Inventory Control Models

Ch 4 (Known Demands)

R. R. Lindeke

IE 3265, Production And Operations Management

Reasons for Holding Inventories

Economies of Scale Uncertainty in delivery lead-times Speculation. Changing Costs Over Time Smoothing: to account for seasonality and/or

Bottlenecks Demand Uncertainty Costs of Maintaining Control System

Characteristics of Inventory Systems Demand

May Be Known or Uncertain May be Changing or Unchanging in Time

Lead Times - time that elapses from placement of order until it’s arrival. Can assume known or unknown.

Review Time. Is system reviewed periodically or is system state known at all times?

Characteristics of Inventory Systems

Treatment of Excess Demand. Backorder all Excess DemandLose all excess demandBackorder some and lose some

Inventory who’s quality changes over timeperishabilityobsolescence

Real Inventory Systems: ABC ideas This was the true basis of Pareto’s Economic

Analysis! In a typical Inventory System most companies

find that their inventory items can be generally classified as: A Items (the 10 - 20% of sku’s) that represent up to 80% of

the inventory value B Items (the 20 – 30%) of the inventory items that

represent nearly all the remaining worth C Items the remaining 50 – 70% of the inventory items

sku’s) stored in small quantities and/or worth very little

Real Inventory Systems: ABC ideas and Control

A Items must be well studied and controlled to minimize expense

C Items tend to be overstocked to ensure no runouts but require only occasional review

See mhia.org – there is an “e-lesson” on the principles of ABC Inventory management – check it out! – do the on-line lesson!

Relevant Inventory Costs Holding Costs - Costs proportional to

the quantity of inventory held. Includes:

1. Physical Cost of Space (3%)

2. Taxes and Insurance (2 %)3. Breakage Spoilage and Deterioration (1%)

4. Opportunity Cost of alternative investment. (18%)

Here these holding issues total: 24% Therefore, in inventory systems, the holding

cost would be taken as: h .24*Cost of product

Lets Try one:

Problem 4, page 193 – cost of inventory Find h first (yearly and monthly) Total holding cost for the given period:

THC = $26666.67 Average Annual Holding Cost

assumes an average monthly inventory of trucks based on on hand data

$3333

Relevant Costs (continued)

Ordering Cost (or Production Cost).Includes both fixed and variable components

slope = c

K C(x) = K + cx for x > 0; 0 for x = 0.

Relevant Costs (continued)

Penalty or Shortage Costs. All costs that accrue when insufficient stock is available to meet demand. These include: Loss of revenue due to lost demand Costs of book-keeping for backordered

demands Loss of goodwill for being unable to satisfy

demands when they occur.

Relevant Costs (continued)

When computing Penalty or Shortage Costs inventory managers generally assume cost is proportional to number of units of excess demand that will go unfulfilled.

The Simple EOQ Model – the most fundamental of all!

Assumptions:

1. Demand is fixed at units per unit time –

typically assumed at an annual rate (use

care).

2. Shortages are not allowed.

3. Orders are received instantaneously. (this

will be relaxed later).

Simple EOQ Model (cont.)Assumptions (cont.):

4. Order quantity is fixed at a value “Q” per

cycle. (we will find this as an optimal value)

5. Cost structure:

a) includes fixed and marginal order

costs (K + cx)

b) includes holding cost at h per unit held per

unit time.

Inventory Levels for the EOQ Model

The Average Annual Cost Function G(Q)

Modeling Inventory:

%

( ) 2: setup cost

c: Unit cost

QT: cycle length T=

h: holding cost of unit cost

K

hQK cQG Q T

Subbing Q/ for T

( )

2

( )2

K cQ hQG Q

Q

K hQG Q c

Q

Finding an Optimal Level of ‘Q’ – the so-called EOQ

Requires us to take derivative of the G(Q) equation with respect to Q

Then, Set derivative equal to Zero:

Now, Solve for Q

'2

( ) 02

K hG Q

Q

Properties of the EOQ (optimal) Solution

2KQ

h

Q is increasing with both K and and decreasing with h

Q changes as the square root of these quantities Q is independent of the proportional order cost, c.

(except as it relates to the value of h = I*c)

Try ONE!

A company sells 145 boxes of BlueMountain BobBons/week (a candy)

Over the past several months, the demand has been steady

The store uses 25% as a ‘holding factor’ Candy costs $8/bx and sells for $12.50/bx Cost of making an order is $35 Determine EOQ (Q*) and how often an order

should be placed

Plugging and chugging:

h = $8*.25 = $2 = 145*52 = 7540

*

*

2*35*7540513.7 514

2

514 .0687540514 3.54145:

Q

QT yr

or wk

In your teams: Compute Pr. 10, pg 201

But, Orders usually take time to arrive!

