1 Inventory Management and Control. 2 Inventory Defined Inventory is the stock of any item or...

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Inventory Management and

Control

2

Inventory Defined

• Inventory is the stock of any item or resource held to meet future demand and can include: raw materials, finished products, component parts, supplies, and work-in-process

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Inventory

Process stage

Demand Type

Number & Value Other

Raw Material WIP

Finished Goods

Independent Dependent

A Items B Items C Items

Maintenance Operating

Inventory Classifications

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E(1)

Independent vs. Dependent Demand

B(4)

E(2)D(1)

C(2)

E(3)B(1)

A

Independent Demand (Demand for the final end-product or demand not related to other items; demand created by

external customers)

Dependent Demand

(Derived demand for component

parts, subassemblies,

raw materials, etc- used to produce final products)

Finishedproduct

Component parts

Independent demand is uncertain Dependent demand is certain

5

Inventory Models

• Independent demand – finished goods, items that are ready to be sold– E.g. a computer

• Dependent demand – components of finished products– E.g. parts that make up the computer

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Types of Inventories (1 of 2)

• Raw materials & purchased parts

• Partially completed goods called work in progress

• Finished-goods inventories (manufacturing firms) or merchandise (retail stores)

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Types of Inventories (2 of 2)

• Replacement parts, tools, & supplies

• Goods-in-transit to warehouses or customers

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The Material Flow Cycle (1 of 2)

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Run time: Job is at machine and being worked onSetup time: Job is at the work station, and the work station is

being "setup."Queue time: Job is where it should be, but is not being

processed because other work precedes it.Move time: The time a job spends in transitWait time: When one process is finished, but the job is waiting

to be moved to the next work area.

The Material Flow Cycle (2 of 2)

WaitTime

MoveTime

QueueTime

SetupTime

RunTimeInput

Cycle Time

Output

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Performance Measures

• Inventory turnover (the ratio of annual cost of goods sold to average inventory investment)

• Days of inventory on hand (expected number of days of sales that can be supplied from existing inventory)

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Functions of Inventory (1 of 2)

1. To “decouple” or separate various parts of the production process, ie. to maintain independence of operations

2. To meet unexpected demand & to provide high levels of customer service

3. To smooth production requirements by meeting seasonal or cyclical variations in demand

4. To protect against stock-outs

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Functions of Inventory (2 of 2)

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

6. To provide a stock of goods that will provide a “selection” for customers

7. To take advantage of economic purchase-order size

8. To take advantage of quantity discounts

9. To hedge against price increases

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• Higher costs– Item cost (if purchased)– Holding (or carrying) cost

• Difficult to control

• Hides production problems

• May decrease flexibility

Disadvantages of Inventory

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Inventory Costs Holding (or carrying) costs

Costs for storage, handling, insurance, etc

Setup (or production change) costs Costs to prepare a machine or process for

manufacturing an order, eg. arranging specific equipment setups, etc

Ordering costs (costs of replenishing inventory) Costs of placing an order and receiving goods

Shortage costs Costs incurred when demand exceeds supply

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Holding (Carrying) Costs

• Obsolescence• Insurance• Extra staffing• Interest• Pilferage• Damage• Warehousing• Etc.

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Inventory Holding Costs(Approximate Ranges)

Category

Housing costs (building rent, depreciation, operating cost, taxes, insurance)

Material handling costs (equipment, lease or depreciation, power, operating cost)

Labor cost from extra handling

Investment costs (borrowing costs, taxes, and insurance on inventory)

Pilferage, scrap, and obsolescence

Overall carrying cost

Cost as a % of Inventory Value

6%(3 - 10%)

3%(1 - 3.5%)

3%(3 - 5%)

11%(6 - 24%)

3% (2 - 5%)

26%

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Ordering Costs

• Supplies

• Forms

• Order processing

• Clerical support

• etc.

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Setup Costs

• Clean-up costs

• Re-tooling costs

• Adjustment costs

• etc.

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Shortage Costs

• Backordering cost

• Cost of lost sales

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Inventory Control System Defined

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

Answers questions as: When to order? How much to order?

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Objective of Inventory Control

To achieve satisfactory levels of customer service while keeping inventory costs within reasonable bounds

Improve the Level of customer service

Reduce the Costs of ordering and carrying inventory

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A system to keep track of inventory

A reliable forecast of demand

Knowledge of lead times

Reasonable estimates of Holding costs

Ordering costs

Shortage costs

A classification system

Requirements of an Effective Inventory Management

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Inventory Counting (Control) Systems• Periodic System

Physical count of items made at periodic intervals; order is placed for a variable amount after fixed passage of time.

