E 1 2 4 8 16 is it easy to find an optimal po2

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Unformatted text preview: ource planning difficult because the schedule is complicated to implement One obvious restriction: “use only integer re-order 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 re-phrase 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.

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