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Instructor’s Solutions Manual for Additional Problems
Operations Management
E I G H T H E D I T I O N
Principles of
Operations Management
S I X T H E D I T I O N
JAY HEIZER Texas Lutheran University
BARRY RENDER Upper Saddle River, New Jersey 07458
Rollins College
VP/Editorial Director: Jeff Shelstad Executive Editor: Mark Pfaltzgraff Senior 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 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, photocop ying, recording, or likewise. For information regarding permission(s), write to: Rights Rights and Permissions Department. Pearson Prentice Hall
TM
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 A-3 Project Management ............................................................... ........................ A-3 A-7 Forecasting ........................................................ ...................................................................................................... .............................................. A-7 Design of Goods and Services ................................................................... ... A-11 A-11 A-15 Managing Quality ....................................................... ......................................................................................... .................................. A-15 Statistical Process Control ...................................................... ............................................................................ ...................... A-18 Process Strategy ........................................................... ................................. A-20 A-20 Capacity Planning ........................................................... .............................. A-23 A-23 A-27 Location Strategies ...................................................... ....................................................................................... ................................. A-27 Layout Strategy ...................................................... ............................................................................................. ....................................... A-30 A-30 Work Measurement ......................................................... ............................. A-34 A-34 A-36 Inventory Management ............................................................ ..................... A-36 A-42 Aggregate Planning ............................................................ .......................... A-42 Materials Requirements Planning (MRP) & ERP ........................................ A-46 A-51 Short-Term Scheduling ........................................................ ................................................................................. ......................... A-51 Just-In-Time and Lean Production Systems ................................................. A-55 Maintenance and Reliability ......................................................... ................ A-57 A-57 Decision Making Tools ........................................................ ................................................................................ ........................ A-59 A-59 A-64 Linear Programming ............................................. ........................................ A-64 Transportation Modeling ....................................................... .............................................................................. ....................... A-70 A-70 Waiting Line Models .................................................. .................................. A-75 A-75 A-79 Learning Curves ............................................... ............................................ A-79 Simulation ....................................................... ..................................................................................................... .............................................. A-80 A-80
iii
1 CHAPTER
1.1
a.
Operations and Productivity
! Last year’s " number of units of output # total factor $ = # $ # productivity $ total dollar value of all inputs used % & =
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
b.
= 0.0546 units dollar
! 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)* =
14,000 units $247,050
c.
= 0.0567 units dollar
! This year’s " ! Last year’s " # total factor $ # total factor $ # $−# $ # productivity $ # productivity $ % & % & ×100% = 0.0567 − 0.0546 × 100% 0.0546 ! Last year’s " # total factor $ # $ # productivity $ % & = + 3.84% ≈ 3.8%
Answer : Total factor productivity increased by 3.798% this year as compared to last year.
A-1
1.2
0.15 = So L =
57,600
(160 )(12 )( L )
, where L = number of laborers employed at the plant.
57,600
(160 )(12 )( 0.15)
= 200
Answer : 200 laborers 1.3
Output = 28,000 customers There are 4 approaches to solving the problem correctly: 1. Input = 7 workers 28,000 Then, = 4,000 customers worker 7 2. Input = 7 ( 40 ) labor weeks Then, 3.
= 100 customers labor week
28,000 7 ( 40 )( 50 )
= 2 customers labor hour
Input = 7 ( 40 )( $250 ) dollars of worker wages Then,
1.4
7 ( 40 )
Input = 7 ( 40 )( 50 ) labor hours Then,
4.
28,000
28,000 7 ( 40 )( $250 )
6,600 Cadillacs
( x )( labor hours )
= 0.40 customers per dollar of labor
= 0.10
x = 66,000 labor hours There are 300 laborers. So, 66,000 labor hours = 220 labor hours laborer 300 laborers 1.5
$ output labor hour
=
52 ( $90 ) + 198 ( $80) 8 ( 45)
=
20,520 360
= $57.00 per labor hour
A-2
3
Project Management
CHAPTER
Gantt Chart
3.1
A
20
B
80
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
! a + 4m + b " ! b− a " Mean: # Variance # $ $ 6 % & % 6 & 120
A:
6 360
B:
6 600
C: D:
6 180 6 120
3.6
! b− a " $ % 6 &
Standard Deviation #
2
= 20
A:
= 60
= 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
= 100 = 30
= 20 6 60 F: = 10 6 120 G: = 20 6 60 H: = 10 6 180 I: = 30 6 E:
( 20 )
2
B:
36 2 ( 60 )
= 11.11
A:
20 6 60
= 3.33
= 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 B:
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 = µ =
502.77 ≅ 22.4 .
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.
3.9
The critical path, listing all critical activities in chronological order: A → B → E → F 1 + 1+ 2 + 2 = 6 ( not CP ) A→C→F 1 + 4 + 2 = 7. This is the CP. The project duration (in weeks): 7 (This is the length of CP.) The slack (in weeks) associated with any and all non-critical paths through the project: Look at the paths that aren’t critical—only 1 here—so from above: A → B → E → F 7 − 6 = 1 week slack.
We have only 1 activity with probabilistic duration. Due date − µ 8 − (1 + 4 + 2) 1 = = = 2 (length of entire path is 7, not 4). For a z = 2 , Z = 0.5 0.5 σ this means P ( Due date < 8) = 97.72% (table lookup) for the path so chance of being OVER 8 weeks is 2.28% (and we know non-CP path will be only 6 weeks)
3.10
Helps to modify the AON with the lowest costs to crash 1. CP is A → C → F ; C is cheapest to crash, so take it to 3 wks at $200. (and $200 < $250) 2. Now both paths through are critical. We would need to shorten A or F, or shorten C and either B/E. This is not worth it, so we would not bother to crash any further.
