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Inventory Management 1

Chapter 12

Inventory Management

McGraw-Hill/Irwin Copyright © 2009 by The McGraw-Hill Companies, Inc. All Rights Reserved.

Inventory

• Inventory– A stock or store of goods– A stock or store of goods

• Independent demand items– Items that are ready to be sold or used

• (Dependent demand -> MRP)

12-2


Types of Inventory

• Raw materials and purchased parts• Work-in-process or Semi-Finished GoodsWork in process or Semi Finished Goods• Finished goods inventories or merchandise• Maintenance and repairs (MRO) inventory, tools and

supplies• Goods-in-transit to warehouses or customers (pipeline

inventory)

12-3

Inventory Functions

• Inventories serve a number of functions such as:1. To meet anticipated customer demandp2. To smooth production requirements3. To decouple operations4. To protect against stockouts5. To take advantage of order cycles6. To hedge against price increases7. To permit operations8. To take advantage of quantity discounts

12-4


Inventory Management

• Management has two basic functions concerning inventory:concerning inventory:1. Establish a system for tracking items in inventory2. Make decisions about

• When to order( ROP,FOI)• How much to order (Q,S)

12-5

Effective Inventory Management

• Requires:1. A system keep track of inventoryy p y2. A reliable forecast of demand3. Knowledge of lead time and lead time variability4. Reasonable estimates of

• holding costs• ordering costs• shortage costs

5. A classification system for inventory items

12-6


Inventory Counting Systems

• Periodic System– Physical count of items in inventory made at– Physical count of items in inventory made at

periodic intervals• Perpetual Inventory System

– System that keeps track of removals from inventory continuously, thus monitoring current levels of each item• Two-bin system

– Two containers of inventory; reorderwhen the first is empty

12-7

A B


Inventory Counting Technologies

• Universal product code (UPC)– Bar code printed on a label that has information about– Bar code printed on a label that has information about

the item to which it is attached• Radio frequency identification (RFID) tags

– A technology that uses radio waves to identify objects, such as goods in supply chains

12-9

Demand Forecasts and Lead Time

• Forecasts– Inventories are necessary to satisfy customer demands, so it is y y

important to have a reliable estimates of the amount and timing of demand

• Lead time– Time interval between ordering and receiving the order

• Point-of-sale (POS) systems– A system that electronically records actual sales– Such demand information is very useful for enhancing

forecasting and inventory management

12-10


ABC Classification System

• A-B-C approach– Classifying inventory according to some measure of importance, and

ll ti t l ff t di lallocating control efforts accordingly– A items (very important)

• 10 to 20 percent of the number of items in inventory and about 60 to 70 percent of the annual dollar value

– B items (moderately important)– C items (least important)

• 50 to 60 percent of the number of items in inventory but only A l

High

AAof items in inventory but only about 10 to 15 percent of theannual dollar value

Annual $ valueof items

LowFew ManyNumber of Items

AA

CCBB

12-11

Item Annual Demand

Unit Cost Annual Dollar Value

Classification

1 2500 330

Example 1 page 563

2 1000 703 1900 5004 1500 1005 3900 7006 1000 9157 200 2108 1000 40009 8000 10

10 9000 211 500 20012 400 300


Item Annual Demand

Unit Cost Annual Dollar Value

Classification

1 2500 330 8250002 1000 70 700003 1900 500 9500004 1500 100 1500005 3900 700 27300006 1000 915 9150007 200 210 420008 1000 4000 40000009 8000 10 80000

10 9000 2 1800011 500 200 10000012 400 300 120000

4500000

150000020000002500000300000035000004000000

Series1

0500000

1000000

1 2 3 4 5 6 7 8 9 10 11 12


ItemAnnual

Demand Unit CostAnnual

Dollar Value Classification

8 1000 4000 4000000 A

5 3900 700 2730000 A

3 1900 500 950000 B6 1000 915 915000 B1 2500 330 825000 B4 1500 100 150000 C

12 400 300 120000 C11 500 200 100000 C9 8000 10 80000 C2 1000 70 70000 C7 200 210 42000 C

10 9000 2 18000 C

Cycle Counting

• Cycle counting– A physical count of items in inventory (reduce discrepancies..)p y y ( p )

• Cycle counting management– How much accuracy is needed? (APICS – recommended)

• A items: ± 0.2 percent• B items: ± 1 percent• C items: ± 5 percent

– When should cycle counting be performed?When should cycle counting be performed?– Who should do it?

