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Inventory Management

JCA Ch. 17STC Ch. 12

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Agenda

Definition of Inventory System

Inventory Costs

Independent vs. Dependent Demand

Single Period Inventory Model

Basic Fixed-Order Quantity Models

Basic Fixed-Time Period Model

Price Break Model & ABC Classification

Other Systems and Issues

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Definition of Inventory System

Inventory is the stock of any item or resource used in an organization.

Types of inventories can be raw materials, finished products, component parts, supplies, and work-in-process.

An inventory system is the set of policies and controls that monitor levels of inventory and determines what levels should be maintained, when stock should be replenished, and how large orders should be.

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Purposes of Inventory

1. To maintain independence of operations.

2. To meet variation in product demand.

3. To allow flexibility in production scheduling.

4. To provide a safeguard for variation in raw material delivery time.

5. To take advantage of economic purchase-order size.

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Inventory Costs

Holding (or carrying) costs. Costs for storage, handling, insurance, etc.

Setup (or production change) costs. Costs for arranging specific equipment setups, etc.

Ordering costs. Costs of someone placing an order, etc.

Shortage costs. Costs of canceling an order, etc.

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E(1

Independent Vs. Dependent Demand

Independent Demand (Demand not related to other items or the final end-product)

Dependent Demand(Derived demand

items for component parts,

subassemblies, raw materials, etc.)

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Classifying Inventory Models

I. Single-Period Inventory Model

II. Fixed-Order Quantity Models (Perpetual Inventory System) Event triggered

III. Fixed-Time Period Models (Periodic System) Time triggered

IV. Miscellaneous models (Price Break & ABC Classification)

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I. Single-Period Inventory Model

Service Level

So

Quantity

Ce Cs

Balance point

Service level =Cs

Cs + CeCs = Shortage cost per unitCe = Excess cost per unit

Service level is the probability that demand will not exceed the stocking level

Stevenson p 587 - 588

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Single Period Model Example

Demand varied between 300 – 500 liters. Ce = $0.20 per unit Cs = $0.60 per unit Service level (SL) = Cs/(Cs+Ce) = .6/(.6+.2) Service level = 0.75

Service Level = 75%

Quantity

Ce CsStock-out risk = 1.00 – 0.75 = 0.25

So, the optimal stocking level (So) = 300 + 0.75 (500 – 300) = 450 liters

Stevenson page 588

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II. Fixed-Order Quantity Models:Model Assumptions

Demand for the product is constant and uniform throughout the period.

Lead time (time from ordering to receipt) is constant.

Price per unit of product is constant.

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II. Fixed-Order Quantity Models:Model Assumptions (Cont’d)

Inventory holding cost is based on average inventory.

Ordering or setup costs are constant.

All demands for the product will be satisfied. (No

back orders are allowed.)

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Basic Fixed-Order Quantity Model and Reorder Point Behavior

R = Reorder pointQ = Economic order quantityL = Lead time

L L

Q QQ

R

Time

Numberof unitson hand

1. You receive an order quantity Q.

2. Your start using them 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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Cost Minimization Goal

Ordering Costs

HoldingCosts

QOPT

Order Quantity (Q)

COST

Annual Cost ofItems (DC)

Total Cost

By adding the item, holding, and ordering costs together, we determine the total cost curve, which in turn is used to find the Qopt inventory order point that minimizes total costs.

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II.1. Basic Fixed-Order Quantity(EOQ) Model: Simple Formula

TC = DC + DQ

S + Q2

H

Total Annual Cost =Annual

PurchaseCost

AnnualOrdering

Cost

AnnualHolding

Cost+ +

TC = Total annual costD = DemandC = Cost per unitQ = Order quantityS = Cost of placing an order or setup costR = Reorder pointL = Lead timeH = Annual holding and storage cost per unit of inventory

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Deriving the EOQ Formula

Using calculus, we take the first derivative of the total cost function with respect to Q, and set the derivative (slope) equal to zero, solving for the

optimized (cost minimized) value of Qopt.

