SOLUTIONS MANUAL for Supply Chain Engineering Models and Applications 2nd Edition Ravi Ravindran, Do by ACADEMIAMILL

SOLUTIONS MANUAL FOR

Supply Chain Engineering: Models and Applications

A. Ravi Ravindran Donald P. Warsing, Jr.

SOLUTIONS MANUAL FOR Supply Chain Engineering: Models and Applications

by A. Ravi Ravindran Donald P. Warsing, Jr.

CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2013 by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Printed in the United States of America on acid-free paper Version Date: 20120802 International Standard Book Number: 978-1-4665-6387-2 (Paperback) This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access www.copyright.com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com

Acknowledgements We are pleased to acknowledge the assistance provided by Madhana Raghavan, Subramanian Pazhani and Nok Kungwalsong, doctoral students in the industrial engineering department at the Pennsylvania State University, in the preparation of the solutions manual.

Table of Contents Chapter 1

Introduction to Supply Chain Engineering…………………………1

Chapter 2

Planning Production in Supply Chains……………………………..3

Chapter 3

Inventory Management Methods and Models……………………30

Chapter 4

Transportation Decisions in Supply Chain Management……….43

Chapter 5

Location and Distribution Decisions in Supply Chains………….54

Chapter 6

Supplier Selection Models and Methods…………………………73

Chapter 7

Managing Risks in Supply Chains………………………….........100

Chapter 8

Global Supply Chain Management……………………………….105

Chapter 1 Introduction to Supply Chain Engineering Solutions to Numerical Exercises only

1.9 Given: Annual demand = 3,000; Inventory turns = 30 (a) Average inventory in the dealer’s lot=

Annual demand 3000 = = 100 cars . Inventory turns 30

(b) The number of days an average car sits in inventory is equal to the days of inventory. Days of inventory =

365 365 = ≈ 12 days Inventory turns 30

If the dealer wants to increase his customer responsiveness, he should stock more items, which results in increased inventory level. As a result of increased inventory level, inventory holding cost increases.  Supply chain cost increases.  Inventory turns decreases.  Average inventory increases.  Days-of-inventory increases

1.10 Given: Annual demand = 5,000; Average Inventory = 500 (a) Inventory Turns =

Annual demand 5000 = = 10 units Average Inventory 500

(b) If XYZ decides to increase its inventory turns: i.

Response time to customers

Increases (Responsiveness decreases)

ii.

Inventory capital

Decreases

iii.

Days of inventory

Decreases

iv.

Return on assets

Increases

Cash – to – Cash cycle

Decreases

vi.

Working capital

Increases

1.11 a) False b) True c) False d) False e) True f) True g) False h) True i) True j) True k) False l) True m) False

Chapter 2 Planning Production in Supply Chains Solutions to Numerical Exercises only

2.6) (a) Last Value method F 13 =F 14 =F 15 = D 12 = 41 (b) Averaging method F 13 =F 14 =F 15 =

34 + 33 + 42 + 34 + 36 + 43 + 34 + 33 + 43 + 31 + 35 + 41 12

= 36.583 ≈ 37

31 + 35 + 41 3

= 35.67 ≈ 36

(d) Exponential smoothing with α = 0.25 Assume F 1 as the average of the 12 months demand = 36.58. The forecasts are shown in Table 1. Table 1. Forecasts (Exponential smoothing)

Month

Sales

1 2 3 4 5 6 7

34.00 33.00 42.00 34.00 36.00 43.00 34.00 3

Forecast (F t ) 36.58 35.94 35.20 36.90 36.18 36.13 37.85

8 9 10 11 12

33.00 43.00 31.00 35.00 41.00

36.89 35.92 37.69 36.01 35.76

F 13 =F 14 =F 15 = (0.25 x 41) + (0.75 x 35.76) = 37.07 ≈ 37 (e) Holt’s method with α = 0.4 and β = 0.5 For illustration, L 1 is assumed to be equal to D 1 and T 1 is 1. Hence, L 1 = 34 and T 1 = 1. The forecasts are shown in Table 2. Table 2. Forecasts (Holts Method)

Month

Sales

Estimate of Level (L t )

1 2 3 4 5 6 7 8 9 10 11 12

34.00 33.00 42.00 34.00 36.00 43.00 34.00 33.00 43.00 31.00 35.00 41.00

34.00 34.60 34.44 37.66 37.25 37.16 39.67 38.25 36.14 38.25 35.66 34.78

Estimate of Trend (T t ) 1.00 0.80 0.32 1.77 0.68 0.30 1.40 -0.01 -1.06 0.52 -1.03 -0.96

For Month 2,

L2 = αD1 + (1 − α )F1 = (0.4) 34 + (0.6) 35 = 34.60 T2 = β (L2 − L1 ) + (1 − β )T1 = (0.5) (34.60 – 34) + (0.7) 1 = 0.80 F 2 = L 2 + T 2 = 35.40 L 13 = (0.4 x 41) + (0.6 x 33.82) = 36.69

Forecast (F t ) 35.00 35.40 34.76 39.42 37.94 37.46 41.08 38.24 35.08 38.77 34.63 33.82

T 13 = 0.5 x (34.78 – 35.66) + (0.5 x -0.96) = 0.48 F 13 = 36.69 + (1 x 0.48) = 37.17 ≈ 37

F 14 = 36.69 + (2 x 0.48) = 37.65 ≈ 38 F 15 = 36.69+ (3 x 0.48) = 38.13 ≈ 38

(f)

Since the demands are higher during the third month of each quarter, there is a definite seasonality pattern present in the demands. Hence, all the methods should be modified to include seasonality.

2.7) Let Dt and Ft be the demand and forecast for period t respectively. Then the forecast error for period t is given by =

Ft-Dt

Further, let the four weights for the four period weighted moving average method be w1 (latest period for which demand data is available, period t), w2 (period t-1), and w3 (period t-2). Joe Kool wishes to minimize the sum of the absolute values of errors (which is a nonlinear objective). To linearize this, let et = et+ − et− , where et+ , et− ≥ 0, ∀t ⇒ et = et+ + et−

The resulting linear program is given as follows: 12

Objective: min Z ∑ (et+ + et− ) = t =4

subject to the constraints: 5200w1+5405w2+5325w3-5510 = e4+ − e4− 5510w1+5200w2+5405w3-5765 = e5+ − e5− 5765w1+5510w2+5200w3-5210 = e6+ − e6− 5210w1+5765w2+5510w3-5375 = e7+ − e7− 5375w1+5210w2+5765w3-5585 = e8+ − e8− 5585w1+5375w2+5210w3-5460 = e9+ − e9− 5

5460w1+5585w2+5375w3-4905 = e10+ − e10− 4905w1+5460w2+5585w3-5755 = e11+ − e11− 5755w1+4905w2+5460w3-6320 = e12+ − e12− w1+w2+w3=1 w1≥w2≥w3≥0 et+ , et− ≥ 0

∀t=4,5,6,...,12

The problem is solved using EXCEL solver. Optimal weights are w1=0.55; w2=0; w3=0.45 Forecast for month 13 = (0.55 x 6320) + (0 x 5755) + (0.45 x 4905) =5683.25 ≈ 5684 2.8) Table 3 gives the forecast and their error for the various methods in Exercise 2.6.

Table 3. Forecast errors for different methods

Et (Averaging)

Et (Threemonth Moving average)

Et (Exponential Smoothing)

Et (Holts method)

Month

Sales

Last Value

Averaging Method

Threemonth Moving average

34.00

36.58

33.00

34.00

35.94

35.4

42.00

33.00

33.50

35.2

34.76

34.00

42.00

36.33

36.9

39.42

8.00

2.33

2.90

5.42

36.00

34.00

35.75

36.33

36.18

37.94

-2.00

-0.25

0.33

0.18

1.94

43.00

36.00

35.80

37.33

36.13

37.46

2.00

-7.20

-5.67

-6.87

-5.54

34.00

43.00

37.00

37.67

37.85

41.08

9.00

3.00

3.67

3.85

7.08

3.57

4.67

3.89

5.24

Exponential Smoothing

Holts method

Et (Last Value)

33.00

34.00

36.57

37.67

36.89

38.24

0.00

43.00

33.00

36.13

36.67

35.92

35.08

-1.00

-6.88

-6.33

-7.08

-7.92

31.00

43.00

36.89

36.67

37.69

38.77

9.00

5.89

5.67

6.69

7.77

35.00

31.00

36.30

35.67

36.01

34.63

-3.00

1.30

0.67

1.01

-0.37

41.00

35.00

36.18

36.33

35.76

33.82

1.00

-4.82

-4.67

-5.24

-7.18

BIAS and MAD values for the methods are given in Table 4.

Table 4. BIAS and MAD for different methods

Method

BIAS 25.00 20.95 24.67 23.33 30.44

Last Value Averaging Three-month Moving average Exponential Smoothing Holts method

MAD 3.67 2.33 2.74 2.59 3.42

Three-month moving average method has the lowest BIAS and MAD values (0.67 and 3.78). So, the three-month moving average method is recommended for forecasting. 2.9) Comparison of forecasting methods

Month 1 2 3 4 5 6

E t (Method 1)

E t (Method 2)

-26 51 -56 -82 60 -20

-37 48 -30 -90 40 -35

(a) Compute MAD, MSE, BIAS 1 n ∑ | et | n t =1 1 n MSE = ∑ et2 n t =1

MAD =

BIAS = ∑ et t =1

Method 1: MAD = (26+51+56+82+60+20)/6= 49.17 MSE = (262+512+562+822+602+202)/6= 2856.17 BIAS=-26+51-56-82+60-20= -73

Method 2: MAD = (37+48+30+90+40+35)/6= 46.67 MSE = (372+482+302+902+402+352)/6=2583.00 BIAS=-37+48-30-90+40-35= -104

(b) Compute tracking signal TS t = BIAS t /MAD t TS 1 = e 1 / | e 1 | TS 2 = (e 1 + e 2 )/ {|e 1 |+ |e 2 |}/2 . . TS 6 = (e 1 + e 2 +…+ e 5 + e 6 )/ {|e 1 |+|e 2 | +…+|e 6 |}/6

Tracking Signal Month 1 2 3 4 5 6

Method 1

Method 2

-1.000 0.649 -0.699 -2.102 -0.964 -1.485

-1.000 0.259 -0.496 -2.127 -1.408 -2.229

(c) Both methods tend to under forecast since the bias is negative. Method 2 appears to be more biased than Method 1. But since: (MAD) Method 2 < (MAD) Method 1 (MSE) Method 2 < (MSE) Method 1 Method 2 is recommended for forecasting purposes.

2.10) Decision Variables: x1

Normal production in week 1 for use in week j for j = 1, 2, 3, 4

Normal production in week 2 for use in week j for j = 2, 3, 4

Overtime production in week 2 for use in week j for j = 2, 3, 4

Normal production in week 3 for use in week j for j = 3, 4

Overtime production in week 3 for use in week j for j = 3, 4

Normal production in week 4 for use in week 4

Inventory at the end of week j, j = 1, 2, 3, 4

Backorder at the end of week j, j = 1, 2, 3, 4

Objective: To minimize the sum of production, inventory and backorder costs. 4

Minimize Z

10 x1 + 10x2 +15x3 +15x4 +15x5 + 15x6 + 3∑ I j + 4∑ b j =j 1 =j 1

Subject to, Inventory balance constraints: The left hand side of the equation is the sum of inventory at the end of week t-1 and the production during week t minus the back order at the end of week t-1. If the sum is less than the demand, backorder (b j ) exists. Else, inventory (I j ) exists. I0 + x1 - b0= 300 - b1+ I1 I1 + x2+ x3 - b1= 700 - b2+ I2 I2 + x4+ x5 - b2 = 900 - b3+ I3 I 3 + x6 - b3 = 800 - b4+ I4 I0 = 0

Initial Inventory constraint: Initial inventory at the beginning of week 1 (end of week 0) is set to be 0. I0 = 0 Initial and Final back order level constraint: Initial back order is assumed to be 0 and the final back order at the end of week 4 is 0, assuming that all the back orders must be filled by the end of the fourth week. b0 =0 b4 =0 Regular time production capacity constraints: Regular production capacity in each week is limited to 700. x1 < 700 x2< 700 x4< 700 x6< 700 Over time production capacity constraints: Over time production capacity in week 2 and 3 is limited to 200. x3 < 200 x5< 200 Non-negativity constraints x1, x2, x3, x4, x5, x6, b1, b2, b3, b4, I1, I2, I3, I4 > 0

2.11)

Week1

Week2

Week3

Week4

Dummy

Week1

700 10

Week2 (Normal)

Week2 (OT)

Week3 (Normal)

Week3 (OT)

0 500

700 200 700 200

Week4

700 300

700

900

800

Variables: x 1j

Normal production in week 1 for use in week j for j = 1, 2, 3, 4

x 2j

Normal production in week 2 for use in week j for j = 1, 2, 3, 4

x 3j

Overtime production in week 2 for use in week j for j = 1, 2, 3, 4

x 4j

Normal production in week 3 for use in week j for j = 1, 2, 3, 4

x 5j

Overtime production in week 3 for use in week j for j = 1, 2, 3, 4

x 6j

Normal production in week 4 for use in week j for j = 1, 2, 3, 4

Note: x 21 , x 31 , x 41 , x 51 , x 61 , x 42 , x 52 , x 62 , x 63 , are production to fill the backorders

2.12) Customer A June

July

Customer B August

June

July

August

Dummy

June RT

100

110

120

100

110

120

June OT

120

130

140

120

130

140

July RT

105

100

110

100

110

July OT

125

120

130

120

130

Aug RT

110

105

100

Aug OT

125

120

2.13) (a) Lt ≤ (0.1) Wt (b) Rt ≤ 50

for t = 1, 2,..., 6

for t =1,2,....,6

This can be linearized as follows: Wt − Wt −1 ≤ 10 Wt −1 − Wt ≤ 10 Note: W1 = 20 (d) I t ≤ 120

for t = 1, 2,..., 6

150

2.14) (a) Let index i =1,...,6 represent the six months. xi yi+ yiIi

Production in month i

Demand for month i

Increase in production in month i Decrease in production in month i Inventory at the end of month i

(b) & (c) The production planning formulation may be written as follows. Objective: 6

Minimize Z = 5∑ yi+ + 3∑ yi− + 3∑ I i =i 1 =i 1 =i 1

Subject to, Demand constraints: The demand in month i should be met. In general, this may be expressed as: Ii-1 + xi = di + Ii. 1000+x1

2500+I1

I1+x2

5000+I2

I2+x3

7500+I3

I3+x4

10000+I4

I4+x5

9000+I5

I5+x6

6000+I6

Inventory constrains: Capacity limits on inventory as well as ending inventory in June. Ii

7000

3000 13

∀ i=1,...,6

Production balance: Production between two successive months is related by the increase or decrease in production. x1

2000 + y1+- y1-

x1 + y2+-y2-

x2 + y3+ - y3-

x3 + y4+ - y4-

x4 + y5+ - y5-

x5 + y6+ - y6-

Non-negativity constraints: xi, yi+,yi-,Ii > 0

∀i=1,...,6

2.15 (Forecasting Case Study) Using the data for the years 2007 – 2010, prepare the initial estimates of the seasonal factors for each quarter Seasonality Index (SI) = -

Average demand during that period Overall averageof demand for all periods

The overall average of quarterly demand using the demand values for 16 quarters

Overall Average = (800+750+600+1500+1700+1100+680+2000+2100+2200+1300+3100+2400+3060+1800+4000 ) / 16 = 1818.125 -

The quarterly average using the four demand values for each quarter is as follows: Quarter 1 average = (800+1700+2100+2400)/4 = 1750 Quarter 2 average = (750+1100+2200+3060)/4 = 1777.5 Quarter 3 average = (600+680+1300+1800)/4 = 1095 Quarter 4 average = (1500+2000+3100+4000)/4 = 2650 14

The seasonality factor for each quarter can then be calculated using the seasonality index formula: Seasonality Index for Quarter 1 = 1750/ 1818.125 = 0. 96253 Seasonality Index for Quarter 2 = 1777.5 / 1818.125 = 0. 977656 Seasonality Index for Quarter 3 = 1095 / 1818.125 = 0. 602269 Seasonality Index for Quarter 4 = 2650/ 1818.125 = 1. 457546 For the six different smoothing constant levels of (α, β) values, forecasts are determined using Holt’s method for the years (2007-2010).

BIAS and STD error measures are

calculated and listed in Table 5. Table 5. BIAS and STD values for different (α, β) values (Alpha, Beta) Level

(Alpha, Beta) values

BIAS

STD

1 2 3 4 5 6

(0.1, 0.1) (0.1, 0.2) (0.1, 0.3) (0.2, 0.2) (0.2, 0.3) (0.3, 0.3)

-6673.39 -5466.121 -4463.673 -2808.392 -1969.469 -1035.496

778.51 763.86 762.05 768.64 781.91 798.29

Based on the two error measures: BIAS and STD, there is no dominant (ALPHA, BETA) value which could be recommended (see Table 5). BIAS is the least for (0.3, 0.3) = -1035.496. But, STD (798.29) is the worst value for this smoothing constant level. STD is comparably good for (0.1, 0.2), (0.1, 0.3) and (0.2, 0.2) with 763.86, 762.05 and 768.64 correspondingly. In order to recommend a best smoothing constant value (ALPHA, BETA), we use a weighted sum approach. The following steps are done: 1) Sum the values of the columns corresponding to BIAS and STD (Table 6). This sum is used to normalize the BIAS and STD values. 15

Table 6. Sum of columns

(Alpha, Beta) Level

(Alpha, Beta) values

BIAS

STD

(0.1, 0.1)

-6673.39

778.51057

(0.1, 0.2)

-5466.121

763.86054

(0.1, 0.3)

-4463.673

762.05034

(0.2, 0.2)

-2808.392

768.63743

(0.2, 0.3)

-1969.469

781.91101

(0.3, 0.3) Total

-1035.496 -22416.541

798.2958 4653.265

2) Calculate the normalized values by dividing each value in column by its corresponding column total (Table 7): Table 7. Normalized values

(Alpha, Beta) Level 1 2 3 4 5 6

(Alpha, Beta) values (0.1, 0.1) (0.1, 0.2) (0.1, 0.3) (0.2, 0.2) (0.2, 0.3) (0.3, 0.3)

BIAS

STD

0.2976994 0.2438432 0.1991241 0.1252821 0.0878578 0.0461934

0.1673041 0.1641558 0.1637668 0.1651824 0.1680349 0.171556

3) Set weights for the error measures. We assume equal weights for both BIAS and STD i.e., 0.5 for STD and 0.5 for BIAS. Table 8 shows the weighted sum. Table 8. Weighted Sum

(Alpha, Beta) Level 1 2 3 4 5 6

(Alpha, Beta) values (0.1, 0.1) (0.1, 0.2) (0.1, 0.3) (0.2, 0.2) (0.2, 0.3) (0.3, 0.3) 16

Weighted Sum 0.232502 0.203999 0.181445 0.145232 0.127946 0.108875

The minimum weighted sum is selected as the best (ALPHA, BETA) levels. (0.3, 0.3) has the minimum weighted sum of 0.108875. So, we recommend using this level for the values of smoothing constants.

c) (0.3, 0.3) F Qi is the forecast for quarter i for the year 2011 L Q1 is the level forecast for quarter 1 for the year 2011 T Q1 is the Trend forecast for quarter 1 for the year 2011 S Qi is the seasonality index for quarter i

L Q1 = 3062.89; T Q1 = 171.96

Period 2011 -Q1 2011 -Q2 2011 -Q3 2011 -Q4

F Q1

= (L Q1 + T Q1 ) SI Q1

= 3113.64

F Q2

= (L Q1 + 2.T Q1 ) SI Q2

= 3330.69

F Q3

= (L Q1 + 3.T Q1 ) SI Q3

= 2155.39

F Q4

= (L Q1 + 4.T Q1 ) SI Q4

= 5466.88

Actual demand

Actual forecast

Errors (e t )

3600 3900 1500 3320 Total

3113.64 3330.69 2155.39 5466.88

-486.36 -569.31 655.39 2146.88 1746.60 MSE STD

236545.1 324112.8 429531.4 4609081 5599270.21 1399817.55 1183.14

Tracking signal (TS k ) -1 -2 -0.702 1.811

BIAS: The method was under forecasting in the years 2007- 2010. However, in 2011 the method is over forecasting. STD: Increased over the 2007-2010 data.