12-16


How Much to Order: EOQ Models

• The basic economic order quantity model• The economic production quantity model• The economic production quantity model• The quantity discount model

12-17

Basic EOQ Model

• The basic EOQ model is used to find a fixed order quantity that will minimize total annual inventory costsq y y

• Assumptions– Only one product is involved– Annual demand requirements are known– Demand is even throughout the year– Lead time does not vary– Each order is received in a single deliveryEach order is received in a single delivery– There are no quantity discounts

12-18


The Inventory Cycle

Profile of Inventory Level Over TimeQ=350 Usage

rate 50Quantityon hand

Reorderpoint = 100units

rate = 50 units/day

s

Receive order

Placeorder

Receiveorder

Placeorder

Receiveorder

Lead time = 2 days

Time

12-19

Total Annual Cost

Cost Ordering AnnualCost Holding AnnualCost Total

+=

+=

SDHQ

yearper unit in usually Demand, unitper cost (carrying) Holding

unitsin quantity Order where

2

===

+=

DHQ

SQ

H

cost Ordering =S

12-20


Goal: Total Cost Minimization

The Total-Cost Curve is U-Shapedua

l Cos

t

SDHQTC +

Ordering Costs

Ann

u

Holding Costs

SQ

HQTC +=2

Order Quantity (Q)

Ordering Costs

Q* (optimal order quantity)

12-21

Deriving EOQ

• Using calculus, we take the derivative of the total cost function and set the derivative (slope)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.

)dd d)(l(cost holdingunit per annualcost)der demand)(orannual(22* ==

HDSQ

12-22


Example 2 page 568

A local distributor for a national tire company expects to sell approximately 9,600 steel belted radial tires of a certain size and t d d i t A l i t i $16 ti dtread design next year. Annual carrying cost is $16 per tire, and ordering cost is $75.The distributor operates 288 days a year.

a. What is the EOQ?b. How many times per year does the store reorder?c. What is the length of an order cycle?d. What is the total annual cost if the EOQ quantity is ordered?

D = 9,600 tires per yearH = $16 per unit per yearS = $75

a. tires30016

75)600,9(220 ===

HDSQ

b. Number of orders per year: D/Q = 9,600 tires/300 tires = 32

c. Length of order cycle: Q/D = 300 tires/ 9,600 tires/yr =1/32 year

= 288*1/32 = 9 workdays

d. TC = Carrying cost + Ordering cost= (Q/2)H + (D/Q)S= (300/2)16 + (9,600/300)75= $2,400 + $ 2,400= $4,800


• Example 3 page 568

Economic Production Quantity (EPQ)

• Assumptions– Only one product is involvedy p– Annual demand requirements are known– Usage rate is constant– Usage occurs continually, but production occurs periodically– The production rate is constant– Lead time does not vary– There are no quantity discountsq y

12-26


EPQ: Inventory Profile

Q

Production Production ProductionUsage Usage

Q*

Imax

and usage and usage and usageonly only

Cumulativeproduction

Amounton hand

Time

12-27

EPQ – Total Cost

Cost SetupCost CarryingTC

⎞⎛

+=

DI

( )

inventory Maximum where

2

max

max

−=

=

+⎟⎠⎞

⎜⎝⎛=

uppQ

I

SQDHI

rate Usage ratedelivery or Production

==

up

p

12-28


EPQ

upp

HDSQp −

=2*

12-29

Example 4 page 571

A toy manufacturer uses 48,000 rubber wheels per year fir its popular dump truck series. The firm makes its own wheels, which it can produce at a rate of 800 per day The toy trucks are assembledproduce at a rate of 800 per day . The toy trucks are assembled uniformly over the entire year .Carrying cost is $1 per wheel a year. Setup cost for a production run of wheels is $45. the firm operates 240 days per year. Determine the

a. Optimal run size.b. Minimum total annual cost for carrying and setup.c. Cycle time for the optimal run size.d. Run time.

