added to balance the problem. The assignment is May—Task 1 Gray—Task 2 Ray—Task 3 Total time = 1 + 1 + 1 = 3 hours 15.2 Convert the minutes into $, Marketing Finance Operations Human Resources Chris $80 $120 $125 $140 Steve $20 $115 $145 $160 Juana $40 $100 $35 $45 Rebecca $65 $35 $25 $75 (Now subtract smallest number in each column from every number.) The Minimum Cost Solution =
Chris Finance $120
Steve Marketing $ 20
Juana Human Resources $ 45
Rebecca Operation $ 25
$210
15.3 The best pairs are assigned as follows: Ajay—Jackie Jack—Barbara Gray—Stella Raul—Dana Total compatibility score (overall) = 90 + 70 + 50 + 20 = 230
A-52
15.4
10 20 30 40
F
D
C
B
E
A
Hour
6
10
20
37
Gantt Chart
13
28
Project Time Due Date Late Days A 9 22 15 B 7 17 3 C 3 16 0 D 4 13 0 E 8 16 12 F 6 9 0
a. 6 10 13 20 28 37 114
Average flow time 19 days6 6
+ + + + += = =
b. total late days 15 3 12
Average lateness 5 daysnumber of jobs 6
+ += = =
c. Maximum lateness 15 days= (for job A—Gantt Table) 15.5 a. The shortest processing time ( )SPT =DBACE
D B A C E Flow Time 0 1 3 6 11 20 = 41 Due Hours 0 12 4 8 6 7 Late Hours 0 — — — 5 13 = 18
Average flow time 41
8.2 hours5
= =
Number of deliveries late C and E 2= =
Average hours late 5 13 18
9 hours2 2
+= = =
A-53
b. The EDD schedule = BCEAD
B C E A D Flow Time 0 2 7 16 19 20 Due Hours 0 4 6 7 8 12 Late Hours 0 — 1 9 11 8
Average flow time 64
12.8 hours5
= =
Number of deliveries late C, E, A, and D 4= =
Average hours late 29
7.25 days4
= =
15.6 5
Cut & sew
Deliver
10 15 20 25
5 10 15 20 25
1
1
2
2
3
3
17114
6 18 23
[123 schedule]
5
Cut & sew
Deliver
10 15 20 25
5 10 15 20 25
3 2 1
17136
6 11 22
[321 schedule]
Answer : The 321 schedule finishes in 22 days, 1 day faster than the 123 schedule, which
finishes in 23 days. 15.7 a. We begin by taking 5 empty slots
b. Next, we find the shortest time in the table. It is product 2678 on Machine B. Since this is on the second machine, this product can be done as late as possible.
2678
c. Then we find the shortest again. It is 2800 on B 2800 2678 d. Then it is 2731 on A. Since it is on the first machine, it is the earliest job.
2731 2800 2678
A-54
e. Finally, we get 2731 2134 2387 2800 2678
5
A
12 19 22
B
2731 2134 2387 2800 2678
7 14 21 25
2731 2134
33 35
2800 26782387 Total time 33 hours=
15.8 a. The jobs should be processed in the sequence: 3, 6, 2, 7, 5, 1, 4
b. Time = 61 hours 15.9 a. Jobs should be processed in the sequence A, B, C, D, E if scheduled by the FCFS
scheduling rule. b. Jobs should be processed in the sequence A, B, C, D, E if scheduled by the EDD
scheduling rule. c. Jobs should be scheduled in the sequence B, E, A, C, D if scheduled by the SPT
scheduling rule. d. Jobs should be processed in the sequence D, C, A, B, E if scheduled by the LPT
scheduling rule. 15.10 Job Due Date Duration (Days) Critical Ratio 103 214 10 1.4 205 223 7 3.3 309 217 11 1.5 410 219 5 3.8 517 217 15 1.1 Jobs should be scheduled in the sequence 517, 103, 309, 205, 412 if scheduled by the
critical ratio scheduling rule.
