inventory control (1)

8/8/2019 inventory control (1)

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1

Inventory Control


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2MULTI-PERIOD MODELS

Fixed-Order Quantity

y Asystem where the order quantity

remains constant but the timebetween orders varies.

Preferred for important or expensive items

because average inventory is lower.

Provides a quicker response to stockouts

Is more expensive to maintain due to

inventory record-keeping costs.


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3

BASIC FIXED-ORDER QUANTITY MODEL AND

REORDER POINT BEHAVIOR

R = Reorder pointQ = Economic order quantityL = Lead time

LL

Q QQ

R

Time

Number

of unitson hand

1. You receive an order quantity Q.

2. You start usingthem up over time. 3. When you reach down to

a level of inventory of R,

you place your next Q

sized order.

4. The cycle then repeats.


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4

FIXED-ORDER-QUANTITY MODEL

ASSUMPTIONS:

y Demand for the product is known, constant, and

uniform throughout the period.

y Lead time (L), which is the time from ordering to

receipt, is constant.

y Price per unit of product is constant (no quantity

discounts).

y Ordering or setup costs are constant

y All demands for the product are known with

certainty, no back orders or stock outs.


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5

COST MINIMIZATION GOAL

Ordering Costs

HoldingCosts

Order Quantity (Q)

COST

Annual Cost of

Items (DC)

Total Cost

QOPT

By adding the item, holding, and ordering costs

together, we determine the total cost curve, which inturn is used to find the Qopt inventory order point thatminimizes total costs


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6

BASIC FIXED-ORDER QUANTITY

(EOQ) MODEL FORMULA

H2Q+S

QD+DC=TC

TotalAnnual =Cost

AnnualPurchase

Cost

AnnualOrdering

Cost

AnnualHolding

Cost+ +

TC=Total annual

cost

D =DemandC =Cost per unit

Q =Order quantity

S =Cost of placing

an order or setup

costR =Reorder point

L =Lead time

H=Annual holding

and storage cost

per unit of inventory


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DERIVING THE EOQ

Using calculus, we take the firstderivative of the total cost function withrespect to Q, and set the derivative(slope) equal to zero, solving for theoptimized (cost minimized) value of Q

opt

Q =2DS

H=

2(Annual Demand)(Order or Setup Cost)

AnnualHolding CostOPT

R eorder p oint, R = d L_

d = average daily demand (constant)

L = Lead time (constant)

_

We also need areorder point totell us when toplace an order


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EOQ EXAMPLE (1) PROBLEM DATA

Annual Demand = 1,000 unitsDays per year considered in average

daily demand = 365Cost to place an order = $10

Holding cost per unit per year = $2.50Lead time = 7 daysCost per unit = $15

Given the information below, what are the EOQ andreorder point?


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EOQ EXAMPLE (1) SOLUTION

Q 2

H

2(1,000 )(10)

2.50 89.443 units orT 9 0 u its

d

1,000 units / year

365days / year 2.7

4 units /d

ay

Reorder point, R d L 2.74units / day (7days) 19 .18 or_

20 u its

In summary, you place an optimal order of 90 units. Inthe course of using the units to meet demand, whenyou only have 20 units left, place the next order of 90units.


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EOQ EXAMPLE (2) PROBLEM DATA

Annual Demand = 10,000 unitsDays per year considered in average dailydemand = 365Cost to place an order = $10

Holding cost per unit per year = 10% of costper unitLead time = 10 daysCost per unit = $15

Determine the economic order quantityand the reorder point given the following


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EOQ EXAMPLE (2) SOLUTION

Q =2D S

H=

2(10,000 )(10)

1.50= 365 .148 units , or OPT 3 6 6 nits

d =

10,000 units / year

365 days / year = 27.397 units / day

nits274or273.97=days)(10units/day27.397=d=_

Place an order for 366 units. When in the course ofusing the inventory you are left with only 274 units,place the next order of 366 units.


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12

FIXED-ORDER QUANTITY MODEL

WITH SAFETY STOCK FORMULA

!

L

1i

2

dL i= WW

LWzdLR !


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13

At Discount Carpet Store, daily demand for

Super Shag Ccarpet stocked by the store is

normally distributed with an average daily

demand of 30 yards and a standard deviation of 5

yards of carpet per day. The lead time for

receiving a new order of carpet is 10 days.

Determine the reorder point and safety stock if

the store wants a service level of 95% with the

probability of a stockout equal to 5 %.

