chapter 4 models for known demandweb.eng.fiu.edu/leet/ein6345inventory/chap4_2013.pdf · chapter 4...

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Chapter 4

Models for Known Demand

Introduction

• EOQ analysis is based on a number of assumptions.

• In the next two chapters we describe some models where

these assumptions are removed.

• This chapter keeps the condition that all variables take

known values (giving deterministic models) while the next

chapter allows some uncertainty (giving probabilistic

models).

• We concentrate on the most widely used models, and use

these to illustrate some general principles.

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Variable Costs

• Unit costs vary with the quantity ordered

• Reordering cost might also vary with order size, particularly

if it includes quality control inspections, or some other

processing of deliveries.

• The transport cost also varies with the amount delivered,

and can have quite complex patterns.

• The holding cost can vary with order size.

• It is common for all of the costs to vary, often in quite

complicated ways.

• We will, therefore, illustrate the general approach with

problems where the unit cost decreases in discrete steps

with increasing order quantity.

Price Discounts from Suppliers

Unit

Cost

Order

Quantity

Lower

Limit

Order

Quantity

Upper

Limit

UC1 0 Qa

UC2 Qa Qb

UC3 Qb Qc

UC4 Qc Qd

UC5 Qd Qe

Order

Quantity

Unit

Cost

Qa Qb Qc Qd

UC1

UC2

UC3

UC4

UC5

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Total Cost Curve using UC1

Order

Quantity

Total

Cost

Price Discounts from Suppliers: Valid Range of the Total Cost Curve using UC1

Order

Quantity

Total

Cost

Qa

Valid

Range

of

Curve

Invalid Range of

Curve

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Order

Quantity

Total

Cost

Qb Qa Qd Qc

Price Discounts from Suppliers:

Finding the Lowest Valid Cost

EOQ:

𝑄𝑜 =2 × 𝑅𝐶 × 𝐷

𝐻𝐶

If the holding cost can be expressed as a proportion, I, of the

unit cost, and for each unit cost UCi, the minimum point of the

cost curve comes with Qoi

𝑄𝑜𝑖 =2 × 𝑅𝐶 × 𝐷

𝐼 × 𝑈𝐶𝑖

For each curve with unit cost UCi this minimum is either ‘valid’

or ‘invalid’.

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Order

Quantity

Total

Cost

Qb Qa Qd Qc

Valid Minimum

Invalid Minimum



• Every set of cost curves will have at least one valid

minimum, and a variable number of invalid minima.

• The valid total cost curve always rises to the left of a valid

minimum. This means that when we search for an overall

minimum cost it is either at the valid minimum or

somewhere to the right of it.

• There are only two possible positions for the overall

minimum cost: it is either at a valid minimum, or else at a

cost break point.

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Worked Example 1:


The annual demand for an item is 2,000 units, each order

costs £10 and the annual holding cost is 40% of unit cost.

The unit cost depends on the quantity ordered as follows:

– £1 for order quantities less than 500

– £0.80 for quantities between 500 and 999

– £0.60 for quantities of 1,000 or more.

What is the optimal order size?

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Worked Example 1:


𝑇𝐶 = 𝑈𝐶 × 𝐷 + 𝑅𝐶 × 𝐷

𝑄+HC × Q

2

𝐷 = 2000 units/yr

𝑅𝐶 = £10 per order

𝐻𝐶= (40%)(UC)

𝑈𝐶= £1.00 (Q<500);

£0.80 (500 ≤ 𝑄 < 999) £0.60 (𝑄 ≥ 1000)

Worked Example 1:


D= 2000 unit/yr

RC= £ 10.00 per order

I= 40% of UC

HC= I×UC

UC= £ 1.00 Q<500

£ 0.80 500<=Q<999

£ 0.60 1000<=Q

Q>=1000 500<=Q<999 Q<500

UC £ 0.60 £ 0.80 £ 1.00

Qo 408.25 353.55 316.23

Invalid Invalid Valid

Order size 1000 500 316.23

TC £ 1,340.00 £ 1,720.00 £ 2,126.49

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Worked Example 2:


During a 50-week year, IH James and Partners notice that

demand for one of its products is more or less constant at 10

units a week. The cost of placing an order, including delivery, is around €150. The company aims for 20% annual return on

assets. The supplier of the item quotes a basic price of €250

a unit, with discounts of 10% on orders of 50 units or more,

15% on orders of 150 units or more and 20% on orders of 500

units or more. What is the optimal order size for the item?