• Perpetual (Continuous) Inventory System System that keeps track of removals from inventory continuously, thus monitoring current levels of each item (constant amount is ordered when inventory declines to a predetermined level)

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Inventory ModelsSingle-Period Inventory Model

One time purchasing decision (Examples: selling t-shirts at a football game, newspapers, fresh bakery products, fresh flowers)

Seeks to balance the costs of inventory over stock 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

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Single-Period Inventory Model

• In a single-period model, items are received in the beginning of a period and sold during the same period. The unsold items are not carried over to the next period.

• The unsold items may be a total waste, or sold at a reduced price, or returned to the producer at some price less than the original purchase price.

• The revenue generated by the unsold items is called the salvage value.

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•Single period model: It is used to handle ordering of perishables (fresh fruits, flowers) and other items with limited useful lives (newspapers, spare parts for specialized equipment).

3

(Newsboy Problem)

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Shortage cost (Cost of Understocking)

• Shortage cost: generally, this cost represents unrealized profit per unit

(Cu=Revenue per unit – Cost per unit)

• If a shortage or stockout cost relates to a spare part for a machine, then shortage cost refers to the actual cost of lost production.

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Excess cost (Cost of Over Stocking)

• Excess cost (Ce): difference between purchase cost and salvage value of items left over at the end of a period.

• If there is a cost associated with disposing of excess items, the salvage cost will be negative.

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Single Period Model

Given the costs of overestimating/underestimating demand and the probabilities of various demand sizes the goal is to identify the order quantity or stocking level that will minimize the long-run excess (overstock)or shortage costs (understock).

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Demand may be discrete or continuous. The demand of computer, newspaper, etc. is usually an integer. Such a demand is discrete. On the other hand, the demand of gasoline is not restricted to integers. Such a demand is continuous. Often, the demand of perishable food items such as fish or meat may also be continuous. •Consider an order quantity Q•Let p = probability (demand<Q)

= probability of not selling the Qth item.•So, (1-p) = probability of selling the Qth item.

Single-Period Models (Demand Distribution)

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• Expected loss from the Qth item = • Expected profit from the Qth item = • So, the Qth item should be ordered if

• Decision Rule (Discrete Demand):– Order maximum quantity Q such that

where p = probability (demand<Q)

se

e

cc

cpor

cppc

s

s

,

)1(

epc

scp1

se

s

cc

cp

Single-Period Models (Discrete Demand)

33

Single-Period Model

es

s

CC

CP

es

s

CC

CP

sold be unit will y that theProbabilit

estimatedunder demand ofunit per Cost C

estimatedover demand ofunit per Cost C

:Where

s

e

P

The service level is the probability that demand will not exceed the stocking level. The service level determines the amount of stocking level to keep.

The service level is the probability that demand will not exceed the stocking level. The service level determines the amount of stocking level to keep.

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Optimal Stocking Level (Choosing optimum Stocking level to minimize these costs is similar to

balancing a seesaw)

Service Level

So

Quantity

Ce Cs

Balance point

Service level =Cs

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

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Service Level

Another way to define ‘Service Level’ is:

• proportion of cycles in which no stock-out occurs

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Service Level

Order Cycle Demand Stock-Outs

1 180 02 75 03 235 454 140 05 180 06 200 107 150 08 90 09 160 010 40 0

Total 1450 55

Since there are two cycles out of ten in which a stockout occurs, service level is 80%. This translates to a 96% fill rate. There are a total of 1,450 units demand and 55 stockouts (which means that 1,395 units of demand are satisfied).

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Single Period Model (Demand is represented by a discrete distribution)

• Unlike the continuous case where the optimal solution is found by determining So which makes the distribution function equal to the critical ratio cs / (cs + ce), in the discrete case, the critical ratio takes place between two values of F(So) or F(Q)

• The optimal So or Q corresponds to the higher value of F( So) or F(Q).

(Note that, in the discrete case, the distribution function increases by jumps)

SEE EXAMPLES 17 & 18 on page 576

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Example : Demand for cookies: Demand Probability of Demand

1,800 dozen 0.052,000 0.102,200 0.202,400 0.302,600 0.202,800 0.103,000 0,05

Selling price=$0.69, cost=$0.49, salvage value=$0.29

What is the optimal number of cookies to make?

c


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Cs= 0.69-0.49=$0.2, Ce= 0.49-0.29=$0.2

Order maximum quantity, Q such that

Demand, Q Probability(demand) Probability(demand<Q), p

1,800 dozen 0.05 0.052,000 0.10 0.152,200 0.20 0.352,400 0.30 0.652,600 0.20 0.852,800 0.10 0.953,000 0,05 1.00

5.0demandyProbabilit

es cc

cQp

s


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Single-Period Models (Continous Demand)

• Often the demand is continuous. Even when the demand is not continuous, continuous distribution may be used because the discrete distribution may be inconvenient.

