Example - Optimal (Q,R) Policy with Service

Download Report

Transcript Example - Optimal (Q,R) Policy with Service

LESSON 19: INVENTORY MODELS (STOCHASTIC)
Q,R SYSTEMS
OPTIMIZATION WITH SERVICE
Outline
• Multi-Period Models
– Lot size-Reorder Point (Q, R) Systems
• Optimization with service
– Procedure for Type 1 Service
– Procedure for Type 2 Service
– Example
Optimization With Service
• In Lesson 18, we discuss the procedure of finding an
optimal Q, R policy without any service constraint and
using a stock-out penalty cost of p per unit.
• Managers often have difficulties to estimate p.
• A substitute for stock-out penalty cost, p. is service
level.
• In this lesson we shall not use stock-out penalty cost,
p. Instead , we shall assume that a service level must
be met. Next slide defines two major types of service
levels.
Optimization With Service
• Type 1 service
– The probability of not stocking out during the lead
time is denoted by . In problems with Type 1
service,  is specified e.g.,  = 0.95
• Type 2 service
– Fill rate, : The proportion of demands that are
met from stock is called filled rate and is denoted
by . In problems with Type 2 service,  is
specified e.g.,  = 0.999
Procedure to Find the Optimal (Q,R) Policy
with Type 1 Service
Goal: Given , , h, K, find (Q,R) to minimize total cost
First, find mean of the lead-time demand,    and
standard deviation of the lead-time demand,    
Step 1: Set Q = EOQ
Step 2: Find z for which area on the left, F(z) =
Step 3: Find R =   z
Procedure to Find the Optimal (Q,R) Policy
with Type 2 Service
Goal: Given , , h, K ,  find (Q,R) to minimize total cost
First, find mean of the lead-time demand,    and
standard deviation of the lead-time demand,    
Step 1: Take a trial value of Q = EOQ
Step 2: Find expected number of shortages per cycle,
n  Q(1   ), standardized loss function, L( z )  n /  ,
and the standard normal variate z from Table A-4, pp.
835-841. Find a trial value of R=   z
Procedure to Find the Optimal (Q,R) Policy
with Type 2 Service
Step 3: Find area on the right, 1-F(z) from Table A-1 or
A-4, pp. 835-841
Step 4: Find the modified
Q  n /(1  F ( z ))  2 K / h  ( n /(1  F ( z ))) 2
Step 5: Find expected number of shortages per cycle,
n  Q(1   ), standardized loss function, L( z )  n /  ,
and the standard normal variate z from Table A-4, pp.
835-841. Find the modified R=   z
Step 6: If any of modified Q and R is different from the
previous value, go to Step 3. Else if none of Q and R
is modified significantly, stop.
Example - Optimal (Q,R) Policy with Service
Annual demand for number 2 pencils at the campus
store is normally distributed with mean 2,000 and
standard deviation 300. The store purchases the
pencils for 10 cents and sells them for 35 cents each.
There is a two-month lead time from the initiation to
the receipt of an order. The store accountant
estimates that the cost in employee time for
performing the necessary paper work to initiate and
receive an order is $20, and recommends a 25
percent annual interest rate for determining holding
cost.
Example - Optimal (Q,R) Policy with Service
a. Find an optimal (Q,R) policy with Type 1 service,
=0.95 and Q=EOQ
b. Find an optimal (Q,R) policy with Type 2 service,
=0.999 and using the iterative procedure
Example - Optimal (Q,R) Policy with Service
Fixed ordering cost, K 
Holding cost, h  Ic 
Mean annual demand,  
Standard deviation of annual demand,  y
Lead time,  
Mean lead - time demand,    
Standard deviation of lead - time demand,    y 

Example - Optimal (Q,R) Policy with Service
a. Type 1 service,  = 0.95
Step 1. Q = EOQ =
Step 2. Find z for which area on the left, F(z) =  = 0.95
Step 3. R =   z 
b. Type 2 service, =0.999
This part is solved with the iterative procedure as
shown next.
Example - Optimal (Q,R) Policy with Service
Iteration 1
Step 1:
2k
Q  EOQ 

h
Step 2:
n  Q (1   ) 
n
L( z )  

z
R    z 
(Table A-4)
Example - Optimal (Q,R) Policy with Service
Step 3: 1  F (z ) 
Step 4:
n
2 K 
n 

Q

 
1  F( z)
h
 1  F( z) 
2

Question: What are the stopping criteria?
Example - Optimal (Q,R) Policy with Service
Step 5:
n  Q (1   ) 
n
L( z )  

z
R    z 
(Table A-4)
Example - Optimal (Q,R) Policy with Service
Iteration 2
Step 3: 1  F (z ) 
Step 4:
n
2 K 
n 

Q

 
1  F( z)
h
 1  F( z) 
2

Question: Do the answers converge?
Example - Optimal (Q,R) Policy with Service
Step 5:
n  Q (1   ) 
n
L( z )  

z
R    z 
(Table A-4)
Fixed cost (K )
Holding cost (h )
Mean annual demand ()
Lead time (in years
Lead time demand parameters:
Mean,
Standard deviation, 
Type 2 service, fill rate, 
EOQ
Step 1 Q =
Q (1   )
Step 2 n =
n /
L(z)=
Table A1/A4, pp. 835 - 41
z=
  z
R=
Table A1/A4, pp. 835 - 41
Step 3 Area on the right=1-F(z )
2
n
/(
1

F
(
z
))

2
K

/
h

(
n
/(
1

F
(
z
)))
Step 4 Modified Q =
Q (1   )
Step 5 n =
n /
L(z)=
Table A1/A4, pp. 835 - 41
z=
  z
R=
Note: K , and h
are input data
input data
input data
<--- computed
input data
input data
Iteration 1 Iteration 2
(Q,R) Systems
Remark
• We solve three versions of the problem of finding an
optimal (Q,R) policy
– No service constraint
– Type 1 service
– Type 2 service
• All these versions may alternatively and more
efficiently solved by Excel Solver. This is discussed
during the tutorial.
Multiproduct Systems
• Pareto effect
– A concept of economics applies to inventory systems
– Rank the items in decreasing order of revenue
generated
– Item group A: top 20% items generate 80% revenue
– Item group B: next 30% items generate 15% revenue
– Item group A: last 50% items generate 5% revenue
Multiproduct Systems
• Exchange curves
– Parameters like K and I are not easy to measure
– Instead of assigning values to such parameters show
the trade off between holding cost and ordering cost
for a large number of values of K/I
– The effect of changing the ratio K/I is shown by plotting
holding cost vs ordering cost
– Similarly, instead of assigning a value to type 2 service
level , one may show the trade off between cost of
safety stock and expected number of stock-outs.
READING AND EXERCISES
Lesson 19
Reading:
Section 5.5, pp. 264-271 (4th Ed.), pp. 255-262 (5th
Ed.)
Section 5.7 (skim) pp. 275-280 (4th Ed.), pp. 265-270
(5th Ed.)
Exercise:
16 and 17, p. 271 (4th Ed.), p. 262 (5th Ed.)