Safety Inventories
Chapter 11 of Chopra
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Why to hold Safety Inventory?
 Desire
for quick product availability
– Ease of search for another supplier
– “I want it now” culture
 Demand
uncertainty
– Short product life cycles
 Safety
inventory
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Measures
 Measures
of demand uncertainty
– Variance of demand
– Ranges for demand
Delivery Lead Time, L
 Measures of product availability

– Stockout, what happens?
» Backorder (patient customer, unique product or big cost advantage) or
Lost sales.
– I. Cycle service level (CSL), % of cycles with no stockout
– II. Product fill rate (fr), % of products sold from the shelf
– Order fill rate, % of orders
» Equivalent to product fill rate if orders contain one product
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Service measures: CSL and fr are different
inventory
CSL is 0%, fill rate is almost 100%
0
time
inventory
0
CSL is 0%, fill rate is almost 0%
time
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Replenishment policies
 When
to reorder?
 How much to reorder?
– Most often these decisions are related.
Continuous Review: Order fixed quantity when total
inventory drops below Reorder Point (ROP).
- ROP meets the demand during the lead time L.
- One has to figure out the ROP.
Information technology facilitates continuous review.
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Demand During Lead time
Di demand in period i.
Mostly Di  Normal( Ri ,  i2 ), or N (mean, variance).
f i , Fi probabilit y density and cumulative density functions for D i
Ri  E ( Di )   Di f i ( Di )dDi
Var ( Di )   i2  E{( Di  Ri ) 2 }   ( Di  Ri ) 2 f i (Di )dDi
cov( Di , D j )   i2, j  E{( Di  Ri )( D j  R j )}
   ( Di  Ri )( D j  R j ) f i , j ( Di , D j )dDi dD j
   i2, j /( i j ) correlatio n coefficien t
L
L
i 1
L
i 1
E ( Di )   Ri by the linearity of integratio n
L
L
L
L
L
Var ( Di )   cov( Di , D j )     cov(Di , D j )
i 1
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i 1 j 1
i 1
2
i
i 1 j 1
j i
6
Normal Density Function
frequency
normdist(x,.,.,1)
normdist(x,.,.,0)
Prob


Mean
x


95.44%
99.74%
Excel statistica l functions :
Density function (pdf) at x : normdist ( x, mean, st _ dev,0)
Cumulative function (cdf) at x : normdist ( x, mean, st _ dev,1)
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7
Cumulative Normal Density
1
prob
normdist(x,mean,st_dev,1)
0
x
norminv(prob,mean,st_dev)
Excel statistica l functions :
Cumulative function (cdf) at x : normdist ( x, mean, st _ dev,1)
Inverse function of cdf at " prob": norminv ( prob, mean, st _ dev)
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Demand During Lead Time Determines ROP
Suppose that demands are identically and independently distributed.
To mean identically and independently distributed, use iid.
L
L
i 1
i 1
E ( Di )  LR and Var ( Di )  L 2
If Di  N ( R,  ) then
2

L
2
D

N
(
LR
,
L

)
 i
i 1



 L

P  Di  a   F a; LR, L  Normdist a, LR, L ,1
 i 1

F is the cumulative density function of the demand in a single period,
say a day. The second equality above holds if demand is Normal.
Coefficien t of variatio n of D : cv  Var( D) / E( D)
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9
Optimal Safety Inventory Levels
inventory
An inventory cycle
Q
ROP
time
Lead Times
Shortage
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I. Cycle Service Level: ROP  CSL
Cycle service level: percentage of cycles with stock out
For example consider 10 cycles :
11 0 111 0 1 0 1
10
CSL  0.7
CSL 
Write 0 if a cycle has stockout, 1 otherwise
CSL  0.7  Probabilit y that a single cycle has sufficient inventory
[Sufficien t inventory]  [Demand during lead time  ROP]
ROP: Reorder point
CSL  Cycle Service Level  F ( ROP ; R  L, L )
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I. Cycle Service Level for Normal Demands
CSL  F ( ROP , R  L, L R )

 P N ( R  L, L
 P N ( R  L, L R )  ROP
2
2
R

Recall N (mean, variance) notation
)  R  L  ROP  R  L

Taking out the mean




2
N ( R  L, L R )  R  L ROP  R  L 

P

Dividing by the StDev

L R
L R 
 

N
(
0
,
1
)



