Transcript Inventory Control Models

Inventory Control
Models
Ch 5 (Uncertainty of Demand)
R. R. Lindeke
IE 3265, Production And
Operations Management
Lets do a ‘QUICK’ Exploration of
Stochastic Inventory Control (Ch 5)
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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)
The Nature of Uncertainty
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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).
Single Period Stochastic
Inventory Models
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These models have the objective of properly
balancing the cost of Underage – having not
ordered enough products vs. Overage – having
ordered more than we can sell
These models apply to problems like:
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Planning initial shipments of ‘High-Fashion’ items
Amount of perishable food products
Item with short shelf life (like the daily newspaper)
Because of this last problem type, this class of
problems is typically called the “Newsboy” problem
The Newsboy Model
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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.
The newsboy must carefully consider these
costs:
co = unit cost of overage
cu = unit cost of underage
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).
Determination of the Optimal
Order Quantity for Newsboy Example
Computing the Critical Fractile:
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We wish to minimize competing costs (Co & Cu):
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G(Q,D) = Co*MAX(0, Q-D) + Cu*MAX(0, D-Q)
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D is actual (potential) Demand
G(Q) = E(G(Q,D)) (an expected value)
Therefore:
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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
Applying Leibniz’s Rule:
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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 as stated earlier
Lets see about this: Prob 5 pg 241
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Observed sales given as a number
purchased during a week (grouped)
Lets assume some data was supplied:
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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
Continuing:
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Compute Critical Ratio:
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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
Continuing:
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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)
Newsboy’s Extensions
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Assuming we have a certain number of parts
on hand, u > 0
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This extends the problem compared to our initial u = 0
assumption for the single period case
This is true only if the product under study
has a shelf life that extends beyond one
period
Here we still compute Q* will order only Q* - u
(or 0 if u > Q*)
Try one (in your Engineering Teams) :
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Do Problem 11a & 11b (pg 249)
Lot Size Reorder Point Systems
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Earlier we considered reorder points (number of
parts on hand when we placed an order) they
were dependent on lead times as a dependent
variable on Q, now we will consider R as an
independent variable just like Q
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)
Lot Size Reorder Point Systems
Additional Assumptions:
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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 cost 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
Describing Demand
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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.
Decision Variables
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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)
Changes in Inventory Over Time
for Continuous-Review (Q, R) System
The Cost Function
The average annual cost is given by:
G (Q, R )  h(Q / 2  R   )  K  / Q  p n( R) / Q
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Interpret n(R) as the expected number of stockouts per
cycle given by the loss integral formula (see Table A-4
(std. values)). And note, the last term is this cost model
is a shortage cost term
The optimal values of (Q,R) that minimizes G(Q,R) can
be shown to be:
2 ( K  pn( R ))
Q
h
1  F ( R )  Qh / p
Solution Procedure
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The optimal solution procedure requires iterating
between the two equations for Q and R until
convergence occurs (which is generally quite fast)
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We consider that the problem has converged if 2
consecutive calculation of Q and R are within 1 unit
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.
Ready to Try one? Lets!
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Try Problem 13a & 13b (pg 261)
Start by computing EOQ and then begin
iterative solution for optimal Q and R values
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:
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1)
2)
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.
Computations
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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.
Comparison of Service
Objectives
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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).
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.
Example: Type 1 Service Pr 5-16
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Desire 95% Type I service Level
F(R) = .95  Z is 1.645 (Table A4)
From Problem 13:  was found to be 172.8
and  was 1400
Therefore: R = Z +  = 172.8*1.645 + 1400
R = 1684.256  1685
Use Q = EOQ = 1265
Example: Type 2 Service Pr 5-17
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Require Iterative Solution:
Q0  EOQ  1295
n( R1 )  1    * Q0  63.25
L  Z1  
n( R1 )
 .3673
172.8
 Z1  .065 & 1  F  R  =.474 from Table A4
Q1 
n( R1 )

 63.25
 n( R1 )

  EOQ   

1 F  R
1

F
R




Q1  63.25
2
.474

 1265  63.25
2
.474

2
 1405
2
Example: Type 2 Service Pr 5-17
(cont.)
n  R2   1    Q1  .05*1405  70.25
L  Z 2   70.25
 0.408
172.8
Z 2  0.02 & 1  F ( R2 )  0.508
R2   Z 2    1397
Q2  70.25
.508

1265
2

 70.25
.508
n  R3   0.05*1411  70.54
L  Z 3   70.54
172.8
 0.4097
Z 3  .02
R3   Z 3    1397 SAME so Stop!
 Q, R  = 1411, 1397 

2
 1411
(s, S) Policies
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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.)