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Chapter 11
Inventory Management
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Types of Inventories
• Raw materials & purchased parts
• Incoming students
• Work in progress
• Current students
• Finished-goods inventories – (manufacturing firms) or merchandise (retail
stores)
• Graduating students
• Replacement parts, tools, & supplies
• Goods-in-transit to warehouses or customers
• Students on leave
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Functions of Inventory
• To meet anticipated demand
• To smooth production requirements
• To decouple components of the production-distribution
• To protect against stock-outs
• To take advantage of order cycles
• To help hedge against price increases or to take advantage of quantity discounts
• To permit operations
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• Inventory level– Low or high
• Customer service levels – Can you deliver what customer wants?
• Right goods, right place, right time, right quantity
• Inventory turnover– Cost of goods sold per year / average inventory investment
• Inventory costs, more will come– Costs of ordering & carrying inventories
Decisions: Order size and time
Inventory performance measures and levers
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Inventory Counting Systems
• A physical count of items in inventory
• Periodic/Cycle Counting System: Physical count of items made at periodic intervals–How much accuracy is needed?
–When should cycle counting be performed?
–Who should do it?
• Continuous Counting System System that keeps track of removals from inventory continuously, thus monitoring current levels of each item
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Inventory Counting Systems (Cont’d)
• Two-Bin System - Two containers of inventory; reorder when the first is empty
• Universal Bar Code - Bar code printed on a label that hasinformation about the item to which it is attached 0
214800 232087768
•RFID: Radio frequency identification
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• Lead time: time interval between ordering and receiving the order, denoted by LT
• Holding (carrying) costs: cost to carry an item in inventory for a length of time, usually a year, denoted by H
• Ordering costs: costs of ordering and receiving inventory, denoted by S
• Shortage costs: costs when demand exceeds supply
Key Inventory Terms
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• A system to keep track of inventory
• A reliable forecast of demand
• Knowledge of lead times
• Reasonable estimates of
– Holding costs
– Ordering costs
– Shortage costs
• A classification system
Effective Inventory Management
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ABC Classification System
Classifying inventory according to some measure of importance and allocating control efforts accordingly.
Importance measure= price*annual sales
AA - very important
BB - mod. important
CC - least important Annual $ volume of items
AA
BB
CC
High
Low
Few ManyNumber of Items
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Inventory Models• Fixed Order Size - Variable Order Interval Models:
– 1. Economic Order Quantity, EOQ– 2. Economic Production Quantity, EPQ– 3. EOQ with quantity discounts
All units quantity discount – 3.1. Constant holding cost
– 3.2. Proportional holding cost
– 4. Reorder point, ROP• Lead time service level• Fill rate
• Fixed Order Interval - Variable Order Size Model– 5. Fixed Order Interval model, FOI
• Single Order Model– 6. Newsboy model
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• Assumptions:– Only one product is involved– Annual demand requirements known– Demand is even throughout the year– Lead time does not vary– Each order is received in a single delivery
• Infinite production capacity– There are no quantity discounts
1. EOQ Model
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The Inventory Cycle
Profile of Inventory Level Over Time
Quantityon hand
Q
Receive order
Placeorder
Receive order
Placeorder
Receive order
Lead time
Reorderpoint
Usage rate
Time
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Average inventory held
• Length of an inventory cycleFrom one order to the next = Q/D
• Inventory held over entire inventory cycleArea under the inventory level = ½ Q (Q/D)
• Average inventory held= Inventory held over a cycle / cycle length
= Q/2
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Total Cost
Annualcarryingcost
Annualorderingcost
Total cost = +
Q2
H DQ
STC = +
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Figure 11-4
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Cost Minimization Goal
Order Quantity (Q)
The Total-Cost Curve is U-Shaped
Ordering Costs
QO
An
nu
al C
os
t
(optimal order quantity)
TCQ
HD
QS
2
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Deriving the EOQ
Using calculus, we take the derivative of the total cost function and set the derivative (slope) equal to zero and solve for Q.
The total cost curve reaches its minimum where the carrying and ordering costs are equal.
Q = 2DS
H =
2(Annual Demand)(Order or Setup Cost)
Annual Holding CostOPT
DSHEOQQ 2)cost( Total
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EOQ example
Demand, D = 12,000 computers per year. Holding cost, H = 100 per item per year. Fixed cost, S =
$4,000/order.Find EOQ, Cycle Inventory, Optimal Reorder Interval and
Optimal Ordering Frequency.
