Transcript Inventory Control Models

6-1
Inventory Control
Models
Ch 4 (Known Demands)
Ch 5 (Uncertainty of Demand)
R. R. Lindeke
IE 3265, Production And Operations
Management
5/27/2016
Chs. 4/5 Inventory Mgt
6-2
Reasons for Holding
Inventories






Economies of Scale
Uncertainty in delivery leadtimes
Speculation. Changing Costs Over Time
Smoothing: seasonality, Bottlenecks
Demand Uncertainty
Costs of Maintaining Control System
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Chs. 4/5 Inventory Mgt
6-3
Characteristics of Inventory Systems

Demand




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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?
Chs. 4/5 Inventory Mgt
6-4
Characteristics of Inventory Systems

Treatment of Excess Demand.




Backorder all Excess Demand
Lose all excess demand
Backorder some and lose some
Inventory who’s quality changes over time


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perishability
obsolescence
Chs. 4/5 Inventory Mgt
6-5
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:



5/27/2016
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
Chs. 4/5 Inventory Mgt
6-6
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 it!
5/27/2016
Chs. 4/5 Inventory Mgt
6-7
Relevant Inverntory Costs
Holding Costs - Costs proportional to
the quantity of inventory held. Includes:

1.
2.
3.
4.

Physical Cost of Space (3%)
Taxes and Insurance (2 %)
Breakage Spoilage and Deterioration (1%)
Opportunity Cost of alternative investment. (18%)
Holding issues total: 24%
Therefore, in inventory systems, the holding
cost is taken as:
 h  .24*Cost of product
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Chs. 4/5 Inventory Mgt
6-8
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


5/27/2016
assumes an average monthly inventory of trucks
based on onhand data
$3333
Chs. 4/5 Inventory Mgt
6-9
Relevant Costs (continued)

Ordering Cost (or Production Cost).
Includes both fixed and variable components
slope = c
K
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C(x) = K + cx for x > 0;
Chs. 4/5 Inventory Mgt
0 for x = 0.
6-10
Relevant Costs (continued)

Penalty or Shortage Costs. All costs
that accrue when insufficient stock is
available to meet demand. These
include:



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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.
Chs. 4/5 Inventory Mgt
6-11
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.
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Chs. 4/5 Inventory Mgt
6-12
The Simple EOQ Model – the
most fundamental of all!

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).
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Chs. 4/5 Inventory Mgt
6-13
Simple EOQ Model (cont.)

Assumptions (cont.):
4. Order quantity is fixed at Q per cycle. (we will
find this is 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.
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Chs. 4/5 Inventory Mgt
6-14
Inventory Levels for the EOQ
Model
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Chs. 4/5 Inventory Mgt
6-15
The Average Annual
Cost Function G(Q)
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Chs. 4/5 Inventory Mgt
6-16
Modeling Inventory:
K  cQ 

G(Q) 
T
hQ 


2
K : setup cost
c: Unit cost

T: cycle length T= Q
l

h: holding cost  % of unit cost 
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Chs. 4/5 Inventory Mgt
6-17
Subbing Q/l for T


K

cQ


hQ


G (Q ) 


 Q
 2
l 

Kl
hQ
 lc 
Q
2
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Chs. 4/5 Inventory Mgt
Finding an Optimal Level of ‘Q’ –
the so-called EOQ


Take derivative of the G(Q)
equation with respect to Q
Set derivative equals Zero:
Kl h
G (Q) 
 0
2
Q
2
'

Now, Solve for Q
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Chs. 4/5 Inventory Mgt
6-18
Properties of the EOQ
(optimal) Solution
2K l
Q
h



Q is increasing with both K and l 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)
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Chs. 4/5 Inventory Mgt
6-19
6-20
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
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Chs. 4/5 Inventory Mgt
6-21
Plugging and chugging:


h = $8*.25 = $2
l = 145*52 = 7540
2 *35* 7540
Q 
 513.7 514
2
*
T  Q  514
 .068 yr
l
7540
or: 514
 3.54wk
145
*
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Chs. 4/5 Inventory Mgt
6-22
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?


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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!
Chs. 4/5 Inventory Mgt
6-23
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
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!!!
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Chs. 4/5 Inventory Mgt
6-24
Sensitivity Analysis
Let G(Q) be the average annual holding and setup cost function given by
G (Q)  K l / Q  hQ / 2
and let G* be the optimal average annual cost.
Then it can be shown that:
G (Q) 1  Q * Q 
 


G*
2  Q Q *
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Chs. 4/5 Inventory Mgt
6-25
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


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that Q* + Q has smaller error than Q* - Q
That is, we tend to have a greater penalty cost
(Error means extra inventory maintenance costs)
if we order too little than too much
Chs. 4/5 Inventory Mgt
6-26
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:
2k l
Q
h'