Worked Example 2:


D= 500 unit/yr

RC= € 150.00 per order

I= 20% of UC

HC= I×UC

UC= € 250.00 Q<50

€ 225.00 50<=Q<150

€ 212.50 150<=Q<500

€ 200.00 500<=Q

500<=Q 150<=Q<500 50<=Q<150 Q<50

UC € 200.00 € 212.50 € 225.00 € 250.00

Qo 61.24 59.41 57.74 54.77

Invalid Invalid Valid Invalid

Order size 500 150 57.74 50

TC € 110,150.00 € 109,937.50 € 115,098.08 € 127,750.00

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Rising Delivery Cost:


• This method for considering unit costs that fall in discrete

steps can be extended to any problem where there is a

discrete change in cost.

• For example, the cost of delivery – and therefore the

reorder cost – rises in steps as more vehicles are needed.

Worked Example 3:

Rising Delivery Cost

Kwok Cheng Ho makes a range of high quality garden

ornaments. On an average day he uses 4 tons of fine grain

sand. This sand costs £20 a ton to buy, and £1.90 a ton to

store for a day. Deliveries are made by modified lorries that

carry up to 15 tons, and each delivery of a load or part load

costs £200. Find the cheapest way to ensure continuous

supplies of sand.

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Worked Example 3:

Rising Delivery Cost

D= 4 unit/day

RC= £ 200.00 per lorries

HC= £ 1.90 per day

UC= £ 20.00

Lorries 1 2 3 4

Q<=15 15<Q<=30 30<Q<=45 45<Q<=60

Order Cost £ 200.00 £ 400.00 £ 600.00 £ 800.00

Qo 29.02 41.04 50.26 58.04

Invalid Invalid Invalid Valid

Order size 15 30 45 58.04

TC £ 147.58 £ 161.83 £ 176.08 £ 190.27

Finite Replenishment Rate

Stock from Production

• If the rate of production is less than demand, each unit arriving is

immediately transferred to a customer and there are no stocks.

• Goods only accumulate when the production rate is greater than

demand.

• The stock level rises at a rate that is the difference between

production and demand. If we call the rate of production P,

stocks will build up at a rate P−D.

• This increase will continue as long as production continues.

• This means that we have to make a decision at some point to

stop production of this item.

• The purpose of this analysis is to find the best time to stop

production, which is equivalent to finding an optimal batch size.

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Produce at rate P per unit time

Transfer to stock

Stock increases at (P−D) per unit time

Meet demand

Meet demand of D per unit time


• We still consider a single item, with demand that is known,

constant and continuous, with costs that are known and

constant, and no shortages allowed.

• We have replenishment at a rate P and demand at a rate D, with

stock growing at a rate P−D. After some time, PT, we decide to

stop production. Then, stock is used to meet demand and

declines at a rate D. After some further time, DT, all stock has

been used and we must start production again.

• We assume there is an optimal value for PT (corresponding to

an optimal batch size) that we always use.

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Time

Stock

level

A

PT DT

(P-D)

D

Finite Replenishment Rate:

Optimal Batch Size

• The overall approach of this analysis is the same as the EOQ:

– find the total cost for one stock cycle,

– divide this total cost by the cycle length to get a cost per

unit time,

– minimize the cost per unit time.

• Consider one cycle of the modified saw-tooth pattern. If we

make a batch of size Q, the maximum stock level with

instantaneous replenishment would also be Q. With a finite

replenishment rate, though, stock never reaches this level, as

units are continuously being removed to meet demand. The

maximum stock level is lower than Q and occurs at the point

where production stops.

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Time

Stock

level

T

A

Q

PT

(P-D)

DT

D


Optimal Batch Size

• Looking at the productive part of the cycle we have:

𝐴 = (𝑃 − 𝐷) × 𝑃𝑇 • We also know that total production during the period is:

𝑄 = 𝑃 × 𝑃𝑇 𝑜𝑟 𝑃𝑇 =𝑄𝑃

• Substituting for PT into the equation for A gives:

𝐴 = 𝑄 ×(𝑃 − 𝐷)