D = 48,000 wheels per yearS = $45H = $1 per wheel per yearp = 800 wheels per dayu = 48,000 wheels per 240 days, or 200 wheels per day


a. wheels2400200800

8001

45)000,48(220 ===up

pHDSQ

b.First compute Imax :

2008001 −−upH

SQDHITC ⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟

⎠⎞

⎜⎝⎛=+=

0

maxmin 2

cost Setup cost Carrying

( ) ( )

800,1$900$900$45$000,481$1,800TC

wheels800,1200800800400,20

max

=+=×+×=

=−=−= upp

QI

c.

d.

,$$$$400,2

$2

days 12dayper wheels200

wheels2,400Q timeCycle 0 ===u

days 3dayper wheels800

wheels400,2 Run time 0 ===p

Q

Quantity Discount Model

• Quantity discount– Price reduction offered to customers for placing large– Price reduction offered to customers for placing large

orders

where2

Cost PurchasingCostOrderingCostCarryingCost Total

++=

++=

PDSQDHQ

priceUnit =P

12-32


Quantity Discounts

12-33

Quantity Discounts

12-34


Example 5 page 575

The maintenance department of a large hospital uses about 816 cases of liquid cleanser annually. Ordering costs are $12, carrying

t $4 d th i h d l i di tcosts are $4 per case a year, and the new price schedule indicates that orders of less than 50 cases will cost $20 per case, 50 to 79 cases will cost $18 per case, 80 to 99 cases will cost $17 per case, and larger orders will cost $16 per case. Determine the optimal order quantity and the total cost. D = 816 cases per year S = $12 H = $4 per case per year

Range PriceRange Price

1 to 49……. $20

50 to 79…… 18

80 to 99…… 17

100 or more…. 16

$20/Case

Range Price

1 to 49……. $20

50 to 79…… 18

80 to 99…… 17

100 or more…. 16

$18/Case

$17/Case

$16/Case

20 60


1. casesHDSEOQ 7097.69

412)816(22

≈===

2. 70 cases can be bought at $18 per case TC70 = Carrying cost + Order cost + Purchase cost

= (Q/2)H + (D/Q0)S + PD= (70/2)4 + (816/70)12 + 18(816) = $14,968

TC50 = (50/2)4 + (816/50)12 + 20(816) = $16,616TC80 = (80/2)4 + (816/80)12 + 17(816) = $14,154TC (100/2)4 (816/100)12 16(816) $13 354 *****TC100 = (100/2)4 + (816/100)12 + 16(816) = $13,354 *****

When to Reorder

• Reorder point– When the quantity on hand of an item drops to this amount, the q y p

item is reordered.– Determinants of the reorder point

1. The rate of demand2. The lead time3. The extent of demand and/or lead time variability4. The degree of stockout risk acceptable to management

12-38


Reorder Point: Under Certainty

LTROP d ×

) as units timesame(in timeLeadLT per week) day,per period,per (units rate Demand

whereLTROP

dd

d

==

×=

12-39

Reorder Point: Under Uncertainty

• Demand or lead time uncertainty creates the possibility that demand will be greater than available supplyg pp y

• To reduce the likelihood of a stockout, it becomes necessary to carry safety stock– Safety stock

• Stock that is held in excess of expected demand due to variable demand and/or lead time

StockSafety timelead duringdemand Expected ROP +=

12-40


Safety Stock

ity

Expected demandduring lead time

Maximum probable demandduring lead timeQ

uant

LT Time

ROP

Safety stock

12-41

Q

ROP

lead timeShortage or back order

0


safety stock ?0

σμ

safety stock ?σ

Safety Stock?