Tracking Signal:

Figure 1. Tracking signal values for Holt’s method 2.5 2

Tracking Signal

1.5 1 0.5 0 -0.5

2011 -Q1

2011 -Q2

2011 -Q3

2011 -Q4

-1 -1.5 -2 -2.5

TS k values are within the limits.

= F= (d) F Q1 Q2

F= Q3

2493.429 + 3129.937 + 2988.699 + 2744.34 = 2839.101 4

F= Q4

Seasonalized forecast (F Qi ) = SQi × FQi F Q1 = 2732.72; F Q2 = 2775.66; F Q3 = 1709.90; F Q4 =4138.12;

Period

Actual demand

Actual forecast

Errors (e t )

et2

2011 -Q1

3600

2732.72

-867.28

752174.6

Tracking signal (TS k ) -1

2011 -Q2

3900

2775.66

-1124.34

1264134.00

-2

2011 -Q3

1500

1709.90

209.90

44058.82

-2.42794

2011 -Q4

3320

4138.12

818.12

669318.5

-1.27644

-963.596

2729686

MSE STD

682421.5 826.0881

Total

Figure 2. Tracking signal values for moving average 0 2011 -Q1

2011 -Q2

2011 -Q3

2011 -Q4

Tracking signal

-0.5 -1 -1.5 -2 -2.5 -3

Tables 9 and 10 show the comparison results of the two methods. Figure 3 shows the tracking signal plot for the methods.

Table 9. BIAS and STD for the moving average and Holt’s

Method Moving Average Holt’s Method

BIAS

STD

-963.596

826.0881

1746.60

1183.14

Table 10. Tracking Signal for the moving average and Holt’s

Moving Average TS 1 TS 2 TS 3 TS 4

-1.000 -2.000 -2.427 -1.276

Holt’s Method -1.000 -2.000 -0.702 1.811

Figure 3. Tracking signal values – Moving Average vs. Holt’s method 1 0.5 0 -0.5

2011 -Q1

2011 -Q2

2011 -Q3

2011 -Q4

-1 -1.5

Holt’s Method

-2

Moving Average

-2.5 -3 -3.5 -4 -4.5

The following are the conclusions from the comparison: 1) Moving average is better than the Holt’s method. From the sales data, it can be observed that the sales for the year 2011 have fluctuated in almost all the quarters. In quarter 1, it has gone up considerably than the trend. In quarter 3 and 4, sales has dropped from its trend till 2010. Moving average method gives a good forecast because of its ability to identify short-term fluctuations quicker than the Holt’s method. Even with a considerably high value of smoothing constant values in the Holt’s method, which can react to the changes in data quickly, resulting forecast errors and variance is larger. It is evident from the results in Table 9. 2) Tracking signal values show that both the forecast methods are within the acceptable value.

2.16 (Aggregate Planning case study) (a) Given below is an example plan for Chase strategy Table 11. Chase Strategy for Exercise 2.16

Month 1 2 3 4 5 6

Workers 35 60 60 80 130 200

RT production 350 600 600 800 1300 2000

Month 7 8 9 10 11 12

Workers 250 300 240 180 150 150

RT production 2500 3000 2400 1800 1500 1500

Note: The above plan (Table 11) assumes no overtime use. There are several other plans possible under the chase strategy.

Table 12. Chase strategy production plan for Exercise 2.16

Month

Regular time production

Over time production

Demand

1 2 3 4 5 6 7 8 9 10 11 12

350 600 600 800 1300 2000 2500 3000 2400 1800 1500 1500

0 0 0 0 0 0 0 0 0 0 0 0

500 600 600 800 1300 2000 2500 3000 2400 1800 1500 1200

Cumulative Inventory at the end of the month t 0 0 0 0 0 0 0 0 0 0 0 300

Table 13. Chase strategy work force analysis for Exercise 2.16

Month 1 2 3 4 5 6 7 8 9 10 11 12

Total workforce 35 60 60 80 130 200 250 300 240 180 150 150

Regular time production 35 60 60 80 130 200 250 300 240 180 150 150

Final Inventory = 300; Final workforce = 150; Cost of the chase strategy = $4,455,000 21

Over time production 0 0 0 0 0 0 0 0 0 0 0 0

Hired

Fired

0 25 0 20 50 70 50 50 0 0 0 0

65 0 0 0 0 0 0 0 60 60 30 0

(b) Given below is an example plan for Level strategy

Average demand / month = 1517. Hence, use a level work force of 150 workers and over time. Table 14. Level Strategy for Exercise 2.16

Month 1 2 3 4 5 6 7 8 9 10 11 12

Initial Inventory 150 1150 2050 2950 3650 3850 3350 2350 850 50 0 0

Production RT OT 1500 0 1500 0 1500 0 1500 0 1500 0 1500 0 1500 0 1500 0 1500 100 1500 250 1500 0 1500 0

Demand 500 600 600 800 1300 2000 2500 3000 2400 1800 1500 1200

Final inventory = 300; Final workers = 150 Note: There are several other plans possible under level strategy also. Table 15. Level strategy production plan for Exercise 2.16

Month

Regular time production

Over time production

Demand

1 2 3 4 5 6 7 8 9 10 11 12

1500 1500 1500 1500 1500 1500 1500 1500 1500 1500 1500 1500

0 0 0 0 0 0 0 0 100 250 0 0

500 600 600 800 1300 2000 2500 3000 2400 1800 1500 1200

Cumulative Inventory at the end of the month t 1150 2050 2950 3650 3850 3350 2350 850 50 0 0 300

Table 16. Level strategy work force analysis for Exercise 2.16

Month 1 2 3 4 5 6 7 8 9 10 11 12

Total workforce 150 150 150 150 150 150 150 150 150 150 150 150

Regular time production 150 150 150 150 150 150 150 150 150 150 150 150

Over time production 0 0 0 0 0 0 0 0 50 125 0 0

Hired

Fired

50 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

Final Inventory = 300; Final workforce = 150; Cost of the Level strategy= $4,243,750 (c) LP Model to determine the optimal production plan for 2012 Decision Variables Wt

Total workers available beginning of month t, after hiring and firing

RP t

Workers assigned to regular time production in month t

OP t

Workers assigned to over time production in month t

Number of workers fired at the beginning of month t

Number of workers hired at the beginning of month t

Cumulative inventory at the end of month t

Number of units produced during regular time production in month t

Number of units produced during over time production in month t

Input Variable Dt

Demand at time period t 23

Objective Function: The objective is to minimize the sum of regular time production, over time production, hiring, firing, and average inventory holding costs

12 12 12 12 12 I Minimize Z = 200∑ X t + 300∑ Yt + 500∑ H t + 3000∑ Ft + 50∑ t =t 1 =t 1 =t 1 =t 1 =t 1 2

Size of the workforce: Total workforce available at the beginning of month t is equal to the total workforce available at the beginning of month t-1 plus the number hired at the beginning month t, minus the number fired at the beginning of month t. Wt = Wt −1 + H t – Ft

for t = 1 , 2, …., 12

Expected number of workers at the beginning: Total workforce available at the beginning of month 0 is equal to 100. W 0 =100 Desired work force at the end of month 12: Total desired workforce during month 12 is equal to 150. W12 ≥ 150 Regular and overtime production capacity constraints: Regular production at month t is equal to the product of regular production capacity of a worker and the number of workers assigned to regular time production in month t = X t 10 RPt

= t for

1,2 …., 12

Overtime production capacity constraints: Overtime production at month t is equal to the product of overtime production capacity of a worker and the number of workers assigned to overtime production in month t = Yt 2 OPt

for = t

1,2 …., 12 24

Demand/Inventory Balance: The left hand side of the equation is the sum of the current regular production (X t ), over time production (Y t ), and the inventory carried over (I t- 1 ). The sum is the total amount available to meet demand in month t. If it exceeds demand (D t ) then we will have an inventory of I t at the end of month t. No shortages are allowed. X t + Yt + I t -1 = Dt + I t

for t = 1,2 …., 12

Initial inventory at the beginning of month 1: Inventory available at the end of time 0 (or beginning of time 1) is 150. I 0 = 150 Desired inventory at the end of month 12: Desired inventory at the end of month 12 is greater than or equal to 100 I 12 > 100 Workforce assignment constraints: The number of workers assigned to regular production in month t is equal to the total workforce available at the beginning of month t after hiring and firing. W = t

RPt

for = t

1,2 …., 12

Overtime production constraints: The number of workers assigned to over time production in month t should be less than the total workforce available at the beginning of month t. OPt ≤ Wt

for t = 1,2 …., 12

Non-negativity constraints Wt , RPt , OPt , H t , Ft , I t , X t ≥ 0 for all t = 1, 2, 3…..,12

(d) The LP model is solved using LINGO and the results are shown in Tables 17 and 18:

Table 17. LP Optimal Production Plan for Exercise 2.16

Month 1 2 3 4 5 6 7 8 9 10 11 12

Regular time production 1000 1000 1000 1000 1836.5 1836.5 1836.5 1836.5 1836.5 1800 1500 1500

Over time production 0 0 0 0 0 0 0 0 367.3 0 0 0

Demand 500 600 600 800 1300 2000 2500 3000 2400 1800 1500 1200

Cumulative Inventory at the end of the month t 650 1050 1450 1650 2186.5 2023.1 1359.6 196.2 0 0 0 300

Table 18. Workforce Analysis for Exercise 2.16

Month 1 2 3 4 5 6 7 8 9 10 11 12

Total workforce 100 100 100 100 183.7 183.7 183.7 183.7 183.7 180 150 150

Regular time production 100 100 100 100 183.7 183.7 183.7 183.7 183.7 180 150 150

Final Inventory = 300; Final workforce =150; Optimal cost of the LP model = $4,121,154 26

Over time production 0 0 0 0 0 0 0 0 183.7 0 0 0

Hired

Fired

0 0 0 0 83.7 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 3.7 30 0

(e) (i)

Monthly Inventory levels: Chase strategy has the least monthly inventory levels. The optimal inventory level given by the LP model is higher than that of the chase strategy. The level strategy has the highest monthly inventory levels among the three. Table 19 shows the monthly inventory levels under the three production plans.

Table 19. Monthly inventory levels under Chase, Level, and LP model (Exercise 2.16)

Month 1 2 3 4 5 6 7 8 9 10 11 12

(ii)

Chase strategy 0 0 0 0 0 0 0 0 0 0 0 300

Level strategy 1150 2050 2950 3650 3850 3350 2350 850 50 0 0 300

LP model 650 1050 1450 1650 2186.5 2023.1 1359.6 196.2 0 0 0 300

Hiring and Firing: Level strategy has the lowest number of hiring and firing. In this problem, under the level strategy, there was hiring of 50 workers in month 1 and no firing. The optimal hiring and firing plan by the LP model had a hiring of 84 workers and firing of 34 workers. The chase strategy had the highest number of hiring and firing (hiring of 265 workers and firing of 215 workers in total). Hiring and firing levels for the three production plans are shown in Table 20.

Table 20. Hiring and Firing under Chase, Level, and LP model (Exercise 2.16)

Chase strategy

Level strategy

LP model

Hiring

Firing

Hiring

Firing

Hiring

Firing

0 25 0 20 50 70 50 50 0 0 0 0

65 0 0 0 0 0 0 0 60 60 30 0

50 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 83.7 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 3.7 30 0

Month 1 2 3 4 5 6 7 8 9 10 11 12

(iii)

Over time use: Under chase strategy, there was 0 over time use. Level strategy used 175 workers in over time. The LP model had the highest over time use with 184 workers. Table 21 shows the over time use for the three production plans. Table 21. Over time use under Chase, Level, and LP model (Exercise 2.16)

Month 1 2 3 4 5 6 7 8 9 10 11 12

Chase strategy 0 0 0 0 0 0 0 0 0 0 0 0

Level strategy 0 0 0 0 0 0 0 0 50 125 0 0 28

LP model 0 0 0 0 0 0 0 0 183.7 0 0 0

(iv)

Total cost: LP model yields a production plan with the lowest cost, followed by the level strategy. The production plan by the chase strategy has the highest total cost. Table 22 shows the total cost for the three production plans.

Table 22. Total cost under Chase, Level, and LP model (Exercise 2.16)

Total cost

Chase

Level

strategy

4455000

4243750

LP model 4121154

Chapter 3 Inventory Management Methods and Models Solutions to Numerical Exercises only 3.8) Note that the data is already sorted by decreasing annual sales (in $). Therefore, to determine reasonable cut-off points for the A, B, and C categories we only need to supplement the sorted sales data with the cumulative percentage of overall sales and the cumulative percentage of overall items. The table below includes this data, along with suggested categorizations of A, B, and C items. Stock-keeping unit (SKU) J-625 Z-454 W-681 J-909 T-988 B-570 M-117 H-033 W-998 C-841 R-596 Y-764 F-496 M-154 M-615 A-620 K-388 B-237 K-778 Y-319 T-670 S-802 T-172 G-676 M-687

Annual Sales ($) $904,366 $838,481 $757,060 $635,764 $596,075 $482,492 $390,553 $262,363 $212,713 $151,413 $125,234 $81,392 $48,131 $37,409 $30,011 $29,830 $27,592 $24,633 $22,551 $19,836 $16,058 $14,996 $10,106 $8,783 $6,624

Cumulative % of SKUs 4.00% 8.00% 12.00% 16.00% 20.00% 24.00% 28.00% 32.00% 36.00% 40.00% 44.00% 48.00% 52.00% 56.00% 60.00% 64.00% 68.00% 72.00% 76.00% 80.00% 84.00% 88.00% 92.00% 96.00% 100.00%

Cumulative % of Sales 15.77% 30.39% 43.59% 54.68% 65.08% 73.49% 80.30% 84.88% 88.58% 91.23% 93.41% 94.83% 95.67% 96.32% 96.84% 97.36% 97.84% 98.27% 98.67% 99.01% 99.29% 99.56% 99.73% 99.88% 100.00%

Class A A A A A A B B B B B B C C C C C C C C C C C C C

Since there are 25 items in total, each item accumulates another 4% (1/25) of the total number of items. Note that the total sales across all items is $5,734,466. Thus, for example, the top-selling item, SKU J-625, accumulates $904,366 $5, 734, 466 = 15.77% of total sales. Similarly, the cumulative percentage of sales across items up to and including SKU W-681 is ( $904,366 + $838, 481 + $757, 060 ) $5, 734, 466 = 43.59%. Note that the AB cut-off occurs before the “80-20” rule of thumb, but reasonably close to those numbers (73.5% of sales and 24% of items), and the B-C cut-off occurs reasonably close to the 9550 rule of thumb. 3.9) As suggested by the hint in the problem statement, denote the item price by p. Also, consistent with the notation in the chapter, let us denote the item cost by c, the item salvage value by s, and the goodwill cost by g. Therefore, according to the problem data, the gross 0.45, which can be solved for c = 0.55 p , and the salvage margin is given by ( p − c ) p = value is s = 0.2c , which can therefore be expressed as s = 0.11 p . (a) Clearly, the problem description implies that a newsvendor model is appropriate. Thus, if we accept Mr. Michaels’ estimate of g = 2 p , the best (optimal) service level is given by the critical ratio,= CR cu ( cu + co ) , or CR =

p−c+ g p−s+g

p − 0.55 p + 2 p p − 0.11 p + 2 p

2.45 p 2.89 p

= 0.848, and therefore the best in-stock probability target is roughly 85%. 31

(b) Using g = 5 p , we obtain CR =

5.45 p 5.89 p

= 0.925, or a target in-stock probability of 92.5%. (c) Clearly, Mr. Michaels’ target of 95% for the in-stock probability level and his estimate of the goodwill cost are inconsistent. If he believes his estimate of g = 2 p , then his instock probability target must be quite a bit lower (approximately 85%). If, however, he believes that 95% service is the right target, then he must accept that the goodwill cost is much higher than his estimate of g = 2 p (or even higher than g = 5 p ). 3.10) The first thing to notice about this problem is that the data gathered appear to indicate that an economic order quantity (EOQ) model is appropriate—i.e., a tradeoff between fixed cost and holding cost, and data that implies that demand transpires at a constant rate over time. (a) Using the TAC expression stated in the chapter, we have

TAC ( Q = 5 ) = A ⋅

D Q + h⋅ Q 2

( 6.25) 

1800  5  + ( 0.25 )(100 )    5  2

= $2, 250.00 + $62.50 = $2,312.50. The best way to assess whether this replenishment policy is good is to compare it to the optimal policy. Without that point of comparison—which will be developed in the answers below—there is no way to say whether this policy is good or bad. (b) Before we even specify the optimal policy, one thing we know for sure is that the Q = 5 policy is not optimal, because if it were the annual costs of ordering and holding would 32

be equal. Clearly, they are not for Q = 5 . Looking at the relative costs, we can see that the Q = 5 policy results in a relatively large ordering cost and a relatively small holding cost. Looking at the TAC expression, if we want Q = 5 to be the optimal policy, this would require one or all of the following: a much smaller fixed cost of ordering, a much lower annual demand, or a much larger cost of holding. Obviously, the rational choice among these three options would be to decrease the ordering cost. Going a little further, we can compute the optimal order quantity using the expression for the EOQ, * QEOQ =

2 AD h

2 ⋅ 6.25 ⋅1800 25 = 30. =

The optimal annual ordering and holding costs are both equal to $375, for a total annual cost of $750. (As an extended exercise, one could also compute the value of the fixed order cost that is required to make Q = 5 the optimal ordering policy. The answer is

A′ = 0.174 , about $0.17.) 3.11) From the original problem data, we have L = 1 day, with no variability (i.e., µ L = 1 day and

σ L = 0 days). In the problem statement, we are also given µ D = 5 monitors and σ D = 1.5 monitors. (a) Using expression (3.13), we have = 5, and using expression (3.14), we have µ DLT µ= D µL

= σ DLT

µ Lσ D2 + µ D2 σ L2 =

1 ⋅ (1.5 ) + 0 = 1.5. Therefore, using expression (3.17), with 2

zCSL = Φ −1 ( CSL ) = Φ −1 ( 0.9 ) = 1.282, we have= R µ DLT + zCSLσ DLT =5 + 1.282 ⋅1.5 =6.92,

which we round up to 7 units. (Note that we round up in order to achieve no less than the in-stock probability target level of 90%, in expectation.) (b) Now, we have µ L = 1.2 days and σ L = 0.4 days, and therefore, = µ DLT µ= D µL =( 5 ) ⋅ (1.2 ) =6 and= σ DLT

µ Lσ D2 + µ D2 σ L2 = 1.2 ⋅ (1.5 ) + ( 5 ) ⋅ ( 0.4 ) = 2.59. Using the 2

6 + 1.282 ⋅ 2.59 = target CSL of 90% results in a reorder point of= R µ DLT + zCSLσ DLT = 9.32. Again, in order to meet or exceed the target CSL, in expectation, we round up to a reorder point of 10 units. (c) If we use a reorder point of R = 7 units when the DLT distribution is actually

DLT  N ( 6, 2.59 ) , the resulting service level can be found by first standardizing this value, giving = z

0.3861. ( R − µ DLT ) σ DLT = ( 7 − 6 ) 2.59 =

This gives an expected

service level of Φ ( 0.3861) = 0.6503, or about 65%, substantially lower than the target CSL. (Note that this can be found by using the Excel function =NORMSDIST(0.3861).)

