1

Inventory (Chapter 16)

What is Inventory?

How Inventory works:two ways of orderingbased on three elements

Inventory models

(to p2)

(to p3)

(to p4)

(to p5)

2

Inventory

- to study methods to deal with

“how much stock of items should be kept on hands that would meet customer demand”

Objectives are to determine:

a) how much to order, and

b) when to order(to p1)

3

Two basic types of Inventory Systems

1) continuous (fixed-order quantity) • an order is placed for the same constant amount

when inventory decreases to a specified level, ie. Re-order point

2) periodic (fixed-time)• an order is placed for a variable amount after a

specified period of time• used in smaller retail stores, drugstores, grocery

stores and offices(to p1)

4

3 basic inventory elements

1.1. Carrying cost, Carrying cost, Cc

• Include facility operating costs, record keeping, interest, etc.

2.2. Ordering cost, Ordering cost, Co

• Include purchase orders, shipping, handling, inspection, etc.

3.3. Shortage (stock out) cost, Shortage (stock out) cost, Cs

• Sometimes peanlties involved; if customer is internal, work delays could result

(to p1)

5

Inventory models

Here, we onlyonly study the following three different models:

1. Basic model8th ed: 1, 15, 19,

9th ed: 1, 10, 13,

2. Model with “discount rate”8th: 24, 26; 9th: 17, 18

3. Model with “re-order points” 8th: 36, 38; 9th: 26,28

(to p6)

(to p23)

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6

1. Basic model

The basic model is known as:

“Economic Order Quantity” (EOQ) Models

Objective is to determine the optimal order size that will minimize total inventory costs

How the objective is being achieved?(to p7)

7

3 Basic EOQ models

Three models to be discussed:

1. Basic EOQ model

2. EOQ model without instantaneous

receipt

3. EOQ model with shortages.

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(to p15)

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8

The Basic EOQ Model

• The optimal order size, Q, is to minimize the sum of carrying costs and ordering costs.

• Assumptions and Restrictions: - Demand is known with certainty and is relatively constant over time. - No shortages are allowed. - Lead time for the receipt of orders is constant. (will consider later) - The order quantity is received all at once and instantaniously.

(to p10)

How to determinethe optimal valueQ*?

9

Determine of Q

We try to• Find the total cost that need to spend for keeping inventory

on hands• = total ordering + stock on hands• Determine its optimal solution by finding its first derivative

with respect to Q

How to get these values?1. Find out the total carrying cost2. Find out the total ordering cost3. Total cost = 1 + 24. d (Total cost) /d Q = 0, and find Q*

(to p10)

10

The Basic EOQ ModelWe assumed that, we will only keep half the inventory over a year then

(to p11)

The total carry cost/yr = Cc x (Q/2). Total order cost = Co x (D/Q)

Then , Total cost = 2QC

QDCTC co Finding optimal Q*

11

The Basic EOQ Model

• EOQ occurs where total cost curve is at minimum value and carrying cost equals ordering cost:

•Where is Q* located in our model?

c

o

co

CDCQ

QCQDCTC

2

2

*

min

(How to obtain this?)Then,

(to p12)

*

12

The Basic EOQ Model

• Total annual inventory cost is sum of ordering and carrying cost:

2QC

QDCTC co

Figure 16.5 The EOQ cost model

To order inventory

To keep inventory

Try to get this value (to p13)Examples

13

The Basic EOQ ModelExample

Consider the following:

days store 62.2 5

311*/

days 311 timecycleOrder

5000,2000,10 :yearper orders ofNumber

500,1$2

)000,2()75.0(000,2000,10)150(

2 :costinventory annual Total

yd 000,2)75.0(

)000,10)(150(22* :sizeorder Optimal

10,000yd D $150, C $0.75, C :parameters Model

*

*min

oc

QD

QD

QCQDCTC

CDCQ

optco

c

o

No. of working days/yr

(to p14)

*

Note: You should pay attention thatall measurement units must be the same

Consider the same example, with yearly

14

The Basic EOQ ModelEOQ Analysis with monthly time frame

$1,500 ($125)(12) cost inventory annual Total

monthper 125$2

)000,2()0625.0(000,2

)3.833()150(2*

* :costinventory monthly Total

yd 000,2)0625.0(

)3.833)(150(22* :sizeorder Optimal

monthper yd 833.3 D order,per $150 C month,per ydper $0.0625 C :parameters Model

min

oc

QCQDCTC

CDCQ

co

c

o

(unit be based on yearly)

12 months a year

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15

The EOQ Model with Noninstantaneous Receipt

Figure 16.6 The EOQ model with noninstantaneous order receiptAlways greater than 0why?