A-6
4
Forecasting
CHAPTER
4.1
Present = period ( week ) 6. So: F7 =
4.2
t 1 2 3 4
1 3
A6 +
1 4
A5 +
1 4
A4 +
1 6
A3 =
1 3
1
1
1
4
4
6
( 52 ) + ( 63) + ( 48) + ( 70 ) = 56.75 patients
At F t 120 — 136 — 114 128 116 125
120 + 136 256 + = = 128, , 2 2 - Checking Data 136 + 114 250 = = 125, ,. 2 2 116 + 114 230 F 5 = = = 115 = Answer 2 2 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-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
= 23,209
i =1
b=
58, 538 − 4 ( 75.75)(191.5) 23, 209 − 4 ( 75.75)
2
=
513.50 256.75
=2
a = 191.5 − 2 ( 75.75) = 40 y ≈ 40 + 2 x x = 85
0 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 ) F t . Let α = F 2 = F 3 = F 4 = F 5 =
1 4 1 4 1 4 1 4
1 4
. Let Monday forecast demand = 88
3
(88 ) + (88 ) = 88 4 3
( 72 ) + (88 ) = 18 + 66 = 84 4 3
( 68 ) + (84 ) = 17 + 63 = 80 4 3
( 48 ) + ( 80 ) = 12 + 60 = 72 4
A-8
4.6
Winter Spring Summer 2001 1, 400 1,500 1, 000 2002 1, 200 1, 400 2,100 2003 1, 000 1,600 2, 000 2004 900 1,500 1,900 4, 500 6,000 7,000
Fall 600 750 650 500 2,500
20,000 = 1,250 16 6,000 Average over spring: = 1,500 4 1,500 Spring index: = 1.2 1, 250 Average over all seasons:
! 5,600 " $ (1.2 ) = 1, 680 sailboats % 4 &
Answer : # 4.7
We need to find the smoothing constant α . We know in general that Ft +1 = α At + (1 − α ) F t , t = 1, 2, 3 . Choose either t = 2 or t = 3 ( t = 1 won’t let us find α because F 2 = 50 = α ( 50 ) + (1− α ) 50 holds for any α ). Let’s pick, e.g., t = 2 . Then F 3 = 48 = α ( 42 ) + (1− α ) 50 . So
48 = 42α + 50 − 50α −2 = − 8α 1 4
= α .
Now we can find F 5 : F 5 = α ( 46 ) + (1 − α ) 50 , with F 5 =
1 3 ( 46 ) + ( 50 ) = 49 ← Answer 4 4
A-9
α =
1 . So 4
4.8
Let X 1 , X 2 ,
!
, X 6 be the prices; Y1 , Y2 ,
!
, Y 6 be the number sold.
6
/ X i
X =
= 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)
i i
i =1
6
/ X
2 i
= 67.1925
(4)
i =1
Then y ≈ a + bx , where y = number sold , x = price , and 6
/ X Y − n ( X )(Y ) i i
b=
i =1
6
/ X
2 i
i =1
− n ( X )
=
2
( 9, 783) − 6 ( 3.25833)( 550) −969.489 = = −277.61395 2 3.49222 67.1925 − 6 ( 3.25833)
a = (Y ) − b ( X ) = 1,454.5578
So at x = 1.80 , y = 1,454.5578 − 277.61395( 1.80) = 954.85270 . Now round to the nearest integer: Answer : 955 dinners n
/ ( A − F ) t
4.9
Tracking Signal = Month
At
May June July August September October November December
100 80 110 115 105 110 125 120
So: MAD : Answer :
87 8 39
t
t =1
MAD F t 100 104 99 101 104 104 105 109
At − F t
( At − F t )
0 24 11 14 1 6 20 11 Sum: 87
0 –24 11 14 1 6 20 11 Sum: 39
= 10.875
10.875
!
1
%
1,000
= 3.586 # to the nearest
A-10
"
th $
&
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
(0.85) 90 of 100 non-defect
$12,500
$4,062.50 (0.15) 75 of 100 non-defect
Use K3
–$43,750
(0.90) 95 of 100 non-defect
–$18,750
$24,375 (0.10) 80 of 100 non-defect
–$75,000
Answer: $27,500—use K1 Outcome Calculations
90 10 1 ( 500 )( 300)( $1.20) − ( 500)( 300 )( $1.30) = ,−$100, 000 + 100 100 2 ,3−$100, 000 + $162, 000 − $19, 500 = $42, 500 70 30 1 (150, 000)( $1.20) − ( 150, 000)( $1.30) = ,−$100, 000 + 100 100 2 ,3−$100, 000 + $126, 000 − $58, 500 = − $32, 500
A-11
90 10 1 (150, 000)( $1.20) − (150, 000)( $1.30) = ,−$130, 000 + 100 100 2 ,3−$130, 000 + $162, 000 − $19, 500 = $12, 500 75 25 1 (150, 000)( $1.20) − (150, 000)( $1.30) = ,−$130, 000 + 100 100 2 ,3−$130, 000 + $35, 000 − $48, 750 = − $43, 750 95 5 1 (150, 000)( $1.20) − (150, 000)( $1.30) = ,−$180, 000 + 100 100 2 ,3−$180, 000 + $171, 000 − $9, 750 = − $18, 750 80 20 1 (150, 000)( $1.20) − (150, 000)( $1.30) = ,−$180, 000 + 100 100 2 ,3−$180, 000 + $144, 000 − $39, 000 = − $75, 000 (0.3) F market
5.2
80.0
66.0 Use D0
(0.7) U market (0.4) F market
84.0
Use D1
60.0
99.0
84.0 (0.6) U market (0.6) F market
Use D2
74.0
89.2
80.2 (0.4) U market
66.7
(All $ figures in millions in tree)
$ Profits : D0 − F : 1,000 ( 80, 000)
= $80, 000, 000
D0 − U : 750 (80, 000)
= $60, 000, 000
D1 − F : 1, 000 (100, 000) − 1, 000, 000 = $99, 000, 000 D1 − U : 750 (100, 000) − 1, 000, 000
= $74, 000, 000
D2 − F : 1, 000 ( 90, 000) − 800, 000
= $89, 200, 000
D2 − U : 750 ( 90, 000) − 800, 000
= $66, 700, 000
Answer : Answer: Design D1 has an expected profit of $84,000,000.