𝑃

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Optimal Batch Size

1. Find the total cost of one stock cycle.

𝑇𝑜𝑡𝑎𝑙 𝑐𝑜𝑠𝑡 𝑝𝑒𝑟 𝑐𝑦𝑐𝑙𝑒= 𝑢𝑛𝑖𝑡 𝑐𝑜𝑠𝑡 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡+ 𝑟𝑒𝑜𝑟𝑑𝑒𝑟 𝑐𝑜𝑠𝑡 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡+ ℎ𝑜𝑙𝑑𝑖𝑛𝑔 𝑐𝑜𝑠𝑡 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡

= UC × Q + RC +HC × 𝐀 × T

2

= UC × Q + RC +HC × 𝐐 × T

2×(𝑃 − 𝐷)

𝑃


Optimal Batch Size

2. Divide this total cost by the cycle length to get a

cost per unit time.

𝑇𝑜𝑡𝑎𝑙 𝑐𝑜𝑠𝑡 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 time

=UC × Q

𝑇+ RC

𝑇+HC × 𝐐

2×(𝑃 − 𝐷)

𝑃

𝑆𝑖𝑛𝑐𝑒 𝑄 = 𝐷 × 𝑇


𝑄+HC × Q

2×(𝑃 − 𝐷)

𝑃

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Optimal Batch Size

3. Minimize this cost per unit time.

𝑑(𝑇𝐶)

𝑑(𝑄)= −𝑅𝐶 × 𝐷

𝑄2+HC

2×(𝑃 − 𝐷)

𝑃= 0

Solve for 𝑄𝑜 (Optimal Batch Size),

𝑄𝑜 =2 × 𝑅𝐶 × 𝐷

𝐻𝐶×

𝑃

(𝑃 − 𝐷)


Optimal Batch Size

Finite Replenishment Rate EOQ

Order

quantity 𝑄𝑜 =2 × 𝑅𝐶 × 𝐷

𝐻𝐶×

𝑃

(𝑃 − 𝐷) 𝑄𝑜 =

2 × 𝑅𝐶 × 𝐷

𝐻𝐶

Cycle

length 𝑇𝑜 =2 × 𝑅𝐶

𝐻𝐶 × 𝐷×

𝑃

(𝑃 − 𝐷) 𝑇𝑜 =

2 × 𝑅𝐶

𝐻𝐶 × 𝐷

Variable

cost

𝑉𝐶𝑜 =

2 × 𝑅𝐶 × 𝐻𝐶 × 𝐷 × (𝑃 − 𝐷)/𝑃

= 𝐻𝐶 × 𝑄𝑜 × (𝑃 − 𝐷) 𝑃

𝑉𝐶𝑜 =

2 × 𝑅𝐶 × 𝐻𝐶 × 𝐷 = 𝐻𝐶 × 𝑄𝑜

Total cost 𝑇𝐶𝑜 = 𝑈𝐶 × 𝐷 + 𝑉𝐶𝑜 𝑇𝐶𝑜 = 𝑈𝐶 × 𝐷 + 𝑉𝐶𝑜

Production

time 𝑃𝑇𝑜 =

𝑄𝑜𝑃

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Worked Example 4:


Demand for an item is constant at 1,800 units a year. The

item can be made at a constant rate of 3,500 units a year.

Unit cost is £50, batch set-up cost is £650, and holding cost is

30% of value a year. What is the optimal batch size for the

item? If production set-up time is 2 weeks, when should this

be started?

Worked Example 5:


Forecasters at Saloman Curtis have estimated the demand for an item that

they import from Taiwan to average 20 units a month. They pay €1,000 for

each unit. Last year the Purchasing and Inward Transport Department

arranged the delivery of 2,000 orders and had running costs of

€5,000,000. The Accounts Section quotes the annual holding costs as

20% of unit cost for capital and opportunity loss, 5% for storage space, 3%

for deterioration and obsolescence and 2% for insurance. All other costs

associated with stocking the item are combined into a fixed annual cost of

€24,000. Calculate the economic order quantity for the item, the time

between orders and the corresponding total cost.

By making the item at a rate of 40 units a month Saloman Curtis could

avoid the fixed cost of €24,000 a year, reduce the unit cost to €900 and

have a batch set-up cost of €1,000. Would it be better for the company to

make the item itself rather than buy it in?