EOQ = 979.79, say 980 computers Cycle inventory = EOQ/2 = 490 unitsOptimal Reorder interval, T = 0.0816 year = 0.98 monthOptimal ordering frequency, n=12.24 orders per year.
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Optimal Quantity is robust
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Total Costs with Purchasing Cost
Annualcarryingcost
PurchasingcostTC = +
Q2
H DQ
STC = +
+Annualorderingcost
PD +
Note that P is the price.Note that P is the price.
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Total Costs with PDC
ost
EOQ
TC with PD
TC without PD
PD
0 Quantity
Adding Purchasing costdoesn’t change EOQ
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• Production done in batches or lots• Capacity to produce a part exceeds the part’s
usage or demand rate• Assumptions of EPQ are similar to EOQ
except orders are received incrementally during production– This corresponds to producing for an order with
finite production capacity
2. Economic Production Quantity
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• Only one item is involved• Annual demand is known• Usage rate is constant• Usage occurs continually
• Production rate p is constant
• Lead time does not vary• No quantity discounts
Economic Production Quantity Assumptions
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Economic Production Quantity
Inventory Level
Usage Usage
Pro
duct
ion
& U
sage
Pro
duct
ion
& U
sage
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Average inventory held
Q/p
Dp-D
Q/DTimeTime
(Q/p)(p-D)(Q/p)(p-D)
Average inventory held=(1/2)(Q/p)(p-D)Average inventory held=(1/2)(Q/p)(p-D)
Total cost=(1/2)(Q/p)(p-D)H+(D/Q)STotal cost=(1/2)(Q/p)(p-D)H+(D/Q)S
Dp
p
H
DSQ
2
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EPQ example
Demand, D = 12,000 computers per year. p=20,000 per year. Holding cost, H = 100 per item per year. Fixed cost, S = $4,000/order.
Find EPQ.
EPQ = EOQ*sqrt(p/(p-D))
=979.79*sqrt(20/8)=1549 computers
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3. All unit quantity discountCost/Unit
$3$2.96
$2.92
Order Quantity
5,000 10,000
Two versionsTwo versionsConstant HConstant H
Proportional HProportional H
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3.1.Total Cost with Constant Carrying Costs
EOQ Quantity
To
tal C
os
t
TCa
TCc
TCbDecreasing Price
Annual demand*discountAnnual demand*discount
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3.1.Total Cost with Constant Carrying Costs
EOQ Quantity
To
tal C
os
t
TCa
TCc
TCbDecreasing Price
Annual demand*discountAnnual demand*discount
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Example Scenario 1
Q*=EOQ Quantity
To
tal C
os
t
aa bb cc
Price a > Price b > Price cPrice a > Price b > Price c
TCa
TCc
TCb
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Example Scenario 2
EOQ Quantity
To
tal C
os
t
aa bb cc
Price a > Price b > Price cPrice a > Price b > Price c
TCa
TCc
TCb
Q*
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Example Scenario 3
EOQ Quantity
To
tal C
os
t
TCa
TCc
TCb
aa bb cc
Price a > Price b > Price cPrice a > Price b > Price c
Q*
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Example Scenario 4
Q*=EOQ Quantity
To
tal C
os
t
TCa
TCc
TCb
aa bb cc
Price a > Price b > Price cPrice a > Price b > Price c
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3.1. Finding Q with all units discount with constant holding cost
Note all the price ranges have the same EOQNote all the price ranges have the same EOQ..Stop if EOQ=Q1 is in the lowest cost range (highest Stop if EOQ=Q1 is in the lowest cost range (highest
quantity range), otherwise continue towards quantity range), otherwise continue towards quantity break points which give lower costsquantity break points which give lower costs
Quantity
To
tal C
os
t
1122
H
DSQ
21
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All-units quantity discountsConstant holding cost
A popular shoe store sells 8000 pairs per year. The fixed cost of ordering shoes from the distribution center is $15 and holding costs are taken as $12.5 per shoe per year. The per unit purchase costs from the distribution center is given as
C3=60, if 0 < Q < 50
C2=55, if 50 <= Q < 150
C1=50, if 150 <= Q
where Q is the order size. Determine the optimal order quantity.