(to p16)

The order quantity is received gradually over time and inventory is drawn on

at the same time it is being replenished.

Example: Let p = production, d = demand, then the total cost (TC) =

16

The EOQ Model with Noninstantaneous ReceiptModel Formulation

)/1(2 :sizeorder Optimal

12

:cost inventory annual Total

demanded isinventory at which ratedaily drate) productionor ( over time received isorder heat which t ratedaily p

*

pdCDCQ

pdQC

QDCTC

c

o

co

Assuming placing an order/yr

(to p17)Example

17

The EOQ Model with Noninstantaneous ReceiptExample

yd 772,1150

2.3218.256,21* levelinventory Maximum

runs 43.48.256,2

000,10*

runs)n (productioyear per orders ofNumber

days 05.15150

8.256,2* length run Production

329,1$150

2.3212

)8.256,2()075(.)8.256,2()000,10()150(1

2*

*min:costinventory annual minimum Total

yd 8.256,2

1502.32175.0

)000,10)(150(2

1

2* :sizeorder Optimal

dayper yd 150 p day,per yd 32.2 10,000/311 year,per yd 10,000 D unit,per $0.75 Cc $150, Co

pdQ

QD

pQ

pdQCc

QDCoTC

pdCc

CoDQ

( to p7)

Let,

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The EOQ Model with Shortages

• In the EOQ model wth shortages, the assumption that shortages cannot exist is relaxed.

• Assumed that unmet demand can be backordered with all demand eventually satisfied.

Shortage = S/Q

On hand = (Q-S)/Q t1 + t2 = S/Q + (Q-S)/Q = 1

Shortage

What we needed

Max level of inventory

Here, we allow Q being shortageshortage, so that we could borrow or replenish the stockslater

Total cost is

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19

The EOQ Model with Shortages

• In the EOQ model wth shortages, the assumption that shortages cannot exist is relaxed.

• Assumed that unmet demand can be backordered with all demand eventually satisfied.

Shortage = S/Q

On hand = (Q-S)/Q

Shortage

What we needed

½*base* height = ½ * (Q-S) * (Q-S)/Q = ½ * (Q-S)2 /Q

Area = ½ * (S/Q) * S = ½ * S2 /Q

Total cost =(to p20)

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The EOQ Model with Shortages

sc

c

s

cs

c

o

ocs

CCCQS

C

CC

C

DCQ

Q

DC

Q

SQC

Q

SCTC

** :level Shortage

2*:quantityorder Optimal

2

)(

2:costinventory Total

cost ordering total costs carrying total costs shortage Total cost Total

22

Example (to p21)

21

The EOQ Model with ShortagesExample

$1,279.20 639.60 465.16 $174.44

2.345,2)000,10)(150(

)2.345,2(2)6.705,1)(75.0(

)2.345,2(2)6.639)(2(

**2*)*(

*2:costinventory Total

yd 6.63975.02

75.02.345,2* :level Shortage

yd 2.345,22

75.0275.0

)000,10)(150(22:quantityorder Optimal

yd 10,000 D yd,per $2 C yd,per $0.75 C $150, C

2222

*

*

sco

QDC

QSQC

QSCTC

CCCQS

CCC

CDCQ

ocs

sc

c

s

cs

c

o

Let,

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22

The EOQ Model with Shortages

Additional Parameters in Example

sday 19.9or year 0.064 10,000639.6

D t shortage a is re which theduring Time

days 53.2or 0.17110,000

639.6-2,345.2 t handon isinventory which during Time

ordersbetween days 0.7326.4

311

orders ofnumber

yearper days t ordersbetween Time

yd 6.705,16.6392.345,2 levelinventory Maximum

yearper orders 26.42.345,2

000,10 orders ofNumber

2

1

SD

SQ

SQ

QD

= Q/D

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23

2. Model with “discount rate”

Price discounts are often offered if a predetermined number of units is ordered or when ordering materials in high volume.

