1

Chapter 11

Inventory Management

2

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

3

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

4

• 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

5

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

6

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

7

• 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

8

• 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

9

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

10

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

12

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

13

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

14

Total Cost

Annualcarryingcost

Annualorderingcost

Total cost = +

Q2

H DQ

STC = +

15

Figure 11-4

16

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

17

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

18

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

20

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.

21

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

25

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

26

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

30

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*

32

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*

33

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

35

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

22

40

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.

41

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.

44

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

46

Inventory and Demand during Lead Time

0

ROP

Demand During LT

LT

InventoryDLT: Demand During LT

0

0

ROPInventory=ROP-DLTUpside

down

47

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

48

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

49

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

50

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

51

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.

52

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

54

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:

55

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

56

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

57

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.

58

Determinants of the Reorder Point

• The rate of demand• The lead time• Demand and/or lead time variability• Stockout risk (safety stock)

59

• 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

61

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

62

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

64

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

65

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)

66

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

67

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)(

*

*

68

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

69

• 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