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Solution• There are three ranges for lot sizes in this problem:
– (0, q2=50), – (q2=50, q1=150) – (q1=150,infinite).
• Holding costs in all there ranges of shoe prices are given as H=12.5, • EOQ is not feasible in the lowest price range because
138.6 < 150.• The order quantity q1=150 is a candidate with cost
TC(150)=8000(50)+8000(15)/150+(12.5)(150)/2 =401,900• Let us go to a higher cost level of (q2=50, q1=150).• EOQ=138.6 is in the appropriate range, so it is another
candidate with costTC(138.6)=8000(55)+8000(15)/138.6+(12.5)(138.6)/2
=441,732• Since TC(150) < TC(132.1), Q=150 is the optimal solution.• Remark: In these computations, we do not need to compute
TC(50), why? Because TC(50) >= TC(132.1).
6.1385.12
)8000)(15(2EOQ
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3.2. Summary of finding Q with all units discount with proportional holding cost
Quantity
To
tal C
os
t
11
11
2
H
DSQ
Note each price range has a different EOQ.Note each price range has a different EOQ.Stop if Q1=“EOQ of the lowest price” feasibleStop if Q1=“EOQ of the lowest price” feasibleOtherwise continue towards higher costs until an EOQ Otherwise continue towards higher costs until an EOQ becomes feasible. becomes feasible.
In each price range, evaluate the lowest costIn each price range, evaluate the lowest cost..Lowest cost is either at an EOQ or price break quantityLowest cost is either at an EOQ or price break quantity
Pick the minimum cost among all evaluatedPick the minimum cost among all evaluated
ExampleExample: Q: Q11
feasible stopfeasible stop
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3.2. Finding Q with all units discount with proportional holding cost
Quantity
To
tal C
os
t
1’1’
11
2
H
DSQ
22
2
H
DSQ
22 ExampleExample: Q: Q1 1 infeasible, Qinfeasible, Q22
feasible, Break point 1 is feasible, Break point 1 is selected since TCselected since TC11 < TC < TC22
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3.2. Finding Q with all units discount with proportional holding cost
Quantity
To
tal C
os
t
1’1’
Stop if 1 is feasible, otherwise Stop if 1 is feasible, otherwise continue towards higher costs until a continue towards higher costs until a
EOQ becomes feasible. EOQ becomes feasible. Evaluate Evaluate cost at all alternativescost at all alternatives
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All-units quantity discountsProportional holding cost
A popular shoe store sells 8000 pairs per year. The fixed cost of ordering shoes from the distribution center is $15 and holding costs are taken as 25% of the shoe costs. The per unit purchase costs from the distribution center is given as
C3=60, if 0 < Q < 50
C2=55, if 50 <= Q < 150
C1=50, if 150 <= Q
where Q is the order size. Determine the optimal order quantity.
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Solution• There are three ranges for lot sizes in this problem:
– (0, q2=50), – (q2=50, q1=150) – (q1=150,infinite).
• Holding costs in there ranges of shoe prices are given as
– H3=(0.25)60=15, – H2 =(0.25)55=13.75 – H1 =(0.25)50=12.5.
• EOQ1 is not feasible because 138.6 < 150.• The order quantity q1=150 is a candidate with cost
TC(150)=8000(50)+8000(15)/150+(0.25)(50)(150)/2 =401,900• Let us go to a higher cost level of (q2=50, q1=150).• EOQ2=132.1 is in the appropriate range, so it is another candidate with
costTC(132.1)=8000(55)+8000(15)/132.1+(0.25)(55)(132.1)/2
=441,800 • Since TC(150) < TC(132.1), Q=150 is the optimal solution.• We do not need to compute TC(50) or EOQ3, why?
6.1385.12
)8000)(15(2
1.13275.13
)8000)(15(2
1
2
EOQ
EOQ
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Types of inventories (stocks) by function
Deterministic demand case• Anticipation stock
– For known future demand
• Cycle stock– For convenience, some operations are performed occasionally and
stock is used at other times• Why to buy eggs in boxes of 12?
• Pipeline stock or Work in Process– Stock in transfer, transformation. Necessary for operations.
