utdallas.edu/~metin 1 safety inventories chapter 11 of chopra
TRANSCRIPT
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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
CSL is 0%, fill rate is almost 100%
CSL is 0%, fill rate is almost 0%
inventory
inventory
time
time
0
0
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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
variance).(mean,or ),,(Mostly
i. periodin demand 2 NRNormalD
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Normal Density Function
Mean
95.44%
99.74%
x
)1,_,,(normdist:at x (cdf)function Cumulative
)0,_,,(normdist:at x (pdf)function Density
:functions lstatistica Excel
devstmeanx
devstmeanx
normdist(x,.,.,0)
Prob
normdist(x,.,.,1)frequency
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Cumulative Normal Density
)_,,(norminv :prob""at cdf offunction Inverse
)1,_,,(normdist:at x (cdf)function Cumulative
:functions lstatistica Excel
devstmeanprob
devstmeanx
0
1
x
normdist(x,mean,st_dev,1)
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.
),( then ),( If
)( and )(
2
1
2
2
11
LLRNDRND
LDVarLRDE
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iii
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1,,,Normdist,;1
LLRaLLRaFaDPL
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)(/)( :D ofn variatiooft Coefficien DEDVarcv
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.
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Optimal Safety Inventory Levels
Lead Times
time
inventory
Shortage
An inventory cycle
ROP
Q
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I. Cycle Service Level: ROP CSL
),;(Level Service Cycle LLRROPFCSL
Cycle service level: percentage of cycles with stock out
ROP] timelead during [Demand inventory]t [Sufficien
inventory sufficient has cycle single ay that Probabilit0.7
7.0
otherwise 1 stockout, has cycle a if 0 Write10
1010111011
:cycles 10consider exampleFor
CSL
CSL
CSL
ROP: Reorder point
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I. Cycle Service Level for Normal Demands
ondistributi normal standard Obtaining )1,0(
StDev by the Dividing ),(
mean out the Taking ),(
),(
notation variance)(mean, Recall ),,(
)1,0(
2
2
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R
R
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R
R
R
R
R
L
LRROPNP
L
LRROP
L
LRLLRNP
LRROPLRLLRNP
ROPLLRNP
NLLRROPFCSL
The last equality is a property of the Normal distribution.
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Example: Finding CSL for given ROP
R = 2,500 /week; = 500L = 4 weeks; Q = 10,000; ROP = 16,000
ss = ROP – L R = Cycle service level,
Average Inventory = Average Flow Time =Average inventory/Thruput=
L timelead during demand of Stdev
F ROP L R L( ; , )
If you wish to compute Average Inventory = Q/2 + ss
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Safety Inventory: CSL ROP
L
LCSLFLCSLF
LRLLRCSLFss
LRROPss
LLRCSLFROPLLRROPFCSL
R
RR
L,0,1)Norminv(CS
)1,0;(),0;(
),,(
:demand ddistributenormally For
:stockSafety
),,(or ),,(
11
1
1
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
Factors driving safety inventory– Replenishment lead time
– Demand uncertainty
ss F CSL L
ROP L R ss
1 0 1( ; , )
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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
ESC E
ESC x ROP f x dxx ROP
(max{
( ) ( )
Demand during lead time - ROP,0})
=
ROPDemand
ROPDemand
if
if
0
ROPDemandShortage
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Inventory and Demand during Lead Time
0
ROP
Demand During LT
LT
InventoryDLT: Demand During LT
0
0
ROPInventory=ROP-DLTUpside
down
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Shortage and Demand during Lead Time
0
ROP
Demand During LT
LTShortage
DL
T: D
eman
d D
urin
g L
T0
0
ROP
Shortage=DLT-ROP
Upsidedown
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Expected shortage per cycle First let us study shortage during the lead time
DLT. of pdf is f where)()(
))max(,0(shortage Expected
D
ROPD
D dDDfROPD
ROPDLTE
Ex:
4
1
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110)}-(11max{0,
4
210)}-(10max{0,
4
110)}-(9max{0,
)()}(d)}(max{0,shortage Expected
Shortage? Expected ,
1/4 prob with 11
2/4 prob with 10
1/4 prob with 9
,10
11
10d
3
1i
33
22
11
dDPROPpROPd
pd
pd
pd
DROP
ii
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Expected shortage per cycle
If demand is normal:
Does ESC decrease or increase with ss, L? Does ESC decrease or increase with expected value of demand?