ROP  R  L 


 P N (0,1) 

L

R


Obtaining standard normal distributi on
The last equality is a property of the Normal distribution.
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Example: Finding CSL for given ROP
R = 2,500 /week; = 500
L = 4 weeks; Q = 10,000; ROP = 16,000
Stdev of demand during lead time   L 
ss = ROP – L R =
Cycle service level, F ( ROP; L  R, L ) 
If you wish to compute Average Inventory = Q/2 + ss
Average Inventory =
Average Flow Time =Average inventory/Thruput=
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Safety Inventory: CSL  ROP
CSL  F ( ROP , R  L, L R ) or ROP  F 1 (CSL, R  L, L R )
Safety stock  ss : ROP  R  L
For normally distribute d demand :
ss  F 1 (CSL, R  L, L R )  R  L
 F 1 (CSL;0, L )  F 1 (CSL;0,1)  L
 Norminv(CS L,0,1)  L
The last two equalities are by properties of the Normal distribution.
Very important remark: Safety inventory is a more general concept.
It exists without lead time. It is the stock held minus the expected demand.
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Finding ROP for given CSL
R = 2,500/week; = 500
L = 4 weeks; Q = 10,000; CSL = 0.90
ss  F 1 (CSL;0,1) L 
ROP  L  R  ss 
Factors driving safety inventory
– Replenishment lead time
– Demand uncertainty
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II. Fill rate:
Expected shortage per cycle

ESC is the expected shortage per cycle

ESC is not a percentage, it is the number of units, also see next page
Demand  ROP if
Shortage  
0
if

Demand  ROP
Demand  ROP
ESC  E (max{Demand during lead time - ROP,0})

ESC =
 ( x  ROP) f ( x)dx
x  ROP
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Inventory and Demand during Lead Time
ROP
0
Inventory
0
ROP
DLT: Demand
During LT
LT
Demand
During LT
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Upside
down
Inventory=
ROP-DLT
0
17
Shortage and Demand during Lead Time
0
DLT: Demand During LT
ROP
Shortage=
DLT-ROP
Upside
down
0
ROP
Shortage
LT
Demand
During LT
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0
18
Expected shortage per cycle

First let us study shortage during the lead time
Expected shortage  E (0, max( DLT  ROP ))

 ( D  ROP ) f

D
( D)dD where f D is pdf of DLT.
D  ROP

Ex:
 d1  9 with prob p1  1/4 


ROP  10, D  d 2  10 with prob p2  2/4 , Expected Shortage?
 d  11 with prob p  1/4 
3
 3

3
Expected shortage   max{0, (d i  ROP )} pi 
i 1
11
 (d  ROP )}P( D  d )
d 10
1
2
1 1
 max{0, (9 - 10)}  max{0, (10 - 10)}  max{0, (11 - 10)} 
4
4
4 4
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Expected shortage per cycle
 Ex:
ROP  10, D  Uniform(6,12), Expected Shortage?
D 12
 1  10 2

1
1  D2
1  12 2
Expected shortage   ( D  10) dD  
 10 D
 
 10(12)   
 10(10) 
6
6 2
6 2
 6 2

D 10
D 10
 172 - 170  2
12
If demand is normal:

 ss

 ss

ESC   ss 1  normdist 
,0,11
,    L  normdist 
,0,1,0






L
L


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Does ESC decrease or increase with ss, L?
Does ESC decrease or increase with expected value of demand?
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Fill Rate


Fill rate: Proportion of customer demand satisfied from stock
Q: Order quantity
ESC
fr  1 
Q
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Finding the Fill Rate
ss  fr
= 500; L = 2 weeks; ss=1000; Q = 10,000;
Fill Rate (fr) = ?

 ss

ESC   ss 1  normdist 
,0,1,1  
 L


 ss

L  normdist 
,0,1,0
 L

ESC  1000(1  normdist (1000 / 707,0,1,1)
 707nomdist (1000 / 707,0,1,0)
ESC  2513
.
fr = (Q - ESC)/Q = (10,000 - 25.13)/10,000 = 0.9975.
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Finding Safety Inventory for a Fill Rate:
fr  ss
If desired fill rate is fr = 0.975, how much safety
inventory should be held?
Clearly ESC = (1 - fr)Q = 250
Try some values of ss or use goal seek of Excel to solve

 ss

 ss

250   ss 1  normdist 
,0,1,1  707normdist 
,0,1,0 
 707

 707


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Evaluating Safety Inventory For Given Fill Rate
Fill Rate
Safety Inventory
97.5%
67
98.0%
183
98.5%
321
99.0%
499
99.5%
767
Safety inventory is very sensitive to fill rate. Is fr=100% possible?
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Factors Affecting Fill Rate
 Safety
inventory: If Safety inventory is up,
– Fill Rate is up
– Cycle Service Level is up.

