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Managing Inventories

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Managing Inventories
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

 DeliveryLead 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

CSL is 0%, fill rate is almost 0%



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 ) correlation coefficient

L L

E ( Di )   Ri by the linearity of integratio n

i 1 i 1

L L L L L L

Var( Di )   cov(Di , D j )    cov(Di , D j ) i

2



i 1 i 1 j 1 i 1 i 1 j 1 6

utdallas.edu/~metin j i

Normal Density Function



frequency

normdist(x,.,.,1) normdist(x,.,.,0)



Prob





  Mean x  



95.44%



99.74%



Excel statistical functions :

Density function (pdf) at x : normdist( x, mean, st _ dev,0)

Cumulative function (cdf)at x : normdist( x, mean, st _ dev,1) 7

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Cumulative Normal Density

1

prob







normdist(x,mean,st_dev,1)





0 x

norminv(prob,mean,st_dev)



Excel statistical functions :

Cumulative function (cdf)at x : normdist( x, mean, st _ dev,1)

Inversefunction of cdf at " prob": norminv( prob, mean, st _ dev) 8

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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

E ( Di )  LR and Var( Di )  L 2

i 1 i 1

L

If Di  N ( R,  ) then  Di  N ( LR, L 2 )

2



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.

Coefficient of variation of D : cv  Var( D) / E( D) 9

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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 consider10 cycles:

11 0 111 0 1 0 1

CSL  Write0 if a cycle has stockout,1 otherwise

10

CSL  0.7

CSL  0.7  Probabilit y that a single cycle has sufficient inventory

[Sufficient 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 ) Recall N (mean,variance) notation



 P N ( R  L, L R )  ROP

2



 P N ( R  L, L )  R  L  ROP  R  L 

2

R Taking out the mean

 

 

 N ( R  L, L R )  R  L  ROP  R  L  Dividing by theStDev

2

P

 L R L R 

  

 N ( 0 ,1) 

 ROP  R  L 

 P N (0,1) 



 Obtaining standard normal distribution

 L R 



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 distributed demand :

ss  F 1 (CSL, R  L, L R )  R  L

 F 1 (CSL;0, L )  F 1 (CSL;0,1)  L

 Norminv(CSL,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 Demand  ROP

Shortage  

 0 if 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=

ROP ROP-DLT

Upside

0 Inventory down

DLT: Demand

During LT

LT

Demand 0

During LT 17

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Shortage and Demand during Lead Time









ROP 0









DLT: Demand During LT

Shortage=

DLT-ROP

Upside ROP

0 down



Shortage

LT

Demand 0

During LT 18

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Expected shortage per cycle

 First let us study shortage during the lead time

Expected shortage  E (0, max(DLT  ROP))



  ( D  ROP) f

D  ROP

D ( D)dD wheref D is pdf of DLT.





 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 11

Expected shortage   max{0,(d i  ROP)} pi   (d  ROP)}P( D  d )

i 1 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  D2 1  122  1  102 

12

1

Expected shortage   ( D  10) dD    10D    10(12)     10(10) 

D 10

6 6 2

 D 10

6 2



 6 2

 





 172 - 170  2



If demand is normal:



  ss   ss 

ESC   ss1  normdist  ,0,11   L  normdist 

, ,0,1,0

  L   L 



 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   ss 

ESC   ss 1  normdist  ,0,11  

, L  normdist  ,0,1,0

  L   L 

ESC  1000(1  normdist (1000 / 707,0,11)

,

 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. s 2  Variance of lead time

L L

E ( Di )  LR Var ( Di )  L   2  R 2  s 2 :  L

2



i 1 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 

  ;0,1,1   L nomdist  ;0,1,0 

ESC   ss 1  normdist   

  L   L  27

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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

Which of two systems provides a higher level of service for a given

safety stock?

Consider locations and demands:

D1  ( R1 , ) D2  ( R2 , )



( R , )

1 2

C C





D3  ( R3 , 3) D4  ( R4 , 4)

K

With k locations centralized, mean and variance of D C   Di

i 1

K K K

  Ri;  )    2 cov(Di , D j )

C C 2 2

R i 1

(

i 1

i

i j

i 1 30

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Sum of Random Variables Are Less Variable

K K

When they are independent, C

cov(Di,Dj)=0   

i 1

i

2

  i

i 1



When they are perfectly positively correlated,

cov(Di,Dj)=σi σj 2

K K

  KK

 C    i2  2  i j     i     i

i 1 i 1  i 1  i 1

i j



When they are perfectly negatively correlated,

cov(Di,Dj)= - σi σj K K K K

C    i2  2  i j 

i 1 i 1

  i2    i

i 1 i 1

i j

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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? 32

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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)

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



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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

For normally distributed 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

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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 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 $15.47

2574200/(1600*20*52)= $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 M1 H1





Left P2 H2 Right

M2





P3 M3 H3 41

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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. 42

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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

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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 Transport Cycle Safety Transit Inventory Total

(Lot size) Cost Inventory Inventory Inventory Cost Cost



AM Rail $78,000 1,000 986 1,644 $108,900 $186,900

(2,000) 120000(0.65) 120000(5/365)

Northeast $90,000 500 658 986 $64,320 $154,320

Trucking

(1,000)

Golden $96,000 250 658 986 $56,820 $152,820

(500) 120000(0.80) 120000(3/365)

Golden $86,400 1,250 658 986 $86,820 $173,220

(2,500)

Golden $78,000 1,500 658 986 $94,320 $172,320

(3,000)

Golden $67,500 2,000 658 986 $109,320 $176,820

(4,000)

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 Option A Option B

Scenario

# 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

Match Supply & Demand



Reduce Buffer Inventory





Supply / Demand Seasonal

Economies of Scale Variability Variability



Cycle Inventory Safety Inventory Seasonal Inventory



•Reduce fixed cost •Quick Response measures

•Aggregate across •Reduce Info Uncertainty

products •Reduce lead time

•Volume discounts •Reduce supply

•Promotion on Sell uncertainty

thru •Accurate Response measures

•Aggregation

•Component commonality 52

utdallas.edu/~metin and postponement

Mass Customization



Mass customization:

– A strategy of producing standardized goods or services,

but incorporating some degree of customization

– Modular design

– Delayed differentiation









53

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Mass Customization I: Customize

Services Around Standardized Products

Warranty for contact lenses: Source: B. Joseph Pine









DEVELOPMENT 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

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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

57

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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

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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 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+LT T



OUL









LT LT 61

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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





R T L

 (T  L) R



 T L

 L T

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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