31. L
å
DL = DL σL = LσD
D
L
=DL s
L
=Ls
D
cv =σ / µ
cv=s/m
DL = DL = 2×2,500
σL = LσD = 2 ×500 = 707
D
L
=DL=2´2,500
s
L
=Ls
D
=2´500=707
ESC = (x – ROP) f (x)dx
x=ROP
34. s
ss
s
L
æ
è
ç
ö
ø
÷
ESC = –ss[1– NORMDIST(ss /σL,0,1,1)]
ESC=–ss[1–NORMDIST(ss/s
L
,0,1,1)]
+σLNORMDIST(ss /σL,0,1,0)
+s
L
NORMDIST(ss/s
L
,0,1,0)
ESC = –1,000[1– NORMDIST(1,000/707,0,1,1)]
ESC=–1,000[1–NORMDIST(1,000/707,0,1,1)]
+707NORMDIST(1,000/707,0,1,0) = 25
+707NORMDIST(1,000/707,0,1,0)=25
35. DL + ss = F
–1(CSL,DL,σL) = NORMINV(CSL,DL,σL)
D
L
+ss=F
–1
(CSL,D
L
,s
L
)=NORMINV(CSL,D
L
,s
L
)
ss = F–1(CSL,DL,σL)– DL = NORMINV(CSL,DL,σL)– DL
ss=F
–1
(CSL,D
L
,s
L
)–D
L
=NORMINV(CSL,D
L
,s
L
)–D
L
36. ss = FS
–1(CSL)×σL = FS
–1(CSL)× LσD
ss=F
S
–1
(CSL)´s
L
=F
S
–1
(CSL)´Ls
D
= NORMSINV(CSL)× LσD
=NORMSINV(CSL)´Ls
D
DL = DL = 2×2,500 = 5,000
σL = LσD = 2 ×500 = 707
D
L
=DL=2´2,500=5,000
s
L
=Ls
D
37. =2´500=707
ss = Fs
–1(CSL)×σL = NORMSINV(CSL)×σL
ss=F
s
–1
(CSL)´s
L
=NORMSINV(CSL)´s
L
= NORMSINV(0.90)×707 = 906
=NORMSINV(0.90)´707=906
ESC = 250 = –ss 1– Fs
ss
σL
⎛
⎝
⎜
⎞
⎠
⎟
⎡
41. ú
+707f
s
ss
707
æ
è
ç
ö
ø
÷
250 = –ss[1– NORMDIST(ss /707,0,1,1)]
250=–ss[1–NORMDIST(ss/707,0,1,1)]
+707NORMDIST(ss /707,0,1,0)
+707NORMDIST(ss/707,0,1,0)
ss = NORMSINV(CSL)× LσD
ss=NORMSINV(CSL)´Ls
D
= NORMSINV(.95)× 9 ×800 = 3,948
=NORMSINV(.95)´9´800=3,948
ss = NORMSINV(.95)× 1×800 =1,316
42. ss=NORMSINV(.95)´1´800=1,316
ss = NORMSINV(.95)× 9 ×400 =1,974
ss=NORMSINV(.95)´9´400=1,974
DL = DL σL = LσD
2 + D2sL
2
D
L
=DL s
L
=Ls
D
2
+D
2
s
L
2
Standard deviation of ddlt σL = LσD
2 + D2sL
2
Standard deviation of ddlt s
L
=Ls
D
43. 2
+D
2
s
L
2
= 7×5002 +2,5002 ×72
=17,500
=7´500
2
+2,500
2
´7
2
=17,500
ss = FS
–1(CSL)×σL = NORMSINV(CSL)×σL
ss=F
S
–1
(CSL)´s
L
=NORMSINV(CSL)´s
L
= NORMSINV(0.90) ×17,500
= 22,491 hard drives
46. (
)
DC = kD σD
C = kσD
D
C
=kD s
D
C
=ks
D
= FS
–1(CSL)× L ×σD
C
i=1
k
∑
=F
S
–1
(CSL)´L´s
D
C
i=1
k
å
47. =
FS
–1(CSL)× L × H
DC
× σ i –σD
C
i=1
k
∑
⎛
⎝
⎜ ⎜
⎞
⎠
⎟ ⎟
=
F
S
–1
(CSL)´L´H
D
C
´s
i
–s
48. D
C
i=1
k
å
æ
è
ç
ç
ö
ø
÷
÷
ss = k × Fs
–1(CSL)× L ×σD
ss=k´F
s
–1
(CSL)´L´s
D
= 4 × Fs
–1(0.9) × 2 × 5
= 4 × NORMSINV(0.9) × 2 × 5 = 36.24 cars
=4´F
s
–1
(0.9)´2´5
=4´NORMSINV(0.9)´2´5=36.24 cars
49. σD
C = 4 ×5 =10
s
D
C
=4´5=10
ss = Fs
–1(0.9)× L ×σD
C = NORMSINV(0.9)× 2 ×10 =18.12
ss=F
s
–1
(0.9)´L´s
D
C
=NORMSINV(0.9)´2´10=18.12
ss = 4× Fs
–1(CSL)× L ×σD
ss=4´F
s
–1
(CSL)´L´s
D
= 4× NORMSINV(0.95)× 4 ×300 = 3,948
50. =4´NORMSINV(0.95)´4´300=3,948
σD
C = 4 ×300 = 600
s
D
C
=4´300=600
ss = Fs
–1(0.95)× L ×σD
C
ss=F
s
–1
(0.95)´L´s
D
C
= NORMSINV(0.95)× 4 ×600 =1,974
=NORMSINV(0.95)´4´600=1,974
= 81× NORMSINV(0.95)× 1×3,000
=81´NORMSINV(0.95)´1´3,000
= 399,699 units
51. =399,699 units
= NORMSINV(0.95)× 1× 9 ×3,000
=NORMSINV(0.95)´1´9´3,000
= 14,804 units
=14,804 units
= 9×14,804 =133,236