A-12
(0.3) Demand rises
5.3
$10,000 Purchase overhead hoist
$14,000
(0.5) Demand stays $10,000 same
(0.2) 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 $0
5.4
Do nothing High demand (0.6) Use A
160K
Low demand (0.4) High demand (0.6)
Use B 180K Use C
Low demand (0.4) High demand (0.6)
$300,000 –$50,000 $300,000 $0 $250,000
302K Low demand (0.4) Upgrade to D
$380,000
380K No upgrade to D
$0
Note: K = $1,000’s
Answer : Use Design C. If demands turns out to be low, upgrade to Design D.
A-13
Support
5.5
Bread & Rolls No support Support Bread & Rolls Pies & Cakes No support Support Full Service No support Support ( p = 0.40)
5.6
$15,000
Bread & Rolls EMV = $12,000 No support p ( = 0.60) Support ( p = 0.40) Bread & Rolls Pies & Cakes
$10,000 $25,000
EMV = $13,000 No support p ( = 0.60) Support ( p = 0.40)
$5,000 $35,000
Full Service EMV = $7,500 No support 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
x
0 b.
Negative
1
2
3
4
5
6
This is a scatter diagram.
A-16
Trips
6.3
a.
40 34
35
36
30
30 24
25 20 15 10
14
10 6
5 0 b.
W
R
I
4 M
2 O
39% of complaints are W, demeaning towards women.
6.4
Manpower
Incorrect measurement
Machines Temperature controls off
Inadequate cleanup
Operator misreads display
Antiquated scales
Variability Technician calculation off
Equipment in disrepair
Inadequate flow controls Incorrect Formulation
Jars mislabeled Incorrect weights
Lack of clear instructions Damaged raw material
Priority miscommunication
Materials
Incorrect maintenance Inadequate instructions
Methods
A-17
6
Statistical Process Control
SUPPLEMENT
S6.1
!
" ! 25 " 420 4 = − $ # $ = 400 . % n& % 25 &
We are given a target of X = 420 . So LCL = X − Z #
!
σ
" ! 25 " 420 4 = + $ # $ = 440 . Thus, % n& % 25 &
UCL = X + Z #
σ
Answer : LCL = 400 calories UCL = 440 calories S6.2
p =
7 250
+
5 250
+"+
=
7 + 5 +"+ 9
30
UCL = p + Z LCL = p − Z
S6.3
9 250
7,500 p (1 − p ) n
p (1 − p ) n
=
300 7,500
= 0.040
= 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 0 Z = 3 (see Normal Table) 2 2 UCL = c + 3 c = 1.5 + 3 1.5 = 5.17 =
A-18
S6.6
!
" $ = answers % n& σ
x ± Z # x =
!
384 24
= 16 lbs.
" ! 0.12 " $ = 2 # 3 $ = 0.08 % & % n&
Z #
σ
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 7,500
= 0.04 , n = 250
( 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 Use mean of 6 weeks of observations = 6 for c , as true c is unknown. 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 = 120X → 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:
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.
24,000 20,000 ? yes! > # #$% $% Anticipated Production Volu me
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 (Millions of dollars)
2
R MC
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 7.3
I
14 12
≤ V ≤ 25,000 .
C
C
10 Cost (millions)
R
2 3
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, 250x 1 , 400 x = 2, 000, 000 I&R , 2 x = 5,000 Intersect , ,1, 000, 000 + 1, 650 ( 5, 000) = $9, 250, 000 3 3, 000, 000 + 1, 250 x = 7, 500, 000 + 650x 1 , 600 x = 4, 500, 000 R&C , 2 x = 7,500 Intersect , ,3, 000, 000 + 1, 250 ( 7, 500) = $12, 375, 000 3 For all V between 5, 000 ≤ V ≤ 7,500
A-21
2 3
units .
7.4
Breakeven points a. R : 21, 000, 000 + 450x = 750x 0 x = 70, 000 C : 26, 250, 000 + 400 x = 750x 0 x = 75, 000 M : 15, 000, 000 + 500x = 750x 0 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 a. Continuous : 2, 400, 000 + 20 x = 80x 0 x = 40, 000 Repetitive : 1, 950, 000 + 30 x = 80x 0 x = 39, 000 Mass Customization : 1, 480, 000 + 40 x = 80x 0 x = 37, 000 Intermittent : 1,800, 000 + 40 x = 80 x 0 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
109 High demand (0.6)
148
135
Additional minor expansion Large expansion
90
135
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
High demand (0.3) Small expansion
140,000 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
$10,000
16,000 Demand up medium (0.6)
Large expand
$90,000
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)
(2) (1)
(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
Actual (or expected) Output =
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 =
! " 200 cars # $ % 38.4 cars per bay & S7.6
a.
