Inventory Control Subject to Deterministic Demand Operations Analysis and Improvement 2015 Spring Dr. Tai-Yue Wang Industrial and Information Management

Inventory Control Subject to Deterministic Demand

Operations Analysis and Improvement

2015 Spring

Dr. Tai-Yue WangIndustrial and Information Management Department

National Cheng Kung University

Dr. Tai-Yue Wang IIM Dept. NCKU

Contents

Introduction Types of Inventories Why Inventory? Characteristics of Inventory Systems Relevant Costs The EOQ Model Extension to a Finite Production Rate Quantity Discount Models


Contents

Quantity Discount Models Resource-constrained Multiple Product

Systems EOQ Models for Production Planning


Overview of Operations Planning Activities


Introduction -- 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?


Breakdown of the total investment in inventories


Treatment of Excess Demand. Backorder all Excess Demand Lose all excess demand Backorder some and lose some

Inventory that changes over time Perishability – 農產品 Obsolescence – 過期之設備備品

Introduction -- Characteristics of Inventory Systems


Introduction -- Purposes

Demand is known Methods to control individual item

inventory


Types of Inventories

Raw material Components Work-in-Process (WIP) Finished goods


Reasons for Holding Inventories

Economies of Scale Uncertainty in delivery leadtimes, supply Speculation-- Changing Costs Over Time Transportation Smoothing Demand Uncertainty Logistics Costs of Maintaining Control System


Relevant Costs

Holding Costs - Costs proportional to the quantity of inventory held. Includes:

a) Physical Cost of Space (3%)

b) Taxes and Insurance (2 %)

c) Breakage Spoilage and Deterioration (1%)

*d) Opportunity Cost of alternative investment. (18%)

Note: Since inventory may be changing on a continuous basis, holding cost is proportional to the area under the inventory curve.


Relevant Costs (continued)

Ordering Cost (or Production Cost).

Includes both fixed and variable components.

slope = c

K

C(x) = K + cx for x > 0 and =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 for lost demand Costs of bookeeping for backordered demands Loss of goodwill for being unable to satisfy

demands when they occur. Generally assume cost is proportional to number of

units of excess demand.


The EOQ Model —The Basic Model

Assumptions:

1. Demand is fixed at l units per unit time.

2. Shortages are not allowed.

3. Orders are received instantaneously. (this will be

relaxed later).



Assumptions:

4. Order quantity is fixed at Q per cycle. (can be proven

optimal.)

5. Cost structure:

a) Fixed and marginal order costs (K + cx)

b) Holding cost at h per unit held per unit time.




Q is the size of the order At t=0, Q is increased instantaneously from 0 to Q The objective is to choose Q to minimize the

average cost per unit time. In each cycle, the total fixed plus proportional

order cost is C(Q)=K+cQ Since the inventory is consumed by the rate of λ,

the cycle length T is computed by Q/ λ



In addition, the average inventory level during one order cycle is Q/2.

Thus, the annual cost, G(Q)

2

22)(

hQc

Q

K

hQQ

cQKhQ

T

cQKQG



Thus

0 0/2)(

2//)(3"

2'

QforQKQG

hQKQG

Since G”(q) > 0, G(Q) is a convex function of Q and G’(0)=- ∞ and G’(∞)=h/2, the curve of G(Q) is in next slide.




The EOQ Model —The Basic Model --Properties of the EOQ 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 = Ic)


The EOQ Model —The Basic Model --Properties of the EOQ Solution

The optimal value of Q occurs where G’(Q)=0

h

KQ

2*

Q* is known as the economic order quantity(EOQ).


The EOQ Model —The Basic Model --Example

2KQ

h

Number 2B pencils at campus bookstore are sold at a rate of 60 per week.

The pencil cost is two cents each and sell for 15 cents each. It cost the bookstore $12 to initiate an order and the holding cost are based on annual interest rate of 25 percent.

Please determine the optimal number of pencils for the bookstore to purchase and the time between placement of orders.


2KQ

h

The annual demand rate λ=(60)(52)=3,120 The holding cost h=(0.25)(0.02)=0.005

870,3005.0

)120,3)(12)(2(2* h

KQ

The cycle time is T=Q/λ = 3,870/3,120 =1.24 years

The EOQ Model —The Basic Model --Example


2KQ

h

In previous example, if the pencils must be ordered four months in advance, we would try to find out when to place order depends on how much inventory on hand.