A-55
16 CHAPTER
Just-In-Time and Lean Production Systems
16.1 a. ( )2
1 dp
D SQ
H
× ×=−
( )2 2
1 dp
D SQ
H
× ×=−
( )( ) ( ) ( )( ) ( )( )( )22 400
5001 1,000 10 1 1,000,000 10 .2 2,000,000
$102 2 100,000 200,000 200,000
dpQ H
SD
− −= = = = =
×
b. 10 1
hr 10 min60 6
⇒ ⇒
Note that the lead time was not needed in this problem. 16.2 10 min. – 2 min. = 8 min. improvement required 16.3 How to improve setups (from Figure 16.4 in Heizer/Render text).
Step 1: Separate setup into preparation and actual setup, doing as much as possible while the machine/process is operating.
Step 2: Move material closer and improve material handling Step 3: Standardize and improve tooling Step 4: Use one-touch system to eliminate adjustments Step 5: Train operators and standardize work procedures
16.4 ( )( )
( )5,00010,000
2 1, 250,000 152 37,500,0003,750,000 1,936 units
1020 11 dp
D SQ
H
× ×× ×= = = = =−−
1,936
per container 19.36 containers100
=
A-56
16.5 Note: 6S = minutes (or 110 hour) $100× shop labor cost .1 $100 $10= × =
( )( )( )( )
( )
( )
50500
2 12,500 102 250,0005,555 74
50 1 451
1EOQ Safety stock Lead time 74 10 500
2
334 oil pumps
dp
D SQ
H
× ×= = = = =−−
= + + = + +
=
334No. of Kanban containers 9.27 9 or 10 containers required
36= = =
16.6 Where: D = annual demand, S = set-up or order cost, H = holding cost, d = daily demand rate, p = daily production rate. Solving for S (set-up cost):
( ) ( ) ( )2 2 1501,0001 150 10 1 22,500 10 1 0.15
2 2 40,000 80,000
191,250$2.39
80,000
$2.39 set-up 60 minute hourSet-up time 2.69 minute set-up
$50 hour
dpQ H
SD
× × − × × − × × −= = =
×
= =
×= =
A-57
17 CHAPTER
Maintenance and Reliability
17.1 Failure Rate Analysis Station No. 1: N = 1,000 F = 22
22
0.0221,000
1 0.022 0.978 97.8%
FR
R
= =
= − = =
Station No. 2: N = 1,000 F = 51
51
0.0511,000
1 0.051 0.949 94.5%
FR
R
= =
= − = =
Answer : Unit #1 is clearly better, with a lower failure rate and better reliability.
17.2 Reliability ( ) 1 2 3s nR R R R R= × × × ×! . So, ( )( )( )5 0.98 0.99 0.96 0.9304R = =
Answer : The reliability is 93.04%
A-58
17.3 Task Errors FR Reliability 1 1 0.001 0.999 2 6 0.006 0.994 3 4 0.004 0.996 4 2 0.002 0.998 5 3 0.003 0.997 6 1 0.001 0.999 7 0 0.000 1.000 8 2 0.002 0.998 9 1 0.001 0.999 10 2 0.002 0.998 11 1 0.001 0.999 12 1 0.001 0.999 13 0 0.000 1.000 14 2 0.002 0.998 15 3 0.003 0.997 Total 29
Overall Reliability ( ) Product of all reliabilities above 0.971 97.1%sR = = =
17.4 Reliability (stick) ( )21 1 0.99 0.9999= − − =
Reliability (cloth) ( )21 1 0.98 0.999998= − − =
The reliability of the snake-charmer’s work 0.9999 0.999998 0.9998 99.98%= × = = 17.5
.96
.96
.96
.99
.99
.93
.93
.93
.90
.90
.90
.90
Mountain test’s reliability ( )31 1 0.96 0.999936= − − =
Bump test’s reliability ( )21 1 0.99 0.999900= − − =
Speed test’s reliability ( )31 1 0.93 0.999657= − − =
Sudden brake test’s reliability ( )41 1 0.90 0.999900= − − =
So, overall reliability ( ) 0.999936 0.999900 0.999657 0.999900 0.9994 99.94%sR = × × × = =
A-59
A MODULE
Decision Making Tools
A.1
$20,000 $30,000 $26,500AssemblyLine
–$10,000 $50,000 $29,000PlantAddition
$0 $0 $0No NewProduct
0.35Competes
0.65Doesn’t Compete
ExpectedValue
Prob.States of Nature
DecisionAlternatives
Answer : ( ) ( )0.35 $20,000 0.65 $50,000 $29,000 $10,500+ − =
A.2
0 2,000 3,000N
–4,000 8,000 9,000M
–10,000 6,000 20,000L
FixedSlight
IncreaseMajor
Increase
States of Nature
DecisionAlternatives
Minimum
0
–4,000
–10,000
–50,000 4,000 40,000O –50,000 Use “maximin: criterion. No floor space (N). A.3 Row Average Increasing capacity $700,000 Using overtime $700,000 Buying equipment $733,333 ← Using equally likely, “Buying equipment” is the best option.