3.12) (a) As in Example 3.14, we start by computing the decentralized optimal order cycle for the DC, given by τ 2* = 2A2 ( Dh2 ) = 2 ⋅ 50 ( 20, 000 ⋅ 30 ) =0.01291 years, or an average of 77.46 orders per year. This is equivalent to Q2* = D ⋅τ 2* = 258 units. The order quantity multiplier, m, for the plant is computed from the condition, m ( m + 1) ≥ ( A1 A2 ) ⋅ ( h2 h1 ) = 6, which we can find by solving min z

subject to m ( m + 1) − z = 6 34

m ≥ 1 and integer z ≥ 0. This small problem is easily set up and solved in Excel, giving m* = 3 , and therefore = τ 1* m= τ 2* 0.03873 years (25.8 orders per year), which is equivalent to Q1* = D ⋅τ 1* =775 units.

Using expression (3.37), we can compute the supply chain cost of this

decentralized solution, which is TC ( m,τ 2 ) = $18,074, broken down across stage 1 and stage 2 as TC1 = $7746 and TC2 = $10,328. (b) For

centralized

solution,

start

computing

( A1 A2 ) ⋅ ( h2 h1 − 1) =

2, and then use this to find m* from ( 200 50 ) ⋅ ( 30 20 − 1) = min z

subject to

m ( m + 1) − z = 2

m ≥ 1 and integer z ≥ 0. This results in m* = 1 , and we use this to compute

τ 2* = =

2 ( A1 m + A2 )  D ( ( m − 1) h1 + h2 )  2 ⋅ ( 200 1 + 50 )  20, 000 ⋅ ( (1 − 1) ⋅ 20 + 30 ) 

= 0.02887 years (34.6 orders per year), or equivalently, Q2* = D ⋅τ 2* = 577 units. Thus,= τ 1* m= τ 2* 0.02887 years (36.4 orders per year), and Q1* = D ⋅τ 1* = 577 units. The supply chain cost of this centralized solution is TC ( m* ,τ 2* ) = $17,321.

(c) The gap between the decentralized solution computed in (a) and the centralized, first-best solution computed in (b) is 4.35% (of the optimal value). The centralized solution achieves this reduction in overall supply chain cost by essentially flipping the cost burden between stage 1 to stage 2 from what it was in the decentralized solution, such that TC1 ( m* ,τ 2* ) = $6,928 and TC2 ( m* ,τ 2* ) = $10,392 (with a small rounding error in the

sum). This is a more dramatic difference than in Example 3.14 regarding how the cost allocation between the supply chain stages changes as we move from a decentralized solution to a centralized solution.

3.13) As in Example 3.15, demand at the field warehouse follows a Poisson distribution with a mean of λ = 10 units/day, and the replenishment lead times at each stage j across this serial supply chain are L j = 2 days ( j = 1, 2,3 ). Therefore lead-time demand at echelon 3 is also Poisson with mean λ3 = 20 units/day. Similarly, the lead-time demand for echelon 2 is Poisson with mean λ2 = 40 units/day, and lead-time demand for echelon 1 is Poisson with mean λ1 = 60 units/day. As in Example 3.15, we approximate these daily demands using normal distributions, such that for echelons 1, 2 and 3, respectively, we use

(

)

(

)

(

)

D1  N 60, 60 , D2  N 40, 40 , and D3  N 20, 20 . In this case, however, the penalty cost incurred at the field warehouse for insufficient supply is b = $10 per unit short per period. Moreover, installation holding costs are h1 = $1.50 per unit per period, h2 = $2.25 per unit per period, and h3 = $3 per unit per period. This means that echelon holding costs H= h= 1 1

$1.50, H 2 = h2 − h1 = $0.75, and H 3 = h3 − h2 = $0.75.

Now, we can

compute the optimal echelon (and installation) base-stock level for echelon 3 to be

b + ∑ i =1 H i 10 + (1.5 + 0.75 ) * −1 * = 0.94231. Since we θ = = Sˆ3* = FDLT θ , where ( ) 3 ,3 3 N 10 + (1.5 + 0.75 + 0.75 ) b + ∑ i =1 H i N −1

are using a normal approximation for the underlying Poisson demand, we can compute * Sˆ= 20 + Φ −1 ( 0.94231) ⋅ ( 20= ) 27.04, which we round to Ŝ3* = 28 to ensure that we meet or 3

exceed the specified fractile of the demand distribution. For stage 2 ( j = 2 ) of this serial

b + ∑ i =1 H i 10 + 1.5 θ lj = = = system, we compute N b + ∑ i =1 H i 10 + (1.5 + 0.75 + 0.75 ) j −1

b + ∑ i =1 H i j −1

b + ∑ i =1 H i j

0.88462 and θ uj =

10 + 1.5 = 0.93878, such that S 2l =  40 + Φ −1 ( 0.88462 ) ⋅ ( 40 )  = 48 10 + (1.5 + 0.75 )

 S 2l + S 2u  = Sˆ2 = and S 2u =  40 + Φ −1 ( 0.93878 ) ⋅ ( 40 )  = 50. Therefore,  49. For stage 1 ( j  2  b + ∑ i =1 H i 10 θ = = 1), the site farthest upstream, = = 0.76923 (note N b + ∑ i =1 H i 10 + (1.5 + 0.75 + 0.75 ) j −1

l j

b + ∑ i =1 H i 10 = = 0.86957, such that ∑ i =1 H i is an empty sum for j = 1 ) and θ = j b + ∑ i =1 H i 10 + 1.5 j −1

j −1

that S1l =

u j

60 + Φ −1 ( 0.76923) ⋅ ( 60 )  = 66 and S1u =

60 + Φ −1 ( 0.86957 ) ⋅ ( 60 )  = 69.

 S1l + S1u  = Sˆ1 = Therefore,  68. As one would expect, the increase in holding cost and  2  decrease in penalty cost relative to Example 3.15 results in lower computed stocking levels. 3.14) * (a) The EOQ value for each item j can be computed from QEOQ, j =

( 2 A D ) h . For each j

item, the annual demand D j is given in Table 3.12, and the values of Aj and h j can be 37

obtained from data that also appears in Table 3.12. The holding costs are given by h j = ic j , where i = 0.25 is the annual holding cost rate, and c j is the unit cost of item j

given in Table 3.12. The fixed ordering cost for item j is = Aj $10 + FC j , where FC j is the freight cost for item j given in Table 3.12, and $10 is the cost of receiving any given item at the ATL DC, also given in Table 3.12. Thus, the resulting EOQ values, expressed in both units and in volume (cu ft) are as stated in the table below. Item 6.5-bu Rear Bagger 3.5-cu ft Tow-behind Broadcast Spreader 12-V Oscillating Fan Canopy Double-bucket holder

EOQ (units) 388 402 1822 988 3114

EOQ volume (cu ft) 12,432 10,857 3,644 1,482 1,869

Clearly, the EOQ values for the rear bagger, broadcast spreader, and oscillating fan are infeasible—i.e., beyond the truckload volume of 3500 cu ft stated in Table 3.12. Therefore, the optimal feasible value for each of these items is the full truckload quantity, or the maximum number of units that results in a total physical volume less than or equal to 3500 cu ft. The updated optimal individual order quantities are as follows.

Item 6.5-bu Rear Bagger 3.5-cu ft Tow-behind Broadcast Spreader 12-V Oscillating Fan Canopy Double-bucket holder

Order quantity (units) 109 (TL) 129 (TL) 1750 (TL) 988 (EOQ) 3114 (EOQ)

Order quantity volume (cu ft) 3,488 3,483 3,500 1,482 1,869

(b) To determine whether a reorder point R j achieves a given fill rate, we use expression (3.16),

S ( Rj )

1− β (Q j , R j ) =

σ DLT , j ⋅ L ( z j )

1− =

By trial and error, we can adjust R j accordingly to bring the value of β ( Q j , R j ) to the required fill rate, in this case 0.98. Otherwise, solving for the appropriate value of R j would require that we compute L−1 ( ⋅ ) , the inverse of the standard normal loss function, or utilize a lookup table. Merely having the ability to compute L ( z j ) , however, from Excel functions NORMSDIST and NORMDIST, as indicated in the chapter, allows us to quickly evaluate the fill rate for a given value of R j , making the trial and error process quite quick. The values of Q j were determined in part (a), above. Since the problem

= σ D , j ⋅ 3 . In addition, since states that the lead time is a consistent three days, σ DLT ,j we are informed that Milo’s uses a coefficient of variation of 0.3 to estimated daily demand uncertainty, we have σ D= 0.3 ⋅ µ D , j , and we know that mean daily demand is ,j given by µ D , j = D j 365 and that µ DLT , j = 3µ D , j . Finally, for a given value of R j , we have = zj

(R − µ j

DLT , j

)σ

DLT , j

Now, since we are informed that R j must be chosen such that the resulting safety stock is non-negative, we can start the “trial-R” process by computing the fill rate for R j = µ DLT , j , which gives zero safety stock. Thus, we can summarize what we have up to this point in the following table.

Item 6.5-bu Rear Bagger 3.5-cu ft Tow-behind Broadcast Spreader 12-V Oscillating Fan Canopy Double-bucket holder

µ DLT , j

σ DLT , j

β ( Q j , R j = µ DLT , j )

109 129 1750 988 3114

49.3 37.0 78.1 42.7 152.1

8.5 6.4 13.5 7.4 26.3

0.9673 0.9802 0.9969 0.9971 0.9966

Note that for each of the items except the rear bagger, β ( Q j , R j = µ DLT , j ) ≥ 0.98, meaning that any larger value of R j will only lead to (further) overshooting the target fill rate. Therefore, we only need to increase R1 , the reorder point of the rear bagger, to a level A value of R1 = 53 results in a fill rate of

that will achieve the target fill rate.

β (= Q1 109,= R1 53 = ) 0.9828.

n = *

(

∑

k j =1

hj Dj

2 A + ∑ j =1 a j k

)

where in this case, A = $776, the cost of truckload shipment containing items j = 1, 2,3 , and a j = 10 , the cost to receive each item in the DC. Therefore, letting the items be denoted as j = 1 for the fan, j = 2 for the canopy, and j = 3 for the bucket holder, we have n* =

( 4.5)( 9,500 ) + ( 7 )( 5, 200 ) + ( 3)(18,500 ) 2 ( 776 + 3 ⋅10 )

= 9.14.

Since this number is an expected value, it is fine for it to be fractional, and we can use it to compute the joint order quantities, Q j = D j n* , giving us the results in the following table. Joint order quantity (units) 1039 569 2024

Item 12-V Oscillating Fan Canopy Double-bucket holder

Joint order quantity volume (cu ft) 2,078 853.4 1,214.4

In total, this comes to a volume of 4,145.8 cu ft, which exceeds the truckload capacity, and is therefore an infeasible solution. Solving this problem in Excel, we can easily set-up an optimization problem using the Excel Solver to

(A+∑

min TAC ( n ) =

3 j =1

∑ a ⋅n + j

)

3 j =1

hj Dj

subject to Dj

∑ ω ⋅ n ≤ 3500, 3

j =1

where ω j is the per-unit volume (cu ft) of item j (see Table 3.12). The resulting optimal solution is n joint = 10.83, which gives the order quantity outcomes in the table below.

Joint order quantity (units) 877 480 1708

Item 12-V Oscillating Fan Canopy Double-bucket holder

Joint order quantity volume (cu ft) 1,754 720 1,024.8

In total, this comes to a volume of 3,498.8 cu ft, which of course is feasible because it was constrained to be feasible. 41

(d) First, we must compute the order frequency implied by the Solver solution from part (c). This is given by computing= τ joint 1= n joint 0.09235 years, or 33.71 days, between orders. In a periodic-review, ( S , T ) system, this amounts to a review period of approximately 33 days (rounding down to ensure that we order no less frequently than what is implied by the order quantities above). Thus, the review period plus lead time amounts to a total period of 36 days. In the table below, we show the resulting values of µ DLTR , j , σ DLTR , j , and S= 1.282. Φ −1 ( 0.9 ) = µ DLTR , j + zCSL ⋅ σ DLTR , j , where zCSL = j Item 12-V Oscillating Fan Canopy Double-bucket holder

µ DLTR , j

σ DLTR , j

936.99 512.88 1824.66

46.85 25.64 91.23

998 546 1942

Therefore, for each of these items, the resulting inventory policy is to review the inventory position x j every 33 days, and place an order for S j − x j units.

Chapter 4 Transportation Decisions in Supply Chain Management Solutions to Numerical Exercises only 4.4) The data in Table 4.12 correspond to a given freight class row in a rate tariff like that specified by Table 4.3 in the chapter. Thus, as indicated in the chapter, the first step is to set j equal to the appropriate value to determine the rates that apply to a shipment of W = 3000 lb. Specifically, we= set j arg { B j : B j ≤ W < B j +1} , meaning that j = 4 since B4 = 2000 and B5 = 5000, and therefore that Ri ,4 = $18.05 and Ri ,5 = $14.21 (where i indicates the freight class, which is not given, but also not needed to answer the question). Thus, we can compute

FCLTL (= 3000 ) max {$83.24, min {$18.05 ⋅ 3000 100,$14.21 ⋅ 5000 100}} = max {$83.24, min {$541.50,$710.50}} = $541.50. cwt In this case, the effective rate is rLTL $18.05/cwt, which is to 100 ⋅ $541.50 3000 = ( 3000 ) =

be expected, since the portion of the expression inside the “min” indicates that a 3,000-lb does not receive a “weight break” rate. For the 4,200-lb shipment, B4 , B5 , Ri ,4 , and Ri ,5 remain the same, and FCLTL (= 4200 ) max {$83.24, min {$18.05 ⋅ 4200 100,$14.21 ⋅ 5000 100}} = max {$83.24, min {$758.10,$710.50}} = $710.50,

cwt which gives an effective rate of rLTL $16.92/cwt. This 100 ⋅ $710.50 4200 = ( 4200 ) =

indicates that the 4,200-lb shipment weight does indeed exceed the weight break and therefore results in an effective rate between Ri ,4 = $18.05 and Ri ,5 = $14.21. 4.7) The full rate tariff is as follows: Origin ZIP

44135

(OH)

Destination ZIP

27709

(NC)

Minimum Charge

$95.39 LTL rate ($/cwt)

Class 500

L5C 357.34

M5C 296.61

M1M 225.12

M2M 182.24

M5M 129.88

M10M 94.51

M20M 48.74

M30M 48.74

M40M 48.74

400

287.98

239.03

181.42

146.87

104.66

76.16

39.34

300

220.48

183.00

138.90

112.44

80.13

58.31

30.14

250

193.23

160.39

121.72

98.54

70.22

51.10

26.42

200

150.49

124.91

94.81

76.75

54.69

39.80

20.55

175

135.63

112.57

85.44

69.17

49.30

35.86

18.50

150

116.43

96.64

73.35

59.38

42.32

30.79

15.95

125

99.09

82.25

62.43

50.53

36.02

26.21

13.60

110

86.09

71.45

54.23

43.90

31.28

22.77

11.94

100

80.52

66.83

50.72

41.06

29.26

21.29

11.15

8.65

7.52

76.18

63.22

47.98

38.85

27.68

20.14

10.76

8.35

7.25

70.60

58.60

44.48

36.00

25.66

18.67

10.38

8.03

6.99

65.65

54.49

41.36

33.48

23.86

17.36

10.08

7.81

6.79

61.93

51.40

39.02

31.58

22.51

16.38

9.79

7.58

6.59

59.39

49.30

37.41

30.29

21.38

15.71

9.69

7.51

6.53

56.86

47.19

35.82

28.99

20.49

15.03

9.59

7.43

6.46

53.76

44.62

33.86

27.41

19.36

14.22

9.50

7.36

6.40

50.66

42.05

31.92

25.83

18.23

13.40

9.40

7.28

6.32

Using the Class 85 row, and applying the 46% discount (to the minimum charge as well), the appropriate rates are as follows. Origin ZIP

44135

(OH)

Destination ZIP

27709

(NC)

Minimum Charge

$51.51 LTL rate ($/cwt)

Class 85

L5C 38.12

M5C 31.64

M1M 24.02

M2M 19.44

M5M 13.86

M10M 10.08

M20M 5.61

M30M 4.34

M40M 3.77

cwt Using this rate data, we can apply the effective rate computation rLTL (W ) =

{

}

100 ⋅ FCLTL (W ) W , where FCLTL (= W ) max MC , min { Rij ⋅ W 100, Ri , j +1 ⋅ B j +1 100} , to build the following table in Excel:

( )

rLTL (W )

ln (W )

ln rLTL

51.51 38.12 38.12 38.12 37.23 35.16 33.31 31.64 31.64 31.64 30.02 28.26 26.69 25.28 24.02 24.02 24.02 24.02 22.87 21.60 20.46 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.44 19.25 18.23

4.6052 5.2983 5.7038 5.9915 6.0521 6.1092 6.1633 6.2146 6.3969 6.5511 6.6846 6.7452 6.8024 6.8565 6.9078 7.0901 7.2442 7.3778 7.4384 7.4955 7.5496 7.6009 7.6962 7.7832 7.8633 7.9374 8.0064 8.0709 8.1315 8.1887 8.2428

3.9418 3.6408 3.6408 3.6408 3.6171 3.5599 3.5058 3.4545 3.4545 3.4545 3.4020 3.3414 3.2842 3.2301 3.1789 3.1789 3.1789 3.1789 3.1299 3.0727 3.0186 2.9673 2.9673 2.9673 2.9673 2.9673 2.9673 2.9673 2.9673 2.9573 2.9032

cwt

W (lb) 100 200 300 400 425 450 475 500 600 700 800 850 900 950 1000 1200 1400 1600 1700 1800 1900 2000 2200 2400 2600 2800 3000 3200 3400 3600 3800

cwt

W (lb) 4000 4200 4400 4600 4800 5000 5200 5400 5600 5800 6000 6200 6400 6600 6800 7000 7200 7400 7600 7800 8000 8200 8400 8600 8800 9000 9200 9400 9600 9800 10000

( )

rLTL (W ) cwt

ln (W )

ln rLTL

17.32 16.50 15.75 15.06 14.43 13.86 13.86 13.86 13.86 13.86 13.86 13.86 13.86 13.86 13.86 13.86 13.86 13.62 13.27 12.93 12.60 12.29 12.00 11.72 11.46 11.20 10.96 10.73 10.50 10.29 10.08

8.2940 8.3428 8.3894 8.4338 8.4764 8.5172 8.5564 8.5942 8.6305 8.6656 8.6995 8.7323 8.7641 8.7948 8.8247 8.8537 8.8818 8.9092 8.9359 8.9619 8.9872 9.0119 9.0360 9.0595 9.0825 9.1050 9.1270 9.1485 9.1695 9.1901 9.2103

2.8519 2.8031 2.7566 2.7121 2.6696 2.6287 2.6287 2.6287 2.6287 2.6287 2.6287 2.6287 2.6287 2.6287 2.6287 2.6287 2.6287 2.6118 2.5852 2.5592 2.5339 2.5092 2.4851 2.4616 2.4386 2.4161 2.3941 2.3726 2.3516 2.3309 2.3107

cwt

Also in Excel, we can apply the INTERCEPT and SLOPE functions to the columns of cwt this table containing the values of lnW and ln ( rLTL ) to obtain the regression estimate cwt ln  rLTL ln W 5.6770 − 0.3500 lnW . Therefore, we can take the “exp” = (W ) ln Cw + b=

cwt of this linear function to specify our rate estimating function, which is rˆLTL (W ) = CwW b

= 292.0666 W −0.35 .