So we want to reorder at inventory on hand, R, the reorder point.

where is the lead timeR

The EOQ Model — Order Lead time


2KQ

h



2KQ

h

If the lead time exceeds one cycle, it is more difficult to determine the reorder point

Let EOQ=25, demand rate = 500/year, lead time = 6 weeks,

Cycle time T = 25/500=0.05 year = 2.6 weeks or Lead time = /T = 2.31 cycles

two cycles + 0.31 cycle 0.0155 year

R=0.0155*500=7.75 8



2KQ

h



2KQ

h

Procedure:1. Form the ratio of /T

2. Get the fractional remainder of the ratio

3. Multiply this fractional remainder by cycle length to convert to year

4. Multiply the result of previous step by the demand rate



The EOQ Model — 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:

2//)( hQQKQG

*

*

* 2

1)(

Q

Q

Q

Q

G

QG


The EOQ Model — Sensitivity Analysis

In general, G(Q) is relatively insensitive to errors in Q

If would results lower average annual cost than a value of

QQQ *

QQQ *


The EOQ Model — Example

2KQ

h

A company produces desks at a rate of 200 per month. Each desk requires 40 screws purchased from a supplier. The screw costs 3 cents each. Fixed delivery charges and cost of receiving and storing equipment of screws amount to $100 per shipment, independently of the size of the shipment.

The firm uses 25 percent interest rate to determine the holding cost

What standing order size should they use?


The EOQ Model — Example

2KQ

h

Solution Annual demand=(200)(12)(40)=96,000 Annual holding cost per screw = 0.25*0.03=0.0075 EOQ

505970075.0

)000,96)(100)(2(2* h

KQ


EOQ With Finite Production Rate

Suppose that items are produced internally at a rate P > λ. The total cost is

Then the optimal production quantity to minimize average annual holding and set up costs has the same form as the EOQ, namely:

)/1)(2/(/)( PhQQKQG

)/1(

2*

Ph

KQ

EOQ With Finite Production Rate --Inventory Levels for Finite Production Rate Model


EOQ With Finite Production Rate — Example

2KQ

h

Example A company produces EPROM for its customers. The demand

rate is 2,500 units per year. The EPROM is manufactured internally at rate of 10,000 units per year. The cost for initiating the production is $50 and each unit costs the company $2 to manufacture. The cost of holding is based on a 30 percent annual interest rate.

Please determine the optimal size of a production run, the length of each production run, and the average annual cost of holding and setup.

What is the maximum level of the on-hand inventory of the EPROM?



2KQ

h

Solution: h=0.3*20.6 per unit per year The modified holding cost h’=0.6*(1-2,500/10,000)=0.45

the length of each production run T=Q/=745/2500=0.298 year the average annual cost of holding and setup

74545.0

)500,2)(50)(2(2'

* h

KQ

41.3352/)/1)((/)( *** PhQQKQG



2KQ

h

Solution: What is the maximum level of the on-hand inventory of the

EPROM?

unitsPQH 559)/1(*


Quantity Discount Models

Two kinds of quantity discount: 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

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.

All-Units Discount Order Cost Function


Incremental Discount Order Cost Function


Quantity Discount Models –all units discount

Trash bag company’s price schedule:

QQ

QQ

QQ

QC

000,1for 28.0

000,1500for 29.0

5000for 30.0

)(


Quantity Discount Models –all units discount

Procedure:1. Starting from the lowest price interval and

determine the largest realizable EOQ value.

2. Compare the value of average annual cost at the largest realizable EOQ and at all the price breakpoints that are greater than the largest realizable EOQ. The optimal one is the one with lowest average annual cost.


Quantity Discount Models –all units discount --example


For c=0.28, Q*=414 X

c=0.29, Q*=406 X

c=0.30, Q*=400 OK

QQ

QQ

QQ

QC

000,1for 28.0

000,1500for 29.0

5000for 30.0

)(


Quantity Discount Models –all units discount --example

G(400)=204

G(500)=198.1

G(1,000)=200.8 Q=500 with lowest average annual cost


Quantity Discount Models –Incremental discount


And G(Q) becomes

QQ

QQ

QQ

QC

000,1for )000,1(28.0295

000,1500for )500(29.0150

5000for 30.0

)(

)2/(//)(

cost holding average

cost/Q setupcost/Q material)(

QhQKQQC

QG


Quantity Discount Models –Incremental discount

Procedure:1. Find C(Q) equation for all price intervals

2. Substitute C(Q) into G(Q), compute the minimum values of Q for each price intervals

3. Determine which minima computed from previous step are realizable, compute the average annual costs at the realizable EOQ values and pick the lowest one.