How do we decide if we should order more to take advantage of the discount being offered? (to p24)

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Quantity Discounts with Constant Carrying CostsAnalysis Approach

Solution method:

1. We first determine the optimal order size, Q*2. We then compare with any lower total cost with a discount price and accept the one has the minimum total cost

Example (to p25)

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Quantity Discounts with Constant Carrying Costs Example(1 of 2)

200 D unit,per $190 C $2,500, C co

The following discount schedule is offered, which size of order should we subscript?

Quantity Price 1- 49 50 – 89 90 +

$1,400 1,100 900

Consider the following example:

5.72190

)200)(500,2(2C

D2CQ

have weThen,

c

o *

Falls in this section, Now we compare the TC of this, toThe next discount class, 90+

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Quantity Discounts with Constant Carrying Costs Example(2 of 2)

- Compute total cost at eligible discount price ($1,100):

- Compare with total cost of with order size of 90 and price of $900:

- Because $194,105 < $233,784, maximum discount price should be taken and 90 units ordered.

784,233$)200)(100,1(2

)5.72()190()5.72(

)200)(500,2(

2*

**

PDQCQ

DCTC co

105,194$)200)(900(2

)90)(190()90(

)200)(500,2(

2*

**

PDQCcQ

CoDTC

Do we always take the offer?

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27

No

• It depends on the carrying cost

• Example:

• Carrying Costs as a Percentage of Price(to p28)

Chapter 16 - Inventory Management

28

Quantity Discounts with Carrying Costs as a Percentage of Price

Example (1 of 2)

- Consider the same example, but with 15% of TC as carrying cost

- Then, the carrying cost for each category is as follows:

- Data: Co = $2,500, D = 200 computers per year

Quantity Price (TC) Carrying Cost

0 - 49 $1,400 1,400(.15) = $210

50 - 89 1,100 1,100(.15) = 165

90 + 900 900(.15) = 135

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29

Quantity Discounts with Carrying Costs as a Percentage of Price

Example (2 of 2)

- Compute optimum order size for purchase price without discount and Cc = $210:

- Compute new order size:

- Compute minimum total cost:

- Compare with cost, discount price of $900, order quantity of 90:

- Optimal order size computed as follows:

- Since this order size is less than 90 units , it is not feasible,thus optimal order size is 77.8 units.

69210

)200)(500,2(22*

c

o

CDCQ

8.77165

)200)(500,2(2* Q

845,232$)200)(100,1(2

)8.77(1658.77

)200)(500,2(2*

* PDQC

QDCTC c

o

630,191$)200)(900(2

)90)(135(90

)200)(500,2( TC

1.86135

)200)(500,2(2* Q (less than 90 as needed)

(note: 69 falls onto 50-89 category)

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30

Quantity Discount Model Solution with QM for Windows

Exhibit 16.4

31

3. Model with “re-order points”

• The reorder point is the inventory level at which a new order is placed.

• Order must be made while there is enough stock in place to cover demand during lead time.

• Formulation: R = dL, where d = demand rate per time period, L = lead time

Then R = dL = (10,000/311)(10) = 321.54

Working days/yr

What wouldhappen?

(to p32)

32

Reorder Point• Inventory level might be depleted at slower or faster rate during lead time.

• When demand is uncertain, safety stock is added as a hedge against stockout.

Two possible scenarios

Safety stock!

No Safetystocks!

We should then ensureSafety stock is secured!

How?

(to p33)

33

Determining Safety Stocks Using Service Levels

• We apply the Z test to secure its safety level,

)( LZLdR d

Reorder point

Safety stock

Average sample demand

How these values are represented in the diagram of normal distribution?

(to p34)

34

Reorder Point with Variable Demand

stocksafety

yprobabilit level service toingcorrespond deviations standard ofnumber demanddaily ofdeviation standard the

timeleaddemanddaily average

pointreorder where

LZ

Z

Ld

R

LZLdR

d

d

d

Example (to p35)

35

Reorder Point with Variable DemandExample

Example: determine reorder point and safety stock for service level of 95%.

26.1. : formulapoint reorder in termsecond isstock Safety

yd 1.3261.26300)10)(5)(65.1()10(30

A)(Appendix 1.65 Zlevel, service 95%For

dayper yd 5 days, 10 L day,per yd 30 d

LZLdR

d

d

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36

Determining the Reorder Point with Excel

Exhibit 16.5