Lot size: If Lot size Q is up,
– Cycle Service Level does not change. Reorder point,
demand during lead time specify Cycle Service
Level.
– Expected shortage per cycle does not change. Safety
stock and the variability of the demand during the
lead time specify the Expected Shortage per Cycle.
Fill rate is up.
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To Cut Down the Safety Inventory
 Reduce
the Supplier Lead Time
– Faster transportation
» Air shipped semiconductors from Taiwan
– Better coordination, information exchange, advance retailer
demand information to prepare the supplier
» Textiles; Obermeyer case
– Space out orders equally as much as possible

Reduce uncertainty of the demand
– Contracts
– Better forecasting to reduce demand variability
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Lead Time Variability
Supplier’s lead time may be uncertain:
L  Average lead time.
L
E ( Di )  LR
i 1
s 2  Variance of lead time
L
Var ( Di )  L   2  R 2  s 2 :  L2
i 1
The formulae do not change:
ss  F 1 (CSL;0,1)   L  F 1 (CSL;0,1) 
L 2  R 2 s 2

 ss

 ss

ESC   ss 1  normdist 
;0,1,1   L nomdist 
;0,1,0 
 L

 L


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27
Impact of Lead Time Variability, s
R = 2,500/day; = 500
L = 7 days; Q = 10,000; CSL = 0.90
StDev of LT
0
1
2
3
4
5
6
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ss
1695
3625
6628
9760
12927
16109
19298
Jump in ss
1930
3003
3132
3167
3182
3189
28
Methods of Accurate Response to Variability

Centralization
– Physical, Laura Ashley
– Information
» Virtual aggregation, Barnes&Nobles stores
– Specialization, what to aggregate

Product substitution

Raw material commonality - postponement
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Centralization: Inventory Pooling
Which of two systems provides a higher level of service for a given
safety stock?
Consider locations and demands:
)
D1  ( R1 ,
1
D3  ( R3 , 3)

D2  ( R2 ,
2
)
( R , )
C
C
D4  ( R4 , 4)
K
With k locations centralized, mean and variance of
D C   Di
i 1
R
C
K
  Ri;
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i 1

(
K
) 
C 2
i 1

2
i
K
 2 cov( Di , D j )
i j
i 1
30
Sum of Random Variables Are Less Variable
When they are independent, C
 
cov(Di,Dj)=0
K

i 1
K
2
i
  i
i 1
When they are perfectly positively correlated,
cov(Di,Dj)=σi σj
2
K


 C   i2  2 i j    i    i
i 1
i 1
i 1
 i 1 
K
K
K
i j
When they are perfectly negatively correlated,
cov(Di,Dj)= - σi σj
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C 
K
K
i 1
i 1
i j
2

 i  2 i j 
K
K
i 1
i 1
2

 i   i
31
Factors Affecting Value of Aggregation

When to aggregate? Statistical checks: Positive correlation and
Coefficient of Variation.
– Aggregation reduces variance almost always except when products are
positively correlated
– Aggregation is not effective when there is little variance to begin with.
When coefficient of variation of demand is relatively small (variance w.r.t.
the mean is small), do not bother to aggregate.