A-60
A.4 ( ) ( ) ( ) ( )3 2 35 15Expected value under certainty 50 20 100 100
10 10 100 10015 4 35 15 69
= + + +
= + + + =
A.5 ( ) ( ) ( ) ( )E A 0.4 40 0.2 100 0.4 60 60= + + =
( ) ( ) ( ) ( )E B 0.4 85 0.2 60 0.4 70 74= + + = ←
( ) ( ) ( ) ( )E C 0.4 60 0.2 70 0.4 70 66= + + =
( ) ( ) ( ) ( )E D 0.4 65 0.2 75 0.4 70 69= + + =
( ) ( ) ( ) ( )E E 0.4 70 0.2 65 0.4 80 73= + + =
Choose Alternative B.
A.6 ( ) ( ) ( ) ( )1 1 3 1Expected value under certainty 50 92 40 64 10 23 12 16 61
5 4 10 4= + + + = + + + =
A.7 Solution Approach: Decision tree, since problem is under risk and has more than one
decision.
Wait 1 day
Buy now $70,000
Buy now $55,000
$0
Avail. (.30)
21Wait 1 day
Unavailable(.70)55
$0
Avail. (.60)
33
33
Buy now $30,000
Wait 1 day
Unavailable(.40)
70
$0
(Numbers in Nodes are in $1,000’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-61
A.8 We use a decision table, since expected marginal value of perfect information is asked for.
Profits (= Payoffs)
D1: Good Market: 80(400) – 25,000 = $7,000 Bad Market: 70(375) – 25,000 = $1,250 D2: Good Market: 85(450) – 30,000 = $8,250 Bad Market: 80(425) – 30,000 = $4,000 D3: Good Market: 90(475) – 33,000 = $9,750 Bad Market: 80(425) – 33,000 = $1,000
Decision Alternatives : D1, D2, D3, N (N = Do Nothing)
States of Nature : Good Market, Bad Market
Decision Table and Solution
7,000 1,250D1
8,250 4,000D2
9,750 1,000D3
Good Market(0.6)
Bad Market(0.4)
States of Nature
DecisionAlternatives
EV
4,700
6,550
6,250
0 0N 0 The car enthusiast should use design D2. Exp. value under certainty: ( )( ) ( )( )0.6 9,750 0.4 4,000+ = $7,450
Minus Max Expected Value: –$6,550 Expected (Marginal) Value $900→
A-62
A.9 More than one decision is involved, and problem is under risk, so use a decision tree approach.
20Modest (0.3)
response
160
160
220Sizable (0.7)response
Advertise
40Don’t Advertise544
800
Low (0.4)demand
High (0.6)demand
544
LargeFacility
270
242
223
270
Don’t expand
Expand
Low (0.4)demand
High (0.6)demand
SmallFacility
200
Note: Payoffs and expected payoffs are in $1000’s.
a. Build the large facility. If demand proves to be low, then advertise to stimulate demand. If demand proves to be high, no advertising option is available (so don’t advertise).
b. Expected Payoff: $544,000. A.10 Approach: Set up and solve via a decision table.
Cost per dozen bagels: Labor: $16.50 hr
$1.65 dozen10 dozen hr
=
Ingredients: $0.85 hr
$2.50 dozen
+
$3.00 if sold during the day
Profit per dozen bagels$1.00 if not
∗
∗∗
= −
$3.00 $5.50 $2.50
$1.00 $1.50 $2.50
∗
∗∗
= −− = −
Suppose: B = dozens baked in A.M. D = dozens sold during the day (= dozens demanded)
Then, ( )$3 $1 iftotal profit
$3 ifD B D B DB B D
− − ≥= ≤
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-63
120 12040
119 12341
118 12242
40(0.20)
41(0.30)
(= States of Nature)
DozensBaked B
EV
120
122.2
123.2
117 12143 122.8
Dozens Demanded D
120 120
123 123
126 126
42(0.35)
43(0.15)
125 129
= DecisionAlternatives
Answer : 42 dozen bagels
A.11 a.