BEP ( $ ) =
F
/ '(1 − (Vi Pi ))* W i
=
450 0.5125
= 878.05
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 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 members $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 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. 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
(Membership 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
Location Strategies
CHAPTER
8.1
C x =
C y = 8.2
8.3
( 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) 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) 68,000
Site A B C D
= 6.67
= 3.02
Total Weighted Score 174 185 187 165
/ Population weights = 5, 000 C x = C y =
( 0 )( 2, 050 ) + ( −1)( 550 ) + ( 2 )(1, 025) + ( 3)( 775) + (− 2 )( 250 )+ (− 2 )( 350 ) 5,000
= 0.525
( 0 )( 2, 050) + (1.5)( 550 ) + (− 1)(1, 025) + ( 3)( 775) + (− 3)( 250)+ (− 1)( 350) 5,000
Coordinates: ( 0.525, 0.205) A
8.4
150 125 $ cost 100 (millions) 75
(0, 25) (1/2, 30)
(2, 60)
50 25 (0, 20) (0, 10) 0
B
C
(1, 35)
V
1
2 3 4 5 6 10,000’s of Autos = V For all V such that 10, 000 ≤ V ≤ 60, 000 . A-27
= 0.205
8.5
Site A B C D
Score 5w + 320 4w + 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
L.A. K.C. Charlotte Char
12 10 TC (millions $)
8
K.C. (20, 7) (30, 9.5)
6 4
L.A.
2
V
0
5
10
15
20
25
30
35 (thousands) cutoff
14,100, 000 + 180V ≤ 1, 000, 000 + 300V , For all V so: 24,100, 000 + 180V ≤ 2, 000, 000 + 250V , 0 ≤ V ≤ 35, 000 3 1 3,100,000 ≤ 120V , ⇔ 22,100, 000 ≤ 70V , 0 ≤ V ≤ 35, 000 3 1 1 ,25,833 ≤ V ⇔ 0 Answer : 30, 000 ≤ V ≤ 35, 000 3 2 ,3 30,000 ≤ V
A-28
8.7
/ weights = 14,000 (or 14 for calculations below) C x = C y =
( 2 )( 3.5 ) + ( 8 )( 7 ) + ( 6 )( 3.5 ) 14 ( 6 )( 3.5) + (1)( 7 ) + ( 2 )( 3.5 ) 14
= 6.0 = 2.5
A-29
9
Layout Strategy
CHAPTER
9.1
9.2
( 60 )( 60 sec ) 3,600
a.
Cycle time =
b.
Theoretical number of work stations =
c.
Yes, it is feasible.
1
D 3 C
=
180 PLAs
180
= 20 sec per PLA
A2 B4
Department pair Weekly Cost AB : AC : AD : BC : BD : CD :
( 8)(800 ) = ( 6 )( 700 ) = ( 4 )( 400 ) = 1 10 )( 300 ) = 2( 1 7 200 ) = 2 ( )( 1 9 600 ) = 2 ( )( 1 2 1 2 1 2
($) 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 i
Cycle time
C.T. =
=
274 ( seconds ) Cycle time ( seconds)
60 ( 60 ) seconds
60 trucks Answer : 5
b.
= 60 seconds ( per truck ) so n =
274 60
= 4.5667 → n = 5 .
Steps 1 and 2: Sample Answer c = 60 seconds +
- From (a)
number of stations is at least 5 . Precedence diagram:
40 30 B
D
20 H
6 E
40 A
30 J
25 50 C
F
18
15
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
Available and Fit A — B, C — B, F, G E, F, G — D, E, G E, G — E, I I, H I — J
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
Answer : Station (Other answers possible, 1 depending upon how ties 2 are broken in above 3 procedure) 4 5 6
9.4
Assigned A — C — B F — D G — E H I — J
(Broke a tie)
(Broke ties) (Broke ties) (Broke a tie)
(Broke a tie)
Tasks A C B, F D, G E, H, I J
c.
n = 6 work stations are in our answer. t i 274 Efficiency = = = 0.7611 n ( C.T.) 6 ( 60 )
a.
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 2
= $15, 200
9.5
Cost of 3 attempts:
a. b.
Attempt 1 2
1 S D
Work Area 2 D S
c.
3
M
D
3 M M
Cost $1,088 $1,142
S
$1,100 ↑
2 '(( 23)(10 ) + ( 32 )( 5) + ( 20 )( 8))* = $1,100
9.6
a.
Theoretical minimum number of stations = Cycle time =
b.
60 5
/ task times cycle time
= 12 minutes . So minimum number of stations =
WS #1
WS #2
10 min
12 min
A
B
48 12
= 4 stations
WS #5 D F
6 min 8 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
S10.1
! Zs " Required sample size = n = # $ where s = 0.15 , x = 0.4 , z = 1.96 (for 95% confidence), % hx & 2
' (1.96 )( 0.15) ) n=4 5 = 54.0225 ≈ 54 0.10 0.4 ( )( ) (4 *5
h = 10% accuracy level
S10.2
a.
b.
S10.3
S10.4
2 hx n ( 0.10 )( 0.40 ) ( 12 ) ! Zs " Thus, n=# Z = = = 0.924 $ 0.15 s % 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 Average observed time = = 0.4484 minutes 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
Average observed time =
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 sum of times
Observed time =
b.
Normal time = ( Observed time) × ( Performance rating factor) = 6.525× 95%
c.
=
1.74
a.
number of cycles
16
= 0.10875 minutes = 6.525 seconds
= 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 Allowance fraction = = = = 0.15 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 1289 →
Average Dollar Volume 400 × 3.75 = 1, 500.00
Percent of Total $ Volume 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 S (Ordering Cost) = $30. So, Q∗ ( Order Quantity ) = 240 =
Thus, TC ( Total Annual Cost ) = PD +
HQ
+
2
SD
2
2 DS
2 ( 4,800 )( 30 ) 2
H
.