2170-172
)10(102
10
6
1)12(10
2
12
6
110
26
1
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1)10(shortage Expected
Shortage? Expected ),12,6( ,10
2212
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D
DD
DD
dDD
UniformDROP
Ex:
ESC ss normdistss
LL normdist
ss
L
1 0 11 0 1 0
, , , , , ,
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Fill Rate
Fill rate: Proportion of customer demand satisfied from stock Q: Order quantity
Q
ESCfr 1
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Finding the Fill Rate ss fr
= 500; L = 2 weeks; ss=1000; Q = 10,000; Fill Rate (fr) = ?
fr = (Q - ESC)/Q = (10,000 - 25.13)/10,000 = 0.9975.
ESC ss normdistss
LL normdist
ss
L
ESC normdist
nomdist
ESC
1 0 11 0 1 0
1000 1 1000 707 0 11
707 1000 707 0 1 0
2513
, , , , , ,
( ( / , , , )
( / , , , )
.
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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
0,1,0,
7077071,1,0,
7071250
ssnormdist
ssnormdistss
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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
2222
11
:)( )( L
L
ii
L
ii sRLDVarLRDE
Supplier’s lead time may be uncertain:
timelead of Variance time.lead Average 2 sL
22211 )1,0;()1,0;( sRLCSLFCSLFss L
0,1,0;1,1,0;1
LL
L
ssnomdist
ssnormdistssESC
The formulae do not change:
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Impact of Lead Time Variability, s
R = 2,500/day; = 500
L = 7 days; Q = 10,000; CSL = 0.90
StDev of LT ss Jump in ss
0 1695 -
1 3625 1930
2 6628 3003
3 9760 3132
4 12927 3167
5 16109 3182
6 19298 3189
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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
K
iji
ji
K
ii
CK
ii
CDDRR
11
22
1
),cov(2)(;
),(111 RD
),( CC
R
Which of two systems provides a higher level of service for a given safety stock? Consider locations and demands:
),(333 RD
),(222 RD
),(444 RD
With k locations centralized, mean and variance of
K
ii
C DD1
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Sum of Random Variables Are Less Variable
K
ii
K
ii
C
11
2
K
ii
K
ii
K
i
K
jii
jiiC
1
2
11 1
2 2
When they are independent, cov(Di,Dj)=0
When they are perfectly positively correlated, cov(Di,Dj)=σi σj
When they are perfectly negatively correlated, cov(Di,Dj)= - σi σj
K
ii
K
ii
K
i
K
jii
jiiC
11
2
1 1
2 2
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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
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Impact of Correlation on Aggregated Safety Inventory (Aggregating 4 outlets)
0
0.1
0.2
0.3
0.4
0.5
0.6
0 0.2 0.4 0.6 0.8 1
Benefit
Benefit=(SS before - SS after) / SS before
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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?
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Use CSL=0.95 Supply lead time L=4 weeks for motors and cleaners
L L,0,1)Norminv(CS ss :demand ddistributenormally For For a single store
Motor safety inventory=Norminv(0.95,0,1) 2 (40)=132Cleaner safety inventory=Norminv(0.95,0,1) 2 (100)=329Value 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 storesStandard deviation of Motor demand=40(40)=1,600Standard deviation of Cleaner demand=40(100)=4,000
For the aggregated storeMotor safety inventory=Norminv(0.95,0,1) 2 (1600)=5,264Cleaner safety inventory=Norminv(0.95,0,1) 2 (4,000)=13,159Value of motor ss=5264(500)=$2,632,000Value 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 Cleaner Mean demand/wk 20 1,000 SD of demand 40 100 Disaggregate cv 2 0.1 Value/Unit $500 $30 Disaggregate ss value $105,600,000 $15,792,000 Aggregate cv 0.05 0.0025 Aggregate ss value $2,632,000 $394,770 Inventory cost savings $102,968,000 $15,397,230 Holding Cost Saving $25,742,000 $3,849,308 Saving / Unit 2574200/(1600*20*52)=$15.47 $0.046
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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.