8 – (6 + 2) = 0
Ad?
5 – (6 + 2) = –3
P(success) = 0.5
Prob (not success) = 0.5
advertise
5 – 6 = –1Advertise
Don’t
EMV = 0.5(0) + 0.5(–3) = –1.5
10 – 6 = 4Rev – Cost
PayoffEMV = 0.6(0.4) + 0.4(–1) = 2
P(high) = 0.6
P(low) = 0.4Build?
$0Don’t Build
Build
2
b. Build studio, but don’t advertise (even if demand is low) c. Expected value = $2 million
A.12 a.
0
10 – 6 = 4
P(high) = 0.6
P(low) = 0.4
Demand
Don’t Build
Build
Build
Build?
Ad?Build
EMV = 0.6(4) +0.4(0) = 2.4
0
We don’t really careabout this part of the tree(which has a –1 EMV anyways)
Don’t Build b. EVPI = EV with PI – EMV = 24 – 2 = 0.4 = $400,000
A-64
B MODULE
Linear Programming
B.1 a. Let T = number of trucks to produce per day
Let C = number of cars to produce per day max z = 300T + 220C
S.T. 1 1
140 60
T C+ ≤
1 1
150 50
T C+ ≤
, 0T C ≥ b. Graph feasible region:
10
20
30
40
50
60
10 20 30 40 50 60
T
C
70 80 90
A
B
C
(1) (2)O
c.
( )( )( )
( ) ( )( )
Point Coordinates valueO 0, 0 0A 0, 50 11,000C 40, 0 12,000B Solve 2 equations in 2 unknowns 12,600
derived from 1 and 2to obtain 20, 30
z
←
d. Produce 20 trucks and 30 cars daily for a profit of $12,600 per day.
A-65
B.2 We solve this problem by the isocost line method:
2
4
6
8
10
12
2 4 6 8 10 12
x
14 16 18
(6, 6)
1
x2
14
16
18
(0, 4)
z = 12z = 4 z = 11–2
Answer : Unique optimal solution is (0, 4) with z = 4
B.3 Feasible region is a line segment AB, where A = (0,0), B = (3, 5). Solution via isoprofit line
method is shown.
2
4
6
8
2 4 6
x
(1)
1
x2
–8
–6
–4
z = 20
–2
z = 0z = 5
(2)
(3)
A
B
(4)
(5)
Answer : Unique optimal solution is ( ) ( )1 2, 3, 5x x = , with objective function value 20.
A-66
B.4 Using the isoprofit line method.
123456
1 2 3 4 5 6 7 8 9
Opt. sol.789
(2)
–1–2–3–4
(5)
(6)
(4)(3)
(1)z = 9
z = –4 = –6z
A
B
Answer : Unique optimal solution is ( ) ( ), 1, 5A B = . It has objective function 9.0.
B.5 Feasible region is same as in Problem B.4. Use Isoprofit line method.
123456
1 2 3 4 5 6 7 8 9
789
–1–2–3–4
z = 1
A
z = 4B
Answer : Problem is feasible and unbounded.