=9´14,804=133,236
ss =100× Fs
–1(CSL)× L ×σD
ss=100´F
s
–1
(CSL)´L´s
D
=100× NORMSINV(0.95)× 2 ×10 = 2,326
=100´NORMSINV(0.95)´2´10=2,326
σD
C = 100 ×10 =100
52. s
D
C
=100´10=100
ss = Fs
–1(CSL)× L ×σD
C = NORMSINV(0.95)× 2 ×100 = 233
ss=F
s
–1
(CSL)´L´s
D
C
=NORMSINV(0.95)´2´100=233
DL = DL
σL = LσD
D
L
=DL
s
L
=Ls
D
ss = FS
–1(CSL)×σL = NORMSINV(CSL)× LσD,ROP = DL + ss
53. ss=F
S
–1
(CSL)´s
L
=NORMSINV(CSL)´Ls
D
,ROP=D
L
+ss
Mean demand during T + L periods, DT+L = (T + L)D
Mean demand during T+L periods, D
T+L
=(T+L)D
Std dev demand during T + L periods, σT+L = T + LσD
Std dev demand during T+L periods, s
T+L
=T+Ls
D
OUL = DT+L + ss
ss = FS
–1(CSL) ×σD+L = NORMSINV(CSL) ×σT+L
Average lot size, Q = DT = DT
OUL=D
54. T+L
+ss
ss=F
S
–1
(CSL)´s
D+L
=NORMSINV(CSL)´s
T+L
Average lot size, Q=D
T
=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
55. (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
56. 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
57. 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
58. schopra
insert some space between the square root and sigma
Mean demand during T + L periods, DT+L = (T + L)D
= (2 + 4)2,500 = 15,000
Mean demand during T+L periods, D
T+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
Std dev demand during T+L periods, s
T+L
=T+Ls
D
=4+2
()
500=1,225
ss = FS
–1(CSL) ×σD+L = NORMSINV(CSL) ×σT+L
= NORMSINV(0.90) ×1,225 = 1,570 boxes
ss=F
S
–1
(CSL)´s
59. D+L
=NORMSINV(CSL)´s
T+L
=NORMSINV(0.90)´1,225=1,570 boxes
OUL = DT+L + ss =15,000+1,570 =16,570
OUL=D
T+L
+ss=15,000+1,570=16,570
Chapter 3: What to Change in an Organization: Frameworks
1
Chapter Overview
Change leaders must understand both the HOW and the WHAT
of change. The focus here is on WHAT needs to change
Open systems organizational frameworks are valuable
assessment tools of what needs to change. Nadler and
Tushman’s Congruence Model is explored in detail
The non-linear and interactive nature of organizations is
explored to make sense of their complexity
Quinn’s competing values model is used to create a bridge
between individual and organizational levels of analysis
Organizational change over time is discussed
72. 21
Evolution stages
Revolution stages
Size of
organization
LARGE
SMALL
YOUNG
Age of Organization
MATURE
1: Growth through
CREATIVITY
1: Crisis of
LEADERSHIP
2: Growth through
DIRECTION
2: Crisis of
AUTONOMY
3: Growth through
DELEGATION
4: Growth through
COORDINATION
5: Growth through
COLLABORATION
3: Crisis of
CONTROL
4: Crisis of
RED TAPE
5: Crisis of ?
PHASE 1
PHASE 2
PHASE 3
PHASE 4
PHASE 5
81. Cycles per Season
Expected
Profit
Number of Order Cycles per Season
FIGURE 13-5 Expected Profit versus Number of Order Cycles
per Season
1. The expected total quantity ordered during the season with
two orders is less than that with