In real life,
– Is the electricity demand in Arlington and Plano are positively or negatively
correlated? Is there an underlying factor which affects both in the same
direction? Note that a big portion of electricity is consumed for
heating/cooling.
– Are the Campbell soup sales over time positively or negatively correlated?
How many soups can you drink per day?
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Impact of Correlation on Aggregated Safety
Inventory (Aggregating 4 outlets)


Safety stocks are proportional to the StDev of the demand.
With four locations, we have total ss proportional to 4*σ

If four locations are all aggregated,
ss proportional to 4*σ with correlation 1
ss proportional to 2*σ with correlation 0

Benefit=SS before - SS after / SS before


33
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Impact of Correlation on Aggregated
Safety Inventory (Aggregating 4 outlets)
Benefit=(SS
before - SS after) / SS before
0.6
0.5
0.4
0.3
Benefit
0.2
0.1
0
0
0.2
0.4
0.6
0.8
1
34
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EX 11.8: W.W. Grainger a supplier of
Maintenance and Repair products






About 1600 stores in the US
Produces large electric motors and
industrial cleaners
Each motor costs $500; Demand is iid
Normal(20,40x40) at each store
Each cleaner costs $30; Demand is iid
Normal(1000,100x100) at each store
Which demand has a larger coefficient of
variation?
How much savings if motors/cleaners
inventoried centrally?
35
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Use CSL=0.95
Supply lead time L=4 weeks for motors and cleaners
For normally distribute d demand : ss  Norminv(CS L,0,1)  L
For a single store
Motor safety inventory=Norminv(0.95,0,1) 2 (40)=132
Cleaner safety inventory=Norminv(0.95,0,1) 2 (100)=329
Value of motor ss=1600(132)(500)=$105,600,000
Value of cleaner ss=1600(329)(30)=$15,792,000
Standard deviation of demands after aggregating 1600 stores
Standard deviation of Motor demand=40(40)=1,600
Standard deviation of Cleaner demand=40(100)=4,000
For the aggregated store
Motor safety inventory=Norminv(0.95,0,1) 2 (1600)=5,264
Cleaner safety inventory=Norminv(0.95,0,1) 2 (4,000)=13,159
Value of motor ss=5264(500)=$2,632,000
36
Value
of
cleaner
ss=13,159(30)=$394,770
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EX. 11.8: Specialization: Impact of cv on
Benefit From 1600-Store Aggregation , h=0.25
Motors
Mean demand/wk
SD of demand
Disaggregate cv
Value/Unit
Disaggregate ss value
Aggregate cv
Aggregate ss value
Inventory cost savings
Holding Cost Saving
Saving / Unit
20
40
2
$500
$105,600,000
0.05
$2,632,000
$102,968,000
$25,742,000
$15.47
2574200/(1600*20*52)=
Cleaner
1,000
100
0.1
$30
$15,792,000
0.0025
$394,770
$15,397,230
$3,849,308
$0.046
37
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Slow vs Fast Moving Items

Low demand = Slow moving items, vice versa.
– Repair parts are typically slow moving items


Slow moving items have high coefficient of variation, vice versa.
Stock slow moving items at a central store
Buying a best seller at Amazon.com vs. a Supply Chain book vs. a Banach spaces
book, which has a shorter delivery time?
- Why cannot I find a “driver-side-door lock cylinder” for my 1994 Toyota
Corolla at Pep Boys?
- Your instructor on March 26 2005.
-
“Case Interview books” are not in our s.k.u. list. You must check with our
central stores.
- Store keeper at Barnes and Nobles at Collin Creek, March 2002.
38
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Product Substitution


Manufacturer driven
Customer driven
Consider: The price of the products substituted for each other and
the demand correlations

One-way substitution
– Army boots. What if your boot is large? Aggregate?

Two-way substitution:
– Grainger motors; water pumps model DN vs IT.
– Similar products, can customer detect specifications.
If products are very similar, why not to eliminate one of them?
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Component Commonality. Ex. 11.9
 Dell
producing 27 products with 3 components
(processor, memory, hard drive)
 No product commonality: A component is used in only 1
product. 27 component versions are required for each
component. A total of 3*27 = 81 distinct components
are required.
 Component commonality allows for component
inventory aggregation.
40
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Max Component Commonality

Only three distinct versions for each component.
– Processors: P1, P2, P3. Memories: M1, M2, M3. Hard drives: H1, H2, H3


Each combination of components is a distinct product. A
component is used in 9 products.
Each way you can go from left to right is a product.
Left
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P1
M1
H1
P2
M2
H2
P3
M3
H3
Right
41
Example 11.9: Value of Component Commonality
in Safety Inventory Reduction
450000
400000
350000
300000
250000
SS
200000
150000
100000
50000
0
1
2
3
4
5
6
7
8
9
# of products a component is used in
Aggregation provides reduction in total standard deviation.
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42
Standardization
Standardization
– Extent to which there is an absence of variety in a product,
service or process
The
degree of Standardization?
Standardized products are immediately available to
customers
Who wants standardization?
– The day we sell standard products is the day we lose a
significant portion of our profit
– A TI manager on November 1, 2005
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Advantages of Standardization

Fewer parts to deal with in inventory & manufacturing
– Less costly to fill orders from inventory

Reduced training costs and time

More routine purchasing, handling, and inspection
procedures

Opportunities for long production runs, automation

Need for fewer parts justifies increased expenditures on
perfecting designs and improving quality control
procedures.
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Disadvantages of Standardization

Decreased variety results in less consumer appeal.