• Students in the class
Stochastic demand case• Safety stock
– Stock against demand variations
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4. When to Reorder with EOQ Ordering
• Reorder Point - When the quantity on hand of an item drops to this amount, the item is reordered. We call it ROP.
• Safety Stock - Stock that is held in excess of expected demand due to variable demand rate and/or lead time. We call it ss.
• (lead time) Service Level - Probability that demand will not exceed supply during lead time. We call this cycle service level, CSL.
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Optimal Safety Inventory Levels
Lead Times
time
inventory
Shortage
An inventory cycle
ROP
Q
45
Safety Stock
LT Time
Expected demandduring lead time
Maximum probable demandduring lead time
ROP
Qu
an
tity
Safety stock
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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
ROP
0
Demand During LT
LTShortage
DL
T: D
eman
d D
urin
g L
T0
0
ROP
Shortage=DLT-ROP
Upsidedown
)}(,0max{ ROPDLTShortage
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Cycle Service Level
Cycle service level: percentage of cycles with shortage
ROP] timelead during [Demand inventory]t [Sufficien
inventory sufficient has cycle single ay that Probabilit0.7
7.0
otherwise 1 shortage, has cycle a if 0 Write10
1010111011
:cycles 10consider exampleFor
CSL
CSL
CSL
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DLT : Demand during lead timeLT and demand may be uncertain.
2222
1
1
)(
timelead during demand theof Variance
))(()(
timelead during demand theof valueExpected
DLTLTD
LT
ii
LT
ii
DLDVar
DLDE
timelead of Variance
periods ofnumber in timelead Average
demand of Variance
periodper demand Average
2
2
LT
D
L
D
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Reorder Point
ROP
Risk ofa stockout
Service level
Probability ofno stockout
Expecteddemand Safety
stock0 z
Quantity
z-scale
ROP = E(DLT) + z ROP = E(DLT) + z σσDLTDLT
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Finding safety stock from cycle service level CSL
SL)normsinv(C
SL)normsinv(C
variablenormal Standard:z )(
)),(()( 2
DLT
DLT
DLT
DLT
ssDLROP
DLROP
DLROPzP
ROPDLNormalPROPDLTPCSL
Note we use normal density for the demand Note we use normal density for the demand during lead time during lead time
The excel function normsinv has default values of 0 and 1 for the mean and standard deviation. Defaults are used unless these values are specified.
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Example: Safety inventory vs. Lead time variability
D = 2,500/day; D = 500L = 7 days; CSL = 0.90
Normsinv(0.9)=1.3, either from table or Excel
If LT=0, DLT=sqrt(7)*500=1323ss=1323*normsinv(0.9)=1719.8
ROP=(D)(L)+ss=17500+1719.8
If LT=1, DLT=sqrt(7*500*500+2500*2500*1)=2828ss=2828*normsinv(0.9)=3676
ROP=(D)(L)+ss=17500+3676
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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
4
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 we assume that DLT is normal,
3.-11 Table use )()),0(max( then Let zEROPDLTELDROP
z DLTDLT
2170-172
)10(102
10
6
1)12(10
2
12
6
110
26
1
6
1)10(shortage Expected
Shortage? Expected ),12,6( ,10
2212
10
212
10
D
DD
DD
dDD
UniformDROP
• Ex:
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Example: Finding expected shortage per cycle
Suppose that the demand during lead time has expected value 100 and stdev 30, find the expected shortage if ROP=120.
z=(120-100)/30=0.66.
E(z)=0.153 from Table 11-3.
Expected shortage = 30*0.153=4.59
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Fill rate
• Fill rate is the percentage of demand filled from the stock
• In a cycle– Fill rate = 1-(Expected shortage during LT) / Q
• For normally distributed demand
Q
zED
D
Q
zE
DLT
DLT
)(
rate) Fill1(*yearper short units ofnumber Expected
)(1
cycleper Demand
cycleper shortage Expected1rate Fill
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Example: computing the fill rate
Suppose that the demand during lead time has expected value 100 and stdev 30, find the expected shortage if ROP=120. Compute the fill rate if order sizes are 200. Compute the annual expected shortages if there are 12 order cycles per year.
Expected shortage per cycle=4.59 from the last example.
Fill rate = 1-4.59/20=0.7705Annual expected shortage=12*4.59=55.08.