This is a realistic relaxation of the EOQ ideas – but it doesn’t change the model

This requires the user to know the order “Lead Time” And then they trigger an order at a point before

the delivery is needed to assure no stock outs In our example, what if lead time is 1 week?

We should place an order when we have 145 boxes in stock (the one week draw down)

Note make sure lead time units match units in T!

But, Orders usually take time to arrive!

What happens when order lead times exceed T? We proceed just as before (but we compute /T) is the lead time is units that match T Here, lets assume = 6 weeks then:

/T = 6/3.545 = 1.69 – in the Blue Mount Bon-Bon case Place order: 1.69 cycles before we need product! Trip Point is then 0.69*Q* = .69*514 = 356 boxes This trip point is not for the next stock out but the one

after that (1.69 T from now!)– be very careful!!!

Sensitivity Analysis

Let G(Q) be the average annual holding and set-up cost function given by

and let G* be the optimal average annual cost. Then it can be shown that:

( ) / / 2G Q K Q hQ

( ) 1 *

* 2 *

G Q Q Q

G Q Q

Sensitivity

We find that this model is quite robust to Q errors if holding costs are relatively low

We find, given a Q – error in ordering quantity that Q* + Q has smaller error than Q* - Q (Error

means extra inventory maintenance costs) That is, we tend to have a greater penalty cost if

we order too little than too much

Inventory Levels for Finite Production Rate Model

EOQ With Finite Production Rate

Suppose that items are produced internally at a rate P (> λ, the consumption rate). Then the optimal production quantity to minimize average annual holding and set up costs has the same form as the EOQ, namely:

Except that h’ is defined as h’= h(1- λ/P)

2

'

kQ

h

This is based on solving:

( )2

( ) 1 /2

1

P is annual production rate

H is maximum on hand quantity

K hHG Q

TK hQ

G Q PQ

H Q P

Lets Try one:

We work for Sam’s Active Suspensions They sell after market kits for car “Tooners” They have an annual demand of 650 units Production rate is 4/day (working at 250 d/y) Setup takes 2 technicians working 45

minutes @$21/hour and requires an expendable tool costing $25

Continuing:

Each kit costs $275 Sam’s uses MARR of 18%, tax at 3%,

insurance at 2% and space cost of 1% Determine h, Q*, H, T and break T down to:

T1 = production time in a cycle (Q*/P)

T2 = non producing time in a cycle (T – T1)

Engineering Teams: (You can) Do It

Quantity Discount Models

All Units Discounts: the discount is applied to ALL of the units in the order. Gives rise to an order cost function such as that pictured in Figure 4-9

All-Units Discount Order Cost Function

Quantity Discount Models

Incremental Discounts: the discount is applied only to the number of units above the breakpoint. Gives rise to an order cost function such as that pictured in Figure 4-10.

Incremental Discount Order Cost Function

Properties of the Optimal Solutions

For all units discounts, the optimal will occur at the bottom of one of the cost curves or at a breakpoint. (It is generally at a breakpoint.). One compares the cost at the largest realizable EOQ and all of the breakpoints beyond it. (See Figure 4-11).

For incremental discounts, the optimal will always occur at a realizable EOQ value. Compare costs at all realizable EOQ’s. (See Figure 4-12).

All-Units Discount Average Annual Cost Function

To Find EOQ in ‘All Units’ discount case:

Compute Q* for each cost levelCheck for Feasibility (the Q computed

is applicable to the range) – we say it is “Realizable”

Compute G(Q*) for each of the realizable Q*’s and the break points.

Chose Q* as the one that has lowest G(Q)

Lets Try one:

Product cost is $6.50 in orders <600, $3.50 above 600.

Organizational I is 34% K is $300 and annual demand is 900

*1

2 300 900495

0.34 6.50Q

*2

2 300 900674

0.34 3.50Q

Lets Try one: Both of these are Realizable (the value is ‘in

range’) Compute G(Q) for both and breakpoint (600) G(Q) = c + (*K)/Q + (h*Q)/2

.34 6.5 495900 300(495) 900 (6.50) 495 2(495) $6942.43

G

G

.34 3.5 674900 300(674) 900 (3.50) 674 2(674) $3951.62

G

G

.34 6.5 600900 300(600) 900 (6.50) 600 2(600) $3957.00

G

G

Order 674 at a time!

Average Annual Cost Function for Incremental Discount Schedule

In an Incremental Case:

Cost is a strictly varying function of Q -- It varies by interval!