! 2 × 240 " ! 30 × 4,800 " $+ # $ % 2 & % 240 &
= ( 4,800× 27 ) + #
= 129, 600 + 240 + 600 = $130, 440 A-36
=
12.3
D (Annual Demand) = 14,558, P (Purchase Price/Unit) = $5, H (Holding Cost/Unit) = $4,
2 DS
S (Ordering Cost/Order) = $22, Q =
=
2 × 14, 558 × 22
= 400 tons per order 4 H HQ SD ! 4 × 400 " ! 22 × 14,558 " TC = PD + + = ( 5 × 14,558) + # $+ # $ = 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, Q=
2 DS
=
2 × 4,800 × 120
= 181.42 = 181 units (rounded off).
35
H
Thus, TC ( Total Cost ) = PD +
HQ
2
+
SD Q
! 35 × 181 " ! 120 × 4,800 " $+ # $ 181 % 2 & % &
= ( 4,800 × 325) + #
= 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.
12.5
Q1 =
2 DS
Q2 =
2 DS
Q3 =
2 DS
=
2 × 4,800 × 120 35
H
=
= 181 units
2 × 4,800 × 120 32.5
H
=
= 188 units
2 × 4,800 × 120
= 196 units 30 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, EOQ = 188 units . Thus, TC (188 units ) = PD +
HQ
2
+
SD Q
! 32.5 × 188 " ! 120 × 4,800 " $+ # $ 2 188 % & % &
= ( 325× 4,800) + #
= 1, 560, 000 + 3, 055 + 3, 064 = $1, 566,119 TC ( 200 units ) = PD +
HQ
2
+
SD Q
! 30 × 200 " ! 120 × 4,800 " $+ # $ 2 200 % & % &
= ( 300 × 4,800) + #
= 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.
A-37
12.6
D = 12,500 year , so d = 50 day , p = 300 day , S = $30 order , H = $2 unit year
2 DS × H
p
Q=
b.
Number of production runs ( N ) =
c.
Maximum inventory level ( I max ) = Q #1 −
d.
Days of demand satisfied by each production run =
p − d
=
2 × 12, 500 × 30 300 × = 612.37 × 1.095 = 671 2 300 − 50
a.
D
=
Q
12,500 = 18.63 671
! %
d "
50 " ! ! 1" $ = 671 #1 − 300 $ = 671# 1− 6 $ = 559 p & % & % & 250 18.63
= 13.42 days in demand
only mode Time in production for each order =
Q
671
= 2.24 days in production for each p 300 order. Total time = 13.42 days per cycle. 2.24 Thus, percent of time in production = = 16.7% . 13.42 12.7
=
! %
H = $2 unit year , S = $10 order , D = 4, 000 # d = Q=
2 DS
=
4,000 " = 16 $ . So, 250 &
2 × 4, 000 × 10
= 200 . ROP (reorder point) = l × d , l (lead time = 5 days). 2 H 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
12.9
300 =
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 90,000
= $4.08
2 ( 6, 000)( 30 ) 2 DS = = 189.74 units 10 H 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% Salt 3 2 $6.00 $312.00 20 0.07% 100.0% $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
2 DS
Q=
b.
ROP for calzone, σ ( demand during lead time) = LT × σ = 40 , Z = 1 (from Table)
H
= 800
calzones . Order every
Q
a.
d
days = 10 days
for 0.1587, ROP = d × LT + Z × σ dLT = 360 c.
d.
12.14
On hand = 85, days left = 1. Since ROP = d × LT + Z × σ dalt , Z =
ROP − d × LT
σ LT
.
Thus, Z of 0.25 gives 0.5987. So there is a 40.13% chance of stockout. Q D = 10 orders term . = 400 = average inventory , 2 Q Q Holding cost term = H = ( 400 )( 0.25) = $100 2 D Order cost term = S = (10 )(10 ) = $100 Q
S = $16 , H = $0.40 calzone term , p = 160 , d = 80 , D = 8,000 for 100 days of the term
2 DS H (1 − d p )
a.
Q=
b.
Cycle =
c.
Run production for
Q d
=
= 1,131.37
1,131.37 80 Q p
= 14.14
days =
1,131
=
160
A-40
7.07 days
12.15 Under present price of $6.40 per box:
Economic Order Quantity: Q∗
2 DS
2 × 5000 × 25
= 395.3 or 395 boxes where 0.25 × 6.40 H D = annual demand, S = set-up or order cost, H = holding cost DS QH Total cost = order cost + holding cost + purchase cost = + + CD 2 Q
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 = + 2 Q =
=
5, 000 × 25
+
+
3, 000
5, 000 × 0.25× 6.00 2
+ 5, 000 × 6.00 =
41.67 + 3, 750.00 + 30, 000
= $33,791.67
Therefore, the old supplier with whom they would incur a total c ost 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 Q∗
=
2 DS H (1 − pd )
=
2 × 10, 000 × 200 50 1.00 (1 − 200 )
=
2,309.4 or 2,309 units
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 Total restaurant demand for the year
= 7,000,000 = 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 Personnel Hotel (Room) Termi- Restaurant TermiQuarter Demand Req. Hire nate Demand Req. Hire nate Winter 800,000 8 8 160,000 2 2 Spring 2,200,000 22 14 800,000 10 8 Summer 3,300,000 33 11 960,000 12 2 Fall 700,000 7 0 26 160,000 2 0 10 Totals 7,000,000 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 TermiHires nations 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) Demand Req. Hire 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
Personnel Quarter from Quarter from Hire Agency Hires 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
13.4
Personnel cost = 36 quarters of labor × 5,000 = Hiring cost = 9 ( = 7 + 2 ) hires @ $1,000 =
$180,000
Termination cost = 0 terminations @ $2,000 = Staffing agency costs = 60 @ 6,500 =
0 390,000 $579,000
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 Plan B: Month Demand Production Mar 1,000 1,100 Apr 1,200 1,100 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
Hire
9,000
Fire 700
Extra Cost 56,000 12,000 8,000 8,000 12,000 16,000
300 200 200 300 200
Inventory 200 100
Sub-Contracting
200 100 400 200
Therefore, Plan B would be preferred.