Using this function, the estimated effective rate of a 8,500-lb

cwt shipment is rˆLTL (8500 ) = 292.0666 ⋅ (8500 )

compute

the

actual

effective

−0.35

=$12.31/cwt.

rate,

first

{

compute

FCLTL ( 8500 ) =

}

$1008, which gives an max $51.51, min {$13.86 ⋅ ( 8500 ) 100,$10.08 ⋅ (10, 000 ) 100} = cwt effective rate of rLTL $11.86/cwt. (8500 ) = 100 ⋅ ( $1008) 8500 =

4.8) We repeat the Class 85 rates from Exercise 4.7, as follows: LTL rate ($/cwt) Class 85

L5C 38.12

M5C 31.64

M1M 24.02

M2M 19.44

M5M 13.86

M10M 10.08

M20M 5.61

M30M 4.34

M40M 3.77

The weight break for the M5C weight group is the point at which a shipment of

500 < W ≤ 1000 lb is charged as if it actually weighed 1000 lb. Therefore, the break occurs at ( $31.64 / cwt= ) ⋅ wM5C

( $24.02 / cwt ) ⋅ (10 cwt ) , or wM5C = 7.59 cwt (759 lb). Thus, all

shipments of weight 759 ≤ W < 1000 lb receive a freight charge of FCLTL = $240.20. ( $24.02 / cwt ) ⋅ (10 cwt ) =

For the M5M weight group, the break occurs at

( $13.86 / cwt ) ⋅ wM5M = ( $10.08 / cwt ) ⋅ (100 cwt ) , or wM5M = 72.73 cwt (7273 lb). 4.9) From the problem data, we are given the shipment density of s = 12.72 lb / cu ft. From Google Maps, we can obtain the shipment distance of d = 558 mi. As in Example 4.13, since s and d are given, we can restate expression (4.22) as

ton-mi rLTL

G1     1 (Wton , s, d ) = PPI LTL   7 ⋅G  , 7  G ⋅W −  3   2 ton 2   46

where = G1 s 2 8 + 14 , G2 = d 15 29 , and G3 = s 2 + 2 s + 14 . For the PPI LTL term, we use the value stated in Example 4.13, or PPI LTL ,2000 = 87.12.

Thus, the resulting LTL rate

estimation function is

ton-mi rLTL

34.2248     1 (Wton s, d , PPI LTL ) = 87.12   7  ⋅ 201.2384  7  26.3435 ⋅ Wton −     2 =

14.8166 . 0.14286 26.3435 ⋅ Wton − 3.5

Recall that this function gives the (circa 2000) LTL rate in units of $/ton-mi. Thus, for the ton-mi shipment in question, = Wton 8500 = 2000 4.25 tons, and we obtain a rate of rLTL ( 4.25) =

14.8166  26.3435 ⋅ ( 4.25 ) 

0.14286

0.51282 $/ton-mi. We multiply this by 558 mi, the − 3.5 = 

length of the shipment lane for which we are estimating the LTL rate, to obtain the rate in cwt $/ton, and then by 1/20 ton/cwt to obtain the rate in $/cwt, giving us rLTL = $14.31/cwt.

This result is 20.7% higher than the actual effective rate computed in Exercise 4.7. Clearly, there are many possible sources of error in this estimating function. The best explanation is perhaps the fact that in the Kay and Warsing (2009) study, the estimating expression is stated to be accurate to an average weighted residual error of about 11%. Therefore, some of the sampled origin-destination pairs in Kay and Warsing’s study must have exhibited smaller error values, and some must have had error values in the 15 – 20% range, or perhaps higher. This O-D pair happens to be one for which the rate estimate is not extremely precise, at least not for a shipment of 8500 lb. Still, the value of the Kay-

Warsing estimate is that, without a rate tariff, we can generate an estimate of the LTL rate (in this case a more approximate estimate). cwt b = 4.10) From Exercise 4.7, we have rˆLTL 292.0666 W −0.35 . Note that this gives the (W ) C= wW

estimated LTL rate in $/cwt. Our goal is to create an equivalent expression that gives the estimated LTL rate in $/unit. Since each unit weighs w pounds, a shipment of Q units cwt weighs W = wQ pounds, and we can obtain the estimated LTL rate as rˆLTL ( wQ ) =

292.0666 ( wQ )

−0.35

cwt $/cwt. Since there are w 100 cwt per unit, we can convert rˆLTL ( wQ ) ,

in $/cwt, to a value in $/unit by multiplying by w 100 , which gives rˆLTL ( wQ ) 292.0666 ( wQ ) =

−0.35

⋅ ( w 100 )

= 2.920666 w0.65Q −0.35 $/unit.

4.11) (a) Expression (4.5) in the chapter provides a broad generalization of the total annual costs relevant to logistics decision making, extending the basic ordering cost – holding cost tradeoff from Chapter 3 to also include transportation costs. In the case of Option 1, since the freight terms are F.O.B. destination, freight prepaid, the relevant costs include only ordering cost and holding costs related to cycle stock and safety stock (i.e., the costs considered in Chapter 3). Note however, that there is no uncertainty implied by the problem, so that there is no need for safety stock and therefore no holding costs related to safety stock. For Option 2, since the freight terms are F.O.B. origin, freight collect, we must extend the list of considered costs to include holding costs related to goods in transit and transportation costs. Finally, note that materials costs are relevant to both options because of the 1.5% discount in the item price under Option 2.

(b) For Option 1, as indicated above, because of the freight terms, we only need to consider ordering costs, cycle stock and safety stock holding costs, and the cost of materials. Moreover, note that the problem states that we are not restricted to a particular shipping schedule, but each shipment must be a full-truckload quantity of Q = 3200 units. This makes the decision simple, and the resulting costs are TAC ( Q ) = OC ( Q ) + HC ( Q ) + MC = A⋅

D Q + h ⋅ + Dc Q 2

=( 25 ) ⋅

26, 000 3200 + ( 0.25 )( 35 ) ⋅ + ( 26, 000 )( 35 ) 3200 2

= $203 + $16,800 + $910, 000 = $927, 003. (c) For Option 2, first we must clearly account for those components of cost that are different from Option 1. First, not only does our company now pay the transportation cost, this cost will also be 10% higher than it was for Option 1, meaning that the per-shipment

= freight charge is FC1

= (1.1 )( $882 ) $970.20. Also, since the supplier no longer has to

pay the transportation bill, the item cost decreases by = 1.5% to c1

0.985 )( $35 ) (=

$34.48. Now, we know that the order quantity decision is the optimal solution to min TAC ( Q ) = OC ( Q ) + HC ( Q ) + TC ( Q ) + MC =A⋅

D Q D + h1 ⋅ + FC1 ⋅ + Dc1 Q 2 Q

= ( A + FC1 ) ⋅ subject to

D Q + h1 ⋅ + Dc1 Q 2

Q ≤ 3200 .

Given the structure of the objective function, the optimal solution is clearly of EOQ form, provided that the resulting EOQ is feasible. Thus, we can compute a potential solution of Q1 =

2 ⋅ ( A + FC1 ) ⋅ D h1 2 ⋅ (10 + 970.20 ) ⋅ ( 26, 000 ) ( 0.25)( 34.48)

= 2237, Clearly, this solution is feasible, and therefore, we can compute the total relevant annual cost to be TAC ( Q ) = A ⋅

D Q D + h1 ⋅ + FC1 ⋅ + Dc1 Q Q 2

= (10 ) ⋅

26, 000 2237 26, 000 + ( 0.25 )( 34.48 ) ⋅ + ( 970.20 ) ⋅ + ( 26, 000 )( 34.48 ) 2237 2 2237

= $291 + $12,184 + $11, 276 + $896, 480 = $920, 231.

(d) Between Options 1 and 2, clearly Option 2 is the better choice. Option 2 saves a total of $6,772 in annual costs, driven by a savings of over $18,000 in holding cost and materials costs, which overcomes the $11,000-plus annual cost of transportation. (e) In this case, we use the order quantity decision from part (c), Q1 = 2237. Knowing this, we must determine the LTL freight cost for a shipment of this size. Based on the problem data, an item weighs w = 12 lb, and therefore a shipment of size Q1 = 2237 units weighs W = w ⋅ Q = 26,844 lb. Therefore, using the methods stated in the chapter, we have (see also Exercise = 4.4) j arg { B j : B j ≤ W < B j +1} , which in this case gives j = 7 since B7 = 20,000 and B8 = 30,000, and therefore Ri ,7 = $7.21 and Ri ,8 = $5.54 (where

again i indicates the freight class, not given, but also not needed to answer the question). With this information, we can compute

{

}

FCLTL ( Q= max $80.33, min {$7.21 ⋅ ( 26,844 ) 100,$5.54 ⋅ ( 30, 000 ) 100} 1) = max {$80.33, min {$1935.45,$1662.00}} = $1662. Thus, the annual transportation cost for this option is TC= ( Q ) FCLTL ⋅

D Q1

(1662 ) ⋅

26, 000 2237

= $19,317.

Clearly, this option is worse than Option 2 above since it would result in a dramatic increase in annual transportation cost over that option with no changes in the other costs. In order to assess this LTL option fairly, however, we clearly need to open up the possibility of a shipment quantity other than Q1 = 2237 units. We could do this if we used the rate tariff to create a function to estimate LTL freight costs and then used, for example, Excel Solver to find the value of Q that minimizes a TAC function that has this LTL cost estimate embedded in it. 4.12) (a) For each item, the inventory cycle length is computed by taking the quotient of the order quantity Q and the annual demand D, as given in Table 4.14, or τ j = Q j D j . (Note that, since D is given in units/year, this computation gives the cycle length in years. To obtain the cycle length in days we must multiply by the number of operating days per year, or 365, also given in Table 4.14.) Thus, the cycle lengths are as stated in the following table. 51

Item HP slimline PC

Annual demand 4000 units

Shipment size 72 units

Inventory cycle

HP LaserJet printer

2600 units

435 units

61.07 days

Lexmark multi-function inkjet printer

6000 units

1022 units

62.17 days

Logitech wireless mouse (case of 10)

1200 cases

48 cases

14.60 days

6.57 days

(b) We can see from the results in (a) that the two printers have very similar inventory cycle lengths, whereas those of the PC and the case of mice differ by a factor of more than 2. Therefore, the most logical candidates to consider for joint shipment are the two printers. (c) To generate a trial joint solution, we can use methods from Chapter 3, particularly the expression for the optimal number of joint shipments, expression (3.29),

n = *

(

∑

k j =1

hj Dj

2 A + ∑ j =1 a j k

)

where A = $2,016, the cost of a joint TL shipment, as given in Table 4.14, and where a j = $20, as given in Table 4.13. Moreover, from the data in Table 4.13, we can compute

h j = ic j , where i = 0.35 is the inventory holding cost rate. Therefore, we have

n* =

( 0.35)(160 )( 2600 ) + ( 0.35)( 60 )( 6000 ) 2 ( 2016 + 2 ⋅ 20 )

= 8.13. This gives a joint shipment inventory cycle of τ = 1= n* 0.123 years, or 44.9 days. joint The resulting total annual of this joint shipment policy would be

(

TAC ( n* ) = A + ∑ j =1 2

= $49, 646.

∑ a ⋅n + j

)

2 j =1

hj Dj

2 n*

(d) It turns out that the trial solution computed in part (c) cannot be optimal because it is infeasible. Using n* = 8.13, we obtain values of Q j = D j n* of Q1 = 320 units and Q2 = 738 units, which have volumes of 1660.8 cu ft and 2309.9 cu ft, respectively. In total, a shipment of this size would have a volume of 3970.7 cu ft, which exceeds the stated truckload capacity of 3200 cu ft. In order to find the optimal joint shipment quantity of these two items, we must solve

(A+∑

min TAC ( n ) =

2 j =1

∑ a ⋅n + j

)

2 j =1

hj Dj

subject to Dj

∑ ω ⋅ n ≤ 3200, 2

j =1

where ω j is the per-unit volume (cu ft) of item j (see Table 4.14). The resulting optimal solution is n joint = 10.09, which gives the order quantity outcomes in the table below.

Joint order quantity (units) 258 594

Item HP LaserJet printer Lexmark multi-function inkjet printer

Joint order quantity volume (cu ft) 1,339 1,859

Interestingly, adding TAC ( n joint ) = $54,350 to the TAC values for the other two items shipped independently, we obtain a total annual cost of $91,651 across all four items. One can easily verify, however, that the original overall TAC of shipping all four items independently was only $84,334.

Chapter 5 Location and Distribution Decisions in Supply Chains Solutions to Numerical Exercises only 5.5 Solution Variables: 𝑥1𝐴 – quantity bought from supplier 1 at $10/unit 𝑥1𝐵 – quantity bought from supplier 2 at $7/unit

𝑥2 – quantity bought from supplier 2

𝑥3𝐴 – quantity bought from supplier 3 at $6/unit

𝑥3𝐵 – quantity bought from supplier 3 at $4/unit

Constraints:

𝑥1𝐴 + 𝑥1𝐵 + 𝑥2 + 𝑥3𝐴 + 𝑥3𝐵 = 2000 𝑥1𝐴 ≤ 300𝛿1𝐴

𝑥1𝐴 ≥ 300𝛿1𝐵

𝑥3𝐴 ≤ 500𝛿3𝐴

𝑥3𝐴 ≥ 500𝛿3𝐵

𝑥1𝐵 ≤ 300𝛿1𝐵

𝑥3𝐵 ≤ 700𝛿3𝐵

𝑥2 ≤ 800𝛿2

𝑥𝑖𝑗 ≥ 0, 𝛿𝑖𝑗 ∈ (0,1), 𝑥2 ≥ 0 for i = 1,3 and j = A, B Objective Function:

Minimize 𝑍 = 100𝛿1𝐴 + 10𝑥1𝐴 + 7𝑥1𝐵 + 500𝛿2 + 2𝑥2 + 300𝛿3𝐴 + 6𝑥3𝐴 + 4𝑥3𝐵

5.6 Solution Variables: 𝑥1𝐴 – quantity bought from supplier 1 at $10/unit 𝑥1𝐵 – quantity bought from supplier 2 at $7/unit

𝑥2 – quantity bought from supplier 2

𝑥3𝐴 – quantity bought from supplier 3 at $6/unit

𝑥3𝐵 – quantity bought from supplier 3 at $4/unit Constraints:

𝑥1𝐴 + 𝑥1𝐵 + 𝑥2 + 𝑥3𝐴 + 𝑥3𝐵 = 2000 𝑥1𝐴 ≤ 300𝛿1𝐴

301𝛿1𝐵 ≤ 𝑥1𝐵 ≤ 600𝛿1𝐵 𝛿1𝐴 + 𝛿1𝐵 ≤ 1 𝑥2 ≤ 800𝛿2

𝑥3𝐴 ≤ 500𝛿3𝐴

501𝛿3𝐵 ≤ 𝑥3𝐵 ≤ 1200𝛿3𝐵 𝛿3𝐴 + 𝛿3𝐵 ≤ 1

𝑥𝑖𝑗 ≥ 0, 𝛿𝑖𝑗 ∈ (0,1), 𝑥2 ≥ 0 for i = 1,3 and j = A, B Objective Function:

Minimize 𝑍 = 100𝛿1𝐴 + 10𝑥1𝐴 + 7𝑥1𝐵 + 500𝛿2 + 2𝑥2 + 300𝛿3𝐴 + 6𝑥3𝐴 + 4𝑥3𝐵

5.7 Solution Let M denote a large positive number. Define ai

No. of units of Product A made in range i, i=1,...,4

No. of units of Product B made in range j, j=1,...,3

No. of units of Product C made in range k, k=1,2

Let, δ ai = 1 if production of Product A is range i, i=1,2,3,4 [ a i >0] ; 0 Otherwise δ bj = 1 if production of Product B is range j, j=1,2,3 [  b j >0] ; 0 Otherwise δ ck = 1 if production of Product C is range k, k=1,2 [c k >0] ; 0 Otherwise For product A, Range 1=[0,40]; range 2=[41,100]; range 3=[101,150]; range 4: ≥ 151. For product B, Range 1=[0,50]; range 2=[51,100]; range 3: ≥ 101 For product C, Range 1=[0,100]; range 2: ≥ 101 Objective: We need to maximize the total profit. (Profit = Selling Price- Cost) max Z = 12(a1+ a2 + a3 + a4) + 9 (b1 + b2+ b3)+ 7(c1 + c2) –(10a1 + 9a2 + 8a3 + 7a4 + 6b1 + 4b2 + 3b3 + 5c1 + 4c2) Subject to the following constraints: Engineering service:

(a1+a2+a3+a4)+2(b1+b2+b3)+(c1+c2) ≤100

Labor:

10(a1+a2+a3+a4)+4(b1+b2+b3)+5(c1+c2) ≤700

Material:

3(a1+a2+a3+a4)+2(b1+b2+b3)+(c1+c2) ≤400

Quantity discount constraints: We need to ensure that the quantity discounts are applied accurately. This is done as follows.