• We shall discuss two distributions:

Uniform distribution

Normal distribution

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Example 2: The J&B Card Shop sells calendars. The once-a-year order for each year’s calendar arrives in September. The calendars cost $1.50 and J&B sells them for $3 each. At the end of July, J&B reduces the calendar price to $1 and can sell all the surplus calendars at this price. How many calendars should J&B order if the September-to-July demand can be approximated by

a. uniform distribution between 150 and 850

Single-Period Models (Continuous Demand)

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Overage cost

ce = Purchase price - Salvage value =1.5-1=$0.5

Underage cost

cs = Selling price - Purchase price =3-1.5=$1.5


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p =0.75

Now, find the Q so that p = probability(demand<Q) =0.75

Q* = a+p(b-a) =150+0.75(850-150)=675

se

s

cc

c

Area =

850150 Demand

Pro

babi

lity

Area=

Q*


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Example 3: The J&B Card Shop sells calendars. The once-a-year order for each year’s calendar arrives in September. The calendars cost $1.50 and J&B sells them for $3 each. At the end of July, J&B reduces the calendar price to $1 and can sell all the surplus calendars at this price. How many calendars should J&B order if the September-to-July demand can be approximated by

b. normal distribution with = 500 and =120.


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Solution to Example 3: ce =$0.50, cu =$1.50 (see Example 2)

p = = 0.750.501.50

1.50

se

s

cc

c


46


Now, find the Q so that p = 0.75

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Single-Period Models (Continuous Demand

Q= So = mean + zσ

= 500 + .68(120)

= 582

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Single Period Example 15 (pg. 574)Demand is uniformly distributed

• Ce = $0.20 per unit

• Cs = $0.60 per unit

• Service level = Cs/(Cs+Ce) = .6/(.6+.2)

• Service level = .75

• Opt. Stock.Level=S0=300+.75(500-300)= 450 liters

Service Level = 75%

Quantity

Ce Cs

Stockout risk = 1.00 – 0.75 = 0.25

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Uniform Distribution[Continuous Dist’n]

• A random variable X is uniformly distributed on the interval (a,b), U(a,b), if its pdf and cdf are:

• Properties– P(x1 < X < x2) is proportional to the length of the interval [F(x2) –

F(x1) = (x2-x1)/(b-a)]

– E(X) = (a+b)/2 V(X) = (b-a)2/12

• U(0,1) provides the means to generate random numbers, from which random variates can be generated.

otherwise ,0

,1

)( bxaabxf

bx

bxaab

axax

xF

,1

,

,0

)(

50

50

Poisson Distribution[Discrete Dist’n]

• Poisson distribution describes many random processes quite well and is mathematically quite simple.– where > 0, pdf and cdf are:

– E(X) = = V(X)

otherwise ,0

,...1,0 ,!)( x

x

exp

x

x

i

i

i

exF

0 !)(

51

51

Normal Distribution [Continuous

Dist’n]

• A normally distributed random variable X has the pdf:

– Mean:

– Variance:

– Denoted as X ~ N(,2)

• Special properties: .– symmetric about .

– The maximum value of the pdf occurs at x = ; the mean and mode are equal.

xx

xf ,2

1exp

2

1)(

2

02

52

52


Dist’n]

• Evaluating the distribution:– Use numerical methods (no closed form)

– Independent of and using the standard normal distribution:

Z ~ N(0,1)

– Transformation of variables: let Z = (X - ) / ,

z t dtez 2/2

2

1)( where,

)()(

2

1

)(

/)(

/)( 2/2

xx

x z

dzz

dze

xZPxXPxF

53

53


Dist’n]

• Example: The time required to load an oceangoing vessel, X, is distributed as N(12,4)– The probability that the vessel is loaded in less than 10 hours:

• Using the symmetry property, (1) is the complement of (-1)

1587.0)1(2

1210)10(

F

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Single Period Model(Demand is represented by a continous distribution)

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 and we can assume that demand for shirts is approximately normally distributed. We make $10 on every shirt we sell at the game, but lose $5 on every shirt not sold. What is the optimal stocking level for shirts?