Designs may be frozen with too many imperfections remaining.

High cost of design changes increases resistance to improvements
– Who likes optimal Keyboards?

Standard systems are more vulnerable to failure
– Epidemics: People with non-standard immune system stop the
plagues.
– Computer security: Computers with non-standard software stop the
dissemination of viruses.

Another reason to stop using Microsoft products!
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Inventory–Transportation Costs:
Eastern Electric Corporation: p.427

Major appliance manufacturer, buys motors from Westview motors in Dallas

Annual demand = 120,000 motors

Cost per motor = $120; Weight per motor 10 lbs.

Current order size = 3,000 motors
» 30,000 pounds = 300 cwt
– 1 cwt = centum weight = 100 pounds; Centum = 100 in Latin.

Lead time = 1 + the number of days in transit

Safety stock carried = 50% of demand during delivery lead time

Holding cost = 25%

Evaluate the mode of transportation for all unit discounting based on shipment
weight
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46
AM Rail proposal:
Over 20,000 lbs at 0.065 per lb in 5 days

For the appliance manufacturer
– No fixed cost of ordering besides the transportation cost
– No reason to transport at larger lots than 2000 motors, which
make 20,000 lbs.
» Cycle inventory=Q/2=1,000
» Safety inventory=(6/2)(120,000/365)=986
» In-transit inventory