• As the amount of safety stock carried increases, the risk of stockout decreases.– This improves customer service level

• Service level– The probability that demand will not exceed supply during lead

time– Service level = 100% - Stockout risk

12-44


How Much Safety Stock?

• The amount of safety stock that is appropriate for a given situation depends upon:for a given situation depends upon:1. The average demand rate and average lead time2. Demand and lead time variability3. The desired service level

timeleadduringdemand Expected ROP += dLTzσ

demand timelead ofdeviation standard Thedeviations standard ofNumber

wheretimeleadduring

==

dLT

dLT

zσ

12-45

Reorder Point

The ROP based on a normalDistribution of lead time demand

ROP

Risk ofstockoutService level

Expected QuantityROPpdemand

Safetystock

0 z

Qua t ty

z-scale

12-46


Reorder Point: Demand Uncertainty

LT ROP zd d+= σ

) as units time(same periodper demand of stddev. Theper week) day,(per periodper demand Average

deviations standard ofNumber where

ddz

d =

=

=

σ

)asunitstime(same timeLead LT d=

LT :Note ddLT σσ =

12-47

Example 8 page 580

Suppose that the manager of a construction supply house determined from historical records that demand for sand during lead time average 50 t i t I dditi th d t i d th t50 metric tons. In addition, suppose the manager determined that demand during lead time could be described by a normal distribution that has a mean of 50 metric tons and a standard deviation of 5 metric tons. Answer these questions, assuming that the manager is willing to accept a stockout risk of no more than 3 percent:

a. What value of z is appropriate?b. How much safety stock should be held?

Wh t d i t h ld b d?c. What reorder point should be used?

Expected lead time demand = 50 metric tonsσdLT = 5 metric tonsRisk = 3 percent


a. From table, using a service level of 1-.03 = 0.9700 obtain z = 1.88b. Safety stock = σdLT =1.88(5) = 9.4 metric tonsc. ROP = Expected lead time demand + Safety stock = 50 + 9.4

= 59.4 metric tons

(order when ROP >= stock position = On hand + On order)

*** Pl d f thi l ( 580) l ******** Please read more for this example (page 580) please *****

Reorder Point: Lead Time Uncertainty

average lead time


Reorder Point: Lead Time Uncertainty

LT ROP LTzdd +×= σ

)asunitstime(sametimeleadAverageLT

) as units time(same timelead of stddev. Theper week) day,(per periodper Demand

deviations standard ofNumber where

LT

d

ddz

===

σ

)asunitstime(same timelead Average LT d=

12-51

Example 9 page 580A restaurant uses an average of 50 jars of a special sauce each week. Weekly

usage of sauce has a standard deviation of 3 jars. the manager is willing to accept no more than a 10 percent risk of stockout lead time, which is two weeks Assume the distribution of usage is normalweeks. Assume the distribution of usage is normal.

a. Determine the value of z.b. determine the ROP.

σd = 3 jars per week acceptable risk = 10 percent, service level = .90

a. from table service level of 0.90 z = 1.28

weeks2 LT per week jars 50 ==d

b.= 105.43

5.43 10021.28(3) 250z LTd ROP d +=+×=+×= LTσ


Shortages and Service Levels

The expected number of units short in each order cycle is:cycle is:

E(n) = E(z)σdLTwhere

E(n) = Expected number of units short per order cycleE(z) = Standardized number of units short obtained from

following tablefollowing tableσdLT = Standard deviation of lead time demand

**Excel formula for E(Z) ......A2 = cell contain value of z

=(1/SQRT(2*PI()))*EXP(-0.5*A2*A2)-A2*(1-NORMSDIST(A2))