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Worked Example 5:


Purchase Production

Qo 63.25 59.63

PTo 0.124225999

To 0.263523138 0.248451997

VCo € 18,973.67 € 8,049.84

TCo € 258,973.67 € 224,049.84

Fixed € 24,000.00 --

Total € 282,973.67 € 224,049.84

Alternatives for Customers when

their Demand cannot be met

Customer demands an item

The item is out of stock

Customer waits for

the next

replenishment-

giving back-orders

Customer keeps all

future business

with the supplier

Customer transfers

some future

business to another

supplier

Customer moves to

another supplier-

giving lost sales

Customer transfers

some future

business to another

supplier

Customer transfers

all future

business to another

supplier

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Back-orders

• A back-order occurs when a customer demands an item that

is out of stock, and then waits to receive the item from the

next delivery from suppliers.

• Back-ordering is more likely when the unit cost is high, there

is a wide range of items, it would be too expensive to hold

stocks of the complete range, lead times from suppliers are

reasonably short, there is limited competition, and customers

are prepared to wait.

• In extreme cases an organization keeps no stock at all and

meets all demand by back orders. This gives ‘make to order’

rather than ‘make for stock’ operations.

Back-orders

• We will look at the more common case where some stock is

kept, but not enough to cover all demand.

• The key question is, how much of the demand should be met

from stock and how much from back-orders?

• There is always some cost associated with back-orders,

including administration, loss of goodwill, some loss of future

orders, emergency orders, expediting, and so on. This cost is

likely to rise with increasing delay. We can, then, define a

shortage cost, SC, which is time-dependent and is a cost per

unit per unit time delayed.

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Stock Cycle with Back-orders

Time

Stock

level

T

T1

T2

Q

S

Q-S

Back-orders

1. Find the total cost of one stock cycle. 𝑇𝑜𝑡𝑎𝑙 𝑐𝑜𝑠𝑡 𝑝𝑒𝑟 𝑐𝑦𝑐𝑙𝑒

= 𝑢𝑛𝑖𝑡 𝑐𝑜𝑠𝑡 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 + 𝑟𝑒𝑜𝑟𝑑𝑒𝑟 𝑐𝑜𝑠𝑡 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡+ ℎ𝑜𝑙𝑑𝑖𝑛𝑔 𝑐𝑜𝑠𝑡 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡+ 𝑠ℎ𝑜𝑟𝑡𝑔𝑒 𝑐𝑜𝑠𝑡 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡

= UC × Q + RC +HC × (Q − S) × 𝑇1

2+𝑆𝐶 × 𝑆 × 𝑇22

Where 𝑇1 = (𝑄 − 𝑆) 𝐷 and 𝑇2 = 𝑆 𝐷 ; then

𝑇𝑜𝑡𝑎𝑙 𝑐𝑜𝑠𝑡 / 𝑐𝑦𝑐𝑙𝑒 = UC × Q + RC +HC × (𝑄 − 𝑆)2

2 × 𝐷+𝑆𝐶 × 𝑆2

2 × 𝐷

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Back-orders

2. Divide this total cost by the cycle length to get a

cost per unit time.

𝑇𝑜𝑡𝑎𝑙 𝑐𝑜𝑠𝑡 / 𝑐𝑦𝑐𝑙𝑒

= UC × Q + RC +HC × (𝑄 − 𝑆)2

2 × 𝐷+𝑆𝐶 × 𝑆2

2 × 𝐷

𝑆𝑖𝑛𝑐𝑒 𝑄 = 𝐷 × 𝑇


𝑄+HC × (𝑄 − 𝑆)2

2 × 𝑄+𝑆𝐶 × 𝑆2

2 × 𝑄

Back-orders


𝑑(𝑇𝐶)

𝑑(𝑄)= −𝑅𝐶 × 𝐷

𝑄2+HC

2−HC × 𝑆2

2 × 𝑄2−SC × 𝑆2

2 × 𝑄2= 0

and 𝑑(𝑇𝐶)

𝑑(𝑆)= −HC +

HC × S

𝑄−SC × S

𝑄= 0

Solving these simultaneous equations,

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Back-orders

𝑄𝑜 =2 × 𝑅𝐶 × 𝐷 × (𝐻𝐶 + 𝑆𝐶)

𝐻𝐶 × 𝑆𝐶

𝑆𝑜 =2 × 𝑅𝐶 × 𝐻𝐶 × 𝐷

𝑆𝐶 × (𝐻𝐶 + 𝑆𝐶)

In addition,

𝑇1 =𝑄𝑜 − 𝑆𝑜𝐷; 𝑇2 =

𝑆𝑜𝐷; 𝑇 = 𝑇1 + 𝑇2

Worked Example 6:

Back-orders

Demand for an item is constant at 100 units a month. Unit

cost is £50, reorder cost is £50, holding cost is 25% of value a

year, shortage cost for back-orders is 40% of value a year.