0 ≤ a1 ≤ 40δa1

a1 ≥ 40δa2 0 ≤ a2 ≤ 60δa2 a2 ≥ 60δa3 0 ≤ a3 ≤ 50δa3 a3 ≥ 50δa4 0≤ a4 ≤Mδa4 0 ≤ b1 ≤50δb1 b1 ≥ 50δb2 0≤ b2 ≤50δb2 b2 ≥ 50δb3 0≤ b3 ≤Mδb3 0≤ c1 ≤100δc1 c1 ≥ 100δc2 0 ≤ c2 ≤ Mδc2 δai, δbj, δck ∈{0,1} ∀ i,j,k 5.8 Solution Let M denote a large positive number. (a) Given 𝑥1 + 𝑥2 ≤ 3

3𝑥1 + 4𝑥2 ≥ 10

Let δ∈{0,1}. Then, rewriting the above equations in ≤ 0 form with all terms and constants on the left hand side, we have the required either-or constraints 𝑥1 + 𝑥2 − 3 ≤ 𝑀𝛿

−3𝑥1 − 4𝑥2 + 10 ≤ 𝑀(1 − 𝛿) 57

δ ∈ {0,1} (b) Given that x2∈{0,4,7,10,12}. Let δi∈{0,1}, ∀i=1,2,3,4 be binary variables. The required set of linear constraints are 𝑥2 = 4𝛿1 + 7𝛿2 + 10𝛿3 + 12𝛿4 𝛿1 + 𝛿2 + 𝛿3 + 𝛿4 ≤ 1

The second equation ensures that at most one of the binary variables equals 1; if all the binary variables are equal to 0, then x3=0 from the first equation. (c) Given that if x2 ≤ 3, then x3 ≥ 6; Else (x2≥4) x3 ≤ 4; [x2 and x3 are integers]. This can be written as a set of Either-Or constraints as below. Let δ∈{0,1}; then 𝑥2 − 3 ≤ 𝑀𝛿

−𝑥3 + 6 ≤ 𝑀𝛿

−𝑥2 + 4 ≤ 𝑀(1 − 𝛿) 𝑥3 − 4 ≤ 𝑀(1 − 𝛿) δ∈ (0,1)

Note that if δ=0, the first two constraints are binding; else if δ=1, the third and fourth constraints are binding. (d) We need at least two out of the five constraints to be satisfied (binding). Let δi∈{0,1} ∀i=1,2,3,4,5 be binary variables. Then, we can model this using the “m out of n type of constraints” as follows. 𝑥1 + 𝑥2 − 7 ≤ 𝑀𝛿1

−𝑥1 + 𝑥2 + 3 ≤ 𝑀𝛿2

−4𝑥1 + 3𝑥2 + 10 ≤ 𝑀𝛿3 𝑥2 − 6 ≤ 𝑀𝛿4

2𝑥1 + 3𝑥2 − 20 ≤ 𝑀𝛿5

𝛿1 + 𝛿2 + 𝛿3 + 𝛿4 + 𝛿5 ≤ 3

The final inequality ensures that at least two of the five constraints are satisfied.

5.9 Solution Write 𝑥1 = 20 𝛿1 + 21 𝛿2 = 𝛿1 + 2𝛿2

𝑥2 = 20 𝛿3 + 21 𝛿4 + 22 𝛿5 = 𝛿3 + 2𝛿4 + 4𝛿5 where 𝛿𝑖 ∈ (0,1)i = 1, 2 … ,5

Now the objective function reduces to

𝑍 = (𝛿1 + 2𝛿2 )2 − (𝛿1 + 2𝛿2 )(𝛿3 + 2𝛿4 + 4𝛿5 ) + 𝛿3 + 2𝛿4 + 4𝛿5

= 𝛿12 + 4𝛿22 + 4𝛿1 𝛿2 − 𝛿1 𝛿3 − 2𝛿1 𝛿4 − 4𝛿1 𝛿5 − 2𝛿2 𝛿3 − 4𝛿2 𝛿4 − 8𝛿2 𝛿5 +𝛿3 + 2𝛿4 + 4𝛿5

Note that 𝛿𝑖2 = 𝛿𝑖 for 𝛿𝑖 ∈ (0,1), while the product terms 𝛿𝑖 𝛿𝑗 can be reduced to linear form as follows:

Let 𝑦𝑖𝑗 = 𝛿𝑖 𝛿𝑗

Add the constraints

𝛿𝑖 + 𝛿𝑗 − 𝑦𝑖𝑗 ≤ 1

𝛿𝑖 + 𝛿𝑗 − 2𝑦𝑖𝑗 ≥ 0

So the given nonlinear integer program reduces to the following linear (0-1) integer program: Min 𝑍 = 𝛿1 + 4𝛿2 + 4𝑦12 − 𝑦13 − 2𝑦14 − 4𝑦15 − 2𝑦23 − 4𝑦24 − 8𝑦25 +𝛿3 + 2𝛿4 + 4𝛿5 Subject to

𝛿1 + 4𝛿2 + 4𝑦12 + 𝑦13 + 2𝑦14 + 4𝑦15 + 2𝑦23 + 4𝑦24 + 8𝑦25 ≤ 8 𝛿1 + 2𝛿2 ≤ 2

𝛿3 + 2𝛿4 + 4𝛿5 ≤ 7 𝛿1 + 𝛿𝑗 − 𝑦1𝑗 ≤ 1

𝛿1 + 𝛿𝑗 − 2𝑦1𝑗 ≥ 0 𝑓𝑜𝑟 𝑗 = 2,3,4,5 𝛿2 + 𝛿𝑗 − 𝑦2𝑗 ≤ 1

𝛿2 + 𝛿𝑗 − 2𝑦2𝑗 ≥ 0 𝑓𝑜𝑟 𝑗 = 3,4,5 𝑦𝑖𝑗 ∈ (0,1), 𝛿𝑗 ∈ (0,1)

5.10 Solution a) Warehouse location and distribution problem: Decision Variables: X ij – No. of units shipped from warehouse site ‘i’ to retailer ‘j’. 1 𝑖𝑓 𝑠𝑖𝑡𝑒 𝑖 𝑖𝑠 𝑠𝑒𝑙𝑒𝑐𝑡𝑒𝑑 𝑡𝑜 𝑏𝑢𝑖𝑙𝑑 𝑎 𝑤𝑎𝑟𝑒ℎ𝑜𝑢𝑠𝑒 δi = � 0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

Constraints:

Capacity constraints: Each warehouse cannot ship more than what it can hold. If a particular site is not chosen to build a warehouse (δ i = 0), then the RHS of the constraint becomes zero and hence all units shipped from that site are also forced to zero since X ij ≥0 for all i, j. 8

� 𝑋𝑖𝑗 ≤ 𝐴𝑖 𝛿𝑖 𝑓𝑜𝑟 𝑖 = 1, 2, 3 (𝐴𝑖 − 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑜𝑓 𝑤𝑎𝑟𝑒ℎ𝑜𝑢𝑠𝑒 𝑠𝑖𝑡𝑒 𝑖) 𝑗=1

Demand constraints: Each retailer’s demand should be satisfied. The sum of all goods received by a retailer from all warehouses must be greater than or equal to that retailer’s demand. 3

� 𝑋𝑖𝑗 ≥ 𝐷𝑗 𝑓𝑜𝑟 𝑗 = 1, 2, … ,8 (𝐷𝑗 − 𝑑𝑒𝑚𝑎𝑛𝑑 𝑜𝑓 𝑟𝑒𝑡𝑎𝑖𝑙𝑒𝑟 𝑗) 𝑖=1

Non-negativity and binary constraints: 𝑋𝑖𝑗 ≥ 0 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑖, 𝑗

𝛿𝑖 ∈ (0,1) 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑖

Objective function:

The objective is to minimize the sum of transportation costs and fixed costs. Transportation cost: This is the sum of the product of number of units transported from site ‘i’ to retailer ‘j’ (X ij ) and the unit cost of shipping (C ij ). 8

𝑇𝐶 = � � 𝐶𝑖𝑗 𝑋𝑖𝑗 = 4𝑋11 + 5𝑋12 + ⋯ + 2.9𝑋37 + 3.5𝑋38 𝑗=1 𝑖=1

Fixed costs: This is given by the sum of the fixed costs (F i ) and the corresponding δ variables.

𝐹𝐶 = � 𝐹𝑖 𝛿𝑖 = 100000𝛿1 + 80000𝛿2 + 110000𝛿3 𝑖=1

The objective function is given by 𝑀𝑖𝑛𝑖𝑚𝑖𝑧𝑒 𝑍 = 𝑇𝐶 + 𝐹𝐶

The optimal solution from Excel solver is as follows:

Table 1 Distribution plan for multiple warehouses formulation (5.10 (a))

Units shipped W1 W2 W3

R1 0 0 20000

R2 0 20000 0

R3 0 20000 0

R4 0 0 20000

R5 0 25000 0

R6 0 0 25000

R7 0 15000 10000

R8 0 0 25000

Table 2 Cost for multiple warehouses formulation (5.10 (a))

Shipping cost Fixed cost Total optimal cost

555500 190000 745500

b) Dedicated warehouse formulation: Decision Variables: 𝛿𝑖 = �

1 𝑖𝑓 𝑠𝑖𝑡𝑒 𝑖 𝑖𝑠 𝑠𝑒𝑙𝑒𝑐𝑡𝑒𝑑 𝑡𝑜 𝑏𝑢𝑖𝑙𝑑 𝑎 𝑤𝑎𝑟𝑒ℎ𝑜𝑢𝑠𝑒 0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

𝑋𝑖𝑗 = �

1 𝑖𝑓 𝑠𝑖𝑡𝑒 𝑖 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑠 𝑟𝑒𝑡𝑎𝑖𝑙𝑒𝑟 ′𝑗′ 0 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

Constraints:

Capacity constraints: 8

� 𝐷𝑗 𝑋𝑖𝑗 ≤ 𝐴𝑖 𝛿𝑖 𝑓𝑜𝑟 𝑖 = 1, 2, 3 𝑗=1

(𝐴𝑖 = 𝑐𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑜𝑓 𝑤𝑎𝑟𝑒ℎ𝑜𝑢𝑠𝑒 𝑠𝑖𝑡𝑒 𝑖 𝑎𝑛𝑑 𝐷𝑗 = 𝑑𝑒𝑚𝑎𝑛𝑑 𝑜𝑓 𝑟𝑒𝑡𝑎𝑖𝑙𝑒𝑟 ′𝑗′)

Dedicated warehouse constraint: Each retailer can be supplied by only one warehouse.

Delta 0 1 1

� 𝑋𝑖𝑗 = 1 𝑓𝑜𝑟 𝑒𝑣𝑒𝑟𝑦 𝑗 𝑖=1

Binary constraints:

𝑋𝑖𝑗 ∈ (0,1) 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑖, 𝑗 𝛿𝑖 ∈ (0,1) 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑖

Objective Function:

The objective function is the sum of transportation costs and the fixed costs. Transportation costs: 8

𝑇𝐶 = � � 𝐶𝑖𝑗 �𝐷𝑗 𝑋𝑖𝑗 � 𝑗=1 𝑖=1

Fixed costs:

= 4 ∗ 20,000 ∗ 𝑋11 + 5 ∗ 20,000 ∗ 𝑋12 + ⋯ + 2.9 ∗ 25,000 ∗ 𝑋37 + 3.5 ∗ 25,000 ∗ 𝑋38 3

𝐹𝐶 = � 𝐹𝑖 𝛿𝑖 = 100000𝛿1 + 80000𝛿2 + 110000𝛿3 𝑖=1

The objective function is given by 𝑀𝑖𝑛𝑖𝑚𝑖𝑧𝑒 𝑍 = 𝑇𝐶 + 𝐹𝐶

The optimal solution from Excel solver is as follows:

Table 3 Distribution plan for dedicated warehouses formulation (5.10 (b))

X ij W1 W2 W3

R1 0 0 20000

R2 0 20000 0

R3 0 20000 0

R4 0 0 20000

R5 0 25000 0

R6 0 0 25000

R7 0 0 25000

Table 4 Cost for dedicated warehouses formulation (5.10 (b))

Transportation costs Fixed costs Total

560000 190000 750000

R8 0 0 25000

Delta 0 1 1

c) In both part (a) and (b), the entire demand is satisfied using warehouses 2 and 3 only; the entire capacity of warehouse 1 remains unused. The only change between the two distribution options is w.r.t. retailer 7 – in part (a) retailer 7’s demand is satisfied by both warehouses 2 and 3, while in part (b) only warehouse 3 is used. The fixed cost is same for both the options and the shipping cost is lower in the multiple warehouses option by $4500 compared to the dedicated warehouse option. 5.11 Solution Let 𝐼1̅ (𝑆𝐶) be the inventory in the SC with just one central warehouse.

(a) Under the current system with 8 warehouses, the total inventory in the supply chain: 𝐼8̅ (𝑆𝐶) = �√8� 𝐼1̅ (𝑆𝐶)

After closing 3 warehouses, the total inventory in the supply chain becomes 𝐼5̅ (𝑆𝐶) = �√5� 𝐼1̅ (𝑆𝐶)

Reduction in the total inventory in the supply chain is

5 𝐼5̅ (𝑆𝐶) = � ≈ 0.79 8 𝐼8̅ (𝑆𝐶)

Hence, the supply chain inventory reduces by 21%. (b) Negative impacts of risk pooling: •

Supply chain risk increases

•

Average delivery time to customers will increase

•

Outbound logistics cost will increase

5.12 Solution a) With one DC: The optimal order quantity 𝑄 ∗ is

2𝐴𝐷 𝑄∗ = � ℎ

where A is the set up cost, D is the annual demand and h is the annual inventory holding cost per unit. 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝐼𝑛𝑣𝑒𝑛𝑡𝑜𝑟𝑦 = 𝐼 ̅ =

which is also the total inventory in the supply chain.

𝑄∗ 2

Under 3 DC option: 𝐷1 =

𝐷 𝐷 , 𝐷2 = 𝐷3 = 2 4

where D 1 , D 2 , D 3 are the annual demands at DC-1, DC-2 and DC-3. Thus, 𝑄∗ 2𝐴(0.5𝐷) 𝑄1∗ = � = �√0.5�𝑄 ∗ = ℎ √2 𝑄2∗ = �√0.25�𝑄 ∗ = 1

𝑄∗ = 𝑄3∗ 2 𝑄∗

Total inventory in the supply chain = 2 [𝑄1∗ + 𝑄2∗ + 𝑄3∗ ] = 2 �

√2

+ 0.5 + 0.5� = 𝐼 ̅ (1.707)

Total inventory in the supply chain increases by more than 70%. (b) Total number of orders: 𝐷

With one DC, number of orders N = 𝑄∗.

With 3 DCs, the total number of orders in the supply chain = 𝑁1 + 𝑁2 + 𝑁3 , where 𝑁1 =

𝐷 (2)

�

𝑄∗

� √2

𝑁

√2

= 0.707𝑁

𝐷 (4) 𝑁2 = ∗ = 0.5 𝑁 = 𝑁3 𝑄 �2�

Thus the total number of orders in the supply chain = N(0.707+0.5+0.5) = N(1.707). In other words, the total number of orders also increases by 70.7%. (c) Frequency of orders decreases by 70.7% in DC-1 and 50% in DC-2 and DC-3 compared to the one DC option. 5.13 Solution (a) Three DC system For service level (SL) = 80%, the safety factor (K SL ) is 0.85 (using Normal tables). Safety stock at DC 1 = SS 1 = K SL σ 1 = (0.85) x 100 = 85 units Safety stock at DC 2 = SS 2 = K SL σ 2 = (0.85) x 50 = 42.5 units Safety stock at DC 3 = SS 3 = K SL σ 3 = (0.85) x 50 = 42.5 units Total safety stock in the SC = 85 + 42.5 + 42.5 = 170 units (b) New option (one DC with SL = 80%) Variance of lead time demand = 1002 + 502 + 502 = 15000 𝜎𝐿𝑇𝐷 = √15000 = 122.474

Safety stock at 80% SL = (0.85)(122.474) ≈ 104 units (c) New option (one DC with SS = 170) 170 = 𝐾𝑆𝐿 (122.474) 𝐾𝑆𝐿 ≅ 1.39

From Normal tables, new SL ≈ 92%.

5.14 Solution (Mighty manufacturing) Scenario A has 4 DCs, while scenario B has 8 DCs, twice the number. Thus, scenario A uses risk pooling and the pros and cons of risk pooling discussed in Section 5.3.3 apply. (a) With respect to “demand fulfillment”, scenario A offers better customer service with a higher fill rate. However, the average delivery time is longer due to fewer DCs. (b) The outbound transportation cost, from DCs to customers will be higher in scenario A. However, the inbound transportation cost from plants to DCs will be lower due to economies of scale. The total transportation cost (inbound + outbound) could be higher or lower depending on the freight rate. (c) Since scenario A carries less inventory in the supply chain, it is in a better position when the forecasts are too high. (d) Scenario B, which carries more inventory in the supply chain, can handle forecasts that are too low. However, a risk pooling strategy (scenario A) can handle demand uncertainty much better compared to scenario B. Without specific details on forecasts and demands, it would not be possible to state which scenario is better. 5.15 Solution (Supplier Selection and Order Allocation Case Study) Variables: X hijk = the quantity of product h delivered to buyer i by supplier j at price level k. Zj =

1, if purchasing product from supplier j 0, otherwise.

D hijk ∈ (0,1)

{ binary variables to control break up points}

Objective Function: Min total costs= Fixed costs + Variable costs

Min Z= 3500𝑍1 + 3600𝑍2 + 180𝑋1111 + 165𝑋1112 + 178𝑋1121 + 166𝑋1122 + 180𝑋1211 + 165𝑋1212 + 178𝑋1221 + 166𝑋1222 + 80𝑋2111 + 70𝑋2112 + 83𝑋2121 + 69𝑋2122 + 80𝑋2211 + 70𝑋2212 + 83𝑋2221 + 69𝑋2222 Constraints:

Break Point constraints: •

For Buyer 1 purchasing Product 1 from Supplier 1. 𝑋1111 ≤ 85𝑑1111 𝑋1111 ≥ 85𝑑1112

•

𝑋1112 ≤ 𝑀𝑑1112 or 𝑋1112 ≤ 65𝑑1112

For Buyer 1 purchasing Product 1 from Supplier 2.

𝑋1121 ≤ 95𝑑1121 𝑋1121 ≥ 95𝑑1122

•

𝑋1122 ≤ 𝑀𝑑1122 or 𝑋1122 ≤ 80𝑑1122

For Buyer 2 purchasing Product 1 from Supplier 1.