Z L(Z) Z L(Z) Z L(Z) Z L(Z) Z L(Z)0.00 0.398942 0.41 0.227011 0.82 0.116028 1.23 0.052737 1.64 0.0211370.01 0.393962 0.42 0.223621 0.83 0.113981 1.24 0.051652 1.65 0.0206370.02 0.389022 0.43 0.220267 0.84 0.111962 1.25 0.050587 1.66 0.0201470.03 0.384122 0.44 0.216949 0.85 0.109972 1.26 0.049539 1.67 0.0196680.04 0.379261 0.45 0.213667 0.86 0.108009 1.27 0.048510 1.68 0.0191980.05 0.374441 0.46 0.210422 0.87 0.106074 1.28 0.047498 1.69 0.0187380.06 0.369660 0.47 0.207212 0.88 0.104166 1.29 0.046504 1.70 0.0182880.07 0.364919 0.48 0.204038 0.89 0.102285 1.3 0.045528 1.71 0.0178470.08 0.360218 0.49 0.200900 0.90 0.100431 1.31 0.044568 1.72 0.0174150.09 0.355557 0.50 0.197797 0.91 0.098604 1.32 0.043626 1.73 0.0169930.10 0.350935 0.51 0.194729 0.92 0.096803 1.33 0.042700 1.74 0.0165790.11 0.346353 0.52 0.191696 0.93 0.095028 1.34 0.041791 1.75 0.0161740 12 0 341811 0 53 0 188698 0 94 0 093279 1 35 0 040897 1 76 0 0157780.12 0.341811 0.53 0.188698 0.94 0.093279 1.35 0.040897 1.76 0.0157780.13 0.337309 0.54 0.185735 0.95 0.091556 1.36 0.040020 1.77 0.0153900.14 0.332846 0.55 0.182806 0.96 0.089858 1.37 0.039159 1.78 0.0150100.15 0.328422 0.56 0.179912 0.97 0.088185 1.38 0.038313 1.79 0.0146390.16 0.324038 0.57 0.177051 0.98 0.086537 1.39 0.037483 1.80 0.0142760.17 0.319693 0.58 0.174225 0.99 0.084914 1.4 0.036668 1.81 0.0139200.18 0.315388 0.59 0.171432 1.00 0.083315 1.41 0.035868 1.82 0.0135730.19 0.311122 0.60 0.168673 1.01 0.081741 1.42 0.035083 1.83 0.0132330.20 0.306895 0.61 0.165947 1.02 0.080190 1.43 0.034312 1.84 0.0129000.21 0.302707 0.62 0.163254 1.03 0.078664 1.44 0.033555 1.85 0.0125750.22 0.298558 0.63 0.160594 1.04 0.077160 1.45 0.032813 1.86 0.0122570.23 0.294448 0.64 0.157967 1.05 0.075680 1.46 0.032085 1.87 0.0119460.24 0.290377 0.65 0.155372 1.06 0.074223 1.47 0.031370 1.88 0.0116420.25 0.286345 0.66 0.152810 1.07 0.072789 1.48 0.030669 1.89 0.0113450.26 0.282351 0.67 0.150280 1.08 0.071377 1.49 0.029981 1.90 0.0110540.27 0.278396 0.68 0.147781 1.09 0.069987 1.5 0.029307 1.91 0.0107710.28 0.274479 0.69 0.145315 1.10 0.068619 1.51 0.028645 1.92 0.0104930.29 0.270601 0.70 0.142879 1.11 0.067274 1.52 0.027996 1.93 0.0102220.30 0.266761 0.71 0.140475 1.12 0.065949 1.53 0.027360 1.94 0.0099570.31 0.262959 0.72 0.138102 1.13 0.064646 1.54 0.026736 1.95 0.0096980.32 0.259196 0.73 0.135760 1.14 0.063365 1.55 0.026124 1.96 0.0094450.33 0.255470 0.74 0.133448 1.15 0.062103 1.56 0.025525 1.97 0.0091980.34 0.251782 0.75 0.131167 1.16 0.060863 1.57 0.024937 1.98 0.0089570.35 0.248131 0.76 0.128916 1.17 0.059643 1.58 0.024360 1.99 0.0087210.36 0.244518 0.77 0.126694 1.18 0.058443 1.59 0.023796 2.00 0.0084910.37 0.240943 0.78 0.124503 1.19 0.057263 1.6 0.023242 2.01 0.0082660.38 0.237404 0.79 0.122340 1.20 0.056102 1.61 0.022700 2.02 0.0080460.39 0.233903 0.80 0.120207 1.21 0.054961 1.62 0.022168 2.03 0.0078320.40 0.230439 0.81 0.118103 1.22 0.053840 1.63 0.021647 2.04 0.007623


