Download Optimal Replenishment Policy for MML Step Sheets

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts

Bayesian inference in marketing wikipedia , lookup

Institute of Cost Accountants of India wikipedia , lookup

Channel coordination wikipedia , lookup

Transcript
Optimal Replenishment Policy for MML Step Sheets
Joe Halo, Gaston Moliva, Jacob Boyle
November 18th, 2014
IE 425: Stochastic Operations Research
Executive Summary (300 to 400 words)
This is a case study on the optimal replenishment policy of inventory of step sheets that will be
inserted into a folder providing detailed information about the program. The information
provided on these sheets will be curriculum, partnership with industry, admission guidelines,
placement statistics, and financial assistance. Ordering of these step sheets must be done in large
quantities in order to be cost efficient, however the tricky part is re-printing after some of the
information in the packets is changed or becomes out dated. ACG also known as Advanced Color
Graphics is the company through which these documents are printed by and with that, they set the prices
and set bundles. As it is stated in the Introduction the cost of ordering an exact reprint is $1,157 for the
initial 500 sets and an additional $174 for every 500 sets. Unfortunately for the university, if the reprint
has changes or updates made to it the price goes up to $1,532 for the initial set of 500, but the price for
500 additional sets stays the same at $174. It is known that the current demand for step sheets is 1000 per
year and the probability that the step sheets require an update at the end of any given academic year is
approximately 0.255. As found in this case study the most economic order quantity (EOQ) that minimizes
the average annual cost is 4000 sets. In order to come to this conclusion transition state diagrams were
built, that can be found in the appendix, and a cost analysis was done on various order quantities ranging
from 1000 to 5000. As part of the cost analysis the probability matrix was formed from the transition state
diagrams and the steady state probabilities were calculated from that matrix. The fact that 4000 sets is the
most economical order quantity gives some insight as to how the university should continue to business
with ACG. Even though the demand is only 1000 per year, orders should only be placed in quantities of
4000 and saved for the next years.
Introduction
The objective of the case study was to determine the most economical way to re order sheets for their
program to give out to prospective students. Every year there is a demand of 1000 sheets with a cost of
$1157 without a redesign and a cost of $1532 with a redesign for the first 500 sheets and any additional
500 sheets was $174 for both scenarios. After every year there is a probability of .255 that there will be a
redesign after each year. Given all these conditions identifying a cost effective solution had to be
calculated. Estimated costs for several scenarios were calculated where the program order different
amounts of sheets such as 1000, 2000, 3000 and so on at the beginning of each year.
Problem statement
The MML program requires sheets to give to their prospective student, at a rate of around 1000 per
year. Due to the demand, the sheets have almost declines to zero and the program has to reorder new
sheets. The objective is to find the economic order quantity that minimizes the cost for the program. At a
rate of $1157 to reorder without changes and $1532 to reorder with changes for 500 sheets and any
additional 500 sheets is $174 for both, the reorder quantity has to be calculated so that the company
orders so that the cheapest alternative is bought. The rate in which the changes occur to the sheets is .255,
this has to be taken into account so the correct amount is ordered.
It is assumed that the MML program orders sheets once a year. With having a demand of 1000
each year it is also assumed that the MML program will only order in quantities of 1000. Starting at 1000
and going as high as they feel necessary. With this model in mind the total cost for each group of orders
(1000, 2000, 3000, 4000, etc) were calculated. This would then be plotted to find where the price dipped
lowest and then began trending back upwards. The lowest part of this U-shaped graph would be the
economic order quantity (EOQ). To calculate these costs a steady state diagram was drawn for each
possible order quantity. From these diagrams the probability matrix were calculated. Having this matrix
allowed for the calculation of the steady state probabilities or the long term probabilities. Once these were
calculated the total costs were calculated using the probabilities for reordering new designed sheets and
reordering sheets of the same design. These were then plotted to find the EOQ.
Solution procedure
The Markov chain model was the same model used throughout the case study for every order
quantity. Each higher order quantity of a 1000 had one more state added but nothing else changed. For all