𝑋1211 ≤ 85𝑑1211 𝑋1211 ≥ 85𝑑1212

•

𝑋1212 ≤ 𝑀𝑑1212 or 𝑋1212 ≤ 65𝑑1212

For Buyer 2 purchasing Product 1 from Supplier 2.

𝑋1221 ≤ 95𝑑1221 𝑋1221 ≥ 95𝑑1222

•

𝑋1222 ≤ 𝑀𝑑1222 or 𝑋1222 ≤ 80𝑑1222

For Buyer 1 purchasing Product 2 from Supplier 1.

𝑋2111 ≤ 120𝑑2111 𝑋2111 ≥ 120𝑑2112

𝑋2112 ≤ 𝑀𝑑2112 or 𝑋2112 ≤ 50𝑑2112 67

•

For Buyer 1 purchasing Product 2 from Supplier 2. 𝑋2121 ≤ 125𝑑2121 𝑋2121 ≥ 125𝑑2122

•

𝑋2122 ≤ 𝑀𝑑2122 𝑋2122 ≤ 35𝑑2122

For Buyer 2 purchasing Product 2 from Supplier 1.

𝑋2211 ≤ 120𝑑2211 𝑋2211 ≥ 120𝑑2212

•

𝑋2212 ≤ 𝑀𝑑2212 or 𝑋2212 ≤ 50𝑑2212

For Buyer 2 purchasing Product 2 from Supplier 2.

𝑋2221 ≤ 125𝑑2221 𝑋2221 ≥ 125𝑑2222

𝑋2222 ≤ 𝑀𝑑2222 or 𝑋2222 ≤ 35𝑑2222

Product demand constraints: •

•

For Buyer 1, Product 1 should meet a demand of 150 𝑋1111 + 𝑋1112 + 𝑋1121 + 𝑋1122 ≥ 150

For Buyer 2, Product 1 should meet a demand of 175

𝑋1211 + 𝑋1212 + 𝑋1221 + 𝑋1222 ≥ 170

For Buyer 1, Product 2 should meet a demand of 200

𝑋2111 + 𝑋2112 + 𝑋2121 + 𝑋2122 ≥ 200

For Buyer 2, Product 2 should meet a demand of 180

𝑋2211 + 𝑋2212 + 𝑋2221 + 𝑋2222 ≥ 180

Supplier capacity constraints: •

For Supplier 1 the capacity of Product 1 should not exceed 400.

•

𝑋1111 + 𝑋1112 + 𝑋1121 + 𝑋1122 ≤ 400𝑧1

For Supplier 2 the capacity of Product 1 should not exceed 450 68

𝑋1211 + 𝑋1212 + 𝑋1221 + 𝑋1222 ≤ 450𝑧2

•

For Supplier 1 the capacity of Product 2 should not exceed 480

•

𝑋2111 + 𝑋2112 + 𝑋2121 + 𝑋2122 ≤ 480𝑧1

For Supplier 2 the capacity of Product 2 should not exceed 460

Lead time constraints: •

𝑋2211 + 𝑋2212 + 𝑋2221 + 𝑋2222 ≤ 460𝑧2

For Buyer 1, lead time for Product 1 should not exceed 12.5 days.

8(𝑋1111 + 𝑋1112 ) + 17(𝑋1121 + 𝑋1122 ) ≤ 12.5((𝑋1111 + 𝑋1112 + 𝑋1121 + 𝑋1122 ) •

•

For Buyer 2, lead time for Product 1 should not exceed 19 days.

14(𝑋1211 + 𝑋1212 ) + 24(𝑋1221 + 𝑋1222 ) ≤ 19((𝑋1211 + 𝑋1212 + 𝑋1221 + 𝑋1222 )

For Buyer 1, lead time for Product 2 should not exceed 18 days.

28(𝑋2111 + 𝑋2112 ) + 8(𝑋2121 + 𝑋2122 ) ≤ 18((𝑋2111 + 𝑋2112 + 𝑋2121 + 𝑋2122 )

For Buyer 2, lead time for Product 2 should not exceed 14 days.

16(𝑋2211 + 𝑋2212 ) + 12(𝑋2221 + 𝑋2222 ) ≤ 14((𝑋2211 + 𝑋2212 + 𝑋2221 + 𝑋2222 )

Quality Constraints: •

Percentage of rejects for Buyer 1.

0.03(𝑋1111 + 𝑋1112 ) + 0.09(𝑋1121 + 𝑋1122 ) + 0.06(𝑋2111 + 𝑋2112 ) + 0.02(𝑋2121 + 𝑋2122 ) ≤ 0.06(𝑋1111 + 𝑋1112 + 𝑋1121 + 𝑋1122 + 𝑋2111 + 𝑋2112 + 𝑋2121 + 𝑋2122 )

•

Percentage of rejects for Buyer2.

0.03(𝑋1211 + 𝑋1212 ) + 0.09(𝑋1221 + 𝑋1222 ) + 0.06(𝑋2211 + 𝑋2212 ) + 0.02(𝑋2221 + 𝑋2222 ) ≤ 0.04(𝑋1211 + 𝑋1212 + 𝑋1221 + 𝑋1222 + 𝑋2211 + 𝑋2212 + 𝑋2221 + 𝑋2222

X hijk ≥ 0, for all h, i, j, k j, k

z 1 , z 2 : binary variables

Optimal Solution by Excel Solver is given in Table 5.

D hijk : binary variables, for all h, i,

Table 5 Optimal Solution to Problem 5.15

Product Buyer Supplier 1 1 1 1 1 1 1 1 2 1 1 2 1 2 1 1 2 1 1 2 2 1 2 2 2 1 1 2 1 1 2 1 2 2 1 2 2 2 1 2 2 1 2 2 2 2 2 2

Break-Level 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2

Quantity 85 65 0 0 85 65 25 0 40 0 125 35 20 0 125 35

150 0 150 25 40 160 20 160

5.16 Solution (Pip Boys: Risk Pooling Case Study) (a) Decentralized supply network with 5 regional DCs Safety stock in region i is given by 𝑆𝑆𝑖 = 𝐾𝑆𝐿 𝜎𝑖 where 𝐾𝑆𝐿 = safety factor = 1.96 and 𝜎𝑖 is the standard deviation of the demand over lead time for region i. Hence, for the steering wheel, Safety stock for Region 1, SS 1 = (1.96) * 180 = 353 units. Similarly, safety stocks for other regions are: SS 2 = 229; SS 3 = 159 units; SS 4 = 194 units; and SS 5 = 318 units. Thus, total safety stock for steering wheel = 1253 units.

Similarly, we calculate the safety stock for the spark plug: SS 1 = 1019; SS 2 = 1176; SS 3 = 666; SS 4 = 862; SS 5 = 745. Total safety stock for spark plug = 4468 units. (b) Centralized supply chain with one central DC in Omaha Since the regional demands are independent, the variance of the combined demand at the Omaha DC over lead time is given by: V(LTD) = Sum of the variances of the regional lead time demands. (i) For steering wheels 𝑉(𝐿𝑇𝐷) = 1802 + 1172 + 812 + 992 + 1622 = 88695

𝜎𝐿𝑇𝐷 = √88695 = 297.82

Safety Stock = SS = 1.96 ∗ 297.82 = 584 units

(ii) For spark plugs

𝜎𝐿𝑇𝐷 = �5202 + 6002 + 3402 + 4402 + 3802 = 1041.13

Safety Stock = SS = 1.96 ∗ 1041.13 = 2041 units (c) Annual Savings

(i) Reduction in total safety stock Steering wheels = 1253 – 584 = 669 units Spark plugs =4468 – 2041 = 2427 units (ii) Safety stock holding cost Steering wheels = (669)($45)(0.2) = $6021 Spark plugs = (2427)($1)(0.2) = $485 (iii) Cost savings as a percentage of annual material cost: 6021

Steering wheels = (6000)(45) = 2.23%

485

Spark plugs = 230000 = 0.21%

(d) Yes, there is a difference. The savings are much higher for steering wheel. Reason: Steering wheel is a slow-moving item (6000 units vs. 230000 for spark plugs). Its coefficient of variation (ratio of STD to mean) is higher compared to spark plugs. Hence steering wheels generate a much larger risk pooling-driven reduction in safety stock relative to its annual demand (669/6000 = 11%), compared to spark plugs (2427/230000 =1%).

Chapter 6 Supplier Selection Models and Methods Solutions to Numerical Exercises only 6.7 Solution Decision Variables: A m – No. of units of part ‘A’ that are made each week A b - No. of units of part ‘A’ that are bought each week B m - No. of units of part ‘B’ that are made each week B b - No. of units of part ‘B’ that are bought each week S I – Idle time per week Constraints: Demand: A m + A b ≥ 100 (Need 100 units of A) B m + B b ≥ 200 (Need 200 units of B) Capacity: A m ≤ 60 B m ≤ 120 Also, 𝐴𝑚

Supply:

𝐵

+ 5𝑚 + S I = 40 3 B b ≤ 150

Am, Bm, Ab, Bb, S I ≥ 0

Non-negativity: Objective:

Minimize Z = A m + 2B m + 1.2A b + 1.5B b + 3S I Optimal Solution: The LP was solved in Excel Solver and the optimal solution is given in Table 1. Table 1 Optimal Solution (Exercise 6.7)

Decision Variable Am Ab Bm Bb SI

Value at optimum 60 40 100 100 0

6.8 Solution to Part (a) L 1 -norm to scale criteria values and L 1 -metric to rank: L 1 -norm of the TCO criteria = 125000 + 95000 + 65000 = 285000 L 1 -norm of the Service criteria = 10 + 5 + 3 = 18 L 1 -norm of Experience = 9 + 6 + 3 = 18 For scaling using L 1 -norm, each of the criteria values is divided by their respective L 1 norms. The scaled values are shown in Table 2. Table 2 Values scaled using L 1 norm (Exercise 6.8)

TCO

Service

Experience

(Min)

(Max)

Supplier A

0.439

0.555

0.5

Supplier B

0.333

0.278

0.333

Supplier C

0.228

0.167

The ideal values are: 0.228 for TCO, 0.555 for Service and 0.5 for Experience. L 1 metric of Supplier A = |0.439 – 0.228| + |0.555 – 0.555| + |0.5 – 0.5| = 0.211 L 1 metric of Supplier B = |0.333 – 0.228| + |0.278 – 0.555| + |0.333 – 0.5| = 0.549 L 1 metric of Supplier C = |0.228 – 0.228| + |0.167 – 0.555| + |0.167 – 0.5| = 0.721 The final ranking is shown in Table 3. Table 3 Supplier ranking using L 1 metric (Exercise 6.8)

TCO

Service

Experience

(Min)

(Max)

Supplier A

0.439

0.555

Supplier B

0.333

Supplier C Ideal values

L 1 metric

Rank

0.5

0.211

0.278

0.333

0.549

0.228

0.167

0.721

0.228

0.555

0.5

The final ranking is Supplier A > Supplier B > Supplier C.

Solution to Part (b) L 2 -norm to scale criteria values and L 2 -metric to rank: L 2 -norm of the TCO criteria = �(1250002 + 950002 + 650002 ) = 169926.455 L 2 -norm of the Service criteria = �(102 + 52 + 32 ) = 11.576 L 2 -norm of Experience = �(92 + 62 + 32 ) = 11.225

For scaling using L 2 -norm, each of the criteria values is divided by their respective L 2 norms. The scaled values are shown in Table 4.

Table 4 Values scaled using L 2 norm (Exercise 6.8)

TCO

Service

Experience

(Min)

(Max)

Supplier A

0.736

0.864

0.802

Supplier B

0.559

0.432

0.535

Supplier C

0.383

0.259

0.267

L 2 metric of Supplier A = �|0.736 − 0.383|2 + |0.864 − 0.864|2 + |0.802 − 0.802|2 = 0.353

L 2 metric of Supplier B = �|0.559 − 0.383|2 + |0.432 − 0.864|2 + |0.535 − 0.802|2 = 0.537

L 2 metric of Supplier C = �|0.383 − 0.383|2 + |0.259 − 0.864|2 + |0.267 − 0.802|2 = 0.808

The final ranking is shown in Table 5. Table 5 Supplier ranking using L 2 metric (Exercise 6.8)

TCO

Service

Experience

(Min)

(Max)

Supplier A

0.736

0.864

Supplier B

0.559

Supplier C Ideal values

L 2 metric

Rank

0.802

0.353

0.432

0.535

0.383

0.259

0.267

0.802

0.383

0.864

0.802

Here too, the final ranking is Supplier A > Supplier B > Supplier C. Thus there are no rank reversals, including the ones in Section 6.3.6.

6.9

Solution a) A pair-wise comparison matrix is constructed as shown below. C1

Total Score

Rank

All the diagonal elements of the matrix are 1. According to this DM, C 2 > C 1 . Therefore the second element of the first row is 0. Similarly, C 1 > C 5 . Therefore the fifth element of the first row is 1. The entire matrix is constructed this way. The row sums are computed. The criterion with the highest score is given the first rank and so on. b) The score for the first criteria is computed as shown. Total score for criteria C 1 = (9 x 5) + (4 x 4) + (4 x 3) + (7 x 2) + (6 x 1) = 93 The scores for all the criteria are calculated in a similar manner. Weight for a criterion is got by dividing the score of that criterion by the total score (= 450). The criterion with the highest weight gets the first rank. The calculations are shown in the following table. There are rank reversals between section (a) and (b). The first three rankings differ while the last 2 are the same.

Rank Criteria

5 Score

(5 pts) (4 pts) (3 pts) (2 pts)

Weights Rank

Rank in 6.9 (a)

(1 pt)

0.207

0.215

0.209

103

0.229

0.14

450 6.10 Solution Scaling the data using the Ideal value method: The ideal value for Price criterion = $140,000 The ideal value for Size criterion = 3600 The ideal value for Quality criterion = 85 The scaled values are shown in Table 6. Table 6 Criteria values scaled using ideal value (Exercise 6.10)

Price

Size

Quality

0.778

0.882

0.875

0.889

0.772

0.941

0.737

0.765

Use the L ∞ metric method to rank the four suppliers: L ∞ metric of Supplier A = Max (|0.778 – 1|, |0.778 – 1|, |0.882 – 1|) = 0.222 L ∞ metric of Supplier B = Max (|0.875 – 1|, |0.889 – 1|, |1 – 1|) = 0.125 L ∞ metric of Supplier C = Max (|1 – 1|, |0.772 – 1|, |0.941 – 1|) = 0.278 L ∞ metric of Supplier D = Max (|0.737 – 1|, |1 – 1|, |0.765 – 1|) = 0.263 The final ranking is shown in Table 7. Table 7 Supplier ranking using L ∞ metric (Exercise 6.10)

Price

Size

Quality

L ∞ metric

Rank

0.778

0.882

0.222

0.875

0.889

0.125

0.772

0.941

0.278

0.737

0.765

0.263

Ideal

The final supplier ranking is Supplier B > Supplier A > Supplier D > Supplier C.

6.11 Solution Decision Variables: x 11 – No. of units bought from supplier 1 at level 1 x 12 – No. of units bought from supplier 1 at level 2 x 2 – No. of units bought from supplier 2 x 31 – No. of units bought from supplier 3 at level 1 x 32 – No. of units bought from supplier 3 at level 2 δ 2 = 1 if supplier 2 is selected; 0 otherwise 79

β ij =

if purchase is made from supplier ‘i’ at level ‘j’

otherwise

i = 1, 3 and j = 1, 2. Constraints: Capacity: The maximum that supplier 2 can supply is 800 units. x 2 ≤ 800 δ 2 Demand: 2000 units of the product are required. x 11 + x 12 + x 2 + x 31 + x 32 ≥ 2000 Quantity Discount for supplier 1: 0 ≤ x 11 ≤ 300 β 11 x 11 ≥ 300 β 12 0 ≤ x 12 ≤ 300 β 12 Quantity Discount for supplier 3: 0 ≤ x 31 ≤ 500 β 31 x 31 ≥ 500 β 32 0 ≤ x 32 ≤ 700 β 32 Non-negativity and integer constraints: x 11 , x 12 ,x 2 ,x 31 ,x 32 ≥ 0 δ 2 ∈ (0,1)

Objective function:

β 11 , β 12 , β 31 , β 32 ∈ (0,1)

Minimize Z = 100 β 11 + 500 δ 2 + 300 β 31 + 10x 11 + 7x 12 + 2x 2 + 6x 31 + 4x 32

6.12 Solution to Case Study-3 (Supplier Ranking): a. Scaling the supplier criteria values using Linear Normalization (Simple Linearization): All supplier data are scaled to be able to compare them meaningfully. Simple linearization uses the following formulation: H j = Max f ij L j = Min f ij where f ij is the value assigned to alternative i for criterion j R ij = Range for criterion j = H j - Lj The scaled criteria value of f ij denoted by r ij is given by 𝑓𝑖𝑗 − 𝐿𝑗

for a larger-the-better criterion

𝐻𝑗 − 𝑓𝑖𝑗

for a smaller-the better criterion

r ij =

𝑅𝑗

Simple linearization converts minimization criteria to maximization. Table 8 gives the scaled values. b. Applying the L 2 metric method to rank the suppliers: The scaled values are used to rank the suppliers using L 2 metric, given by the formula, 2

||L 2 || = �∑𝑛𝑗=1( |𝑥𝑗 − 𝑦𝑗 | )2 The best supplier is Supplier 7 with a L2 metric score of 2.17 and the worst is Supplier 21 with a score of 7.00. Supplier 21 is the worst with regard to all individual criteria. Ranks of other suppliers are as shown in Table 9.

Table 8 Scaled values using simple linearization (Exercise 6.12)

Supplier Price

C pk

Defective Flexibility

Service

Distance

Lead time

0.78 0.11 0.89 0.56 1.00 0.56 0.44 0.33 0.89 0.33 0.22 0.44 0.11 0.22 0.33 0.33 0.00 0.44 0.67 0.11 0.00 1

0.10 1.00 0.00 0.15 0.29 0.57 0.43 0.57 0.15 0.36 0.00 0.15 0.43 0.27 0.43 0.19 0.36 0.10 0.00 0.36 0.00 1

0.33 1.00 0.00 0.58 0.86 0.99 0.96 0.99 0.58 0.92 0.00 0.58 0.96 0.86 0.96 0.72 0.92 0.33 0.00 0.92 0.00 1

0.50 1.00 0.30 0.70 0.80 0.98 1.00 0.00 0.60 0.70 0.50 0.60 0.70 0.74 0.72 0.30 0.40 0.54 0.78 0.70 0.00 1

0.97 0.91 1.00 0.71 0.46 0.59 1.00 0.14 0.57 0.29 0.92 0.62 0.71 0.09 0.03 0.89 0.90 0.90 0.86 0.29 0.00 1

0.92 0.74 0.95 0.63 0.55 0.79 1.00 0.26 0.13 0.00 0.39 0.45 0.53 0.58 0.55 0.26 0.29 0.58 0.66 0.26 0.00 1

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 Ideal

0.40 0.00 1.00 0.60 0.72 0.20 0.00 0.00 0.20 0.40 0.60 0.00 0.00 0.08 0.20 0.00 0.20 0.00 0.40 0.40 0.00 1

c. Rating method: V

The weight for criterion ‘j’ is given by w j = ∑𝑝 j V . This implies that ∑𝑗 w j = 1. 𝑗=1

Sub-criteria of Quality and Delivery are handled separately. Weights are obtained for the first level in the hierarchy of criteria without considering the sub-criteria. Each set of subcriteria is then rated separately. The final weights of the sub-criteria are obtained by multiplying its weight with its corresponding upper level criteria weight. Thus, the criteria weight is split among its sub-criteria.

