Discussion 2

All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher.

Printed in the United States of America.

12-‹#›

Copyright ©2013 Pearson Education, Inc. publishing as Prentice Hall.

DL = Di σL = σ i 2 + 2 ρijσ iσ j

i> j ∑

i=1

L

∑ i=1

L

∑

DL = DL σL = LσD

cv =σ / µ

DL = DL = 2×2,500

σL = LσD = 2 ×500 = 707

ESC = (x – ROP) f (x)dx x=ROP

∞

∫

ESC = –ss 1– Fs ss σL

⎛

⎝ ⎜

⎞

⎠ ⎟

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥ +σL fs

ss σL

⎛

⎝ ⎜

⎞

⎠ ⎟

ESC = –ss[1– NORMDIST(ss /σL,0,1,1)]

+σLNORMDIST(ss /σL,0,1,0)

ESC = –1,000[1– NORMDIST(1,000/707,0,1,1)]

+707NORMDIST(1,000/707,0,1,0) = 25

DL + ss = F –1(CSL,DL,σL) = NORMINV(CSL,DL,σL)

ss = F–1(CSL,DL,σL)– DL = NORMINV(CSL,DL,σL)– DL

ss = FS –1(CSL)×σL = FS

–1(CSL)× LσD

= NORMSINV(CSL)× LσD

DL = DL = 2×2,500 = 5,000

σL = LσD = 2 ×500 = 707

ss = Fs –1(CSL)×σL = NORMSINV(CSL)×σL

= NORMSINV(0.90)×707 = 906

ESC = 250 = –ss 1– Fs ss σL

⎛

⎝ ⎜

⎞

⎠ ⎟

⎡

⎣ ⎢ ⎢

⎤

⎦ ⎥ ⎥ +σL fs

ss σL

⎛

⎝ ⎜

⎞

⎠ ⎟

= –ss 1– Fs ss 707 ⎛

⎝ ⎜

⎞

⎠ ⎟

⎡

⎣ ⎢

⎤

⎦ ⎥+707 fs

ss 707 ⎛

⎝ ⎜

⎞

⎠ ⎟

250 = –ss[1– NORMDIST(ss /707,0,1,1)]

+707NORMDIST(ss /707,0,1,0)

ss = NORMSINV(CSL)× LσD

= NORMSINV(.95)× 9 ×800 = 3,948

ss = NORMSINV(.95)× 1×800 =1,316

ss = NORMSINV(.95)× 9 ×400 =1,974

DL = DL σL = LσD 2 + D2sL

2

Standard deviation of ddlt σL = LσD 2 + D2sL

2

= 7×5002 +2,5002 ×72

=17,500

ss = FS –1(CSL)×σL = NORMSINV(CSL)×σL

= NORMSINV(0.90) ×17,500 = 22,491 hard drives

= FS –1(CSL)× L ×σ i

i=1

k

∑

DC = Dii=1 k

∑ ; var DC( ) = σ i2 + 2 ρijσ iσ j i> j ∑

i=1

k

∑ ;

σD C = var DC( )

DC = kD σD C = kσD

= FS –1(CSL)× L ×σD

C

i=1

k

∑

= FS –1(CSL)× L × H

DC × σ i –σD

C

i=1

k

∑ ⎛

⎝ ⎜⎜

⎞

⎠ ⎟⎟

ss = k × Fs –1(CSL)× L ×σD

= 4 × Fs –1(0.9) × 2 × 5

= 4 × NORMSINV(0.9) × 2 × 5 = 36.24 cars

σD C = 4 ×5 =10

ss = Fs –1(0.9)× L ×σD

C = NORMSINV(0.9)× 2 ×10 =18.12

ss = 4× Fs –1(CSL)× L ×σD

= 4× NORMSINV(0.95)× 4 ×300 = 3,948

σD C = 4 ×300 = 600

ss = Fs –1(0.95)× L ×σD

C

= NORMSINV(0.95)× 4 ×600 =1,974

= 81× NORMSINV(0.95)× 1×3,000

= 399,699 units

= NORMSINV(0.95)× 1× 9 ×3,000

= 14,804 units

= 9×14,804 =133,236

ss =100× Fs –1(CSL)× L ×σD

=100× NORMSINV(0.95)× 2 ×10 = 2,326

σD C = 100 ×10 =100

ss = Fs –1(CSL)× L ×σD

C = NORMSINV(0.95)× 2 ×100 = 233

DL = DL

σL = LσD

ss = FS –1(CSL)×σL = NORMSINV(CSL)× LσD,ROP = DL + ss

Mean demand during T + L periods, DT+L = (T + L)D

Std dev demand during T + L periods, σT+L = T + LσD

OUL = DT+L + ss

ss = FS –1(CSL) ×σD+L = NORMSINV(CSL) ×σT+L

Average lot size, Q = DT = DT

Chapter 12 • Managing Uncertainty in a Supply Chain: Safety Inventory 343

The next step is to evaluate the distribution of demand during the time interval T ! L. Using Equation 12.2, demand during the time interval T ! L is normally distributed, with

The safety inventory in this case is the quantity in excess of DT+L carried by Wal-Mart over the time interval T ! L. The OUL and the safety inventory ss are related as follows:

(12.17)

Given the desired CSL, the safety inventory (ss) required is given by

(12.18)

The average lot size equals the average demand during the review period T and is given as

(12.19)

In Figure 12-6, we show the inventory profile for a periodic review policy with lead time L " 4 and reorder interval T " 7. Observe that on day 7, the company places an order that determines available inventory until day 18 (as illustrated in the line from point 1 and point 2). As a result, the safety inventory must be sufficient to buffer demand variability over T ! L " 7 ! 4 " 11 days.

We illustrate the periodic review policy for Wal-Mart in Example 12-13.

Evaluation Safety Inventory for a Periodic Review Policy

Weekly demand for Legos at a Wal-Mart store is normally distributed, with a mean of 2,500 boxes and a standard deviation of 500. The replenishment lead time is two weeks, and the store manager has decided to review inventory every four weeks. Assuming a periodic-review replenishment policy, evaluate the safety inventory that the store should carry to provide a CSL of 90 percent. Evaluate the OUL for such a policy.

Analysis: In this case, we have

Average demand per period, D " 2,500

Standard deviation of demand per period, sD = 500

EXAMPLE 12-13

Average lot size, Q = DT = DT

ss = FS-11CSL2 * sT+L = NORMSINV1CSL2 * sT+L OUL = DT+L + ss

Standard deviation of demand during T + L periods, sT+L = 1T + LsD Mean demand during T + L periods, DT+L = 1T + L2D

5

OUL

DT

DL Safety Inventory

T = 7

L = 4 W

ar eh

ou se

In ve

nt or

y

10

L

T L

SS 0

Review Point 0

Review Point 1

Review Point 2

Review Point 3

15 Days

20 25

2

1

FIGURE 12-6 Inventory Profile for Periodic Review Policy with L " 4, T " 7

M12_CHOP3952_05_SE_C12.QXD 11/15/11 6:52 PM Page 343

Mean demand during T + L periods, DT+L = (T + L)D = (2 + 4)2,500 = 15,000

Std dev demand during T + L periods, σT+L = T + LσD = 4 + 2( )500 = 1,225

ss = FS –1(CSL) ×σD+L = NORMSINV(CSL) ×σT+L

= NORMSINV(0.90) ×1,225 = 1,570 boxes

OUL = DT+L + ss =15,000+1,570 =16,570