So =mean + zσCs = $10 and Ce = $5; P ≤ $10 / ($10 + $5) = .667

Z.667 = .432 therefore we need 2,400 + .432(350) = 2,551 shirts

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Multi-Period Inventory Models

Fixed-Order Quantity Models (Types of)Economic Order Quantity Model (EOQ)Economic Production Order Quantity (Economic Lot Size) Model (EPQ)Economic Order Quantity Model with Quantity Discounts

Fixed Time Period (Fixed Order Interval) Models

56

Fixed Order Quantity Models:Economic Order Quantity Model

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Economic Order Quantity Model Assumptions (1 of 2):

• Demand for the product is known with certainty, it is constant and uniform throughout the period

• Lead time (time from ordering to receipt) is known and constant

• Price per unit of product is constant (no quantity discounts). So it is not included in the total cost.

• Inventory holding cost is based on average inventory

58

Economic Order Quantity Model Assumptions (2 of 2):

• Ordering or setup costs are constant

• All demands for the product will be satisfied (no backorders are allowed)

• No stockouts (shortages) are allowed

• The order quantity is received all at once. (Instantaneous receipt of material in a single lot)The goal is to calculate the order quantitiy that

minimizes total cost

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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(Inv. Level)

1. You receive an order quantity Q.

2. You 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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EOQ Model

Reorder Point

(ROP)

Time

Inventory LevelAverageInventory

(Q/2)

Lead Time

Order Quantity

(Q)

Demand rate

Order placed Order received

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EOQ Cost Model: How Much to Order?

By adding the holding and ordering costs together, we determine the total cost curve, which in turn is used to find the optimal order quantity that minimizes total costs

Slope = 0Slope = 0

Total CostTotal Cost

Order Quantity, Order Quantity, QQ

Annual Annual cost ($)cost ($)

Minimum Minimum total costtotal cost

Optimal orderOptimal order QQoptopt

Carrying Cost =Carrying Cost =HHQQ

22

Ordering Cost =Ordering Cost =SSDD

QQ

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• More units must be stored if more are ordered

Purchase OrderDescription Qty.Microwave 1

Order quantity


Order quantity

Why Holding Costs Increase?

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Cost is spread over more units

Example: You need 1000 microwave ovens





1000 Order (Postage $ 0.33) 1 Order (Postage $330)

Order quantity


Why Ordering Costs Decrease ?

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Basic Fixed-Order Quantity (EOQ) Model Formula

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 =Annual 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 =Annual 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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EOQ Cost Model

Annual ordering cost =Annual ordering cost =S S DD

QQ

AnnualAnnual carrying costcarrying cost = =HHQQ

22

Total cost = +Total cost = +S S DD

QQH H QQ

22

TC = +S D

Q

H Q

2

=- +S D

Q2

H

2TC

Q

0 =- +S D

Q2

H

2

Qopt =2SD

H

Deriving Qopt Proving equality of costs at optimal point

=S D

Q

H Q

2

Q2 =2S D

H

Qopt =2 S D

H

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

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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

Reorder point, R = d L_

Reorder point, R = d L_

(constant) timeLead = L

(constant) demanddaily average = d_

We also need a reorder point to tell us when to place an order

We also need a reorder point to tell us when to place an order

1) How much to order?

2) When to order?

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Optimal Order Quantity

Expected Number of Orders

Expected Time Between Orders Working Days / Year

Working Days / Year

= =× ×

= =

= =

=

= ×

Q*D SH

ND

Q*

TN

dD

ROP d L

2

EOQ Model Equations

68

EOQ Example 1 (1 of 3)

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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EOQ Example 1(2 of 3)

Q = 2DS

H =

2(1,000 )(10)

2.50 = 89.443 units or OPT 90 unitsQ =

2DS

H =

2(1,000 )(10)

2.50 = 89.443 units or OPT 90 units

d = 1,000 units / year

365 days / year = 2.74 units / dayd =

1,000 units / year

365 days / year = 2.74 units / day

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

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

20 units

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

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

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EOQ Example I(3 of 3)

TCTCminmin = =SSDD

QQHHQQ

22

TCTCminmin = = (10)(1,000)(10)(1,000)

9090((2,52,5)()(990)0)

22

TCTCminmin = $ = $ 111 111 + $ + $111111 = = 22 22 $$

Orders per year =Orders per year = DD//QQoptopt

== 1000/1000/9900

== 1111 orders/year orders/year

Order cycle timeOrder cycle time== 365/(365/(DD//QQoptopt))