A-44
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
CD Case
Period (week) 1
2
3
4
5
6
7
550
400
500
50 500 500 400
400 400 500
500 500
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
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 Gross Requirements Scheduled Receipts CD Projected On Hand Top Net Requirements Planned Order Receipts Planned Order Releases
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
500
500
100
100
8
100
14.3
Lot Size
Lot for Lot
Lot for Lot
Lot for Lot
Lot for Lot
Lot for Lot
Lot for Lot
Lot for Lot
Lot for Lot
Lead On Safety AlloTime Hand Stock cated
1
1
3
1
1
1
1
1
—
—
—
—
—
3,000
3,000
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
LowLevel Code
0
1
1
Item ID
Ball Point Pens
Period (day) 1
2
3
Gross Requirements Scheduled Receipts Projected On Hand Net Requirements Planned Order Receipts Planned Order Releases
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 Net Requirements Point Planned Order Receipts Planned Order Releases Gross Requirements Fine Scheduled Receipts Point Projected On Hand Ball Net Requirements Point 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
Lot Size
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
Lead On Safety AlloTime Hand Stock cated
1
1
1
1
1
1
1
1
1
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
—
LowLevel Code
0
1
1
1
1
2
2
2
3
Item ID
Ball Point Pens
Period (week, day) 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
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
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
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-ha nd 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-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
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-51
Gantt Chart
15.4
F
6
D
10
C
13
B
20
E
28
A
37 10
Project A B C D E F
15.5
20 Hour
30
Time 9 7 3 4 8 6
40
Due Date 22 17 16 13 16 9
6 + 10 + 13 + 20 + 28 + 37
Late Days 15 3 0 0 12 0 114
a.
Average flow time =
b.
Average lateness =
c.
Maximum lateness = 15 days (for job A—Gantt Table)
a.
The shortest processing time ( SPT ) =DBACE
6 total late days
Flow Time Due Hours Late Hours
6 =
15 + 3 + 12
number of jobs
D 0 0 0
Average flow time
6
B 1 12 —
=5
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 =
=
A-52
= 19
days
days
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 Average hours late = = 7.25 days 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 23 25
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.
Then we find the shortest again. It is 2800 on B
d.
Then it is 2731 on A. Since it is on the first machine, it is the earliest job. 2731 2800 2678
A-53
2800 2678
e.
Finally, we get 2731 2134
2387 5
2800 12
2678
19
22
A 2731 2134 2387 2800 2678 B
2731
7
14
2134
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= Q2
S =
10
=
2 × D × S H (1 − pd )
2 × D × S H (1 − pd )
Q 2 ( H ) (1 − pd )
2 D
=
=
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.
!
2
400 (1, 000 ) (10 ) (1 − 500 ) (1, 000, 000)( 10)( .2) 2,000,000
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=
1,936 100
2 × D × S H (1 − pd )
=
2 × (1, 250, 000) × 15 5,000 20 (1 − 10,000 )
=
37, 500, 000 10
per container = 19.36 containers
A-55
=
3,750,000 = 1,936 units
16.5
Note: S = 6 minutes (or Q=
=
2 × D × S H (1 −
d p
)
=
1 10
hour) ×$100 shop labor cost
( 2 )(12, 500)(10) 50 50 (1 − 500 )
250, 000 45
=
$10
5, 555 = 74
"1# EOQ + Safety stock + Lead time = 74 + 10 + $ % ( 500) &2'
= 334
oil pumps
No. of Kanban containers = 16.6
=
= .1× $100 =
334 36
=
9.27 = 9 or 10 containers required
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 = =
Set-up time =
Q 2 × H × (1 − pd )
=
2 D
150 1502 × 10 × (1− 1,000 )
2 × 40, 000
=
22, 500 × 10 × (1− 0.15) 80, 000
191,250 = $2.39 80,000 $2.39 set-up× 60 minute hour = 2.69 minute set-up $50 hour
A-56
17 CHAPTER
17.1
Maintenance and Reliability
Failure Rate Analysis Station No. 1: N = 1,000 F = 22 22 FR = 1,000
=
0.022
R = 1 − 0.022 = 0.978 = 97.8%
Station No. 2: N = 1,000 F = 51 51 FR = 1,000
=
0.051
R = 1 − 0.051 = 0.949 = 94.5% Answer : Unit #1 is clearly better, with a lower failure rate and better reliability. 17.2
Reliability ( R s ) = R1 × R2 × R3 × !× Rn . So, R5 Answer : The reliability is 93.04%
A-57
=
( 0.98)( 0.99)( 0.96) = 0.9304
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
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
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
Overall Reliability ( R s ) = Product of all reliabilities above = 0.971 = 97.1% 17.4
2
Reliability (stick)
= 1−
(1 − 0.99)
Reliability (cloth)
= 1−
(1 − 0.98)
2
=
0.9999
=
0.999998
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)
Bump test’s reliability
= 1−
(1 − 0.99)
Speed test’s reliability
= 1−
(1 − 0.93)
Sudden brake test’s reliability
2
3
= 1−
3
=
0.999936
=
0.999900
=
0.999657
(1 − 0.90)
4
=
0.999900
So, overall reliability ( R s ) = 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.