The results of the rating method are given in Table 10. Quality and Delivery will not have final weights as they are split among their corresponding sub-criteria. Table 9 Ranking of Suppliers using L 2 metric (Exercise 6.12)

Supplier Price

C pk

Defective Flexibility

Service

Distance

Lead time L2 Metric Rank

0.78 0.11 0.89 0.56 1.00 0.56 0.44 0.33 0.89 0.33 0.22 0.44 0.11 0.22 0.33 0.33 0.00 0.44 0.67 0.11 0.00 1

0.10 1.00 0.00 0.15 0.29 0.57 0.43 0.57 0.15 0.36 0.00 0.15 0.43 0.27 0.43 0.19 0.36 0.10 0.00 0.36 0.00 1

0.33 1.00 0.00 0.58 0.86 0.99 0.96 0.99 0.58 0.92 0.00 0.58 0.96 0.86 0.96 0.72 0.92 0.33 0.00 0.92 0.00 1

0.50 1.00 0.30 0.70 0.80 0.98 1.00 0.00 0.60 0.70 0.50 0.60 0.70 0.74 0.72 0.30 0.40 0.54 0.78 0.70 0.00 1

0.97 0.91 1.00 0.71 0.46 0.59 1.00 0.14 0.57 0.29 0.92 0.62 0.71 0.09 0.03 0.89 0.90 0.90 0.86 0.29 0.00 1

0.92 0.74 0.95 0.63 0.55 0.79 1.00 0.26 0.13 0.00 0.39 0.45 0.53 0.58 0.55 0.26 0.29 0.58 0.66 0.26 0.00 1

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 Ideal

0.40 0.00 1.00 0.60 0.72 0.20 0.00 0.00 0.20 0.40 0.60 0.00 0.00 0.08 0.20 0.00 0.20 0.00 0.40 0.40 0.00 1

1.38 1.37 1.58 1.25 1.07 1.11 1.28 1.98 1.63 1.68 1.84 1.68 1.59 1.75 1.62 1.80 1.71 1.72 1.63 1.65 2.65

6 5 7 3 1 2 4 20 10 13 19 14 8 17 9 18 15 16 11 12 21

Table 10 Ranking criteria using the Rating method (Exercise 6.12)

Quality

Rating Criteria Weights Final Weights Criteria Rank

Defective Flexibility Service

Deliver y

Delivery Distanc Lead e time

Price

Quality

C pk

0.241

0.310

0.5

0.138

0.103

0.207

0.615

0.385

0.241

0.155

0.138

0.103

0.127

0.08

Pair wise comparison of criteria/sub-criteria and Borda Count: Pair-wise comparison of all criteria is carried out first ignoring the sub-criteria as shown in Table 11. Table 11 Pair-wise comparison of criteria (Exercise 6.12)

Quality Price Quality C pk Price Quality C pk Quality Defective Flexibility Service Delivery Distance Delivery Lead time

1 1

0 1 1 1

0 0 0

Delivery Lead Defective Flexibility Service Delivery Distance time 1 1 1 1 1 1 1 1 1 1 0 0 1 0 1 1 1 1 1 0 1

Criteria are ranked using the results of this comparison and assigned weights by the Borda Count method. The weights of the criteria are given in Table 12. 84

Table 12 Borda Count and Criteria Weights (Exercise 6.12)

Criteria Sum Price Quality C pk Quality Defective Flexibility Service Delivery Distance Delivery Lead time

4 5

Borda Count Rank 2 1

Subcriteria Sum

2 2 2 1 3

4 5 3 2 1

Weights 0.267 0.333 0.5 0.5 0.133 0.067 0.2 0.667 0.333

Final Weights 0.267 0.1675 0.1675 0.133 0.067 0.133 0.067

AHP: The criteria and sub-criteria level comparison is given below in Table 13. Table 13 AHP Pair-wise comparison matrix (Exercise 6.12)

Quality

Delivery

Price Quality C pk Defective Flexibility Service Delivery Distance Lead time

Price

Quality C pk

1 2

0.5 1 1 1

0.167 0.125 0.25

Delivery

Defective Flexibility Service

Delivery

6 7

8 9

4 4

1 0.333 5

3 1 6

0.2 0.167 1

Dist ance

Lead time

1 0.5

2 1

1 1

0.143 0.111 0.25

The first level of the hierarchy alone is considered first and its normalized A matrix (Table 14) is computed to determine the individual criteria weights.

Table 14 Normalized pair-wise comparison matrix (Exercise 6.12)

Price Quality Flexibility Service Delivery Sum

Price

Quality

Flexibility

Service

Delivery

0.282 0.565 0.047 0.035 0.071 1

0.250 0.499 0.071 0.055 0.125 1

0.310 0.362 0.052 0.017 0.259 1

0.296 0.333 0.111 0.037 0.222 1

0.427 0.427 0.021 0.018 0.107 1

Average (Weights) 0.313 0.437 0.061 0.033 0.157

The above weights and the original A matrix are then used to perform the consistency for the first level of hierarchy. AW = λ max W 1

1 2 ⎛2 1 ⎜1 1 AW = ⎜ 6 7 1 1 ⎜ 8 1

⎝4

6 8 4 7 9 4⎞ 1 1 3 5⎟ ⎟ 1 1 1 6⎟ 3

9 1

5 6 1⎠

1.781621

0.3131091 ⎛ 0.4372329 ⎞ ⎜ 0.0605067⎟ 0.0325623 ⎝ 0.156589 ⎠

2.406415

0.304158

1.781621 ⎛2.406415⎞ = ⎜0.304158⎟ 0.166549 ⎝0.842082⎠ 0.166549

0.842082

λ max = Average �0.3131091 , 0.4372329 , 0.0605067 , 0.0325623 , 0.156589 � = 5.333868466

CI =

(λmax – n ) (n – 1)

(5.333868466−5)

RI for n = 5 is 1.11

(5 −1)

= 0.08347

𝐶𝐼

CR = 𝑅𝐼 = 0.0752 < 0.1

The consistency check results are summarized in Table 15.

Table 15 Consistency check results of the pair-wise comparison matrix (Exercise 6.12)

λ max Consistency Index Random Index

5.333868 0.083467

Consistency Ratio

0.075196

1.11

The Consistency Ratio obtained is 0.075 which is below the threshold 0.1. Hence we have been acceptably consistent in assigning weights to the various criteria. Consistency check for the sub-criteria level is not required as there are only two criteria and there is no scope for inconsistency to arise. The final criteria weights computed using AHP are shown in Table 16. Table 16 Criteria weights using AHP (Exercise 6.12)

Criteria/Subcriteria Price Cpk Defective Flexibility Service Distance Lead time

Weights 0.313 0.219 0.219 0.061 0.033 0.104 0.052

c. Using the weights obtained in part (c) and the scaled supplier data in part (a), to rank the suppliers: Supplier ranking is computed using the scaled values calculated in part (a) and the weights in part (c) under each method. The results of all three methods are summarized in Table 17.

Table 17 Ranking using different methods (Exercise 6.12)

Supplier

L2 Metric

Rank

Rating Score

Rating Rank

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

1.38 1.37 1.58 1.25 1.07 1.11 1.28 1.98 1.63 1.68 1.84 1.68 1.59 1.75 1.62 1.80 1.71 1.72 1.63 1.65 2.65

6 5 7 3 1 2 4 20 10 13 19 14 8 17 9 18 15 16 11 12 21

0.560 0.616 0.586 0.543 0.704 0.644 0.633 0.363 0.500 0.444 0.337 0.396 0.448 0.374 0.446 0.387 0.406 0.392 0.458 0.411 0.000

6 4 5 7 1 2 3 19 8 12 20 15 10 18 11 17 14 16 9 13 21

Borda Count Score 0.558 0.601 0.587 0.533 0.705 0.632 0.617 0.387 0.510 0.441 0.322 0.392 0.438 0.358 0.436 0.397 0.406 0.386 0.441 0.399 0.000

Borda Count Rank 6 4 5 7 1 2 3 17 8 10 20 16 11 19 12 15 13 18 9 14 21

AHP Score

AHP Rank

0.529 0.639 0.502 0.499 0.711 0.663 0.632 0.475 0.535 0.462 0.239 0.405 0.464 0.385 0.476 0.420 0.415 0.376 0.382 0.406 0.000

6 3 7 8 1 2 4 10 5 12 20 16 11 17 9 13 14 19 18 15 21

d. Comparison of rankings: Supplier 5 has been consistently ranked as the best through all the methods. There are minor differences in supplier rankings under different methods. This is because of the involvement of the decision maker and the varying methodology giving rise to different weights for the criteria/sub-criteria in different methods. The only glaring exception is the rank reversal of supplier 8 and 19 under the AHP method.

6.13 Solution to Case Study-4 (Supplier Selection and Order Allocation) a. Formulation: Variables: Table 18 Number of variables for the MILP problem

Variable X ijk

Description Type No of units of product ‘i’ sent Continuous positive from supplier ‘k’ to buyer ‘j’ integer

δ1

Includes fixed cost when orders Binary integer are allocated to supplier 1 Includes fixed cost when orders Binary integer are allocated to supplier 1

δ2

Number 2 x 2 x 2 = 8 (2 products, 2 buyers and 2 suppliers) 1 1 Total = 10

Constraints: Demand constraint: The total units received by buyer ‘j’ for product ‘i’ from both the suppliers must be enough to meet the demands. For instance, consider buyer 1 (j=1) product 1 (i=1). The demand is 150. 2

� 𝑋11𝑘 = 150

𝑘=1

Similarly,

Suppliers’ capacity constraints:

𝑋111 + 𝑋112 = 150

(1)

𝑋121 + 𝑋122 = 175

(2)

𝑋211 + 𝑋212 = 200

(3)

𝑋221 + 𝑋222 = 180

(4)

𝑋111 + 𝑋121 ≤ 300 𝛿1

(5)

𝑋111 + 𝑋122 ≤ 175 𝛿2

(6)

𝑋211 + 𝑋221 ≤ 280 𝛿1

(7)

𝑋212 + 𝑋222 ≤ 175 𝛿2

(8)

Here δ 1 and δ 2 are binary integer variables. When supplier 1 is used, δ 1 is forced to be 1 and when not selected, δ 1 = 0. Non-negativity constraints: Units of products are non-negative integer quantities. 𝑋𝑖𝑗𝑘 ∈ Z*, the set of all non-negative integers.

But for all practical purposes and computational simplicity, it can be assumed that the X ijk ’s are continuous, non-negative variables. Thus, 𝑋𝑖𝑗𝑘 ≥ 0

Objective functions: 1. Cost criteria:

(9)

Fixed-cost terms

Minimize total cost Min Z 1 = 1200 δ 1 + 1325 δ 2 + 190 X 111 + 175 X 112 + 194 X 121 + 200 X 122 + 335 X 211 +

360

X 212

(10)

370

X 221

365

X 222

Variable-cost terms

2. Lead time criteria: Minimize weighted average lead time 𝑀𝑖𝑛 𝑍2 =

15 X111 + 17 X112 + 19 X121 + 18 X122 + 24X211 + 21X212 + 11X221 + 12 X222 705 (11)

3. Quality criteria: Minimize weighted average quality 𝑀𝑖𝑛 𝑍3 =

0.07 (X111 + X121 ) + 0.09 (X112 + X122 ) + 0.06 (X211 + X221 )+ 0.05 (X212 + X222 ) 705 (12)

Thus for the three criteria MILP problem, Equations (1) – (9) are constraints (Nine constraints) and Equations (10) – (12) are objective functions. b. Obtaining the Ideal Values: Three single-objective problems (with objectives shown in Equations (10), (11) and (12)) were solved separately with the same set of constraints (Equations (1) – (9)). The results are as shown in Table 19. Table 19 Results of the three single-objective problems (Exercise 6.13)

Cost

Lead time

Quality

Minimize Cost

195425

18.205674

0.0663121

Minimize Lead Time

204625

16.425532

0.0667376

Minimize Quality

201800

17.085106

0.0603128

Ideal Value

195425

16.425532

0.0603128

The bounds on the criteria are as follows: 195425 ≤ Cost ≤ 204625 16.425532 ≤ Lead time ≤ 18.205674 0.0603128 ≤ Quality ≤ 0.0667376

Goal Programming Models using the four approaches: The target values are given in Table 20. Table 20 Ideal values and targets for the three criteria

Criteria

Ideal Value

Target = 1.1 * Ideal value

Cost

195425

214967.5

Lead time

16.425532

18.0680852

Quality

0.0603128

0.06634408

i. Pre-emptive Goal Programming: Cost as a goal constraint: 1200 δ 1 + 1325 δ 2 + 190 X 111 + 175 X 112 + 194 X 121 + 200 X 122 + 335 X 211 + 360 X 212 + 370 X 221

365

X 222

d1-

d1+

214967.5

(13) Lead time as a goal constraint: 15 X111 + 17 X112 + 19 X121 + 18 X122 + 24X211 + 21X212 + 11X221 + 12 X222 + d2− 705 − d2+ = 18.0680852

(14)

Quality as a goal constraint: 0.07 (X111 + X121 ) + 0.09 (X112 + X122 ) + 0.06 (X211 + X221 )+ 0.05 (X212 + X222 ) + d2− 705 − d2+ = 0.6634408

(15)

Real constraints include constraints shown in Equations (1) – (9) and non-negativity of deviational variables. Goal constraints are Equations (13) – (15). 92

The objective function is Minimize Z = P 1 d 1 + + P 2 d 2 + + P 3 d 3 + (16) The problem was solved in three stages using Excel Solver. The solution is as shown in Table 21. Table 21 Solution to the Preemptive GP (Exercise 6.13)

Priority

Criteria

Values

di+

di-

Cost

214967.5

Weighted average lead time 18.068085

Weighted average quality

0.0662341

The objective function resulted in a value of zero, indicating there is actually scope for improvement on the objective values obtained. ii. Non-Preemptive Goal Programming: The goal constraints are scaled by dividing the coefficients of the terms by the corresponding target values. Scaled cost constraint: 1200

1325

190

175

194

200

335

δ 1 + 214967.5 δ 2 + 214967.5 X 111 + 214967.5 X 112 + 214967.5 X 121 + 214967.5 X 122 + 214967.5 214967.5 360

370

365

214967.5

X 211 + 214967.5X 212 + 214967.5 X 221 + 214967.5 X 222 + d 1 - - d 1 + = 214967.5

5.58 x 10-3 δ 1 + 6.16 x 10-3δ 2 + 8.84 x 10-4 X 111 + 8.14 x 10-4 X 112 + 9.02 x 10-4 X 121 + 9.30 x 10-4 X 122 + 1.55 x 10-3 X 211 + 1.67 x 10-3 X 212 + 1.72 x 10-3 X 221 + 1.7 x 10-3 X 222 + d 1 - - d 1 + = 1 (17)

Similar scaling is done for the other two goal constraints. Scaled lead time constraint 15

X 111 + 18.0680852 x 705 X 112 + 18.0680852 x 705 X 121 + 18.0680852 x 705 X 122 +

18.0680852 x 705 24

X 211 + 18.0680852 x 705 X 212 + 18.0680852 x 705 X 221 + 18.0680852 x 705 X 222

18.0680852 x 705 18.0680852

18.0680852

1.18 x 10-3 X 111 + 1.33 x 10-3 X 112 + 1.49 x 10-3 X 121 + 1.41 x 10-3 X 122 + 1.88 x 10-3 X 211 + 1.65

10-3

X 212

8.64

10-4

X 221

9.42

10-4

X 222

(18)

Scaled quality constraint 0.07

0.09

0.06

(X 111 + X 121 ) + 0.06634408 x 705 (X 112 + X 122 ) + 0.06634408 x 705 (X 211 + X 221 ) + 0.06634408 x 705 0.05

0.06634408

0.06634408 x 705

(X 212 + X 222 ) + d 3 - - d 3 + = 0.06634408

(X 212

1.497 x 10-3 (X 111 + X 121 ) + 1.924 x 10-3 (X 112 + X 122 ) + 1.28 x 10-3 (X 211 + X 221 ) + 1.07 x 10-3 X 222 )

d3-

d3+

(19) The objective function is Minimize Z = w 1 d 1 + + w 2 d 2 + + w 3 d 3 + = 0.5 d 1 + + 0.3 d 2 + + 0.2 d 3 + (20) Real constraints are Equations (1) – (9) and non-negativity of deviational variables. Goal constraints are Equations (17) – (19). The problem was solved in excel solver. Solution is as shown below in Table 22. 94

Table 22 Solution of the Non-Preemptive GP

Criteria

Value

di+

di-

Cost

214967.5

Weighted average lead time

18.068085

Weighted average quality

0.0662341

All the targets have been satisfied (d i +’s = zero) and the objective function has value equal to zero. iii. Min-Max Goal Programming The linearized formulation of the Min-Max goal programming is as follows. M = Max (d 1 +, d 2 +, d 3 +) This is brought about by the constraints: M ≥ d 1 +; M ≥ d 2 +; M ≥ d 3 +; (21) Apart from Equation (21), the other constraints include the scaled goal constraints in Equations (17) – (19) and the real constraints in Equations (1) – (9) and non-negativity of deviational variables. The objective function is to Minimize M

(22)

The solution is shown in Table 23. Table 23 Solution for Tchebycheff’s min-max goal programming (Exercise 6.13)

Criteria

Value

di+

di-

Cost

214967.5

Weighted average lead time

18.068085

Weighted average quality

0.0662341

All the targets have been satisfied and all the deviational variables take the value zero. Hence the objective function has value zero. iv. Fuzzy Goal Programming (use the bounds obtained in part (b)) 95

If U 1 is the upper bound and L 1 is the lower bound for the cost objective, from the bounds obtained in part (b), we know that U 1 = 204625

L1 = 195425

Similarly for the other 2 objectives the upper and lower bounds are defined as, U 2 = 18.205674

L2 = 16.425532

U 3 = 0.0667376

L3 = 0.0603128

The three objective functions Z 1 , Z 2 , and Z 3 (Equations (10) – (12)) are minimization functions. Hence the lower bound is the ideal value in each case. For each of the objectives, the quantity 𝑍 − 𝐿𝑖

� 𝑖

𝑈𝑖 − 𝐿𝑖

� is computed. The MILP objective is to minimize the maximum of these quantities. The

objectives are automatically scaled in this expression. Therefore the fuzzy goal programming objective is, 𝑍 − 195425

𝑍 − 16.425532

𝑍 − 0.0603128

1 2 1 Min Max��204625− � , �18.205674− � , �0.0667376− �� (23) 195425 16.425532 0.0603128

The problem is formulated in a similar fashion to Tchebycheff’s model where M takes the maximum of the three values. Minimize M 𝑍 − 195425

𝑍 − 16.425532

(24) 𝑍 − 0.0603128

1 2 1 Where M = Max ��204625− � , �18.205674− � , �0.0667376− �� 195425 16.425532 0.0603128

This is achieved through the following constraints:

𝑍 − 195425

(25)

𝑍 − 16.425532

(26)

𝑍 − 0.0603128

(27)

1 M ≥ �204625− � 195425

2 M ≥ �18.205674− � 16.425532 1 M ≥ �0.0667376− � 0.0603128

M≥0

(28)

Thus the objective function for the fuzzy GP is (24) and the constraints are Equations (25) – (28) and Equations (1) – (9). The solution for the fuzzy GP is shown in Table 24.