== 336565//1111 == 33.133.1daysdays

++

+

71


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 given the following…

Determine the economic order quantity and the reorder point given the following…

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Q =2DS

H=

2(10,000 )(10)

1.50= 365.148 units, or OPT 366 unitsQ =

2DS

H=

2(10,000 )(10)

1.50= 365.148 units, or OPT 366 units

d =10,000 units / year

365 days / year= 27.397 units / dayd =

10,000 units / year

365 days / year= 27.397 units / day

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

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

274 units

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

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

73

EOQ Example 3

HH = $0.75 per yard = $0.75 per yard SS = $150 = $150 DD = 10,000 yards = 10,000 yards

QQoptopt = =22 S S DD

HH

QQoptopt = =2(150)(10,000)2(150)(10,000)

(0.75)(0.75)

QQoptopt = 2,000 yards = 2,000 yards

TCTCminmin = + = +S S DD

QQH H QQ

22

TCTCminmin = + = +((150)(10,000)150)(10,000)

2,0002,000(0.75)(2,000)(0.75)(2,000)

22

TCTCminmin = $750 + $750 = $1,500 = $750 + $750 = $1,500

Orders per year = D/Qopt

= 10,000/2,000

= 5 orders/year

Order cycle time =311 days/(Order cycle time =311 days/(DD//QQoptopt))

== 311/5311/5

== 62.2 store days62.2 store days

74

When to Reorder with EOQ Ordering ?• Reorder Point – is the level of inventory at which a

new order is placed

ROP = d . L

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

• Service Level - Probability that demand will not exceed supply during lead time (probability that inventory available during the lead time will meet the demand) 1 - Probability of stockout

75

Reorder Point Example

Demand = 10,000 yards/year

Store open 311 days/year

Daily demand = 10,000 / 311 = 32.154 yards/day

Lead time = L = 10 days

R = dL = (32.154)(10) = 321.54 yards

76

Determinants of the Reorder Point

• The rate of demand

• The lead time

• Demand and/or lead time variability

• Stockout risk (safety stock)

77

Answer how much & when to order Allow demand and lead time to vary

Follows normal distribution Other EOQ assumptions apply

Consider service level & safety stock Service level = 1 - Probability of stockout Higher service level means more safety stock More safety stock means higher ROP

Probabilistic Models

78

Safety Stock

LT Time

Expected demandduring lead time

Maximum probable demandduring lead time

ROP

Qu

an

tity

Safety stock

Safety stock reduces risk ofstockout during lead time

79

Reorder Point With Variable Demand

Reorderpoint, R

Q

LTLT

TimeTimeLTLT

Inve

nto

ry le

vel

0

80

Reorder Point with a Safety Stock

Reorderpoint, R

QQ

LT

Time

LT

Inve

nto

ry le

vel

0

Safety Stock

81

Reorder Point With Variable Demand and Constant Lead Time

R = dL + zd Lwhere

d = average daily demandL = lead time

d = the standard deviation of daily demand

z = number of standard deviationscorresponding to the service levelprobability

zd L = safety stock

82

Reorder Point for Service Level

Probability of meeting demand during lead time = service level

Probability of a stockout

R

Safety stock

dLExpected Demand

zd L

The reorder point based on a normal distribution of LT demand

83

Reorder Point for Variable Demand (Example)

The carpet store wants a reorder point with a 95% service level and a 5% stockout probability

d = 30 yards per day, (demand is normally distributed) d= 5 yards per day

For a 95% service level, z = 1.65

R = dL + z d L

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

= 326.1 yards

Safety stock = z d L

= (1.65)(5)( 10)

= 26.1 yards

L= 10 days

84

Shortages and Service Levels

It is also important to specify:

1) Expected number of units short per order cycle

E(n) =E(z) σdLT

where E(z) is standardized number of units short obtained from Table 12.3, pg. 569.

2) Expected number of units short per year

E(N) =E (n) (D/Q)

3) Annual Service Level

SLannual = 1- E(N)/D that is percentage of demand filled directly from inventory, known also as FILL RATE.

85

85

Example 10 (pg. 568)– shortages and service levels

Suppose standard deviation of lead time demand is known to be 20 units. Lead time demand is approximately normal.

(a) For lead time service level of 90 percent, determine the expected number of units short for any other cycle.

(b) What lead time service level would imply an expected shortage of 2 units?

86

86

Answer – shortage and service levels

(a) For lead time service level of 90 percent, determine the expected number of units short for any other cycle.

