Transcript Chapter 4 Inventory Control Subject to Known Demand McGraw-Hill/Irwin

Chapter 4
Inventory Control Subject to
Known Demand
McGraw-Hill/Irwin
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
5/27/2016
Chs. 4/5 Inventory Mgt
6-3
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?
5/27/2016
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
that changes over time
perishability
obsolescence
5/27/2016
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:

(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 20 – 30% of the inventory items
sku’s) stored in small quantities and/or worth very little
 A Items
5/27/2016
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 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%)
(Total: 24%)
 h  .24*Cost of product
Note: Since inventory may be changing on a continuous basis, holding cost is
proportional to the area under the inventory curve.
5/27/2016
Chs. 4/5 Inventory Mgt
6-8
Lets Try one:
4, page 193 – cost of inventory
 Find h first (yearly and monthly)
 Total holding cost for the given period:
 Problem
 THC
= $26666.67
 Average Annual
Holding Cost
 assumes
an average monthly inventory of
trucks based on on hand data
 $3333
5/27/2016
Chs. 4/5 Inventory Mgt
6-9
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.
5/27/2016
Chs. 4/5 Inventory Mgt
6-10
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 book-keeping for backordered
demands
5/27/2016
Chs. 4/5 Inventory Mgt
6-11
Relevant Costs (continued)
Penalty
or Shortage Costs. All costs
that accrue when insufficient stock is
available to meet demand. These
include:
Loss
of goodwill for being unable to satisfy
demands when they occur.
Generally assume cost is proportional to
number of units of excess demand.
5/27/2016
Chs. 4/5 Inventory Mgt
6-12
The Simple EOQ 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).
5/27/2016
Chs. 4/5 Inventory Mgt
6-13
Simple EOQ Model (cont.)
Assumptions
(cont.):
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.
5/27/2016
Chs. 4/5 Inventory Mgt
Inventory Levels for the EOQ
Model
5/27/2016
Chs. 4/5 Inventory Mgt
6-14
6-15
The Average Annual
Cost Function G(Q)
5/27/2016
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 
5/27/2016
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
5/27/2016
Chs. 4/5 Inventory Mgt
6-18
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
'

Solve for Q
5/27/2016
Chs. 4/5 Inventory Mgt
6-19
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 = Ic)

5/27/2016
Chs. 4/5 Inventory Mgt
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 ans sells for $12.50/bx
 Cost of ordering is $35
 Determine EOQ (Q*)
5/27/2016
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
*
Q
T
 514
 .068 yr
l
7540
or: 514
 3.54wk
145
*
5/27/2016
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 trigger an order at a point before the
delivery is needed to not have 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 units of lead time match units in T!
5/27/2016
Chs. 4/5 Inventory Mgt
6-23
But, Orders usually take time to
arrive!


What happens when lead time exceeds T?
It is just as before (but we compute /T)
 is the lead time is similar units as T

Here, in weeks  = 6 weeks then:
/T = 6/3.545 = 1.69
 Place order 1.69 cycles before we need product
 Trip Point is then .69*Q* = .69*514 = 356 boxes
 This trip point is not for the next stock out but the one
after that (1.7 T from now!) – be very careful!!!
5/27/2016
Chs. 4/5 Inventory Mgt
6-24
Sensitivity Analysis
Let G(Q) be the average annual holding and set-up
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 *
5/27/2016
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
Q* + Q has smaller error than Q* - Q
 Error here mean storage costs penalty
 that
5/27/2016
Chs. 4/5 Inventory Mgt
6-26
EOQ With Finite Production Rate
Suppose that items are produced internally at a rate
P > λ. 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'
Except that h’ is defined as h’= h(1- λ/P)
5/27/2016
Chs. 4/5 Inventory Mgt
6-27
This is based on solving
K hH
G (Q )  
2
T
K l  hQ

 1  l / P 

G (Q ) 
2
Q
P is annual production rate
H is max on hand quantity

H  Q  1 l
5/27/2016
P

Chs. 4/5 Inventory Mgt
6-28
Inventory Levels for Finite
Production Rate Model
5/27/2016
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 (250 d/y)
 Setup takes 2 techs working 45 minutes
@$21/hour and requires an expendible tool
costing $25
5/27/2016
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:
= production time in a cycle (Q*/P)
 T2 = non producing time in a cycle (T – T1)
 T1
5/27/2016
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.
5/27/2016
Chs. 4/5 Inventory Mgt
All-Units Discount
Order Cost Function
5/27/2016
Chs. 4/5 Inventory Mgt
6-32
6-33
Incremental Discount
Order Cost Function
5/27/2016
Chs. 4/5 Inventory Mgt
6-34
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).
5/27/2016
Chs. 4/5 Inventory Mgt
All-Units Discount Average
Annual Cost Function
5/27/2016
Chs. 4/5 Inventory Mgt
6-35
6-36
To Find EOQ in ‘All Units’
discount case:
 Compute
Q* for each cost type
 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)
5/27/2016
Chs. 4/5 Inventory Mgt
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

Q1* 
Q2* 
5/27/2016
 2  300  900 
 2  300  900 
 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  6.5 674 

 .34  6.5 600 
G (495)  $6942.43
G (674)  900  (6.50) 
 900  300 
674
G (674)  $3951.62
G (600)  900  (6.50) 
 900  300 
600
G (600)  $3957.00
5/27/2016
Chs. 4/5 Inventory Mgt
2
2
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 but as
Incremental Case:
 Cost
Function: Basically states that we pay
6.50 for each up to 600 then 3.50 for any
more:
= 6.5(Q), Q  600
 C(Q) = 3.5(Q – 600) + 3900, Q > 600
 C(Q)/Q = 6.5, Q  600
 C(Q)/Q = 3.5 + (1800/Q), Q > 600
 C(Q)
5/27/2016
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)
 Finding Q* is a process of writing a G(Q)
equation for this range and then differentiation :