Find an optimal inventory policy for

the item.

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Lost Sales

• An initial stock of Q runs out after a time Q/D, and all

subsequent demand is lost until the next replenishment

arrives.

• There is an unsatisfied demand of D× T− Q.

• When shortages are allowed, the aim of minimizing cost is no

longer the same as maximizing revenue.

• If shortages are allowed we might, in some circumstances,

minimize costs by holding no stock at all – but this would

certainly not maximize revenue.

• In this analysis we will maximize the net revenue, which is

defined as the gross revenue minus costs. For this we have

to define SP as the selling price per unit.

Lost Sales

• We also need to look at the cost of lost sales, which has two

parts.

• First, there is a loss of profit, which is a notional cost that we

can define as SP − UC per unit of sales lost.

• Second, there is a direct cost, which includes loss of goodwill,

remedial action, emergency procedures, and so on. We will

define this as DC per unit of sales lost.

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Stock Level with Lost Sale

Time

Stock

level

Q

Q/D

T

D

Lost Sales

1. Find the total cost of one stock cycle. 𝑇𝑜𝑡𝑎𝑙 𝑐𝑜𝑠𝑡 𝑝𝑒𝑟 𝑐𝑦𝑐𝑙𝑒

= 𝑢𝑛𝑖𝑡 𝑐𝑜𝑠𝑡 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 + 𝑟𝑒𝑜𝑟𝑑𝑒𝑟 𝑐𝑜𝑠𝑡 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡+ ℎ𝑜𝑙𝑑𝑖𝑛𝑔 𝑐𝑜𝑠𝑡 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡+ 𝑙𝑜𝑠𝑡 𝑠𝑎𝑙𝑒𝑠 𝑐𝑜𝑠𝑡 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡

= UC × Q + RC +HC × Q

2×𝑄

𝐷+ 𝐷𝐶 × (𝐷 × 𝑇 − 𝑄)

then

𝑁𝑒𝑡 𝑅𝑒𝑣𝑒𝑛𝑢𝑒

= SP × 𝑄 − UC × Q − RC +HC × 𝑄2

2 × 𝐷+ 𝐷𝐶 × (𝐷 × 𝑇 − 𝑄)

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Lost Sales

2. Divide the net revenue by the cycle length to get

the net revenue per unit time.

𝑅 =1

𝑇× [𝑄 × 𝐷𝐶 + 𝑆𝑃 − 𝑈𝐶 − 𝑅𝐶 −

HC × 𝑄2

2 × 𝐷− 𝐷𝐶 × 𝐷 × 𝑇]

Let LC = cost of each unit of lost sales including loss of profits

= DC + SP – UC; and

Z = proportion of demand satisfied = 𝑄/(𝐷 × 𝑇)

𝑅 = 𝑍 × [𝐷 × 𝐿𝐶 −𝑅𝐶 × 𝐷

𝑄−HC × 𝑄

2− 𝐷𝐶 × 𝐷]

Lost Sales


𝑑(𝑅)

𝑑(𝑄)=𝑍 × 𝑅𝐶 × 𝐷

𝑄2−Z × 𝐻C

2= 0

Solving this equation,

𝑄𝑜 =2 × 𝑅𝐶 × 𝐷

𝐻𝐶

𝑅𝑜 = 𝑍 × 𝐷 × 𝐿𝐶 − 𝑄𝑜 × 𝐻𝐶

= 𝑍 × 𝐷 × 𝐿𝐶 − 2 × 𝑅𝐶 × 𝐻𝐶 × 𝐷

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Lost Sales

• If 𝐷 × 𝐿𝐶 > 2 × 𝑅𝐶 × 𝐻𝐶 × 𝐷; set 𝑍 = 1no shortage

• If 𝐷 × 𝐿𝐶 < 2 × 𝑅𝐶 × 𝐻𝐶 × 𝐷; set 𝑍 = 0Don’t stock

• If 𝐷 × 𝐿𝐶 < 2 × 𝑅𝐶 × 𝐻𝐶 × 𝐷; 𝑍 = any value

Worked Example 7:

Lost Sales

Find the best ordering policy for three items with the following

costs.