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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.
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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.
P1
P2
P3
M1
M2
M3
H1
H2
H3
Left Right
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Example 11.9: Value of Component Commonalityin Safety Inventory Reduction
0
50000
100000
150000
200000
250000
300000
350000
400000
450000
1 2 3 4 5 6 7 8 9
SS
# of products a component is used in
Aggregation provides reduction in total standard deviation.
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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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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 (Lot size)
Transport Cost
Cycle Inventory
Safety Inventory
Transit Inventory
Inventory Cost
Total Cost
AM Rail (2,000)
$78,000 120000(0.65)
1,000 986 1,644 120000(5/365)
$108,900 $186,900
Northeast Trucking (1,000)
$90,000 500 658 986 $64,320 $154,320
Golden (500)
$96,000 120000(0.80)
250 658 986 120000(3/365)
$56,820 $152,820
Golden (2,500)
$86,400 1,250 658 986 $86,820 $173,220
Golden (3,000)
$78,000 1,500 658 986 $94,320 $172,320
Golden (4,000)
$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
Current Scenario
Option A Option B
# Locations 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
Reduce Buffer Inventory
Economies of ScaleSupply / Demand
VariabilitySeasonal
Variability
Cycle Inventory Safety Inventory Seasonal Inventory
Match Supply & Demand
•Reduce fixed cost•Aggregate across products•Volume discounts•Promotion on Sell thru
•Quick Response measures•Reduce Info Uncertainty•Reduce lead time•Reduce supply uncertainty
•Accurate Response measures•Aggregation•Component commonality and postponement
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Mass customization:– A strategy of producing standardized goods or services,
but incorporating some degree of customization– Modular design– Delayed differentiation
Mass Customization
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Mass Customization I: Customize Services Around Standardized Products
DEVELOPMENT PRODUCTION MARKETING DELIVERY
Deliver customized services aswell as standardized productsand services
Market customized services with standardizedproducts or services
Continue producing standardized products or services
Continue developing standardized products or services
Source: B. Joseph PineWarranty for contact lenses:
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Mass Customization II: Create Customizable Products and Services
DEVELOPMENT PRODUCTION MARKETING DELIVERY
Deliver standard (but customizable) productsor services
Market customizable products or services
Produce standard (but customizable) products or services
Develop customizable products or services
Customizing the look of screen with windows operating system Gillette sensor adjusting to the contours of the face
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Mass Customization III: Provide Quick Response Throughout Value Chain
DEVELOPMENT PRODUCTION MARKETING DELIVERY
Reduce Delivery Cycle Times
Reduce selection and order processing cycle times
Reduce Production cycle time
Reduce development cycle time
Skiing parkas manufactured abroad vs. in the U.S.A.:
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Mass Customization IV: Provide Point of Delivery Customization
DEVELOPMENT PRODUCTION MARKETING DELIVERY
Deliver standardize portion
Market customized products or services
Produce standardized portion centrally
Develop products where point of delivery customization is feasible
Point of deliverycustomization
Paint mixingLenscrafters for glasses.
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Mass Customization V: Modularize Components to Customize End Products
DEVELOPMENT PRODUCTION MARKETING DELIVERY
Deliver customized product
Market customized products or services
Produce modularized components
Develop modularized products
Computer industry, Dell computers:
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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 ReviewOrder at fixed time intervals (T apart) to raise total inventory
(on hand + on order) to Order up to Level (OUL)
Inventory
OUL
LT
T
LT
T+LT
OUL must cover the Demand during
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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
ssOUL
CSLFss
TL
RLT
R
R
LT
LT
LT
LT
)1,0;(
)(
1
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Example: Periodic Review Policy
R = 2,500/week; R = 500L = 2 weeks; T = 4 weeks; CSL = 0.90What is the required safety inventory?
Factors driving safety inventory– Demand uncertainty– Replenishment lead time– Reorder interval
1570)1,0;(1
LTCSLFss
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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