A-67
B.6 Let S = number of standard bags to produce per week Let D = number of deluxe bags to produce per week
( )
( )
maximum: 10 8
1300
22
3603, 0
z S D
S D A
S D B
S D
= +
+ ≤
+ ≤
≥
100
200
300
100 200 300 400 500 600
400
S
D
(A)
(B)
(0, 540)
(240, 180)
(360, 0)
z = $3,840
z = 2,000
z = 1,000
Extreme (Corner) Points Point Profit (0, 0) $0 (0, 300) $2,400 (360, 0) $3,600
(240, 180) $3,840 ← optimal solution and answer
A-68
B.7
1
2
3
1 2 3 4 5 6
4
x
y
(–2, 4)
–1–2
(–2, 3)
(2)(3)
(4)
(1)
(4, 1)
(–1, 3 1/2)
(5)
5
( )( )( )( )( )
4 line 1
2 line 22 6 line 3
2 8 line 40 line 5
x
x
x y
x y
y
≤− ≤
+ ≤− + ≤
≥
There are 5 corner (extreme) points. B.8
Foods Cost/
Serving Calories/ Serving
Percent Protein
Percent Carbs
Percent Fat
Fruit/ Vegetable
Apple Sauce $0.30 100 0% 100% 0% 1 Canned Corn $0.40 150 20% 80% 0% 1 Fried Chicken $0.90 250 55% 5% 40% 0 French Fries $0.20 400 5% 35% 60% 0 Mac & Cheese $0.50 430 20% 30% 50% 0 Turkey Breast $1.50 300 67% 0% 33% 0 Garden Salad $0.90 100 15% 40% 45% 1 AS CC FC FF MC TB GS Cost 0.3 0.4 0.9 0.2 0.5 1.5 0.9 Servings 0 1.333333 0.457143 0 1.129568 0 0 $1.51 Constraints AS CC FC FF MC TB GS LHS RHS Cals min 100 150 250 400 430 300 100 800 500 Cals max 100 150 250 400 430 300 100 800 800 Protein min 0 30 137.5 20 86 201 15 200 200 Carb min 100 120 12.5 140 129 0 40 311.4286 200 Fat max 0 0 100 240 215 99 45 288.5714 400 Fruit/Veg Min 100 150 0 0 0 0 100 200 200
A-69
Target Cell (Min) Answer Report (Relevant Section) Cell Name Original Value Final Value $I$14 servings $2.91 $1.51 Adjustable Cells Cell Name Original Value Final Value $B$14 serving A 1.50 0.00 $C$14 serving C 0.00 1.33 $D$14 serving FC 1.33 0.46 $E$14 serving FF 0.00 0.00 $F$14 serving M 0.00 1.13 $G$14 serving T 0.00 0.00 $H$14 serving G 1.40 0.00 Constraints Cell Name Cell Value Formula Status Slack $I$17 Cals min LI 800 $I$17 ≥ $J$1 Not Binding 300 $I$18 Cals max L 800 $I$18 ≤ $J$1 Binding 0 $I$19 Protein min 200 $I$19 ≥ $J$1 Binding 0 $I$20 Carb min L 311.4286 $I$20 ≥ $J$2 Not Binding 111.4286 $I$21 Fat max LI 288.5714 $I$21 ≤ $J$2 Not Binding 111.4286 $I$22 Fruit + Veg I 200 $I$22 ≥ $J$2 Binding 0
Adjustable Cells Sensitivity Report (Relevant Section)
Cell
Name Final Value
Reduced Cost
Objective Coefficient
Allowable Increase
Allowable Decrease
$B$14 serving A 0 0.172602 0.3 1E + 30 0.172602 $C$14 serving C 1.333333 0 0.4 0.258904 0.225581 $D$14 serving FC 0.457143 0 0.9 0.105072 0.100581 $E$14 serving FF 0 0.152691 0.2 1E + 30 0.152891 $F$14 serving M 1.129568 0 0.5 0.062909 0.707833 $G$14 serving T 0 0.169316 1.5 1E + 30 0.169318 $H$14 serving G 0 0.666151 0.9 1E + 30 0.688151 Constraints
Cell
Name Final Value
Shadow Price
Constraint R.H. Side
Allowable Increase
Allowable Decrease
$I$17 Cals min LI 800 0 500 300 1E + 30 $I$18 Cals max L 800 –0.00023 800 200 251.6129 $I$19 Protein min 200 0.008983 200 155 40 $I$20 Carb min L 311.4286 0 200 111.4285 1E + 30 $I$21 Fat max LI 288.5714 0 400 1E + 30 111.4286 $I$22 Fruit + Veg I 200 0.001504 200 485.7143 200
A-70
C MODULE
Transportation Modeling
C.1
105
93
8 7
1 2 3 4 SupplyDestination
A
Source 46
31
2 1B
Demand
5
5 4 6 53XX
X X
8 5 0
12 7 1 X
Answer :
a. Final Solution: Shown in the final tableau above.
b. ( ) ( ) ( ) ( ) ( )Total Cost : 5 10 3 9 1 3 6 2 5 1 $97+ + + + =
c. Perform optimality test: Result :
Cell
A-3
A-4
A-1
if “Opened Up”
0
0
0
Change in Cost
Since these numbers
are all nonnegative,
the solution is optimal. Answer : Yes
A-71
C.2 First, apply the optimality test:
Cell Change in Cost if Route
is “Opened Up” C-MI +3 C-J +4 H-L +2 B-MI –1 B-D +4 Resulting new basic feasible solution:
745
255
4 5
Miami Denver LincolnSan
Supply
Destination
CHI.
