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Unformatted text preview: ource planning difficult because the schedule is
complicated to implement
One obvious restriction: “use only integer reorder intervals”
An even tighter restriction “use only intervals that are an integer power of
two” (i.e. 1, 2, 4, 8, 16, . . . )
Is it easy to find an optimal Po2 policy? How much do our costs increase? J. G. Carlsson, U of MN ISyE Lecture 10: Inventory Models November 7, 2013 30 / 55 Synchronizing schedules
An advantage of Po2 policies: suppose we have a production system where
three products, A, B, and C, share a machine
Currently, A, B, and C are produced each week, every three weeks, and every
five weeks, respectively
This means that the production schedule repeats every fifteen weeks: A
B
C 1
x
x
x 2
x 3
x 4
x
x 5
x 6
x 7
x
x 8
x 9
x 10
x
x x 11
x
x 12
x 13
x
x 14
x 15
x 16
x
x
x Note that, once every fifteen weeks, all three items are produced at once (this
may cause overload) J. G. Carlsson, U of MN ISyE Lecture 10: Inventory Models November 7, 2013 31 / 55 Synchronizing schedules If we shift B and C to be produced every two weeks and every four weeks
(instead of three and five), we can work around this problem: A
B
C 1
x 2
x
x 3
x x 4
x
x 5
x
x 6
x
x 7
x 8
x
x 9
x 10
x
x x 11
x 12
x
x 13
x 14
x
x 15
x 16
x
x x The big picture: restricting ourselves to a power of 2 makes scheduling easier J. G. Carlsson, U of MN ISyE Lecture 10: Inventory Models November 7, 2013 32 / 55 EOQ in a Po2 policy
Recall the basic EOQ model without shortages and zero lead time:
minimize
Q≥0 aK
hQ
+ ac +
Q
2 As we discussed earlier, the purchasing cost ac is the least interesting because it
doesn’t depend on Q, so we can get rid of it:
minimize
Q≥0 aK hQ
+
Q
2 We always have a cycle time of T = Q/a, so we can rephrase the problem in
terms of T:
K haT
minimize +
T≥0
T
2
Finally, let’s simplify things even more by defining g = ah/2 (don’t bother
thinking of what this represents):
minimize
T≥0 J. G. Carlsson, U of MN ISyE K
+ gT
T Lecture 10: Inventory Models November 7, 2013 33 / 55 EOQ in a Po2 policy
The solution to the problem
K
+ gT
T≥0
T
is easily obtained by setting derivatives to zero, which gives
(
)
√
dK
K
+ gT = g − 2 = 0 =⇒ T∗ = K/g
dT T
T
and the optimal cost is
√
√
K
+ gT √ = 2 Kg = 2Kah
T
minimize T= K/g as before
Let’s say that the planning periods need to be multiples of some base planning
period TL :
For example, if T is expressed in years but we want our planning periods to be
multiples of weeks, then we’d write TL = 1/52
Generally it would make sense to use the same units for T and TL (i.e. TL = 1), but
we’ll play it safe
J. G. Carlsson, U of MN ISyE Lecture 10: Inventory Models November 7, 2013 34 / 55 EOQ in a Po2 policy
We’re now going to change the constraint “T ≥ 0” to “T = 2i × TL for some
i = {0, 1, 2, . . . }” Costs vs T with K = 250, g = 1
J. G. Carlsson, U of MN ISyE Lecture 10: Inventory Models November 7, 2013 35 / 55 EOQ in a Po2 policy The objective function, f(T) = K/T + gT, is convex: it decreases, and then
increases
∗
Thus we can find the optimal length, T∗ = 2i · TL , by looking for a point that is
better than both of its neighbors:
f(2i ∗ −...
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This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.
 Spring '14

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