σdLT= 20 units

lead time service level is 0.90 from z table (lead time), E(z)= 0.048 (page 569, table 12.3)

E(n) =E(z) σdLT = (0.048) (20)= 0.96 or about 1 unit.

(b) What lead time service level would an expected shortage of 2 units imply?

E(n) = 2 E(n) =E(z) σdLT or E(z) = E(n) / σdLT =(2)/(20)= 0.100 from the

table, lead time service level is 81.06 percent or 81.7%

87

Shortages and Service Levels

• Expected number of units short per year

See example 11, page 568

• Annual Service Level

See example 12, page 570

Note that annual service level will usually be grater than the cycle service level

88

Fixed Order Quantity Models:-Noninstantaneous Receipt-Production Order Quantity

(Economic Lot Size) Model

89

Production done in batches or lotsCapacity to produce a part exceeds that part’s

usage or demand rateAllows partial receipt of material

Other EOQ assumptions apply

Suited for production environment Material produced, used immediately Provides production lot size

Lower holding cost than EOQ modelAnswers how much to order and when to order

Production Order Quantity Model

90

POQ Model Inventory Levels (1 of 2)Inventory Level

TimeSupply Begins

Supply Ends

Production portion of cycle

Demand portion of cycle with no supply

Maximum inventory level

91

POQ Model Inventory Levels (2 of 2)

Time

Inventory Level

Production Portion of

Cycle

Max. Inventory Q/p·(p- u)Q*Q*

Supply Begins

Supply Ends

Inventory level with no demand

Demand portion of cycle with no supply

Average inventory Q/2(1- u/p)

92

D = Demand per year

S = Setup cost

H = Holding cost

d = Demand per day

p = Production per day

POQ Model Equations

Setup Cost

Holding Cost

= *

= *

=

Q

D

QS

(

1/2 * H * Q -u

p1

)-u

p1

( )

Maximum inventory level

93

Production Order Quantity Example (1 of 2)

H = $0.75 per yard S = $150 D = 10,000 yards

u = 10,000/311 = 32.2 yards per day p = 150 yards per day

POQopt = = = 2,256.8 yards

2 S D

H 1 - up

2(150)(10,000)

0.75 1 - - 32.2150

TC = + 1 - = $1,329up

S DQ

H Q2

Production run = = = 15.05 days per orderQp

2,256.8150

94

Production Quantity Example (2 of 2)

H = $0.75 per yard S = $150 D = 10,000 yards

u= 10,000/311 = 32.2 yards per day p = 150 yards per day

QQoptopt = = = 2,256.8 yards = = = 2,256.8 yards

22CCooDD

CCcc 1 - 1 - ddpp

2(150)(10,000)2(150)(10,000)

0.75 1 - 0.75 1 - 32.232.2150150

TCTC = + 1 - = $1,329 = + 1 - = $1,329ddpp

CCooDD

QQ

CCccQQ

22

Production run = = = 15.05 days per orderQp

2,256.8150

Number of production runs = = = 4.43 runs/yearDQ

10,0002,256.8

Maximum inventory level = Q 1 - = 2,256.8 1 -

= 1,772 yards

up

32.2150

95

Fixed-Order Quantity Models:Economic Order Quantity Model

with Quantity Discounts

96

• Answers how much to order & when to order

• Allows quantity discounts

– Price per unit decreases as order quantity increases

– Other EOQ assumptions apply

• Trade-off is between lower price & increased holding cost

Quantity Discount Model

TC = + + PDS D

Q

iP QQ

22Where P: Unit Price

Total cost with purchasing cost

97

Total Costs with Purchasing CostC

ost

EOQ

TC with PD

TC without PD

PD

0 Quantity

Adding Purchasing costdoesn’t change EOQ

98

Quantity Discount Models

• There are two general cases of quantity discount models:

1.Carrying costs are constant (e.g. $2 per unit).

2.Carrying costs are stated as a percentage off purchase price (20% of unit price)

99

1) Total Cost with Constant Carrying Costs(Compute the Common Optimal Order Quantity

OC

EOQ Quantity

Tot

al C

ost TCa

TCc

TCbDecreasing Price

CC a,b,c

1002) Total Cost with Variable Carrying Cost (Compute Optimal Order Quantity for each price

range)

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

101

Quantity Discount – How Much to Order?