Z L(Z) Z L(Z) Z L(Z) Z L(Z) Z L(Z)0.00 0.398942 0.41 0.227011 0.82 0.116028 1.23 0.052737 1.64 0.0211370.01 0.393962 0.42 0.223621 0.83 0.113981 1.24 0.051652 1.65 0.0206370.02 0.389022 0.43 0.220267 0.84 0.111962 1.25 0.050587 1.66 0.0201470.03 0.384122 0.44 0.216949 0.85 0.109972 1.26 0.049539 1.67 0.0196680.04 0.379261 0.45 0.213667 0.86 0.108009 1.27 0.048510 1.68 0.0191980.05 0.374441 0.46 0.210422 0.87 0.106074 1.28 0.047498 1.69 0.0187380.06 0.369660 0.47 0.207212 0.88 0.104166 1.29 0.046504 1.70 0.0182880.07 0.364919 0.48 0.204038 0.89 0.102285 1.3 0.045528 1.71 0.0178470.08 0.360218 0.49 0.200900 0.90 0.100431 1.31 0.044568 1.72 0.0174150.09 0.355557 0.50 0.197797 0.91 0.098604 1.32 0.043626 1.73 0.0169930.10 0.350935 0.51 0.194729 0.92 0.096803 1.33 0.042700 1.74 0.0165790.11 0.346353 0.52 0.191696 0.93 0.095028 1.34 0.041791 1.75 0.0161740 12 0 341811 0 53 0 188698 0 94 0 093279 1 35 0 040897 1 76 0 0157780.12 0.341811 0.53 0.188698 0.94 0.093279 1.35 0.040897 1.76 0.0157780.13 0.337309 0.54 0.185735 0.95 0.091556 1.36 0.040020 1.77 0.0153900.14 0.332846 0.55 0.182806 0.96 0.089858 1.37 0.039159 1.78 0.0150100.15 0.328422 0.56 0.179912 0.97 0.088185 1.38 0.038313 1.79 0.0146390.16 0.324038 0.57 0.177051 0.98 0.086537 1.39 0.037483 1.80 0.0142760.17 0.319693 0.58 0.174225 0.99 0.084914 1.4 0.036668 1.81 0.0139200.18 0.315388 0.59 0.171432 1.00 0.083315 1.41 0.035868 1.82 0.0135730.19 0.311122 0.60 0.168673 1.01 0.081741 1.42 0.035083 1.83 0.0132330.20 0.306895 0.61 0.165947 1.02 0.080190 1.43 0.034312 1.84 0.0129000.21 0.302707 0.62 0.163254 1.03 0.078664 1.44 0.033555 1.85 0.0125750.22 0.298558 0.63 0.160594 1.04 0.077160 1.45 0.032813 1.86 0.0122570.23 0.294448 0.64 0.157967 1.05 0.075680 1.46 0.032085 1.87 0.0119460.24 0.290377 0.65 0.155372 1.06 0.074223 1.47 0.031370 1.88 0.0116420.25 0.286345 0.66 0.152810 1.07 0.072789 1.48 0.030669 1.89 0.0113450.26 0.282351 0.67 0.150280 1.08 0.071377 1.49 0.029981 1.90 0.0110540.27 0.278396 0.68 0.147781 1.09 0.069987 1.5 0.029307 1.91 0.0107710.28 0.274479 0.69 0.145315 1.10 0.068619 1.51 0.028645 1.92 0.0104930.29 0.270601 0.70 0.142879 1.11 0.067274 1.52 0.027996 1.93 0.0102220.30 0.266761 0.71 0.140475 1.12 0.065949 1.53 0.027360 1.94 0.0099570.31 0.262959 0.72 0.138102 1.13 0.064646 1.54 0.026736 1.95 0.0096980.32 0.259196 0.73 0.135760 1.14 0.063365 1.55 0.026124 1.96 0.0094450.33 0.255470 0.74 0.133448 1.15 0.062103 1.56 0.025525 1.97 0.0091980.34 0.251782 0.75 0.131167 1.16 0.060863 1.57 0.024937 1.98 0.0089570.35 0.248131 0.76 0.128916 1.17 0.059643 1.58 0.024360 1.99 0.0087210.36 0.244518 0.77 0.126694 1.18 0.058443 1.59 0.023796 2.00 0.0084910.37 0.240943 0.78 0.124503 1.19 0.057263 1.6 0.023242 2.01 0.0082660.38 0.237404 0.79 0.122340 1.20 0.056102 1.61 0.022700 2.02 0.0080460.39 0.233903 0.80 0.120207 1.21 0.054961 1.62 0.022168 2.03 0.0078320.40 0.230439 0.81 0.118103 1.22 0.053840 1.63 0.021647 2.04 0.007623