Item D RC HC DC SP UC

1 50 150 80 20 110 90

2 100 400 200 10 200 170

3 50 500 400 30 350 320

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Worked Example 7:

Lost Sales

Item1 Item2 Item3

D= 50 100 50

RC= £ 150.00 £ 400.00 £ 500.00

HC= £ 80.00 £ 200.00 £ 400.00

DC= £ 20.00 £ 10.00 £ 30.00

SP= £ 110.00 £ 200.00 £ 350.00

UC= £ 90.00 £ 170.00 £ 320.00

LC= £ 40.00 £ 40.00 £ 60.00

D×LC= £ 2,000.00 £ 4,000.00 £ 3,000.00

SQRT(2×RC×HC×D) £ 1,095.45 £ 4,000.00 £ 4,472.14

Z= 1 0.5 0

Ro= £ 904.55 £ - £ -

Qo= 13.69306394 20 11.18033989

Do Not Stock

Constraints on Stock

• If each inventory item is completely independent, then we can

calculate an optimal order policy for each item in isolation and

there is no need to consider interactions with other items.

• There are situations where, although demand for each item is

independent, we may have to look at these interactions. – When several items are ordered from the same supplier; and

we can reduce delivery costs by combining orders for

different items in a single delivery.

– When there are constraints on some operations, such as

limited warehouse space or a maximum acceptable

investment in stock.

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Constraints on Stock

• Problems of this type can become rather complicated, so we

will illustrate some broader principles by looking at problems

with constraints on stock levels.

• These constraints could include limited storage space, a

maximum acceptable investment in stock, a maximum

number of deliveries that can be accepted, maximum size of

delivery that can be handled, a maximum number of orders

that can be placed, etc.

• Most problems with constraints can be tackled using the same

methods.

• The first shows an intuitive approach to problems where there

is a limited amount of storage space; the second is a more

formal analysis for constrained investment.

Constraints on Storage Space

• If an organization uses the economic order quantity for all

items in an inventory, the resulting total stock might exceed

the available capacity.

• Then we need some way of reducing the stock until it is within

acceptable limits.

• One approach puts an additional cost on space used. Then

the holding cost is in two parts: the original holding cost, HC,

and an additional cost, AC, related to the storage area (or

volume) used by each unit of the item. Then the total holding

cost per unit per unit time becomes HC + AC × Si, where Si is

the amount of space occupied by one unit of item i.

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• The economic order quantity becomes:

𝑄𝑖 =2 × 𝑅𝐶𝑖 × 𝐷𝑖𝐻𝐶𝑖 + 𝐴𝐶 × 𝑆𝑖

• If space is constrained, we can give AC a positive value and

order quantities of all items are reduced from 𝑄𝑜𝑖 to 𝑄𝑖. • As the average stock level is 𝑄𝑖/2, this automatically reduces

the amount of stock.

• The reduction necessary – and the consequent value of AC –

depend on the severity of space constraints. A reasonable

way of finding solutions is iteratively to adjust AC until the

space required exactly matches available capacity.

Worked Example 8:


The Brendell Work Centre is concerned with the storage space

occupied by three items that have the following characteristics.

The reorder cost for the items is constant at $1000, and holding

cost is 20% cent of value a year. If the company wants to

allocate an average space of 300 m3 to the items, what would be

the best ordering policy? How much does the constraint on

space increase variable inventory cost?

Item UC($) D Space (m3/unit)

1 100 500 1

2 200 400 2

3 300 200 3

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Worked Example 8:


Item UC D Space RC HC(1) A HC Qo Space

Req. VCo

1 $ 100.00 500 1 $ 1,000.00 $ 20.00 0 $ 20.00 223.61 111.80 $ 4,472.14

2 $ 200.00 400 2 $ 1,000.00 $ 40.00 0 $ 40.00 141.42 141.42 $ 5,656.85

3 $ 300.00 200 3 $ 1,000.00 $ 60.00 0 $ 60.00 81.65 122.47 $ 4,898.98

375.70 $ 15,027.97

Item UC D Space RC HC(1) A HC Q Space

Req. VC

1 $ 100.00 500 1 $ 1,000.00 $ 20.00 11.3666 $ 31.37 178.55 89.28 $ 4,585.82

2 $ 200.00 400 2 $ 1,000.00 $ 40.00 11.3666 $ 62.73 112.93 112.93 $ 5,800.66

3 $ 300.00 200 3 $ 1,000.00 $ 60.00 11.3666 $ 94.10 65.20 97.80 $ 5,023.52

300.00 $ 15,409.99

Goal Seek:

Item UC D Space RC HC(1) A HC Q Space

Req. VC

1 $ 100.00 500 1 $ 1,000.00 $ 20.00 11.3667 $ 31.37 178.55 89.28 $ 4,585.82

2 $ 200.00 400 2 $ 1,000.00 $ 40.00 11.3667 $ 62.73 112.93 112.93 $ 5,800.66

3 $ 300.00 200 3 $ 1,000.00 $ 60.00 11.3667 $ 94.10 65.20 97.80 $ 5,023.52

300.00 $ 15,410.00

Solver:

Constraints on Average Investment

in Stock

• Suppose an organization stocks N items and has an upper

limit, UL, on the total average investment. The calculated

EOQ for each item 𝑖 is 𝑄𝑜𝑖, but again we need some means of

calculating the best, lower amount 𝑄𝑖 which allows for the

constraint.

• The average stock of item i is 𝑄𝑖/2 and the average

investment in this item is 𝑈𝐶 × 𝑄𝑖/2. The problem then

becomes one of constrained optimization – minimize the total

variable cost subject to the constraint of an upper limit on

average investment.

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in Stock

𝑀𝑖𝑛𝑖𝑚𝑖𝑧𝑒 𝑉𝐶 = 𝑅𝐶𝑖 × 𝐷𝑖𝑄𝑖+𝐻𝐶𝑖 × 𝑄𝑖2

𝑁

𝑖=1

𝑆𝑢𝑏𝑗𝑒𝑐𝑡 𝑡𝑜 𝑈𝐶𝑖 × 𝑄𝑖2≤ 𝑈𝐿

𝑁

𝑖=1


in Stock

• The usual way of solving problems of this type is to eliminate

the constraint by introducing a Lagrange multiplier and

incorporating this into a revised objective function.

• In particular, we can adjust the constraint to make it less than

or equal to zero, and the result is multiplied by the Lagrange

multiplier, 𝐿𝑀. This is incorporated in the revised objective

function:

𝑀𝑖𝑛𝑖𝑚𝑖𝑧𝑒 𝑉𝐶

= 𝑅𝐶𝑖 × 𝐷𝑖𝑄𝑖+𝐻𝐶𝑖 × 𝑄𝑖2

𝑁

𝑖=1

+ 𝐿𝑀 × 𝑈𝐶𝑖 × 𝑄𝑖2− 𝑈𝐿

𝑁

𝑖=1

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in Stock

• To solve this, we differentiate with respect to both 𝑄𝑖 and 𝐿𝑀,

and set both of these derivatives to zero to give two

simultaneous equations. We can then eliminate the Lagrange

multiplier to give the overall result:

𝑄𝑖 = 𝑄𝑜𝑖 ×2 × 𝑈𝐿 × 𝐻𝐶𝑖

𝑈𝐶𝑖 × 𝑉𝐶𝑜𝑖𝑁𝑖=1

• When the holding cost is a fixed proportion of the unit cost,

the best order quantity for each item is a fixed proportion of

the EOQ.

Worked Example 9: Constraints on

Average Investment in Stock

A company is concerned with the stock of three items with the

following demand and unit cost.

The items have a reorder cost of £100 and a holding cost of 20%

of value a year. What is the best ordering policy if the company

wants to limit average investment in these item to £4,000?

Item UC($) D

1 100 100

2 60 300

3 40 200

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Worked Example 9: Constraints on

Average Investment in Stock

Item UC D RC HC Qo VCo Avg. Inv.

1 £ 100.00 100 £ 100.00 £ 20.00 31.62 £ 632.46 £ 1,581.14

2 £ 60.00 300 £ 100.00 £ 12.00 70.71 £ 848.53 £ 2,121.32

3 £ 40.00 200 £ 100.00 £ 8.00 70.71 £ 565.69 £ 1,414.21

£ 2,046.67 £ 5,116.67

Item Qo VCo Ratio Q Avg. Inv'

1 31.62 £ 632.46 0.781758 24.72136 £ 1,236.07

2 70.71 £ 848.53 0.781758 55.27864 £ 1,658.36

3 70.71 £ 565.69 0.781758 55.27864 £ 1,105.57

£ 2,046.67 £ 4,000.00

Discrete, Variable Demand

• Often, an organization has to meet demand for an integer

number of units every period, e.g., a car showroom, electrical

retailer, furniture warehouse or computer store.