Source3 1
355 2
HOU.
Demand
40
Jose
6 950
7 4BUF. 30
70 90 45 50
100
75
80
TOTAL COST: ( ) ( ) ( ) ( ) ( ) ( )30 6 40 3 35 1 55 2 45 4 50 4 $825 Answer+ + + + + = =
A-72
C.3 Total supply = 450 < 500 = Total demand.
6
75
4200
9
A B C Supply
Destination
W
Source
10 5100
8X
Demand
0
12 7 6Y 75
250 100 150
200
175
75
0 0 0Z 50 50Dummy =
Optimal solution and meaning :
Ship 200 tons of grain from W to A Ship 100 tons of grain from X to B Ship 75 tons of grain from X to C Ship 75 tons of grain from Y to C.
Results in demands being met at destinations B and C, but in a shortfall of 50 tons at destination A. Total cost of optimal shipping plan: $2,750.
C.4
4
1
821
6
A B C
1
5 2
2
12
7 9 33 25
9 8 94
17
9
D
4
3
2
1
131
Cost $84 16 34 9 75 9 26 $253= + + + + + + = . This is also optimal.
A-73
C.5
128
84
5 10
A B C
Destination
1
Source: 6 1112
3 72 2
D
4
9
E
4 2
Cost $48 64 40 8 12 36 $208= + + + + + = C.6 The only cell with a negative cost improvement index is Houston–Miami. It achieves a “–1”. Allocate 10 to that cell. The result is: Denver Yuma Miami Houston 0 0 10 St. Louis 20 0 0 Chicago 0 20 10 Total cost = $170 C.7 2
44 5 9
6 71
8 101
14
33
11 123
Cost $3 1 4 3 4 5 1 8 1 10 3 12 $89= × + × + × + × + × + × =
A-74
C.8 Cell Improvement Index B-2 –9 C-1 +6 Result:
2 86
1 2
Destinations
A
Sources:9 6
3B 1
7 7C
6
4
22
57 Cost $12 9 18 14 $53= + + + =
A-75
D MODULE
Waiting Line Models
D.1 CURRENT MACHINE: 40λ = , 60µ =
1. Utilization ( ) 4067%
60ρ = =
2. Average number of customers waiting ( ) ( )( )
240 1
160 60 40 3qL = =
− customers
3. Average number of customers in system ( ) 402
60 40sL = =−
customers
4. Average time waiting ( ) ( )40
0.033 hours 2 minutes60 60 40qW = = =
−
5. Average time in system ( ) 1 1hours 3 minutes
60 40 20sW = = =−
PROPOSED MACHINE: 40λ = , 90µ =
1. Utilization ( ) 4044%
90ρ = =
2. Average number of customers waiting ( ) ( )( )
240
0.35690 90 40qL = =
− customers
3. Average number of customers in system ( ) 400.8
90 40sL = =−
customers
4. Average time waiting ( ) ( )40
0.0089 hours 0.533 minutes90 90 40qW = = =
−
5. Average time in system ( ) 1 1hours 1.2 minutes
90 40 50sW = = =−
Answer : Thus, we observe that the proposed machine will give better results with a
decrease in queries as well as time in system.
A-76
D.2 Two machine system
a. 40 1
2 60 3M
λρµ
= = =×
. Each machine is busy, on average, 33 percent of the time.
b. qL = average number of customers waiting in line = 0.081
c. 0.081 0.667 0.748s qL Lλµ
= + = + = customers in the system
d. 0.081
0.002 hours 0.1215 minutes40
7.3 seconds average waiting time in the queue
LW
λ= = = =
=
e. 1 1
0.002 0.0187 hours 1.12 minutes60
67.3 seconds average time in the system
s qW Wµ
= + = + = =
=
The 2-machine system seems to be the best overall, with even further reduction in queues and waiting times.