 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
dQ
Q
2


Set equal to 0 and solve for Q2
 900  2100 
2
2
Q

2
Q 
5/27/2016
 0.595  0
 900  2100 
0.595
Chs. 4/5 Inventory Mgt
 1783
Realizable!
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 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:
c1Q1  c2Q2  ...  cnQn  C
5/27/2016
Chs. 4/5 Inventory Mgt
6-47
Resource Constrained Multi-Product 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
6-48
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 we must have that
n
lj
P
j 1
5/27/2016
 1.
j
Chs. 4/5 Inventory Mgt
6-49
Issues:
 We
are interested in the Family
MAKESPAN (we wish to produce all
products within the chosen cycle time)
 Underlying Assumptions:
 Setup
Cost are not Sequence Dependent (this
assumption is not accurate as we will later see)
 Plant uses a “Rotation” Policy that produces a
single ‘batch’ of each product each cycle
5/27/2016
Chs. 4/5 Inventory Mgt
6-50
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
6-51
Working forward:
 This
last statement means:
 (sj+(Qj/Pj)
T
 Since: Qj = lj*T
 Subbing:
 (sj+((lj*T )/Pj)  T
 T(sj/(1- lj/Pj) = Tmin
 We
must Choose
T(planned cycle time) = MAX(T*,Tmin)
5/27/2016
Chs. 4/5 Inventory Mgt
6-52
Lets Try Problem 30
ITEM
Mon
Reqr
Daily
Reqr
h=
.2*c
l/P
h’
Setup
Time
Setup
Cost
U.
Cost
Daily
Pr.
Rate
Mon.
Pr.
Rate
J55R
125
6.25
4
.045
3.82
1.2
$120
$20
140
2800
H223
140
7
7
.032
6.78
0.8
$68
$35
220
4400
K18R
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
6-53
Compute:
n


2Kj 
j 1

T   n
'
l

h

 j j
j 1
n
TMin 
s
j 1
j
  n lj
1   

  j 1 Pj
Topt  T  , TMin 

 

MAX
Lot Size: Topt  Demand Rate
Uptime; Drawdown Time; Utilization
5/27/2016
Chs. 4/5 Inventory Mgt
6-54
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)

5/27/2016
Chs. 4/5 Inventory Mgt
6-55
The Nature of Uncertainty
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).
5/27/2016
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
It can be shown(*see over) that the optimal number
of papers to purchase is the fractile of the demand
distribution given by F(Q*) = cu / (cu + co).
5/27/2016
Chs. 4/5 Inventory Mgt
Determination of the Optimal
Order Quantity for Newsboy
Example
5/27/2016
Chs. 4/5 Inventory Mgt
6-57
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)) ---- expected value


Therefore:
G (Q)  Co  MAX (0, Q  x) f  x dx  Cu  MAX (0, x  Q) f  x dx
0
0
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 the ordering
5/27/2016
Chs. 4/5 Inventory Mgt
6-59
Applying Leibniz’s Rule:
= CoF(Q) – Cu(1 – F(Q))
 F(Q) is a cumulative Prob. Density Function
(as earlier)
 G’(Q*) = (Cu)/(Co + Cu)
 This is the critical fractile
 d(G(Q))/dQ
5/27/2016
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
5/27/2016
Chs. 4/5 Inventory Mgt
6-61
Continuing:
 Compute
 CR
Critical Ratio:
= Cu/(Co + Cu) = 2.25/(.45 + 2.25) = .8333
 If
we assume a continuous Pr. D. Function
(lets choose a normal function):
 0.967 when F(Z) = .8333 (from Std.
Normal Tables!)
 Z = (Q* - )/ (we compute mean = 9856;
StDev = 4813.5)
 Z(CR)
5/27/2016
Chs. 4/5 Inventory Mgt
6-62
Continuing:
= 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
 Q*
5/27/2016
Chs. 4/5 Inventory Mgt
6-63
Lot Size Reorder Point Systems
Assumptions
 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)
 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:
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

5/27/2016
Chs. 4/5 Inventory Mgt
6-64
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.
5/27/2016
Chs. 4/5 Inventory Mgt
6-65
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. In this
case, 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)
5/27/2016
Chs. 4/5 Inventory Mgt
Changes in Inventory Over Time
for Continuous-Review (Q, R)
System
5/27/2016
Chs. 4/5 Inventory Mgt
6-66
6-67
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. The
standardized loss integral values appear in Table A-4.
The optimal values of (Q,R) that minimizes G(Q,R)
can be shown to be:
2l ( K  pn( R ))
Q
h
5/27/2016
Chs. 4/5
1  F ( R )  Qh
/ pInventory
l Mgt
6-68
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)
where σ is the standard deviation of lead
time demand when demand variance is
high).
5/27/2016
Chs. 4/5 Inventory Mgt
6-69
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) Type 1 service: Choose R so that the probability
of not stocking out in the lead time is equal to a
specified value.
2) Type 2 service. Choose both Q and R so that the
proportion of demands satisfied from stock
equals a specified value.
5/27/2016
Chs. 4/5 Inventory Mgt
6-70
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
A-4) will generally give good results.
5/27/2016
Chs. 4/5 Inventory Mgt
6-71
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).
5/27/2016
Chs. 4/5 Inventory Mgt
6-72
Comparison (continued)
Order Cycle
1
2
3
4
5
6
7
8
9
10
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.
5/27/2016
Chs. 4/5 Inventory Mgt
6-73
(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.)
5/27/2016
Chs. 4/5 Inventory Mgt