Q = 2DS

H =

2(Annual Demand)(Order or Setup Cost)Annual Holding CostOPT

Reorder point, R = d L_

d = average daily demand (constant)

L = Lead time (constant)

_We also need a reorder point to tell us when to place 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 = $10Holding cost per unit per year = $2.50Lead time = 7 daysCost per unit = $15

Given the information below, what are the EOQ and reorder point?

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EOQ Example (1) Solution

Q = 2DS

H =

2(1,000 )(10)2.50

= 89.443 units or OPT 90 units

d = 1,000 units / year365 days / year

= 2.74 units / day

Reorder point, R = d L = 2.74units / day (7days) = 19.18 or _

20 units

In summary, when you only have 20 units left, place the next order of 90 units.

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EOQ Example (2) Problem Data

Annual Demand = 10,000 unitsDays per year considered in average daily demand = 365Cost to place an order = $10Holding cost per unit per year = 10% of cost per unitLead time = 10 daysCost per unit = $15

Determine the economic order quantity and the reorder point.

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EOQ Example (2) Solution

Q =2D S

H=

2(10,000 )(10)1.50

= 365.148 units, or O PT 366 units

d =10,000 units / year

365 days / year= 27.397 units / day

R = d L = 27.397 units / day (10 days) = 273.97 or _

274 units

When in the course of using the inventory you are left with only 274 units, place the next order of 366 units.

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II.2. Fixed-Order Quantity Model with Usage During Production Time

Also known as: Economic Production Quantity (EPQ), or

Gradual Replacement Model

In the previous model, it is assumed that the quantity ordered would be received in one lot

In many situations, production of an inventory item and usage of that item take place simultaneously

Characteristics:- Demand being withdrawn while production is underway- No stock-outs- Constant and known demand, lead time and unit cost

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For this case: Replenishment/production rate (p) > withdrawal/usage rate (u)

TC = (D/Qo)(S) + (Imax/2)(H)

p Qo = EOQ = (2DS / H) ( --------- ) p - u

p - uImax = Qo (--------) Reorder Point (R) = d LT p

II.2. Fixed-Order Quantity Model with Usage During Production Time (Cont’d)

Stevenson page 569 - 571

Imax = maximum inventory TC = total costu = usage rate Qo = EOQ = Q optimump = production rate

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Product X is a standard item in a firm’s inventory. Final assembly of the product is performed on an assembly line that is in operation every day.

One component of product (i.e. component X1) is produced in another department. This dept, when it produces X1, does so at the rate of 100 units per day. Assume there are 250 working days for a year

u = 40 units/day D = 40 x 250 = 10 000 units / yrp = 100 units H = $0.50 per unitS = $50 C of X1 =$7 eachLT = 7 days

Problem Example

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EOQ = ????? R = ??????

EOQ = {2(10000)(50) / (0.50)} {100 / (100-40)}

= 1826 units R = u x LT = 40 (7) = 280 units

Therefore: An order for 1826 units of component X1 should be placed

when the stock drops to 280 units

Problem Example (Cont’d)

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II.3. Fixed-order Quantity Model with Demand Variation (Safety Stock)

The danger of this model occurs during the lead time, due to demand variation

The amount of safety stock depends on the service level desired

The optimal quantity to be ordered:

EOQ = (2DS/H)

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LT Time

Expected demandduring lead time

Maximum probable demandduring lead time

ROP

Qu

an

tity

Safety stockSafety stock reduces risk ofstockout during lead time

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The Reorder Point: R = d L + z d LT

Where : Safety stock = z d LT _ diThe average daily demand: d = ----------- n

_

( di – d )2

The std dev of the daily demand: d = ------------------- n - 1 (Use n - 1 for a sample standard deviation)

II.3: Fixed-order Quantity Model with Safety Stock (Cont’d)

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Given information:

D = 1000 units / yr and Q = 200 units

Desired service level = P = 95%

Std dev during lead time = L = 25 units

L = 15 days and assumed 250 workdays / year

Question: R = ???????