D.3 This model is, again, the M/M/1 model. The service time of 30 seconds means that the
service rate, µ, is 120 per hour.
a. 40 1
120 3λρµ
= = = . In this example and in Problem D.2, the system utilization is the
same.
b. ( ) ( )2 240 1
0.167120 120 40 6qL
λµ µ λ
= = = =− −
customers. The average number of
customers waiting in line is twice as large as the average number (0.081) found in Problem D.2.
c. 40 1
120 40 2sLλ
µ λ= = =
− − customer. On the average, the number of customers in
the system is 1
2, rather than the
3
4 (0.748) found in Problem D.2.
d. ( ) ( )40 1 1
hour minutes 15 seconds120 120 40 240 4qW
λµ µ λ
= = = = =− −
, which
exceeds qW in Problem D.2.
e. 1 1 1 3
hour minutes 45 seconds120 40 80 4sW
µ λ= = = = =
− −. Thus, average time
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.
A-77
D.4 EMPLOYEE SYSTEM: The current system is actually an M/D/1 system, because service times are constant.
a. 10
0.8312
λρµ
= = = . The toll collector is busy 83 percent of the time.
b. ( ) ( )( )2 100 100
2.082 2 12 12 10 48qL
λµ µ λ
= = = =− −
drivers. The average number of
drivers waiting to pay the toll is 2.08.
c. 2.08 0.83 2.91s qL Lλµ
= + = + = drivers. The average number of drivers at any one
toll booth is 2.91.
d. ( ) ( ) ( )10 10
0.208 minutes 12.5 seconds2 2 12 12 10 48qW
λµ µ λ
= = = = =− −
. The
average time drivers spend waiting is 12.5 seconds. e. waiting time service time 12.5 seconds 5 seconds 17.5 secondssW = + = + =
AUTOMATED SYSTEM: The proposed system, although automated, is an M/M/1 system.
a. 10
0.8312
λρµ
= = = . The utilization of the new system is the same as that of the old
system.
b. ( ) ( )2 100 100
4.1612 12 10 24qL
λµ µ λ
= = = =− −
drivers. Twice as many drivers are in
line under the new, automated system.
c. 5s qL Lλµ
= + = drivers. On average, 5 customers are in the system, which is 67
percent more customers than the current system allows.
d. ( ) ( )10 10
0.417 minutes 25 seconds12 12 10 24qW
λµ µ λ
= = = = =− −
. The waiting time
will double with the new system.
e. 1
30s qW Wµ
= + = seconds. The time in the system will rise by 67 percent.
The employee system seems to be better overall.
A-78
D.5 a. Each server handles 60 registrants hour , so it takes 4 servers to handle
200 arrivals hour b., c. 200λ = , 60µ = , 4M = servers yields: 3.29qL = , 0.0164qW = 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, 0.65qL = , 0.0033qW =
( )( )( )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. d. Server utilization rate 83.33%= .
D.6 a. The optimal number of servers is again 4.
b. The wait cost is now $50 person hour q qL W× × (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 15 hourλ = , 20 hourµ =
a. 0.075 hours 4.5 minutesqW = =
b. 1.125qL = people
D.8 10 hourλ = , 30 hourµ =
a. 1
0.0083 hours minute2qW = =
b. 0.083qL = people
A-79
E MODULE
Learning Curves
E.1 In order to estimate the learning curve rate, we take the ratios of the units that have
doubled.
report 2 560.848
report 1 66
report 4 490.860
report 2 56
report 6 450.849
report 3 53
report 8 420.857
report 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 1NT T C= where 0.402C ≈ (from Table E.3);
( )( )48 66 0.402 26.5T = = minutes.
E.3 1NT T C= where 0.400C = (from Table E.3), ( )( )50 66 0.400 26.4T = = minutes
E.4 1NT T C= where 25.513C = and NT is now cumulative time.