models state 1 was the state where the step sheets were redesigned and another order of quantity n (n =
1000, 2000, 3000, 4000, 5000) was ordered. State 2 was also the same for each Markov chain. This state
was the state where the inventory level was at zero and the quantity n was reordered without any changes.
Every other state (3-6) was the previous years inventory level minus 1000 sheets. There were two
transition probabilities for each model. Each time a state went to state 1 the probability was always 0.225.
Every other transition in the diagram had the probability of 0.745. For the full state transition diagrams
and steady state probabilities refer to the appendix.
Figure 1. Order Amount vs. Cost
Computational results
The MML program is a two semester academic program held at Penn State University. This
program focuses on marketing to current Penn State Juniors and Seniors. They find that these groups are
where a large number of their applicants come from. The MML program created a new program brochure
two years ago. The MML program estimates there is a 25.5% chance that the brochure will change after a
year of use. With new brochure sets costing $1532 per 500 sheets and reusing the old brochure costing
$1157 per 500 sheets the MML program wants to know the economic order quantity. After creating a
Markov chain model for the situation and finding the steady state probabilities for each possible order
quantity it was found that the MLL program should order 4000 sets of sheets at a time. With this order
quantity the yearly cost of sheets will be $ 968.26 cents. This is cheaper than 3000 sheets costing $
976.92 per year and 5000 sheets costing $995.70 per year.
Q = 1000 sets
P=
.255 .745
.255 .745
steady state probabilities
(π1 π2 ) = (π1 π2 ) X P
EQ1: π1 = .255π1 + .255π2
EQ2: π2 = .745π1 + .745π2
π1 = .34228188π2
π1 + π2 = 1
.34228π2 + π2 = 1
π2 = .745
π1 = .255
Cost analysis for Q = 1000
re-order with redesign
C(1000) = 1532 + 1X174 = 1706
re-order without redesign
C(1000) = 1157 + 1X174 = 1331
C(1000) = 2054Xπ1 + 1679Xπ2 => 1706(.255) + 1331(.745)
C(1000) = 1426.625
Q = 2000 sets
P=
.255 0
.745
.255 0
.745
.255 .745 0
steady state probabilities
(π1 π2 π3) = (π1 π2 π3) X P
EQ1: π1 = .255π1 + .255π2 + .255π3
EQ2: π2 = .745π3
EQ3: π3 = .745π1 + .745π2
.745π1 = .255(.745π3) + .255π3 => π1 = .59728188π3
π2 = .745π3
π1 + π2 + π3 = 1
.59728π3 + .745π3 + π3 = 1
π3 = .4269
π1 = .2251
π2 = .318
Cost analysis for Q = 2000
re-order with redesign
C(2000) = 1532 + 3X174 = 2054
re-order without redesign
C(2000) = 1157 + 3X174 =1679
C(2000) = 2054Xπ1 + 1679Xπ2 => 2054(.2551) + 1679(.318)
C(2000) = 1057.897
Q = 3000 sets
P=
.255 0
.745 0
.255 0
.745 0
.255 0
0
.255 .745 0
.745
0
steady state probabilities
(π1 π2 π3 π4) = (π1 π2 π3 π4) X P
EQ1: π1 = .255π1 + .255π2 + .255π3 + .255π4
EQ2: π2 = .745π4
EQ3: π3 = .745π1 + .745π2
EQ4: π4 = .745π3
π3 = .745π1 + .745(.745π4) => .745π1 + .555025π4
π3 = .7733π4 + .555025π4
π3 = 1.3283π4
.745π1 = .255(.745π4) + .255( .745π1 + .555025π4) + .255π4
.565025π1 = .5865π4 => π1 = 1.038π4
π1 + π2 +π3 + π4 = 1
1.038π4 + .745π4 + 1.3283π4 + π4 = 1
π4 = .2413
π1 = .2550
π2 = .1798
π3 = .3240
Cost analysis for Q = 3000
re-order with redesign
C(3000) = 1532 + 5X174 = 2402
re-order without redesign
C(3000) = 1157 + 5X174 =2027
C(3000) = 2402Xπ1 + 2027Xπ2 => 2402(.2524) + 2027(.1814)
C(3000) = 976.9198
Q = 4000 sets
P=
.255 0
.745 0
0
.255 0
.745 0
0
.255 0
0
.745 0
.255 0
0
0
.745
.255 .745 0
0
0
steady state probabilities
(π1 π2 π3 π4 π5) = (π1 π2 π3 π4 π5) X P
EQ1: π1 = .255π1 + .255π2 + .255π3 + .255π4 + .255π5
EQ2: π2 = .745π5
EQ3: π3 = .745π1 + .745π2
EQ4: π4 = .745π3
EQ5: π5 = .745π4
π4 = π4
π5 = .745π4
π3 = 1.342π4
π2 = .555π4
π1 = .255π1 + .255(.555π4) + .255(1.342π4) + .255π4 + .255(.745π4) = 1.24π4
π1 + π2 +π3 + π4 + π5 = 1
1.24π4 + .555π4 + 1.342π4 + π4 + .745π4 = 1
π1 = 0.2539
π2 = 0.1137
π3 = 0.2748
π4 = 0.2048
π5 = 0.1526
Cost analysis for Q = 3000
re-order with redesign
C(4000) = 1532 + 7X174 = 2750
re-order without redesign
C(4000) = 1157 + 7X174 =2375
C(4000) = 2750Xπ1 + 2375Xπ2 => 2750(.2539) + 2375(.1137)
C(4000) = 968.26
Q = 5000 sets
P=
.255 0
.745 0
0
0
.255 0
.745 0
0
0
.255 0
0
.745 0
0
.255 0
0
0
.745 0
.255 0
0
0
0
.745
.255 .745 0
0
0
0
steady state probabilities
(π1 π2 π3 π4 π5 π6) = (π1 π2 π3 π4 π5 π6) X P
EQ1: π1= .255π1 + .255π2 + .255π3 + .255π4 + .255π5 + .255π6
EQ2: π2 = .745π6
EQ3: π3= .745π1 + .745π2
EQ4: π4= .745π3
EQ5: π5= .745π4
EQ6: π6= .745π5
π6 = 1.342 π2
π6 = 0.754π5 → 1.342 π2 = 0.754π5 → π5 = 1.801π2
π5 = 0.754π4 → 1.801 π2 = 0.754π4 → π4 = 2.417π2
π4 = 0.754π3 → 1.342 π2 = 0.754π3 → π3 = 3.244π2
π3 = 0.754π1 + 0.754π2 → 3.244 π2 = 0.754π1 + 0.754π2 → π1 = 3.35π2
1 = π1 + π2 + π3 + π4 + π5 + π6
1= 3.35π2 + π2 + 3.244π2 + 2.417π2 + 1.801π2 + 1.342π2
π1 = .2546
π2 = .076
π3 = .247
π4 = .184
π5 = .137
π6 = .102
Cost analysis for Q = 5000
re-order with redesign
C(5000) = 1532 + 9X174 = 3098
re-order without redesign
C(5000) = 1157 + 9X174 =2723
C(5000) = 3098Xπ1 + 2723Xπ2 => 3098(.2546) + 2723(.076)
C(5000) = 995.70
Appendix
Q = 1000 sets
Q = 2000 sets
Q = 3000 sets
Q = 4000 sets
Q = 5000 sets