Decision Alternatives
0.35 Competes
0.65 Doesn’t Compete
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
A.2
Fixed
Slight Increase
Major Increase
Minimum
N
0
2,000
3,000
0
M
–4,000
8,000
9,000
–4,000
L
–10,000
6,000
20,000
–10,000
O –50,000 4,000 Use “maximin: criterion. No floor space (N).
40,000
–50,000
Decision Alternatives
A.3
Increasing capacity Using overtime Buying equipment
Row Average $700,000 $700,000 $733,333
←
Using equally likely, “Buying equipment” is the best option.
A-59
3
2
( 50) +
A.4
Expected value under certainty =
A.5
E ( A ) = 0.4 ( 40 ) + 0.2 (100) + 0.4 ( 60) = 60
35
( 20) +
10 10 100 = 15 + 4 + 35 + 15 = 69
( 100) +
15 100
( 100)
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
4
10
( 50) + ( 92) +
1
( 40) + ( 64) = 10+ 23+ 12+ 16= 61
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
5
4
$30,000
Unavailable (.40)
Wait 1 day 33
$0 Buy now
Avail. (.60) 55
$55,000
Unavailable (.70)
Wait 1 day 21
$0 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
Decision Alternatives
Good Market (0.6)
Bad Market (0.4)
EV
D1
7,000
1,250
4,700
D2
8,250
4,000
6,550
D3
9,750
1,000
6,250
N
0
0
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: Expected (Marginal) Value
–$6,550 → $900
A-61
A.9
More than one decision is involved, and problem is under risk, so use a decision tree approach. Low (0.4) demand
200
242 Small Facility
Expand High (0.6) demand
270
270
Don’t expand Modest (0.3) response
544
Advertise Low (0.4) demand
Large Facility
223
20
160
160
Sizable (0.7) response
544 Don’t Advertise 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 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. $16.50 hr Cost per dozen bagels: Labor: = $1.65 dozen 10 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) ($3 D − $1( B − D ) if B ≥D Then, total profit = * if B ≤ D +$3 B 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)
Dozens Baked B = Decision Alternatives
40 (0.20)
41 (0.30)
42 (0.35)
43 (0.15)
EV
40
120
120
120
120
120
41
119
123
123
123
122.2
42
118
122
126
126
123.2
43
117
121
125
129
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 advertise 2
Build?
P (low) = 0.4
Prob (not success) = 0.5
5 – (6 + 2) = –3
Ad? Don’t Advertise
5 – 6 = –1
Don’t Build
b. c. A.12
$0
Build studio, but don’t advertise (even if demand is low) Expected value = $2 million
a.
Build 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
Linear Programming
MODULE
B.1
a.
b.
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)
Point O
Coordinates ( 0, 0 )
z value 0
A
( 0, 50 ) ( 40, 0 )
11, 000
C B
d.
C
12, 000 Solve 2 equations in 2 unknowns 12,600 ← derived from (1) and ( 2) to obtain ( 20, 30)
Produce 20 trucks and 30 cars daily for a profit of $12,600 per day.
A-64
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 = 12 z = 11
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
1
–1 –2 –3 –4
2
3
4
z = 9
(1)
z = –4 z = –6
5
6
7
8
9
(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
9 8 7 6 5 4 3 2 1 –1 –2 –3 –4
z = 4
z = 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 = 10S + 8D 1 2
S + D ≤ 300
2
D ≤ 360 3 S, D ≥ 0
S+
( A) ( B)
D
(0, 540)
z = $3,840
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 (240, 180)
$3,840
← optimal
solution and answer
A-67
S
500
600
B.7
y
(–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
2 x + 2 y ≤ 6 − x + 2 y ≤ 8 y ≥ 0 − x ≤
( 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 800 800 200 311.4286 288.5714 200
RHS 500 800 200 200 400 200
Target Cell (Min) 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
Answer Report (Relevant Section) Original Value Final Value $2.91 $1.51
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 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
Status Not Binding Binding Binding Not Binding Not Binding Binding
Sensitivity Report (Relevant Section) Final Reduced Objective Allowable Value Cost Coefficient Increase 0 0.172602 0.3 1E + 30 1.333333 0 0.4 0.258904 0.457143 0 0.9 0.105072 0 0.152691 0.2 1E + 30 1.129568 0 0.5 0.062909 0 0.169316 1.5 1E + 30 0 0.666151 0.9 1E + 30
Slack 300 0 0 111.4286 111.4286 0
Allowable Decrease 0.172602 0.225581 0.100581 0.152891 0.707833 0.169318 0.688151
Constraints Final Value 800 800 200 311.4286 288.5714 200
A-69
Shadow Price 0 –0.00023 0.008983 0 0 0.001504
Constraint R.H. Side 500 800 200 200 400 200
Allowable Increase 300 200 155 111.4285 1E + 30 485.7143
Allowable Decrease 1E + 30 251.6129 40 1E + 30 111.4286 200
C
Transportation Modeling
MODULE
Destination 2 3
C.1
1 A
10 5
Source
8
Supply 7
3 4
B Demand
9
4
5 X
8 5 0 3
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
Change in Cost if “Opened Up”
A-3
0
Since these numbers
A-4
0
are all nonnegative,
A-1
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
Demand
5
45 1
100 5
2
35 6
BUF.
4
75 9
7
30
4 50
70
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. Destination A W X Source
B
C
6 200
4
10
5
0
8 75
7
Y
175 6
75 0
Dummy = Z
9 200
100 12
Supply
0
75 0
50
Demand 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
2
8
6
9
2
1
4
9
3
3
9
2
17 7 1
1 9
4
D
2 5
3
C
25 8 1
13
Cost = $84 + 16 + 34 + 9 + 75 + 9 + 26 = $253. This is also optimal.