REORDER POINT FOR

VARIABLE DEMAND (EXAMPLE)


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SOLUTION

The carpet store wants a reorder point with aThe carpet store wants a reorder point with a

95% service level and a 5%95% service level and a 5% stockoutstockout probabilityprobability

dd = 30 yards per day= 30 yards per day

LL = 10 days= 10 days

WWdd = 5 yards per day= 5 yards per day

For a 95% service level,For a 95% service level, zz = 1.65= 1.65

RR == dLdL ++ zz WWdd LL

= 30(10) + (1.65)(5)( 10)= 30(10) + (1.65)(5)( 10)

= 326.1 yards= 326.1 yards

Safety stockSafety stock== zz WWdd LL

= (1.65)(5)( 10)= (1.65)(5)( 10)

= 26.1 yards= 26.1 yards


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FIXED-TIME-PERIOD MODEL

y Inventory is counted at fixed intervals.

y Ceiling (par) inventory is established.

y Safety stock level is established.

y Order quantity to return inventory to ceiling level

varies based on on-hand inventory less safety stock at

time inventory is counted.


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FIXED-TIME PERIOD INVENTORYMODEL

The

McG

raw-

Hill

Com

panie

s,

Inc.,2003


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FIXED-TIME PERIOD MODEL WITH SAFETY

STOCK FORMULA

order)onitems(includeslevelinventorycurrentI

timeleadandrevieover thedemandodeviationstandard

yprobabilitservicespeci iedaordeviationsstandardonumberthez

deman

ddailyaverageorecast

d

daysintimelead

revie sbet eendaysonumbertheT

orderedbetoquantitiyq

:Where

I-Z)(Tdq

T

T

W

W

q = Average demand + Safety stock Inventory currently on hand


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MULTI-PERIOD MODELS: FIXED-TIME PERIOD MODEL:

DETERMINING THE VALUE OF 7T+L

The standard deviation of a sequenceof random events equals the square

root of the sum of the variances

W W

W

W W

T di 1

T

d

T d

2

ince each day is independent and is constant,

(T )

i

2

!


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EXAMPLE OF THE FIXED-TIME PERIOD MODEL

Average daily demand for a prod ct is

20 nits. The review period is 30 days,and lead time is 10 days. Management

has set a policy of satisfying 96 percent

of demand from items in stock.A

t thebeginning of the review period there are

200 nits in inventory. The daily

demand standard deviation is 4 nits.

Given the information below, how many nitssho ld be ordered?


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EXAMPLE OF THE FIXED-TIME PERIOD

MODEL: SOLUTION (PART 1)

W WT+ L d2 2

= (T + L) = 30 + 10 4 = 25.298

FromAppendix D , z = 1.75


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EXAMPLE OF THE FIXED-TIME PERIOD

MODEL: SOLUTION (PART 2)

or644.272,200-44.272800q

200-298)(1.75)(25.10)20(30q

I-Z)(Tdq T

u its645

W

So, to satisfy 96 perce t of the dema d,

you should place a order of645 u its at

this review period


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The corner drug store stocks a popular brand of

sunscreen. The average demand for the

sunscreen is 6 bottles per day, with a standard

deviation of 1.2 bottles. Avendor for the

sunscreen producer checks the drugstore stock

every 60 days. During one visit the drugstore had

8 bottles in stock. The lead time to receive an

order is 5 days. Determine the order size for this

order period that will enable the drug store tomaintain a 95% service level.

EXAMPLE OF THE FIXED-TIME

PERIOD MODEL


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dd= 6 bottles per day= 6 bottles per day

WWdd = 1.2 bottles= 1.2 bottles

TT = 60 days= 60 days

LL = 5 days= 5 days

II= 8 bottles= 8 bottleszz= 1.65 (for a 95% service level)= 1.65 (for a 95% service level)

QQ == dd(T +(T +LL) +) + zzWWdd T +T +LL -- II

= (6)(60 + 5) + (1.65)(1.2) 60 + 5= (6)(60 + 5) + (1.65)(1.2) 60 + 5 -- 88

= 397.96 bottles= 397.96 bottles

EXAMPLE OF THE FIXED-TIME

PERIOD MODEL: SOLUTION


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PRICE-BREAK MODEL FORMULA

ostHoldingnnual

ost)etuporderemand)( r2( nnual

i

2

QOPT

Based on the same assumptions as the EOQ model,the price-break model has a similar Qopt formula:

i = percentage of unit cost attributed to carrying inventoryC = cost per unit

Since C changes for each price-break, the formulaabove will have to be used with each price-break costvalue


