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1 EMBA-2, BUP EO - 702 Inventory Control. EO - 702 M. AsadEMBA-2 Inventory Control Inventory – The...

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1 EMBA-2, BUP EO - 702 Inventory Control
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Page 1: 1 EMBA-2, BUP EO - 702 Inventory Control. EO - 702 M. AsadEMBA-2 Inventory Control Inventory – The longer it sits, the harder it is to move Purposes of.

1

EMBA-2, BUP

EO - 702

Inventory Control

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Inventory Control

• Inventory – The longer it sits, the harder it is to move

• Purposes of Inventory1. To maintain independence of operations2. To meet variation in product demand3. To allow flexibility in production scheduling4. To provide a safeguard for variation in raw

material delivery time5. To take advantage of economic purchase-order

size

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Inventory Control

• Inventory Cost– Holding (or carrying) costs– Setup (or production change) costs– Ordering costs– Shortage costs

Independent vs. Dependent Demand

Independent Demand -Demand for the final end-product or demand not related to other items

Dependent Demand -Derived demand items for component parts, subassemblies, raw materials, etc

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Inventory Systems

• Single-Period Inventory Model– Seeks to balance the costs of inventory overstock

and under stock• Multi-Period Inventory Models

– Fixed-Order Quantity Models-Event triggered (Example: running out of stock)

– Fixed-Time Period Models -Time triggered (Example: Monthly sales call by sales representative)

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Single-Period Inventory Model

uo

u

CC

CP

uo

u

CC

CP

This model states that we should continue to increase the size of the inventory so long as the probability of selling the last unit added is equal to or less than the ratio of: Cu/Co+Cu

This model states that we should continue to increase the size of the inventory so long as the probability of selling the last unit added is equal to or less than the ratio of: Cu/Co+Cu

sold be unit will y that theProbabilit

estimatedunder demand ofunit per Cost C

estimatedover demand ofunit per Cost C

u

o

P

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Single Period Model Example

• Our college basketball team is playing in a tournament game this weekend. Based on our past experience we sell on average 2,400 shirts with a standard deviation of 350. We make $10 on every shirt we sell at the game, but lose $5 on every shirt not sold. How many shirts should we make for the game?

Cu = $10 and Co = $5; P ≤ $10 / ($10 + $5) = .667

Z.667 = .432 (use Cum or Std Nor Dist (.667)therefore we need 2,400 + .432(350) = 2,551 shirts

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Multi-Period Models:

Fixed-Order Quantity Model Vs

Fixed-Time Period Model

1.Order Quantity 2.Recordkeeping 3.Time to maintain 4.When to place Order5.Size of Inventory6.Types of items

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Multi-Period Models:

Fixed-Order Quantity Model 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 • 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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Deriving the EOQ

Q = 2DS

H =

2(Annual D em and)(Order or Setup Cost)

Annual Holding CostOPTQ =

2DS

H =

2(Annual D em and)(Order or Setup Cost)

Annual Holding CostOPT

H 2

Q + S

Q

D + DC = TC H

2

Q + S

Q

D + DC = TC

Total Annual =Cost

AnnualPurchase

Cost

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

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

Cost Minimization Goal : Qopt inventory order point that minimizes total costs

Cost Minimization Goal : Qopt inventory order point that minimizes total costs

Reorder point, R = d L_

Reorder point, R = d L_

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EOQ Example 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?

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

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Fixed-Order Quantity Model with Safety Stock

timelead during demand ofdeviation standard =

yprobabilit service specified afor deviations standard ofnumber the= z

demanddaily averageforecast = d

daysin timelead = L

in UnitsPoint Recorder = R

:Where

L timelead during demand ofdeviation standard =

yprobabilit service specified afor deviations standard ofnumber the= z

demanddaily averageforecast = d

daysin timelead = L

in UnitsPoint Recorder = R

:Where

L

Establishing Safety Stock Level

z + Ld = R point,Reorder L

z + Ld = R point,Reorder L

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Fixed-Time Period Model with Safety Stock Formula

order)on items (includes levelinventory current = I

timelead and review over the demand ofdeviation standard =

yprobabilit service specified afor deviations standard ofnumber the= z

demanddaily averageforecast = d

daysin timelead = L

reviewsbetween days ofnumber the= T

ordered be toquantitiy = q

:Where

I - Z+ L)+(Td = q

L+T

L+T

order)on items (includes levelinventory current = I

timelead and review over the demand ofdeviation standard =

yprobabilit service specified afor deviations standard ofnumber the= z

demanddaily averageforecast = d

daysin timelead = L

reviewsbetween days ofnumber the= T

ordered be toquantitiy = q

:Where

I - Z+ L)+(Td = q

L+T

L+T

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

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Multi-Period Models: Fixed-Time Period Model: Determining the Value of T+L

T+ L di 1

T+ L

d

T+ L d2

=

Since each day is independent and is constant,

= (T + L)

i

2

T+ L di 1

T+ L

d

T+ L d2

=

Since each day is independent and is constant,

= (T + L)

i

2

• The standard deviation of a sequence of random events equals the square root of the sum of the variances

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Example of the Fixed-Time Period Model

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 percent of 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?

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

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Price-Break Model Formula

Cost Holding Annual

Cost) 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

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?

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

units 1,826 = 0.02(1.20)

4)2(10,000)( =

iC

2DS = QOPT

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 = QOPT

units 2,020 = 0.02(0.98)

4)2(10,000)( =

iC

2DS = QOPT

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

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.

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.

0 1826 2500 4000 Order Quantity

Total annual costs

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

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

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

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

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Price-Break Example Solution

iC 2

Q + S

Q

D + DC = TC 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

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) = $12,043.82 (order quantity = 1826)TC(2500-3999)= $10,041 (order quantity = 2500)TC(4000&more)= $9,849.20 (order quantity = 4000)

TC(0-2499)=(10000*1.20)+(10000/1826)*4+(1826/2)(0.02*1.20) = $12,043.82 (order quantity = 1826)TC(2500-3999)= $10,041 (order quantity = 2500)TC(4000&more)= $9,849.20 (order quantity = 4000)

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

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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• 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 safety stock

Operations Strategy


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