5/27/2016
Except that h’ is defined as h’= h(1- λ/P)
Chs. 4/5 Inventory Mgt
6-27
This is based on solving:
K hH
G (Q)  
T
2
K l  hQ

G (Q) 

 1  l / P 
Q
2
P is annual production rate
H is maximum on hand quantity

H  Q  1 l
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P

Chs. 4/5 Inventory Mgt
6-28
Inventory Levels for Finite
Production Rate Model
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Chs. 4/5 Inventory Mgt
6-29
Lets Try one:





We work for Sam’s Active Suspensions
They sell after market kits for car “Pimpers”
They have an annual demand of 650 units
Production rate is 4/day (wotking at 250 d/y)
Setup takes 2 technicians working 45
minutes @$21/hour and requires an
expendable tool costing $25
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Chs. 4/5 Inventory Mgt
6-30
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:


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T1 = production time in a cycle (Q*/P)
T2 = non producing time in a cycle (T – T1)
Chs. 4/5 Inventory Mgt
6-31
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

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.
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Chs. 4/5 Inventory Mgt
All-Units Discount
Order Cost Function
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Chs. 4/5 Inventory Mgt
6-32
6-33
Incremental Discount
Order Cost Function
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Chs. 4/5 Inventory Mgt
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).
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Chs. 4/5 Inventory Mgt
6-34
All-Units Discount Average
Annual Cost Function
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Chs. 4/5 Inventory Mgt
6-35
To Find EOQ in ‘All Units’ discount
case:
Compute Q* for each cost level
 Check for Feasibility (the Q computed
is applicable to the range) –
“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)

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Chs. 4/5 Inventory Mgt
6-36
6-37
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
Q 
 2  300  900 
Q 
 2  300  900 
*
1
*
2
5/27/2016
 0.34  6.50 
 495
 0.34  3.50 
 674
Chs. 4/5 Inventory Mgt
6-38
Lets Try one:



Both of these are Realizable (the value is ‘in
range’)
Compute G(Q) for both and breakpoint (600)
G(Q) = cl + (l*K)/Q + (h*Q)/2
G (495)  900  (6.50) 
 900  300 
495

 .34  6.5 495

 .34  3.5 674 
2
G (495)  $6942.43
G (674)  900  (3.50) 
 900  300 
674
2
G (674)  $3951.62
G (600)  900  (6.50) 
G (600)  $3957.00
5/27/2016
 900  300 
600

 .34  6.5 600 
Chs. 4/5 Inventory Mgt
2
Order 674
at a time!
6-39
Average Annual Cost Function
for Incremental Discount Schedule
5/27/2016
Chs. 4/5 Inventory Mgt
6-40
In an Incremental Case:







Cost is strictly a 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 Equation
Check if “Realizable”
Compute G(Q*) for realizable Q*’s
5/27/2016
Chs. 4/5 Inventory Mgt
6-41
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:




5/27/2016
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 = 3.5 + (1800)/Q (orders beyond 601)
Chs. 4/5 Inventory Mgt
6-42
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) = cl + (l*K)/Q + (h*Q)/2]


 1800     300  900 
 1800    Q2
G2 (Q2* )  900   3.5  


0.34

3.5



 2



Q
Q
Q
2 
2
2 




 900  2100   Q   .34  3.5   .34  900
G2 (Q2* )  900  3.5 
2 
Q2
2 

5/27/2016
Chs. 4/5 Inventory Mgt
6-43
Differentiating G2(Q)
d  G2  Q2  
   900  2100   .34  3.5 
0
0

2
dQ2
Q2
2


Set equal to 0 and solve for Q2
 900  2100 
2
2
Q

2
Q 
5/27/2016
 0.595  0
Realizable!
 900  2100 
0.595
Chs. 4/5 Inventory Mgt
 1783
6-44
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
5/27/2016
Chs. 4/5 Inventory Mgt
Lowest cost –
purchase 1783
about every 2
years!
6-45
Properties of the
Optimal Solutions
Lets jump back into our
teams and do some!
5/27/2016
Chs. 4/5 Inventory Mgt
6-46
Resource Constrained MultiProduct 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:
c1Q1  c2Q2  ...  cnQn  C
5/27/2016
Chs. 4/5 Inventory Mgt
6-47
Resource Constrained MultiProduct Systems

When the condition that:
c1 / h1  c2 / h2  ...  cn / hn

is met, the solution procedure is
straightforward. If the condition is not met,
one must use an iterative procedure involving
Lagrange Multipliers.
5/27/2016
Chs. 4/5 Inventory Mgt
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:
n
lj
P
j 1
5/27/2016
1
j
Chs. 4/5 Inventory Mgt
6-48
6-49
Issues:


We are interested in controlling Family
MAKESPAN (we wish to produce all products
within our chosen cycle time)
Underlying Assumptions:


5/27/2016
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
Chs. 4/5 Inventory Mgt
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:
n
T* 
2 K j
j 1
n
h
j 1

j
'lj
We must assure that this time allows for all setups
and of production times.
5/27/2016
Chs. 4/5 Inventory Mgt
6-50
6-51
Working forward:

This last statement means:





(sj+(Qj/Pj)  T
Of course: Qj = lj*T
So – with substitution: (sj+((lj*T )/Pj)  T
Or: T(sj/(1- lj/Pj) = Tmin
Finally, we must Choose
T(actual cycle time) = MAX(T*,Tmin)
5/27/2016
Chs. 4/5 Inventory Mgt
6-52
Lets Try Problem 30
ITEM
Mon
Req’r
Daily
Req’r
h=
.2*c
l/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
5/27/2016
Chs. 4/5 Inventory Mgt
Compute the Following – in
teams!:
n


2Kj 
j 1



T  n
  l j  h'j 
j 1
n
TMin 
s
j 1
j
  n lj
1   

  j 1 Pj
Topt  T  , TMin 

 

MAX
Lot Size: Topt  Demand Rate
5/27/2016
Uptime; Drawdown
Time; Utilization
Chs. 4/5 Inventory Mgt
6-53
Lets do a ‘QUICK’ Exploration of
Stochastic Inventory Control (Ch 5)





We will examine underlying ideas –
We base our approaches on Probability Density
Functions (means & std. Deviations)
We are concerned with two competing ideas: Q
and R
Q (as earlier) an order quantity and R a
stochastic estimate of reordering time and level
Finally we are concerned with Servicing ideas –
how often can we supply vs. not supply a
demand (adds stockout costs to simple EOQ
models)
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Chs. 4/5 Inventory Mgt
6-54
6-55
The Nature of Uncertainty



5/27/2016
Suppose that we represent demand as:
 D = Ddeterministic + Drandom
If the random component is small compared to
the deterministic component, the models of
chapter 4 will be accurate. If not, randomness
must be explicitly accounted for in the model.
In this chapter, assume that demand is a random
variable with cumulative probability distribution
F(t) and probability density function f(t).
Chs. 4/5 Inventory Mgt
6-56
The Newsboy Model
At the start of each day, a newsboy must decide
on the number of papers to purchase. Daily sales
cannot be predicted exactly, and are represented
by the random variable, D.
Costs:
co = unit cost of overage
cu = unit cost of underage


5/27/2016
It can be shown that the optimal number of
papers to purchase is the fractile of the demand
distribution given by F(Q*) = cu / (cu + co).
Chs. 4/5 Inventory Mgt
6-57
Determination of the Optimal
Order Quantity for Newsboy Example
5/27/2016
Chs. 4/5 Inventory Mgt
6-58
Computing the Critical Fractile:

We wish to minimize competing costs (Co & Cu):



G(Q,D) = Co*MAX(0, Q-D) + Cu*MAX(0, D-Q) ---- D is actual
(potential) Demand
G(Q) = E(G(Q,D)) an expected value
Therefore:


0
0
G (Q)  Co  MAX (0, Q  x) f  x dx  Cu  MAX (0, x  Q) f  x dx
Q
Q
0
0
G (Q)  Co   Q  x)  f  x  dx  Cu   x  Q)  f  x  dx
Here: f(x) is a probability density function
controlling the behavior of ordering
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Chs. 4/5 Inventory Mgt
6-59
Applying Leibniz’s Rule:




d(G(Q))/dQ = CoF(Q) – Cu(1 – F(Q))
F(Q) is a cumulative Prob. Density Function
(as earlier of the quantity ordered)
Thus: G’(Q*) = (Cu)/(Co + Cu)
This is the critical fractile for the order
variable
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Chs. 4/5 Inventory Mgt
6-60
Lets see about this: Prob 5 pg 241


Observed sales given as a number
purchased during a week (grouped)
Lets assume some data was supplied:





Make Cost: $1.25
Selling Price: $3.50
Salvageable Parts: $0.80
Co = overage cost = $1.25 - $0.80 = $0.45
Cu = underage cost = $3.50 - $1.25 = $2.25
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Chs. 4/5 Inventory Mgt
6-61
Continuing:

Compute Critical Ratio:


CR = Cu/(Co + Cu) = 2.25/(.45 + 2.25) = .8333
If we assume a continuous Probability
Density Function (lets choose a normal
distribution):



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Z(CR)  0.967 when F(Z) = .8333 (from Std.
Normal Tables!)
Z = (Q* - )/)
From the problem data set, we compute mean =
9856 & St.Dev. = 4813.5)
Chs. 4/5 Inventory Mgt
6-62
Continuing:



Q* = Z +  = 4813.5*.967 + 9856 = 14511
Our best guess economic order quantity is
14511
(We really should have done it as a Discrete
problem -- Taking this approach we would
find that Q* is only 12898)
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Chs. 4/5 Inventory Mgt
6-63
Lot Size Reorder Point Systems
Assumptions:


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Inventory levels are reviewed continuously
(the level of on-hand inventory is known at
all times)
Demand is random but the mean and
variance of demand are constant.
(stationary demand)
Chs. 4/5 Inventory Mgt
6-64
Lot Size Reorder Point Systems
Assumptions:


There is a positive leadtime, τ. This is the time
that elapses from the time an order is placed
until it arrives.
The costs are:




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Set-up each time an order is placed at $K per order
Unit order cost at $C for each unit ordered
Holding at $H per unit held per unit time ( i. e., per year)
Penalty cost of $P per unit of unsatisfied demand
Chs. 4/5 Inventory Mgt
6-65
Describing Demand



The response time of the system (in this case) is the
time that elapses from the point an order is placed
until it arrives. Hence,
The uncertainty that must be protected against is
the uncertainty of demand during the lead time.
We assume that D represents the demand during
the lead time and has probability distribution F(t).
Although the theory applies to any form of F(t), we
assume that it follows a normal distribution for
calculation purposes.
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Chs. 4/5 Inventory Mgt
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Decision Variables




For the basic EOQ model discussed in Chapter 4,
there was only the single decision variable Q.
The value of the reorder level, R, was determined
by Q.
Now we treat Q and R as independent decision
variables.
Essentially, R is chosen to protect against
uncertainty of demand during the lead time, and Q
is chosen to balance the holding and set-up costs.
(Refer to Figure 5-5)
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Chs. 4/5 Inventory Mgt
6-67
Changes in Inventory Over Time
for Continuous-Review (Q, R) System
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Chs. 4/5 Inventory Mgt
6-68
The Cost Function
The average annual cost is given by:
G (Q, R )  h(Q / 2  R   )  K l / Q  pl n( R ) / Q


Interpret n(R) as the expected number of stockouts
per cycle given by the loss integral formula see
Table A-4 (std. values).
The optimal values of (Q,R) that minimizes G(Q,R)
can be shown to be:
2l ( K  pn( R ))
Q
h
1  F ( R )  Qh / pl
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Chs. 4/5 Inventory Mgt
6-69
Solution Procedure



The optimal solution procedure requires iterating
between the two equations for Q and R until
convergence occurs (which is generally quite
fast).
A cost effective approximation is to set Q=EOQ
and find R from the second equation.
A slightly better approximation is to set Q =
max(EOQ,σ)

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where σ is the standard deviation of lead time demand
when demand variance is high.
Chs. 4/5 Inventory Mgt
6-70
Service Levels in (Q,R) Systems
In many circumstances, the penalty cost, p, is
difficult to estimate. For this reason, it is
common business practice to set inventory
levels to meet a specified service objective
instead. The two most common service
objectives are:

1)
2)
5/27/2016
Type 1 service: Choose R so that the probability of
not stocking out in the lead time is equal to a
specified value.
Type 2 service. Choose both Q and R so that the
proportion of demands satisfied from stock equals a
specified value.
Chs. 4/5 Inventory Mgt
6-71
Computations


For type 1 service, if the desired service level
is α then one finds R from F(R)= α and
Q=EOQ.
Type 2 service requires a complex interative
solution procedure to find the best Q and R.
However, setting Q=EOQ and finding R to
satisfy n(R) = (1-β)Q (which requires Table A4) will generally give good results.
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Comparison of Service
Objectives



Although the calculations are far easier for
type 1 service, type 2 service is generally the
accepted definition of service.
Note that type 1 service might be referred to
as lead time service, and type 2 service is
generally referred to as the fill rate.
Refer to the example in section 5-5 to see the
difference between these objectives in
practice (on the next slide).
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Comparison (continued)
Order Cycle
1
2
3
4
5
6
7
8
9
10

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Demand
180
75
235
140
180
200
150
90
160
40
Stock-Outs
0
0
45
0
0
10
0
0
0
0
For a type 1 service objective there are two cycles out of ten in
which a stockout occurs, so the type 1 service level is 80%. For
type 2 service, there are a total of 1,450 units demand and 55
stockouts (which means that 1,395 demand are satisfied). This
translates to a 96% fill rate.
Chs. 4/5 Inventory Mgt
6-74
(s, S) Policies

The (Q,R) policy is appropriate when inventory levels
are reviewed continuously. In the case of periodic
review, a slight alteration of this policy is required.
Define two levels, s < S, and let u be the starting
inventory at the beginning of a period. Then
If u  s, order S  u
If u  s, don't order
(In general, computing the optimal values of s and S is much
more difficult than computing Q and R.)
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Chs. 4/5 Inventory Mgt