a single order for the same cycle service level. In other words,
it is possible to provide the
same level of product availability to the customer with less
inventory if a second, follow-
up order is allowed after observing some sales.
2. The average overstock to be disposed of at the end of the
sales season is less if a follow-up
order is allowed after observing some sales.
3. The profits are higher when a follow-up order is allowed
during the sales season.
In other words, as the total quantity for the season is broken up
into multiple smaller orders
with the size of each order based on some observed sales, the
buyer is better able to match supply
and demand and increase profitability for Saks. These
relationships are shown in Figures 13-4
and 13-5.
We now consider the case in which the buyer improves her
forecast accuracy for the
second order after observing some of the season’s demand. As a
82. result, the standard deviation of
weekly demand forecast drops from 15 to 3 for the second
seven-week period. In this setting, the
first order stays at 195 shawls as discussed earlier. For the
second order, however, we must
M13_CHOP3952_05_SE_C13.QXD 11/14/11 8:04 PM Page
376
376 Chapter 13 • Determining the Optimal Level of Product
Availability
Unsold
Inventory at
End of Season
Number of Order Cycles per Season
FIGURE 13-4 Leftover Inventory versus Number of Order
Cycles per Season
Expected
Profit
Number of Order Cycles per Season
FIGURE 13-5 Expected Profit versus Number of Order Cycles
per Season
1. The expected total quantity ordered during the season with
two orders is less than that with
a single order for the same cycle service level. In other words,
it is possible to provide the
same level of product availability to the customer with less
83. inventory if a second, follow-
up order is allowed after observing some sales.
2. The average overstock to be disposed of at the end of the
sales season is less if a follow-up
order is allowed after observing some sales.
3. The profits are higher when a follow-up order is allowed
during the sales season.
In other words, as the total quantity for the season is broken up
into multiple smaller orders
with the size of each order based on some observed sales, the
buyer is better able to match supply
and demand and increase profitability for Saks. These
relationships are shown in Figures 13-4
and 13-5.
We now consider the case in which the buyer improves her
forecast accuracy for the
second order after observing some of the season’s demand. As a
result, the standard deviation of
weekly demand forecast drops from 15 to 3 for the second
seven-week period. In this setting, the
first order stays at 195 shawls as discussed earlier. For the
second order, however, we must
M13_CHOP3952_05_SE_C13.QXD 11/14/11 8:04 PM Page
376