A-72
Destination
C.5
A
B 12
1
4
2
2
Source:
C 8
D 5
8
E 10
4
6
11
3
4 2
7
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
8
10
4 6
7 1
1 3
3 4
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: Destinations 1
2 2
A
6
Sources: B
1
8 6
9
6 3
7 C
Cost = $12 + 9 + 18 + 14 = $53
A-74
7 2
7
4
5
2
D MODULE
D.1
Waiting Line Models
CURRENT MACHINE: λ = 40 , µ = 60 40 1. Utilization ( ρ ) = = 67% 60 2
2.
( 40 ) 1 Average number of customers waiting ( Lq ) = = 1 customers 60 ( 60 − 40 ) 3
3.
Average number of customers in system ( L s ) =
4.
Average time waiting (W q ) =
5.
Average time in system (W s ) =
40 60 ( 60 − 40 ) 1
=
60 − 40
40 = 2 customers 60 − 40
0.033 hours = 2 minutes
=
1 20
hours = 3 minutes
PROPOSED MACHINE: λ = 40 , µ = 90 40 1. Utilization ( ρ ) = = 44% 90 2
2.
( 40 ) Average number of customers waiting ( Lq ) = = 0.356 customers 90 ( 90 − 40 )
3.
Average number of customers in system ( L s ) =
4.
Average time waiting (W q ) =
5.
Average time in system (W s ) =
40 90 ( 90 − 40 ) 1
=
1
40 = 0.8 customers 90 − 40
0.0089 hours = 0.533 minutes
hours = 1.2 minutes 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.
A-75
=
D.2
Two machine system λ 40 1 a. . Each machine is busy, on average, 33 percent of the time. = = ρ = M µ 2 × 60 3 b. Lq = average number of customers waiting in line = 0.081
λ = 0.081 + 0.667 = 0.748 customers in the system µ
c.
L s
d.
W q
=
e.
W s
= W q +
=
Lq + Lq
0.081
= 0.002 hours = 0.1215 minutes 40 = 7.3 seconds average waiting time in the queue
λ
=
1
µ
=
0.002 +
1 60
=
0.0187 hours = 1.12 minutes
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. =
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. λ 40 1 ρ = = a. . In this example and in Problem D.2, the system utilization is the = µ 120 3 same. λ 2 402 1 b. Lq = = = = 0.167 customers. The average number of µ ( µ − λ ) 120 (120 − 40 ) 6
c.
d.
customers waiting in line is twice as large as the average number (0.081) found in Problem D.2. λ 40 1 customer. On the average, the number of customers in L s = = = 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 hour = minutes = 15 seconds , which = = W q = 4 µ ( µ − λ ) 120 (120 − 40 ) 240 exceeds W q in Problem D.2.
e.
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. W s
=
1
=
A-76
D.4
EMPLOYEE SYSTEM: The current system is actually an M / D/1 system, because service times are constant. λ 10 ρ = = a. = 0.83 . The toll collector is busy 83 percent of the time. µ 12 b.
Lq
=
λ 2 100 100 = = = 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.
L s
=
Lq +
λ = 2.08 + 0.83 = 2.91 drivers. The average number of drivers at any one µ
toll booth is 2.91. 10 10 λ = = = 0.208 minutes = 12.5 seconds . The 2 µ ( µ − λ ) 2 (12 )(12 − 10 ) 48
d.
W q
e.
average time drivers spend waiting is 12.5 seconds. W s = 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 = = a. = 0.83 . The utilization of the new system is the same as that of the old µ 12 system. 100 100 λ 2 b. Lq = = = = 4.16 drivers. Twice as many drivers are in µ ( µ − λ ) 12 (12 − 10 ) 24 line under the new, automated system. c.
d.
e.
L s
=
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 W q = = = = 0.417 minutes = 25 seconds . The waiting time µ ( µ − λ ) 12 (12 − 10 ) 24 will double with the new system. 1 W s = W q + = 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.
Each server server handles handles 60 60 registrants registrants hour , so it takes 4 servers servers to handle 200 200 arri arriva vals ls hour hour
b., c.
λ = 200 ,
=
60 , M = 4 servers yields: Lq
=
3.29 , W q
=
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 , W q
=
0.0033
Wait cost = ( 0.65 )( ) ( 0.0033) ( $100) = $0.21, Server cost = 5 × 15 = $75
D.6
d.
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 × W q (which are the same as in Problem D.5). Entert Entertain ainmen mentt cost cost = $15 hou hourr , 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.
W q
=
0.07 0.075 5 hour hourss = 4.5 4.5 minu minute tess
b.
Lq
= 1.125 people
λ = 10 hour , µ = 30 hour a.
W q
=
b.
Lq
=
0.0083 hours =
1 2
minute
0.083 people
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 2
=
report 1 report 4
=
0.848
49
=
0.860
=
0.849
=
0.857
56 =
report 3 report 8
=
66
report 2 report 6
56
45 53
=
report 4
42 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 T N = T1C where C ≈ 0.402 (from Table E.3); T 48
=
( 66 ) ( 0.402) = 26.5 minutes. ( 66 ) ( 0.400) = 26.4 minutes
E.3
T N = T1C where C = 0.400 (from Table E.3), T 50
E.4
T N = T1C where C = 25.513 and T N is now cumulative time. T 50
=
=
683.85 m miinutes = 28.06 h ho ours ( 66 ) ( 25.513) = 1, 68
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. corresponding to random number 52 is 3 minutes. Column 6 : Teller time corresponding 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
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