Problem Example 1

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Solution:_d = (1000 / 250) = 4 units / day

P = 95% = 0.95 z = 1.645 (Appendix B – Stevenson p 843)

R = 4(15) + 1.645(25) = 102 units

Therefore: When the stock on hand drops to 102 units, place a new order of 200 units more

Problem Example 1 (Cont’d)

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Daily demand mean = d = 60 units (normally distributed)

Std dev = d = 7 units/dayLead time = LT = 6 daysOrdering cost = S = $ 10 per orderHolding cost = H= $0.50 per unit/yearAssumed: 365 workdays / yearDesired service level = P = 95%

Questions: EOQ = ???R = ?????

Problem Example 2

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Solution:EOQ = {2(60)(365)(10) / 0.50} = 936 unitsP = 95% = 0.95 Z = 1.645 (Appendix B)

Standard deviation during LT = d LT = 6 x (7) = 17.2

R = (60)(6) + 1.645 (17.2) = 388.29 = 389 units

Therefore: When the inventory drops to 389 units, order 936 units more

Problem Example 2 (Cont’d)

II.4: Fixed-order Quantity Model with Demand Variation and Lead Time VariationThe optimal quantity to be ordered:

EOQ = (2DS/H)

Reorder Point (R): R = d LT + ZLT d

+d 2LT

Safety Stock = ZLT d+d2

LT

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Stevenson page 580

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III. Fixed-Time Period Model with

Safety Stock Formula

q = Average demand + Safety stock – Inventory currently on hand

Amount to order:

Q = d (OI + LT) + Z d OI + LT - A

Where: OI = order interval A = amount on hand at reorder time LT = lead time

d = standard deviation of demand

Stevenson page 586

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Problem Example

Average daily demand for a product is 20 units.The review period is 30 days, and lead time is 10 days. Management has set a policy of satisfying 96 percentof demand from items in stock. At the beginning of the review period there are 200 units in inventory. The daily demand standard deviation is 4 units.

Given the information below, how many units should be ordered?

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Problem Example (Cont’d)

Service level = P = 96% (stock-out risk = 4%)

From the Appendix B Z = 1.75

Q = d (OI + LT) + Z d OI + LT - A

= 20 (30 + 10) + (1.75)(4)

= 645 units

Solution:

So, to satisfy 96 percent of the demand, you should place an order of 645 units at this review period.

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IV.1. Price-Break Model Formula

Cost Holding AnnualCost) Setupor der Demand)(Or 2(Annual

= iC

2DS = 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 formula above will have to be used with each price-break cost value.

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Price-Break Example Problem Data (Part 1)

A company has a chance to reduce their inventory ordering costs by placing larger quantity orders using the price-break order quantity schedule below. What should their optimal order quantity be if this company purchases 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 annual demand of 10,000 units?

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)

units 1,826 = 0.02(1.20)

4)2(10,000)( =

iC

2DS = Q 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”.

units 2,000 = 0.02(1.00)

4)2(10,000)( =

iC

2DS = Q OPT

units 2,020 = 0.02(0.98)

4)2(10,000)( =

iC

2DS = Q OPT

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

Interval from 0 to 2499, the Qopt value is feasible.

Interval from 2500-3999, the Qopt value is not feasible.

Interval from 4000 & more, the Qopt value is not feasible.

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

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Price-Break Example Solution (Part 3)

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

0 1826 2500 4000 Order Quantity

Total annual costs So the candidates

for the price-breaks are 1826, 2500, and 4000 units.

Because the total annual cost function is a “u” shaped function.

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Price-Break Example Solution (Part 4)

iC 2

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 under each 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 this problem occurs in the 4000 & more interval. In summary, our optimal order quantity is 4000 units.

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IV.2. ABC Classification System

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

dollars invested

profit potential

sales or usage volume

stock-out penalties

0

30

60

30

60

AB

C

% of $ Value

% of Use

So, identify inventory items based on percentage of total dollar value, where “A” items are roughly top 15 %, “B” items as next 35 %, and the

lower 65% are the “C” items.

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Inventory Accuracy and Cycle CountingDefined

Inventory accuracy refers to how well the inventory records agree with physical count.

Cycle Counting is a physical inventory-taking technique in which inventory is counted on a frequent basis rather than once or twice a year.