• In this section, we will look at a model for discrete, variable

demand. If we know exactly what these demands are going

to be, we can build a deterministic model and find an optimal

ordering policy.

• The way to approach this is to assume there is some optimal

number of period’s demand that we should combine into a

single order.

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• Consider one order for an item, where we buy enough to

satisfy demand for the next N periods. The demand is

discrete and varies, so we can define the demand in period 𝑖 of a cycle as 𝐷𝑖, and an order to cover all demand in the next

𝑁 periods will be:

𝐴 = 𝐷𝑖

𝑁

𝑖=1

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• The average variable cost of stocking the item per period:

𝑉𝐶𝑁 =𝑅𝐶

𝑁+𝐻𝐶 × 𝐷𝑖

𝑁𝑖=1

2

• Starting with 𝑁 = 1, increasing 𝑁 will follow the graph

downward until costs reach a minimum and then start to rise.

• If it is cheaper to order for 𝑁 + 1 periods than for 𝑁 periods,

we are on the left-hand side of the graph.

• If we keep increasing the value of 𝑁, there comes a point

where it is more expensive to order for 𝑁 + 1 periods than 𝑁.

• At this point, we have found the optimal value for 𝑁 and if we

increase it any further, costs will continue to rise.

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• The average variable cost of stocking the item over 𝑁 periods:

𝑉𝐶𝑁 =𝑅𝐶


𝑁𝑖=1

2

• The average variable cost of stocking the item over 𝑁 + 1 periods:

𝑉𝐶𝑁+1 =𝑅𝐶

𝑁 + 1+𝐻𝐶 × 𝐷𝑖

𝑁+1𝑖=1

2

• If 𝑉𝐶𝑁+1 > 𝑉𝐶𝑁, 𝑅𝐶

𝑁 + 1+𝐻𝐶 × 𝐷𝑖

𝑁+1𝑖=1

2>𝑅𝐶


𝑁𝑖=1

2; or

𝑁 × 𝑁 + 1 × 𝐷𝑁+1 >2 × 𝑅𝐶

𝐻𝐶


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• This method usually gives good results, but it does not

guarantee optimal ones.

• It is sometimes possible for the variable cost to be higher

for orders covering 𝑁 + 1 periods rather than 𝑁, but then

fall again for 𝑁 + 2 periods.

• If 𝑉𝐶𝑁+2 > 𝑉𝐶𝑁, 𝑅𝐶

𝑁 + 2+𝐻𝐶 × 𝐷𝑖

𝑁+2𝑖=1

2>𝑅𝐶


𝑁𝑖=1

2; or

𝑁 × 𝑁 + 2 × [𝐷𝑁+1+𝐷𝑁+2] >4 × 𝑅𝐶

𝐻𝐶


• When we find a turning point and we know it is cheaper to

order for 𝑁 periods than for 𝑁 + 1, we simply see if it is also

cheaper to order for 𝑁 + 2 periods.

• If the inequality above is valid, we stop the process and

accept the solution.

• If the inequality is invalid we continue the process until we

find another turning point.

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Worked Example 10:


K C Low calculated her total cost of placing an order for an item

and having it delivered at $1,200, with a holding cost of $50 a

unit a week. She had a fixed production schedule and needed

the following units over the next 11 weeks.

Find an ordering pattern that gives a reasonable cost for the

item.

Week 1 2 3 4 5 6 7 8 9 10 11

Demand 2 4 6 9 9 6 3 4 4 6 9

Worked Example 10:


Week 1 2 3 4 5 6 7 8 9 10 11

Demand 2 4 6 9 9 6 3 4 4 6 9

N= 1 2 3 1 2 3 4 1 2 3

N(N+1)D 8 36 108 18 36 36 80 8 36 108

less less greater less less less greater less less greater

N(N+2)(D) 225 168 225

valid Valid Valid

VC 1250 750 700 1425 1050 1000 975 1300 800 750

chapter 4 models for known demandweb.eng.fiu.edu/leet/ein6345inventory/chap4_2013.pdf · chapter 4...

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