Quantity Discount Models –Incremental discount --example

Trash bag company’s price schedule:1.

2.

QQ

QQ

QQ

QC

000,1for )000,1(28.0295

000,1500for )500(29.0150

5000for 30.0

)(

)2/](/)(*[//)(

)2/(//)()(

QQQCIQKQQC

QhQKQQCQG



2. 2/*3.0*2.0/600*83.0*600)(0 QQQG

OK -- 4000 Q

)2/(*)/529.0(*2.0

/600*8)/529.0(*600)(1

QQ

QQQG

OK-- 5191 Q

)2/(*)/1528.0(*2.0

/600*8)/1528.0(*600)(2

QQ

QQQG

XXX-- 7022 Q



3. Compare G0 and G1

204)(0 QG

58.204)(1 QG


Quantity Discount Models --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 succeeding 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


Average Annual Cost Function for Incremental Discount Schedule


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:

EOQ:

CQcQcQc nn 2211

i

iii h

KQ

2*



When the condition that

is met, the solution procedure is straightforward.

EOQ

nn hchchc /// 2211

n

iii

ii

EOQcCm

mEOQQ

1

*

/



If the condition

is not met, one must use an iterative procedure involving Lagrange Multipliers.

nn hchchc /// 2211


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. j = demand rate for product j Pj= production rate for product j hj = holding cost per unit per unit time for product j Kj= cost of setup the production facility for product j



The goal is to determine the optimal procedure for producing n products on the machine to minimize the cost of holding and setups, and to guarantee that no stock-outs occur during the production cycle.

For the problem to be feasible we must have that

1/1

n

jjj P



We also assume that rotation cycle policy is used. That is, in each cycle, there is exactly one setup for each product, and the products are produced in the same sequence in each production cycle.

Let T be the cycle time, and during time T, exactly one cycle of each product are produced.



So the lot size for product j during time T is

And the average annual cost for product j is

For all products

TQ jj

)2/(/)( 'jjjjjj QhQKQG

n

jjjjjj

n

jj QhQKQG

1

'

1

)2/(/)(



Since So

The goal is to find optimal cycle time to minimize G(T)

TQ jj

n

jjjj ThTKTG

1

' )2//()(

jjQT /

0)(

dT

TdG



So

However, if setup time is a factor, one needs to check if having enough time for setup and production

n

jjj

n

jj

h

K

T

1

'

1*

2



Let sj be the setup time for product j So

And

So we choose the cycle time T =max(T*, Tmin)

TPTsPQsn

jjjj

n

jjjj

11

)/()/(

min

1

1

/1T

P

s

T n

jjj

n

jj


EOQ Models for Production Planning -- Example

A machine serves as a cutting machine for different products. The rotation policy is used and setup cost is proportion to the setup time. Data are followed.

Products Annual Demand(units/year)

Production Rate (units/year)

Setup time(hours)

Variable costs($/unit)

A 4,520 35,800 3.2 40

B 6,600 62,600 2.5 26

C 2,340 41,000 4.4 52

D 2,600 71,000 1.8 18

E 8,800 46,800 5.1 38

F 6,200 71,200 3.1 28

G 5,200 56,000 4.4 31



The firm estimates that the setup costs amount to an average of $110 per hour, based on the cost of worker time and the cost of forced machine idle time during setups. Holding costs are based on a 22 percent annual interest rate charge.

Please find the optimal cycle time for those products.



Solution:1. Verify if is valid.

2. Compute the setup costs and modified holding costs

Setup cost K1=$110*3.2=$352, … etc.

Modified holding cost

1/1

n

jjj P

169335.0000,56/200,5...800,35/520,4/1

n

jjj P

69.7)800,35/520,41(*22.0'1 h



Setup costs(Kj) Modified Holding Costs

352 7.69

275 5.12

484 10.79

198 3.81

561 6.79

341 5.62

484 6.19

Total=2,695

'jh


1529.0

200,5*19.6...600,6*12.5520,4*69.7

695,2*22

1

'

1*

n

jjj

n

jj

h

K

T

3. So

4. Assuming 250 working days for one year, that is, the production repeats at roughly 38 working days.



4. So optimal lots sizes are:

Products Optimal lot sizes for each production run

A 691

B 1,009

C 358

D 398

E 1,346

F 948

G 795


Documents

Inventory Control Subject to Deterministic Demand Operations Analysis and Improvement 2015 Spring Dr. Tai-Yue Wang Industrial and Information Management