All motors shipped 5 days ago are still in-transit
5-days demand=(120,000/365)5=1,644
– Total inventory held over an average day=3,630 motors
– Annual holding cost=3,630*120*0.25=$108,900
– Annual transportation cost=120,000(10)(0.065)=$78,000
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Inventory–Transportation trade off: Eastern
Electric Corporation, see p.426-8 for details
Alternative Transport
(Lot size)
Cost
AM Rail
(2,000)
Northeast
Trucking
(1,000)
Golden
(500)
Golden
(2,500)
Golden
(3,000)
Golden
(4,000)
Cycle
Safety
Inventory Inventory
Transit
Inventory
Inventory Total
Cost
Cost
$78,000
120000(0.65)
$90,000
1,000
500
1,644
120000(5/365)
658
986
$96,000
120000(0.80)
$86,400
250
658
986
$108,900
$186,900
$64,320
$154,320
$56,820
$152,820
1,250
986
120000(3/365)
658
986
$86,820
$173,220
$78,000
1,500
658
986
$94,320
$172,320
$67,500
2,000
658
986
$109,320
$176,820
If fast transportation not justified cost-wise, need to consider rapid response
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Physical Inventory Aggregation:
Inventory vs. Transportation cost: p.428
 HighMed
Inc. producer of medical equipment
sold directly to doctors
 Located in Wisconsin serves 24 regions in USA
 As a result of physical aggregation
– Inventory costs decrease
– Inbound transportation cost decreases
» Inbound lots are larger
– Outbound transportation cost increases
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Inventory Aggregation at HighMed
Highval ($200, .1 lbs/unit) demand in each of 24 territories
– H = 2, H = 5
Lowval ($30/unit, 0.04 lbs/unit) demand in each territory
– L = 20, L = 5
UPS rate: $0.66 + 0.26x {for replenishments}
FedEx rate: $5.53 + 0.53x {for customer shipping}
Customers order 1 H + 10 L
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Inventory Aggregation at HighMed
Option A
Option B
# Locations
Current
Scenario
24
24
1
Reorder Interval
4 weeks
1 week
1 week
Inventory Cost
$54,366
$29,795
$8,474
Shipment Size
8 H + 80 L
2 H + 20 L
1 H + 10 L
Transport Cost
$530
$1,148
$14,464
Total Cost
$54,896
$30,943
$22,938
If shipment size to customer is 0.5H + 5L, total cost of option B
increases to $36,729.
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Summary of Cycle and Safety Inventory
Match Supply & Demand
Reduce Buffer Inventory
Economies of Scale
Supply / Demand
Variability
Cycle Inventory
Safety Inventory
•Reduce fixed cost
•Aggregate across
products
•Volume discounts
•Promotion on Sell
thru
utdallas.edu/~metin
Seasonal
Variability
Seasonal Inventory
•Quick Response measures
•Reduce Info Uncertainty
•Reduce lead time
•Reduce supply
uncertainty
•Accurate Response measures
•Aggregation
•Component commonality
and postponement
52
Mass Customization
Mass customization:
– A strategy of producing standardized goods or services,
but incorporating some degree of customization
– Modular design
– Delayed differentiation
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Mass Customization I: Customize
Services Around Standardized Products
Warranty for contact lenses:
DEVELOPMENT
Source: B. Joseph Pine
PRODUCTION
MARKETING
DELIVERY
Deliver customized services as
well as standardized products
and services
Market customized services with standardized
products or services
Continue producing standardized products or services
Continue developing standardized products or services
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Mass Customization II: Create
Customizable Products and Services
Customizing the look of screen with windows operating system
Gillette sensor adjusting to the contours of the face
DEVELOPMENT
PRODUCTION
MARKETING
DELIVERY
Deliver standard (but
customizable) products
or services
Market customizable products or services
Produce standard (but customizable) products or services
Develop customizable products or services
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Mass Customization III: Provide Quick
Response Throughout Value Chain
Skiing parkas manufactured abroad vs. in the U.S.A.:
DEVELOPMENT
PRODUCTION
MARKETING
DELIVERY
Reduce Delivery Cycle Times
Reduce selection and order processing cycle
times
Reduce Production cycle time
Reduce development cycle time
utdallas.edu/~metin
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Mass Customization IV: Provide Point of
Delivery Customization
Paint mixing
Lenscrafters for glasses.
DEVELOPMENT
PRODUCTION
MARKETING
DELIVERY
Point of delivery
customization
Deliver standardize portion
Market customized products or services
Produce standardized portion centrally
Develop products where point of delivery customization is feasible
utdallas.edu/~metin
57
Mass Customization V: Modularize
Components to Customize End Products
Computer industry, Dell computers:
DEVELOPMENT
PRODUCTION
MARKETING
DELIVERY
Deliver customized product
Market customized products or services
Produce modularized components
Develop modularized products
utdallas.edu/~metin
58
Modular Design
Modular design is a form of standardization in which component parts are
subdivided into modules that are easily replaced or interchanged.
– Good example: Dell uses same components to assemble various
computers.
– Bad example: Earlier Ford SUVs shared the lower body with Ford cars.
– Ugly example:
It allows:
–
easier diagnosis and remedy of failures
–
easier repair and replacement
–
simplification of manufacturing and assembly
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Types of Modularity for Mass
Customization
Component Sharing Modularity, Dell
Cut-to-Fit Modularity,
Gutters that do not require
seams
Bus Modularity, E-books
+
=
Mix Modularity, Paints
Sectional Modularity, LEGO
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Periodic Review
Order at fixed time intervals (T apart) to raise total inventory
(on hand + on order) to Order up to Level (OUL)
Inventory
OUL must cover
the Demand during
T
T+LT
OUL
utdallas.edu/~metin
LT
LT
61
Periodic Review Policy: Safety Inventory
T: Reorder interval
R: Standard deviation of demand per unit time
L+T: Standard deviation of demand during L+T periods
OUL: Order up to level
T L
 (T  L) R
T L
 L T
R

ss  F 1 (CSL;0,1)   T  L
OUL 
R
T L
 ss
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Example: Periodic Review Policy
R = 2,500/week; R = 500
L = 2 weeks; T = 4 weeks; CSL = 0.90
What is the required safety inventory?
ss  F 1 (CSL;0,1)   T  L  1570
Factors driving safety inventory
– Demand uncertainty
– Replenishment lead time
– Reorder interval
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Periodic vs Continuous Review
 Periodic
review ss covers the uncertainty over
[0,T+L], T periods more than ss in continuous case.
 Periodic review ss is larger.
 Continuous review is harder to implement, use it for
high-sales-value per time products
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