Calculate a C(Q) for the applied schedule Divide by Q to convert it to a “unit cost” function Build G(Q) equations for each interval Find Q* from each G(Q) Equation Check if “Realizable” Compute G(Q*) for realizable Q*’s

Trying the previous problem (but as Incremental Case):

Cost Function: Basically states that we pay 6.50 for each unit up to 600 then 3.50 for each unit ordered beyond 601: C(Q) = 6.5(Q), Q < 600 C(Q) = 3.5(Q – 600) + 6.5*600, Q 600 C(Q)/Q = 6.5, Q < 600 (order up to 600)

C(Q)/Q = 3.5 + ((3900 – 2100)/Q), Q 600 or C(Q)/Q = 3.5 + (1800)/Q (orders beyond 601)

Trying the previous but as Incremental Case:

For the First Interval: Q* = [(2*300*900)/(.34*6.50)] = 495 (realizable)

For order > 600, find Q* by writing a G(Q) equation and then optimizing:

[G(Q) = c + (*K)/Q + (h*Q)/2]

* 22 2

2 2 2

*2 2 2

2

1800 300 900 1800( ) 900 3.5 0.34 3.5 2

900 2100 .34 3.5( ) 900 3.5 .34 9002

QG Q Q Q Q

G Q QQ

Differentiating G2(Q)

2 2

22 2

900 2100 .34 3.50 0

2

2Set equal to 0 and solve for Q

d G Q

dQ Q

22

2

900 21000.595 0

900 210017830.595

Q

Q

Realizable!

Now Compute G(Q) for both and “cusp”

G(495) = 900*6.5 + (300*900)/495 + .34((6.5*495)/2) = $6942.43

G(600) = 900*6.5 +(300*900/600) + .34((6.5*600)/2) = $6963.00

G(1763) = 900*(3.5 +(1800/1783)) + (300*900)/1783 + .34*(3.5 +(1800/1783))*(1783/2) = $5590.67

Lowest cost – purchase 1783 about every 2 years!

Properties of the Optimal Solutions

Lets jump back into our teams and do some!

TRY 21b and 22b on Pg 211

Resource Constrained Multi-Product Systems

Consider an inventory system of n items in which the total amount available to spend is C and items cost respectively c1, c2, . . ., cn. Then this imposes the following constraint on the system:

1 1 2 2 ... n nc Q c Q c Q C

Resource Constrained Multi-Product Systems When the condition that:

is met, the solution procedure is straightforward. If the condition is not met, one must use an iterative procedure involving Lagrange Multipliers.

1 1 2 2/ / ... /n nc h c h c h

EOQ Models for Production Planning

Consider n items with known demand rates, production rates, holding costs, and set-up costs. The objective is to produce each item once in a production cycle. For the problem to be feasible the following equation must be true:

1

1n

j

j jP

Issues:

We are interested in controlling Family MAKESPAN (we wish to produce all products within our chosen cycle time)

Underlying Assumptions: Setup Cost (times) are not Sequence Dependent

(this assumption is not always accurate as we will later see)

Plants uses a “Rotation” Policy that produces a single ‘batch’ of each product each cycle – a mixed line balance assumption

1

1

2

*'

n

jj

n

j jj

K

Th

EOQ Models for Production Planning The method of solution is to express the average

annual cost function in terms of the cycle time, T. The optimal cycle time has the following mathematical form:

We must assure that this time allows for all setups and of production times.

Working forward:

This last statement means: (sj+(Qj/Pj)) T Of course: Qj = j*T So – with substitution: (sj+((j*T )/Pj) T Or: T((sj/(1- j/Pj)) = Tmin

Finally, we must Choose T(actual cycle time) = MAX(T*,Tmin)

Lets Try Problem 30ITEM Mon

Req’rDaily Req’r

h = .2*c

/P h’ Setup Time

Setup Cost

Unit Cost

Daily Pr. Rate

Mon. Pr. Rate

J55R 125 6.25 4 .045 3.82 1.2 $102 $20 140 2800

H223 140 7 7 .032 6.78 0.8 $68 $35 220 4400

K-18R 45 2.25 0.6 .023 0.586 2.2 $187 $12 100 2000

Z-344 240 12 9 .073 8.34 3.1 $263.5 $45 165 3300

Given: 20 days/month and 12 month/year; $85/hr for setup

Compute the Following – in teams!:

1

'

1

1

1

2

1

,

Lot Size: Demand Rate

Uptime; Drawdown Time; Utilization

n

jj

n

j jj

n

jj

Minn

j

j j

opt Min MAX

opt

K

Th

s

T

P

T T T

T