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PRICE-BREAK EXAMPLE PROBLEM DATA

(PART 1)

A company has a chance to red ce their inventory

ordering costs by placing larger q antity orders sing

the price-break order q antity sched le below. What

sho ld their optimal order q antity be if this company

p rchases this single inventory item with an e-mail

ordering cost of $4, a carrying cost rate of 2% of the

inventory cost of the item, and an ann al demand of

10

,000

nits?Order Quantity(units) Price/unit($)

0 to 2,499 $1.202,500 to 3,999 1.004,000 or more .98


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PRICE-BREAK EXAMPLE SOLUTION (PART 2)

units2,0200.02(.98)

4)2(10,000)(

i

2Q

OPT

Annual Demand (D)= 10,000 unitsCost to place an order (S)= $4

First, plug data into formula for each price-break value of C

units2,0000.02(1.00)

4)2(10,000)(i

2Q OPT

units1,8260.02(1.20)

4)2(10,000)(

i

2Q

OPT

Carrying cost % of total cost (i)= 2%Cost per unit (C) = $1.20, $1.00, $0.98

Interval from 4000 and more,the Qopt value is not feasible

Interval from 2500-3999, the

Qopt value is not feasible

Interval from 0 to 2499, theQopt value is feasible

Next, determine if the computed Qopt values are feasible or not


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Since the feasible solution occurred in the first price-break, it means that all the other true Qopt values occurat the beginnings of each price-break interval.

0 1826 2500 4000 Order Quantity

Totalannualcosts So the candidates

for the price-

breaks are 1826,2500, and 4000units

Because the total annual cost function isa u shaped function


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iC2

Q+S

Q

D+DC=TC

Next, we plug the true Qopt values into the total cost

annual cost function to determine the total cost undereach price-break

TC(0-2499)=(10000*1.20)+(10000/1826)*4+(1826/2)(0.02*1.20)

= $12,043.82TC(2500-3999)= $10,041TC(4000&more)= $9,949.20

Finally, we select the least costly Qopt, which is thisproblem occurs in the 4000 & more interval. In summary,our optimal order quantity is 4000 units


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The maintenance department of alarge hospital uses about 816 cases of

liquid cleanser annually. Ordering

costs are Rs. 12, carrying costs are Rs.

4 per case a year and the new priceschedule indicates that the orders of

less than 50 cases will cost Rs 20 per

case, 50-79 cases will cost Rs 18 per

case, 80-99 cases will cost Rs. 17 per

case, and larger orders will cost Rs.

16 per case. Determine the optimum

order quantity and the total cost.

PRICE-BREAK EXAMPLE PROBLEM DATA


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Maximum Inventory Level, M

MISCELLANEOUS SYSTEMS:

OPTIONAL REPLENISHMENT SYSTEM

MActual Inventory Level, I

q = M - I

I

Q = minimum acceptable order quantity

If q > Q, order q, otherwise do not order any.


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MISCELLANEOUS SYSTEMS:

BIN

SYSTEMSTwo-Bin System

Full Empty

Order One Bin ofInventory

One-Bin System

Periodic Check

Order Enough toRefill Bin


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ABC CLASSIFICATION SYSTEM

Items kept in inventory are not ofequal importance in terms of:

y dollars invested

y profit potential

y sales or usage volume

y stock-out penalties


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ABC CLASSIFICATION

Class A

y 5 15 % of units

y 70 80 % of value

Class B

y 30 % of units

y 15 % of value

Class Cy 50 60 % of units

y 5 10 % of value


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ABC CLASSIFICATION: EXAMPLE

11 $ 60$ 60 9090

22 350350 4040

33 3030 13013044 8080 6060

55 3030 100100

66 2020 180180

77 10

10

170

170

88 320320 5050

99 510510 6060

1010 2020 120120

PARTPART UNIT COSTUNIT COST ANNUAL USAGEANNUAL USAGE


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INVENTORY ACCURACY AND CYCLE COUNTING

Inventory accuracy refers to how well the

inventory records agree with physical count

Cycle Counting is a physical inventory-

taking technique in which inventory iscounted on a frequent basis rather than once

or twice a year


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EN

D OF CHA

PTER

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