102

Price-Break Example 1 (1 of 3)

ORDER SIZE PRICE

0 - 99 $10

100 - 199 8 (d1)

200+ 6 (d2)

For this problem holding cost is given as a constant value, not as a percentage of price, so the optimal order quantity is the same for each of the price ranges. (see the figure 12.9)

103

Price Break Example 1 (2 of 3)

QQoptopt

Carrying cost Carrying cost

Ordering cost Ordering cost

Inve

nto

ry c

ost

($)

Inve

nto

ry c

ost

($)

QQ((dd1 1 ) = 100) = 100 QQ((dd2 2 ) = 200) = 200

TC TC ((dd2 2 = $6 ) = $6 )

TCTC ( (dd1 1 = $8 )= $8 )

TC TC = ($10 )= ($10 )

104

Price Break Example 1 (3 of 3)

QQoptopt

Carrying cost Carrying cost

Ordering cost Ordering cost

Inve

nto

ry c

ost

($)

Inve

nto

ry c

ost

($)

QQ((dd1 1 ) = 100) = 100 QQ((dd2 2 ) = 200) = 200

TC TC ((dd2 2 = $6 ) = $6 )

TCTC ( (dd1 1 = $8 )= $8 )

TC TC = ($10 )= ($10 )

The lowest total cost is at the second price break

105

Price Break Example 2

QUANTITYQUANTITY PRICEPRICE

1 - 491 - 49 $1,400$1,400

50 - 8950 - 89 1,1001,100

90+90+ 900900

SS = = $2,500 $2,500

HH = = $190 per computer $190 per computer

DD = = 200200

QQoptopt = = = 72.5 PCs = = = 72.5 PCs22SSDD

HH2(2500)(200)2(2500)(200)

190190

TCTC = + + = + + PD PD = $233,784 = $233,784 SSDD

QQoptopt

H H QQoptopt

22

For For QQ = 72.5 = 72.5

TCTC = + + = + + PD PD = $194,105= $194,105SSDD

QQ

H H QQ

22

For For QQ = 90 = 90

106


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 with a rate of 2% of the unit price, 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 with a rate of 2% of the unit price, 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

107

Price-Break Example (2 of 4)

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

108


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?

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

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

109


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

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

110

Multi-period Inventory Models:Fixed Time Period

(Fixed-Order- Interval) Models

111

Orders are placed at fixed time intervals

Order quantity for next interval? (inventory is brought up to target amount, amount ordered varies)

Suppliers might encourage fixed intervals

Requires only periodic checks of inventory levels (no continous monitoring is required)

Risk of stockout between intervals

Fixed-Order-Interval Model

112

Inventory Level in a Fixed Period System

Various amounts (Qi) are ordered at regular time intervals (p) based on the quantity necessary to bring inventory up to

target maximum

pp pp pp

QQ11 QQ22

QQ33

QQ44

Target maximum

TimeTime

d In

vent

ory

d In

vent

ory

113

Tight control of inventory items

Items from same supplier may yield savings in:

Ordering

Packing

Shipping costs

May be practical when inventories cannot be closely monitored

Fixed-Interval Benefits

114

Requires a larger safety stock Increases carrying cost Costs of periodic reviews

Fixed-Interval Disadvantages

115

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

116

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

117

Order Quantity for a Periodic Inventory System

Q = d(tb + L) + zd T + L - Iwhere

d = average demand rateT = the fixed time between ordersL = lead time

d = standard deviation of demand

zd T + L = safety stockI = inventory level

z = the number of standard deviations for a specified service level

118

Fixed-Period Model with Variable Demand (Example 1)

d = 6 bottles per dayd = 1.2 bottlesT = 60 daysL = 5 daysI = 8 bottlesz = 1.65 (for a 95% service level)

Q = d(T + L) + zd T + L - I

= (6)(60 + 5) + (1.65)(1.2) 60 + 5 - 8

= 397.96 bottles

119

Fixed-Time Period Model withVariable Demand (Example 2)(1 of 3)

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 standard deviation of daily demand is 4 units.

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

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

120

Fixed-Time Period Model with Variable Demand (Example 2)(2 of 3)

T+ L d2 2 = (T + L) = 30 + 10 4 = 25.298 T+ L d

2 2 = (T + L) = 30 + 10 4 = 25.298

So, by looking at the value from the Table, we have a probability of 0.9599, which is given by a z = 1.75

So, by looking at the value from the Table, we have a probability of 0.9599, which is given by a z = 1.75

121

Fixed-Time Period Model with Variable Demand (Example 2) (3 of 3)

or 644.272, = 200 - 44.272 800 = q

200- 298)(1.75)(25. + 10)+20(30 = q

I - Z+ L)+(Td = q L+T

units 645

or 644.272, = 200 - 44.272 800 = q

200- 298)(1.75)(25. + 10)+20(30 = q

I - Z+ L)+(Td = q L+T

units 645

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

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

122

ABC Classification System

• Demand volume and value of items vary

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

– dollars invested

– profit potential

– sales or usage volume

– stock-out penalties

123

ABC Classification System

Classifying inventory according to some measure of importance and allocating control efforts accordingly.