Table 24 Solution to the Fuzzy GP problem

Criteria

Value

Fuzzy Objective Value

Cost

199998.814

0.4971537

Weighted Average Lead Time

17.310536

0.4971537

Weighted Average Quality

0.06345661

0.4971537

d. Comparison of the four GP solutions using the Value Path Approach: Table 25 shows the values for cost, weighted average quality and weighted average lead time obtained from the various methods. Table 25 Solutions from different GP methods (Exercise 6.13)

Fuzzy Total Cost Lead time Quality

Tchebycheff

Preemptive

199998.8

214967.5

Nonpreemptive 214967.5

17.31054

18.06808511

18.068085

17.31054

0.063457

0.066234043

0.06623408

0.0662341

0.063457

Best value 199998.8

In order to compare the solutions using the value path approach, the values have to be scaled. Scaling is done by dividing the best value by the actual criterion value so that the best criterion value will be scaled to one and the others will have values less than one. The scaled values are shown in Table 26.

Table 26 Solutions from different GP methods – Scaled (Exercise 6.13)

Total Cost Lead time Quality

Fuzzy

Tchebycheff

Preemptive

0.930367679

0.93036768

Nonpreemptive 0.9303677

0.958072531

0.95807254

0.9580725

0.958066324

0.95806578

0.9580658

The value path graph is shown in Figure 1.

Comparison using Value Path approach 1.02 1 0.98

Fuzzy

0.96

Tchebycheff

0.94

Preemptive

0.92

Non preemptive

0.9 0.88 Total Cost

Lead time

Quality

Figure 1 Comparison of different GP solutions using Value path approach

The graphs indicate that the solution obtained using Fuzzy goal programming is the nondominated solution. The other three solutions are dominated by Fuzzy. They all have the same values (target values) and hence over lap each other. The Fuzzy GP’s solution dominates all others.

Chapter 7 Managing Risks in Supply Chain Solutions to Numerical Exercises only 7.13 Solution (a) From Table 7.9, we note the following: Prob(Loss ≤ 2,800,572) = 0.93 Prob(Loss ≤ 3,235,638) = 0.94 Hence, probability that the annual loss doesn’t exceed $3 million ≈ 0.935 (93.5%) (b) Maximum loss at 80% confidence level = $2,006,971 (c) Probability that the loss exceeds $4 million < 3% (d) Probability that the loss is between $2 and $3 million = Prob (Loss ≤ $3M) – Prob (Loss ≤ 2M) =0.935 – 0.8 =0.135 (13.5%)

7.14 Solution (a) From Equation 7.14, cost of missing delivery window for α between 48.41% and 84.58% equals 𝑅(𝛼) = 4000(tan(𝛼 − 0.5) 𝜋 + 0.1)2 + 5000

Hence, for α = 0.8, we get R(α) = $13,718.81

100

(b) Solve the following equation for α. 4000(tan(𝛼 − 0.5) 𝜋 + 0.1)2 + 5000 = 12,000 𝛼 = 0.7818

Hence, the probability that the cost ≤ 12000 = α = 0.7818 (78.18%) (c) Solve the following equation for α. 4000(tan(𝛼 − 0.5) 𝜋 + 0.1)2 + 5000 = 18,000 𝛼 = 0.831

Hence the probability that the cost exceeds $18,000 = 1 – α = 0.169 (16.9%) 7.15 Solution To determine the supplier rankings, the weighted score of each supplier is obtained by multiplying their attributes (row elements of Table 7.17) by their respective weights (column elements of Table 7.16). Note that the attributes have been scaled so that they all have to be maximized. (a) For the Rating method, the weights given under the ‘Rating’ column of Table 7.16 are used to compute the weighted score of each supplier. For example, for supplier 1, the weighted score S 1 is given by: S 1 = (0.059*80) + (0.052*40)+ (0.059*70) +...+(0.066*80) + (0.047*60) + (0.151*45) = 52.467 The suppliers are then ranked using the weighted scores.

101

\ The weighted scores of the 20 suppliers and their rankings, using the Rating Method, are given in Table 1: Table 1 Supplier Ranking Using Weights from Rating Method

Supplier 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Score 52.467 58.002 50.847 61.665 56.885 47.175 52.974 64.053 52.898 58.522 59.52 64.413 46.441 56.522 52.197 42.863 61.595 57.331 72.751 42.858

Rank 14 8 16 4 10 17 12 3 13 7 6 2 18 11 15 19 5 9 1 20

b) Using the Borda Count weights (given in Table 7.16), the weighted scores of the suppliers are calculated. The supplier scores and their rankings, using the Borda Count method are shown in Table 2.

102

Table 2 Supplier Ranking Using Weights from Borda Count Method

Supplier 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Score 50.814 61.258 51.674 63.059 55.08 48.66 52.5 63.465 55.394 56.995 59.62 65.396 44.092 59.393 48.209 48.645 59.603 54.496 74.224 38.919

Rank 15 5 14 4 11 16 13 3 10 9 6 2 19 8 18 17 7 12 1 20

c) For AHP, the weights under the ‘AHP’ column (Table 7.16) are used to calculate the weighted scores. The supplier rankings, based on AHP, are shown in Table 3.

103

Table 3 Supplier Ranking Using Weights from AHP

Supplier 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Score 49.415 53.746 54.88 57.093 51.785 46.778 55.054 55.77 49.689 54.028 56.323 57.832 48.466 55.166 50.172 44.864 54.934 59.108 63.84 46.8

Rank 16 12 10 4 13 19 8 6 15 11 5 3 17 7 14 20 9 2 1 18

d) In all the three rankings, supplier 19 is always ranked first. Suppliers 4 and 12 are always in the top 5. Supplier 8 is almost in top 5. Hence, as minimum, suppliers 19, 4, 12 and 8 should be pre-qualified and included in the ‘short list’ of suppliers.

104

Chapter 8 Global Supply Chain Management Solutions to Numerical Exercises only 8.8 a) Criteria weights by rating method:

Criteria

Ratings

Weights

10 = 27% 37

24%

16%

Total

100%

16%

By Borda Count the criterion ranked first (GP) gets 5 points, ranked second (DF) gets 4 points and so on. Thus, the weights by Borda Count are: Criteria

Ratings

5 = 33% 15

27%

13%

Total

100%

105

Weights

20%

8.9 a) Following the discussion of AHP in Chapter 6, Section 6.3.6, we compute the criteria weights using the strength of preference information given in Table 8.11 as follows: Pair-wise comparison Matrix (A)

1 6

1 7 1

1 7

1 5

1 3

Sum

1.9

14.5

6.54

1 5

1 3

1 5

16.2

Normalized Pair-wise comparison Matrix (A norm )

Weights

0.53

0.41

0.76

0.19

0.26

43%

0.09

0.07

0.02

0.43

0.16

15%

0.11

0.48

0.15

0.31

0.26

26%

0.18

0.01

0.03

0.06

0.26

11%

0.18

0.01

0.03

0.06

0.26

Note: The criteria weights are obtained by averaging across each row of the A norm matrix.

106

(b) To check the consistency of the pair-wise comparison matrix A, compute the vector X = Aw where w is the vector of criteria weights. 3.21 ⎡1.15⎤ ⎢ ⎥ 𝐗 = 𝐴𝑤 = ⎢2.19⎥ ⎢0.55⎥ ⎣0.26⎦

Then λmax = Average �

3.21 1.15 2.19 0.55 0.26 , , , , � = 6.824 . 43 . 15 . 26 . 11 . 04

The consistency index CI is given by CI =

λmax − 5 6.824 − 5 = = 0.456 5−1 4

For n = 5, the random index (RI) = 1.11

Hence, the consistency ratio CR is given by CR =

CI = 0.411 > 0.1 1.11

Hence the matrix is not consistent. c)

Comparison of criteria weights by Rating, Borda count and AHP. Weights (%) Criteria Rating Method

Borda Count

AHP

107

Comments: •

For all 3 methods, GP is ranked first, followed by DF criterion.

•

Rating method is not able to distinguish among the remaining three criteria – LT, RD and SF. They are all tied for the third place. This is a drawback of the simple rating method. It usually generates several ties among criteria.

•

The top 3 rankings (GP, DF, LT) are the same by both Borda Count and AHP. There is a rank reversal between RD and SF criteria. Such rank reversals are common when using different methods for ranking.

8.10

a) By the Averaging method, the group weight for each criterion is obtained by averaging its weights from the eleven DMs. For example, for criterion GP, Average Weight =

22 + 36 + 22 … … … … . . +44 + 33 + 33 = 34% 11

Similarly, the average weights for the criteria are LT = 15%, DF = 21%, RD = 10%, SF = 19% Thus the group rankings are GP > DF > SF > LT > RD. b) Using the criteria weights (Table 8.12) determine the ranking of the 5 criteria by each DM as given below Criteria weights (%) GP

DM-1

DM-2

DM-3

DM-4

108

DM-5

DM-6

DM-7

DM-8

DM-9

DM-10

DM-11

Thus, GP has 10 first place votes and one second place vote. Its Borda Count points are = 10(5) + (1) 4 = 54 Similarly for the others: LT = 1(4) + 6(3) + 3(2) + 1(1) = 29 DF = 7(4) + 1(3) + 3(2) = 37 RD = 1(3) + 2(2) + 8(1) = 15 SF= 1(5) + 2(4) + 3(3) + 3(2) + 2(1) = 30 The group weights are computed as follows: Weight for GP =

54 54 = = 0.327 54 + 29 + 37 + 15 + 30 165

Weight for LT =

Weight for DF = Weight for RD = 109

29 = 0.176 165

37 = 0.224 165

15 = 0.091 165

Weight for SF =

30 = 0.182 165

The group rankings are GP > DF > SF > LT > RD. c)

Even though the actual weights are different, the relative rankings are the same.

8.11

a) Index set for decision variables: i – set of suppliers (i = 1, 2, 3) j – set of plants (j = 1, 2, 3) k – set of warehouse sites (k = 1, 2, 3, 4) c – set of customer zones (c = 1, 2, 3) t – time periods ( t = 1, 2, 3,…, 15) Decision Variables: 𝑤𝑖𝑗𝑡 – Quantity of raw material shipped from supplier i to plant j in period t

𝑥𝑗𝑡 – Production quantity at plant j in period t

𝑦𝑗𝑘𝑡 – Quantity shipped from plant j to warehouse k in period t

𝑧𝑘𝑐𝑡 – Quantity shipped from warehouse k to customer c in period t

𝑑𝑐𝑡 – Demand fulfilled for customer c in period t

𝑅𝑀𝑗𝑡 – Raw material inventory at plant j at the end of period t

𝐹𝐺𝑗𝑡 – Finished goods inventory at plant j at the end of period t 𝐷𝐼𝑘𝑡 – Inventory level at warehouse k at the end of period t 110

𝛼𝑗 = 1 if plant j is chosen; 0 otherwise

𝛽𝑘 = 1 if warehouse k is chosen; 0 otherwise Objective Function:

The objective is to maximize Gross Profit (GP) given by: GP = Revenue from sales − raw material cost

− plant production cost − plant fixed cost

− warehouse fixed cost − shipping cost from plants to warehouses − shipping cost from warehouses to customer zones

− inventory holding cost at the plant, warehouses and customer zones 15

Revenue from sales = �(20 d1t + 20 d2t + 21d3t ) Raw material cost

t=1

= �[(3 w11t + 2 w21t + 4 w31t ) + (3.5 w12t + 3.5 w22t + 2 w32t ) t=1

+ (3.5 w13t + 3.5 w23t + 2 w33t )] 15

Plant Production Cost = �(7 x1t + 6 x2t + 6 x3t ) t=1

Plant fixed cost = 2,100,000 α1 + 1,900,000 α2 + 1,800,000 α3

Warehouse fixed cost = 800,000 β1 + 750,000 β2 + 700,000 β3 + 750,000 β4

111

Shipping cost from plants to warehouses 15

= �[(3 y11t + 4.5 y21t + 4.5 y31t ) + (3.5 y12t + 4 y22t + 4 y32t ) t=1

+ (4.5 y13t + 4 y23t + 4.5 y33t ) + (4.5 y14t + 4 y24t + 4 y34t )]

Shipping cost from warehouses to customer zones 15

= �[(3 z11t + 3 z21t + 4.5 z31t + 4.5 z41t ) t=1

+ (3 z12t + 3 z22t + 4.5 z32t + 4.5 z42t )

+ (4.5 z13t + 3 z23t + 4.5 z33t + 4.5 z43t )]

Inventory holding cost at the plants 15

= �[(0.02) (RM1t + FG1t ) + (0.02) (RM2t + FG1t ) t=1

+ (0.03) (RM3t + FG3t )]

Inventory holding cost at the warehouses 15

= �[(0.05) DI1t + (0.05) DI2t + (0.01)DI3t + (0.03)DI4t ] t=1

Inventory holding cost at customer zones 15

= �[(0.05) RI1t + (0.06) RI2t + (0.07)RI3t ] t=1

Model Constraints: 1. At the plants •

Production Capacity 112

x1t ≤ 100000α1 x2t ≤ 50000α2

•

Storage Capacity

x3t ≤ 50000α3 RMjt + FGjt ≤ 500000 αj

•

for j = 1, 2, 3 and all t

Raw material flow balance o Plant 1: RM1,t−1 + W1,1,t−2 + W2,1,t−1 + W3,1,t−3 = x1t + RM1t

o Plant 2: RM2,t−1 + W1,2,t−5 + W2,2,t−3 + W3,2,t−1 = x2t + RM2t o Plant 3: RM3,t−1 + W1,3,t−5 + W2,3,t−3 + W3,3,t−1 = x3t + RM3t

for all t = 1, 2,…,15

Note: RM j,0 = 0 and W i,j,t = 0 for all t ≤ 0 •

Finished goods flow balance 4

FGj,t−1 + xjt = � yjkt + FGj,t k=1

for j = 1, 2, 3 and t = 1,…,15 Note: FG j,0 = 0 for all j. 2. At the warehouses •

Flow balance o Warehouse 1: DI1,t−1 + y1,1,t−1 + y2,1,t−4 + y3,1,t−4 = ∑3c=1 z1ct + DI1,t 113

o Warehouse 2: DI2,t−1 + y1,2,t−2 + y2,2,t−3 + y3,2,t−3 = ∑3c=1 z2ct + DI2,t o Warehouse 3: DI3,t−1 + y1,3,t−3 + y2,3,t−4 + y3,3,t−4 = ∑3c=1 z3ct + DI3,t o Warehouse 4: DI4,t−1 + y1,4,t−3 + y2,4,t−4 + y3,4,t−4 = ∑3c=1 z4ct + DI4,t

for all t = 1,…,15

Note: DI k,0 = 0; yjkt = 0 for all t ≤ 0 •

Storage capacity DIkt ≤ 500000βk

•

for k = 1, 2, 3, 4 and t = 1,…,15

Warehouse inflow 15

� � yjkt ≤ Mβk t=1 j=1

for k = 1, 2, 3, 4 Note: (i) If β k = 0, there will be no shipping to warehouse k from any plant (ii) M is a large positive number •

Warehouse outflow 15

� � zkct ≤ Mβk t=1 c=1

for k = 1, 2, 3, 4 Note: If β k = 0, there will be no shipping from warehouse k to any of the customers. 3. At customer zones 114

•

Maximum sales dct ≤ Dct

for c = 1, 2, 3 and t = 1,…,15

where D ct = forecast demands at customer c in period t (given in Table 8.16). (e.g. D 11 = 104000; D 21 = 33000; D 31 = 101000, etc.) •

Flow balance o Customer 1: RI1,t−1 + z1,1,t−1 + z2,1,t−1 + z3,1,t−3 + z4,1,t−3 = d1t + RI1t

o Customer 2: RI2,t−1 + z1,2,t−1 + z2,2,t−1 + z3,2,t−3 + z4,2,t−3 = d2t + RI2t o Customer 3: RI3,t−1 + z1,3,t−3 + z2,3,t−3 + z3,3,t−1 + z4,3,t−1 = d3t + RI3t

for all t = 1,…,15

Note: RI1,0 = 250,000; RI 2,0 = 100,000 and RI3,0 = 250,000. Also z kct = 0 for all t ≤ 0. All the variables w ijt , x jt , y jkt , z kct , RM jt , FG jt , DIkt and RIct are non-negative and α j and β k are binary variables. Results: The MILP model resulted in 798 variables (7 binary) and 410 constraints. The problem was solved using the optimization software LINGO 13.0. It took 1910 iterations and the run time was about 5 seconds. The gross profit was $15,316,120. The optimal network used only the manufacturing plant 1 and warehouse 1 in country 2. Plants 2 and 3 in country 3 were not selected. Warehouse 2 in country 2 and warehouse 3 and 4 in country 4 were not selected.

115

All the raw materials for plant 1 were supplied by supplier S2 in country 2. Supplier 1 in country 1 and supplier 3 in country 3 were not used. Plant 1 produces to its maximum capacity of 100,000 units in periods 2 through 13, namely 12 out of the 15 periods. The demand fulfillments at the customer zones are given below: Customer

Total demand

Demand fulfilled

Percentage

1,459,000

1,215,000

83%

332,000

325,000

98%

1,033,000

260,000

25%

Total

2,824,000

1,800,000

64%

Note that customers 1 and 2 are in the same country where the plant and warehouse are located. Finished goods shipped from warehouse 1 to the customer zones are as follows (in 1000’s): Period 1

100 200

105

130 140 150

965

225

116

15 Total

From the above tables it is clear that customer 1’s demand is fulfilled using 250,000 units of initial inventory and 965000 units of new shipments from the warehouse. Similarly, customer 2’s demand is fulfilled with 100,000 units of initial inventory and 225,000 of additional warehouse shipments. Customer 3 receives only 10000 additional units from the warehouse; the rest comes from its inventory of 250,000 units. Since customer 3 is in country 4, its demand fulfillment is very low (25%), primarily due to high shipping cost, even though its sales price is $1 higher. Finally, the route from supplier 2 to plant 1 to warehouse 1 is used to ship 1,200,000 units of product to customers. These sites provide the least expensive route because supplier 2, plant 1 and warehouse 1 are all in the same country.

117

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