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Determinants of the Reorder Point
• The rate of demand• The lead time• Demand and/or lead time variability• Stockout risk (safety stock)
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• Orders are placed at fixed time intervals• Order quantity for next interval?• Suppliers might encourage fixed intervals• May require only periodic checks of inventory levels• Items from same supplier may yield savings in:
– Ordering– Packing– Shipping costs
• May be practical when inventories cannot be closely monitored
5. Fixed-Order-Interval Model
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A single order must cover the demand until the next order arrives. Exposure to random demand during not only lead time but also before.
Requires higher safety stock than variable order interval models.
May provide savings in set up / ordering costs.
FOI compared against variable order interval models
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6. How many to order for the winter? Parkas at L.L. Bean
Demand Di
Probabability pi
Cumulative Probability of demand being this size or less, F()
Probability of demand greater than this size, 1-F()
4 .01 .01 .99 5 .02 .03 .97 6 .04 .07 .93 7 .08 .15 .85 8 .09 .24 .76 9 .11 .35 .65
10 .16 .51 .49 11 .20 .71 .29 12 .11 .82 .18 13 .10 .92 .08 14 .04 .96 .04 15 .02 .98 .02 16 .01 .99 .01 17 .01 1.00 .00
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Parkas at L.L. Bean
• Cost per parka = c = $45• Sale price per parka = p = $100• Discount price per parka = $50• Holding and transportation cost = $10
• Salvage value per parka = s = $40
• Profit from selling parka = p-c = 100-45 = $55
Underage cost=$55• Cost of overstocking = c-s = 45-40 = $5
Overage cost=$5
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• Single period model: model for ordering of perishables and other items with limited useful lives
• Shortage cost: generally the unrealized profits per unit, $55 for L.L.Bean. We call this underage.
• Excess cost: difference between purchase cost and salvage value of items left over at the end of a period, $5 for L.L.Bean. We call this overage.
6. Single Period Model
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Parkas at L.L. Bean
• Expected demand = 10 (‘00) parkas• Expected profit from ordering 10 (‘00) parkas = $499
The underage and overage probabilities after ordering 1100 parkas
P(D>1100):Probability of underage
P(D<1100):Probability of overage
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Optimal level of product availability
p = sale price; s = outlet or salvage price; c = purchase price
CSL = Probability that demand will be at or below reorder point
At optimal order size,
Expected Marginal Benefit from raising order size =
=P(Demand is above stock)*(Profit from sales)=(1-CSL*)(p - c)
Expected Marginal Cost =
=P(Demand is below stock)*(Loss from discounting)=CSL*(c - s).
Let Co= c-s; Cu=p-c, then the optimality condition is
(1-CSL*)Cu = CSL* Co,
CSL* = Cu / (Cu + Co)
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Parkas at L.L. Bean
Additional100s
ExpectedMarginal Benefit
ExpectedMarginal Cost
Expected MarginalContribution
11th 5500.49 = 2695 500.51 = 255 2695-255 = 2440
12th 5500.29 = 1595 500.71 = 355 1595-355 = 1240
13th 5500.18 = 990 500.82 = 410 990-410 = 580
14th 5500.08 = 440 500.92 = 460 440-460 = -20
15th 5500.04 = 220 500.96 = 480 220-480 = -260
16th 5500.02 = 110 500.98 = 490 110-490 = -380
17th 5500.01 = 55 500.99 = 495 55-495 = -440
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Order Quantity for a Single Order
Co = Cost of overstocking = $5
Cu = Cost of understocking = $55
Q* = Optimal order size
demandstdevdemandmeanCC
CQ
CC
CQDemandPCSL
ou
u
ou
u
_,_,norminv
d,distributenormally is demand theIf
slide.next theSee
917.0555
55)(
*
*
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Optimal Order Quantity without Normal Demands
0
0.2
0.4
0.6
0.8
1
1.2
CumulativeProbability
Optimal Order Quantity = 13(‘00)
0.917
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• Too much inventory– Tends to hide problems– Easier to live with problems than to eliminate them– Costly to maintain
• Wise strategy– Reduce lot sizes– Reduce set ups– Reduce safety stock– Aggregate negatively correlated demands
• Remember component commonality• Delayed postponement
Operations Strategy