AA - very important

BB - mod. important

CC - least important Annual $ value of items

AA

BB

CC

High

Low

Low HighPercentage of Items

124

Classify inventory into 3 categories typically on the basis of the dollar value to the firm $ volume = Annual demand x Unit cost

A class, B class, C class Policies based on ABC analysis– Develop class A suppliers more carefully – Give tighter physical control of A items– Forecast A items more carefully

ABC Analysis

125

% of Inventory Items

Classifying Items as ABC

0

20

40

60

80

100

0 50 100

% Annual $ Usage

AABB

CC

Class % $ Vol % ItemsA 70-80 5-15B 15 30C 5-10 50-60

126

ABC Classification

11 $ 60$ 60 909022 350350 404033 3030 13013044 8080 606055 3030 10010066 2020 18018077 1010 17017088 320320 505099 510510 6060

1010 2020 120120

PARTPART UNIT COSTUNIT COST ANNUAL USAGEANNUAL USAGE

127

ABC Classification

11 $ 60$ 60 909022 350350 404033 3030 13013044 8080 606055 3030 10010066 2020 18018077 1010 17017088 320320 505099 510510 6060

1010 2020 120120

PARTPART UNIT COSTUNIT COST ANNUAL USAGEANNUAL USAGETOTAL % OF TOTAL % OF TOTALPART VALUE VALUE QUANTITY % CUMMULATIVE

9 $30,600 35.9 6.0 6.08 16,000 18.7 5.0 11.02 14,000 16.4 4.0 15.01 5,400 6.3 9.0 24.04 4,800 5.6 6.0 30.03 3,900 4.6 10.0 40.06 3,600 4.2 18.0 58.05 3,000 3.5 13.0 71.0

10 2,400 2.8 12.0 83.07 1,700 2.0 17.0 100.0

$85,400

128

ABC Classification

11 $ 60$ 60 909022 350350 404033 3030 13013044 8080 606055 3030 10010066 2020 18018077 1010 17017088 320320 505099 510510 6060

1010 2020 120120


9 $30,600 35.9 6.0 6.08 16,000 18.7 5.0 11.02 14,000 16.4 4.0 15.01 5,400 6.3 9.0 24.04 4,800 5.6 6.0 30.03 3,900 4.6 10.0 40.06 3,600 4.2 18.0 58.05 3,000 3.5 13.0 71.0

10 2,400 2.8 12.0 83.07 1,700 2.0 17.0 100.0

$85,400

AA

BB

CC

129

ABC Classification

11 $ 60$ 60 909022 350350 404033 3030 13013044 8080 606055 3030 10010066 2020 18018077 1010 17017088 320320 505099 510510 6060

1010 2020 120120


9 $30,600 35.9 6.0 6.08 16,000 18.7 5.0 11.02 14,000 16.4 4.0 15.01 5,400 6.3 9.0 24.04 4,800 5.6 6.0 30.03 3,900 4.6 10.0 40.06 3,600 4.2 18.0 58.05 3,000 3.5 13.0 71.0

10 2,400 2.8 12.0 83.07 1,700 2.0 17.0 100.0

$85,400

AA

BB

CC

% OF TOTAL % OF TOTALCLASS ITEMS VALUE QUANTITY

A 9, 8, 2 71.0 15.0B 1, 4, 3 16.5 25.0C 6, 5, 10, 7 12.5 60.0

130

ABC Classification

100 100 –

80 80 –

60 60 –

40 40 –

20 20 –

0 0 –| | | | | |00 2020 4040 6060 8080 100100

% of Quantity% of Quantity

% o

f V

alu

e%

of

Val

ue

AA

BBCC

131

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

• Cycle counting refers to Physical Count of items in inventory.

• Used often with ABC classification– While A items are counted most often (e.g.,

daily), C items are counted the least frequently.

Inventory Accuracy and Cycle Counting

132

Last Words

Inventories have certain functions.

But too much inventory

- Tends to hide problems

- Costly to maintain

So it is desired

• Reduce lot sizes

• Reduce safety stocks

Date post:	29-Mar-2015
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