!!! See Example 10 page 582 and!!! See Example 10 page 582 andExample 11 page 582

and Example 12 page 584


How Much to Order: FOI

• Fixed-order-interval (FOI) model– Orders are placed at fixed time intervalsp

• Reasons for using the FOI model– Supplier’s policy may encourage its use– Grouping orders from the same supplier can produce savings in

shipping costs– Some circumstances do not lend themselves to continuously

monitoring inventory position

12-57

Fixed-Quantity vs. Fixed-Interval Ordering

12-58


FOI Model

OrderofTimeat PositionInventory

LevelInventoryTarget

Order toAmount −=

order of at timeposition Inventory levelinventory Target

order Amount towhere

===

−=

IPTQ

IPTQ

12-59

FOI Model

( ) LTOILT OI +++= dzdT σ

orders)between timeof(length intervalOrder OI where

=

*Q interestofframe-Time* ×=DQOI

OI* represents the optimal time between orders. Time-frame of interest is an appropriate period (e.g., days or weeks). This is usually based on the time-frame expressed by the average demand rate, d-bar.

12-60


Example 13 page 586

Given the following information, determine the amount to order..percent 99 level service Desired day per units 30 ==d

z = 2.33 for 99 percent service level

days 7 OI days 2 LTunits 71 mereorder tiat handon Amount day per units 3 d

====σ

A - LTOIz LT)(OId order Amount to d +++= σ

units 22071272.33(3)2)30(7 =−+++=

Single-Period Model

• Single-period model– Model for ordering perishables and other items with limited useful g p

lives– Shortage cost

• Generally, the unrealized profit per unit• Cshortage = Cs = Revenue per unit – Cost per unit

– Excess cost• Different between purchase cost and salvage value of items

left over at the end of the periodleft over at the end of the period• Cexcess = Ce = Cost per unit – Salvage value per unit

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Single-Period Model

• The goal of the single-period model is to identify the order quantity that will minimize the long-run excess and q y gshortage costs

• Two categories of problem:– Demand can be characterized by a continuous distribution– Demand can be characterized by a discrete distribution

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Stocking Levels

level Service+

= s

CCC

Cs Ce

unitper cost excess unitper cost shortage

where

==

+

e

s

es

CC

CC

Service level

SoBalance Point

Quantity

So =OptimumStocking Quantity

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chapter 12webstaff.kmutt.ac.th/~charoen.soon/pre485/chap012inventory2010.pdf · inventory...

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