( ) ( )50 66 25.513 1,683.85 minutes 28.06 hoursT = = =
A-80
F MODULE
Simulation
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 Cumulative Probability
RN Interval
Demand
0.20 0.20 01–20 0 0.40 0.60 21–60 1 0.20 0.80 61–80 2 0.15 0.95 81–95 3 0.05 1.00 96–00 4
Probability Cumulative Probability
RN Interval
Lead Time
0.15 0.15 01–15 1 0.35 0.50 16–50 2 0.50 1.00 51–00 3 The results are: Units
Received Begin Inv.
RN
Demand
End Inv.
Lost Sales
Order?
RN
Lead time
5 52 1 4 0 4 37 1 3 0 3 82 3 0 0 Yes 06 1 0 69 2 0 2 10 10 98 4 6 0 6 96 4 2 0 Yes 63 3 2 33 1 1 0 1 50 1 0 0 0 88 3 0 3 10 10 90 3 7 0 Total 23 5 The total stock out cost = 5($40) = $200. The total holding cost = 23($1) = $23.
A-81
F.2 If the reorder point Problem F.1 is changed to 4 units, we have:
Units Received
Begin Inv.
RN
Demand
End Inv.
Lost Sales
Order?
RN
Lead time
5 52 1 4 0 Yes 6 1 4 37 1 3 0 10 13 82 3 10 0 10 69 2 8 0 8 98 4 4 0 Yes 63 3 4 96 4 0 0 0 33 1 0 1 0 50 1 0 1 10 10 88 3 7 0 7 90 3 4 0 Yes 57 3 Total 40 2
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
(2) Random Number
(3) Interval to
Arrival
(4) Time of Arrival
(5) Random Number
(6) Service Time
(7) Start
Service
(8) End
Service
(9) Wait Time
(10) Idle
Time 1 50 2 9:02 52 3 9:02 9:05 0 2 2 28 1 9:03 37 2 9:05 9:07 2 0 3 68 2 9:05 82 3 9:07 9:10 2 0 4 36 1 9:06 69 3 9:10 9:13 4 0 5 90 4 9:10 98 4 9:13 9:17 3 0 6 62 2 9:12 96 4 9:17 9:21 5 0 7 27 1 9:13 33 2 9:21 9:23 8 0 8 50 2 9:15 50 3 9:23 9:26 8 0 9 18 1 9:16 88 4 9:26 9:30 10 0 10 36 1 9:17 90 4 9:30 9:34 13 0 11 61 2 9:19 50 3 9:34 9:37 15 0 12 21 1 9:20 27 2 9:37 9:39 17 0 13 46 2 9:22 45 2 9:39 9:41 17 0 14 01 0 9:22 81 3 9:41 9:44 19 0 15 14 1 9:23 66 3 9:44 9:47 21 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-82
F.4 Time Until Next Arrival
(Minutes)
Probability
Cumulative Probability
Random Number Range
0 0.1 0.1 00–09 1 0.2 0.3 10–29 2 0.3 0.6 30–59 3 0.4 1.00 60–99 Service
Time (Minutes)
Probability
Cumulative Probability
Random Number Range
1 0.1 0.1 00–09 3 0.5 0.6 10–59 5 0.2 0.8 60–79 10 0.2 1.00 80–99 F.5 Simulating employment of one detailer: Detailer 1 Random*
Number
IAT Random** Number
Service Time
Arrival Time
Time In
Time Out
Wait Time
52 2 06 1 2 2 3 37 3 63 3 5 5 8 82 2 57 2 7 8 10 1 69 0 02 1 7 10 11 3 98 3 94 4 10 11 15 1 96 3 52 2 13 15 17 2 33 2 69 3 15 17 20 2 50 2 33 2 17 20 22 3 27 1 32 2 18 22 24 4 45 2 30 2 20 24 26 4 81 3 48 2 23 26 28 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-83
F.6 Simulating employment of two detailers: Detailer 1 Detailer 2 Random
Number
IAT Random Number
ServiceTime
ArrivalTime
Time In
Time Out
Time In
Time Out
Wait Time
52 2 06 1 2 2 3 37 3 63 3 5 5 8 82 2 57 2 7 7 9 69 0 02 1 7 8 9 1 98 3 94 4 10 10 14 96 3 52 2 13 13 15 33 2 69 3 15 15 18 50 2 33 2 17 17 19 27 1 32 2 18 18 20 45 2 30 2 20 20 22 81 3 48 2 23 23 25 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.