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4QA3 F12 Week 10 Lecture Notes

# 4qa3 f12 a gandomi 29 principle of optimality

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Unformatted text preview: est path) in the network to go from node 1 to node T+1, where T is the number of periods. 4QA3 F12 A. Gandomi 29 ●  Principle of Optimality: if we have n stages and there are k<n stages remaining, the optimal policy for the remaining stages is independent of policy adopted in the previous stage. ●  Let fk be minimum cost starting at node k, assuming that an order is placed in period k. We have: f k = min(ckj + f j ) for k = 1,, n j >k f n+1 = 0. 4QA3 F12 A. Gandomi 30 ●  Instead of solving the problem optimally, we could use a heuristic (a rule) that leads to reasonably good solutions but not necessarily optimal solution. ●  The advantage of heuristics is ease of implementation and lower computational effort to reach a solution. 4QA3 F12 A. Gandomi 31 yi =ri for each period i. ●  One could apply the EOQ formula by de^ining: n 1 ●  λ= ●  ∑r n i i =1 The method requires computing the average cost for an order horizon of j periods for j = 1, 2, 3, etc. and stopping at the ^irst instance when the average cost function increases. The average cost for a production quantity spanning j periods, C(j), is given by: 4QA3 F12 ( C ( j ) = K + hr2 + 2hr3 + ... + ( j − 1)hrj A. Gandomi ) j 32 ●  Here one minimizes the average cost per unit of demand (as opposed to the average cost per period as is done in the Silver Meal heuristic). The average cost per unit of demand over j periods is given by: ( C ( j ) = K + hr2 + 2hr3 + ... + ( j − 1)hrj ) (r + r + ... + r ) 1 2 j ●  Here one chooses the order horizon to most closely balance the total holding cost with the set- up cost. 4QA3 F12 A. Gandomi 33 ●  The time- phased net requirements for the base assembly in a table lamp over the next six weeks are Week 1 2 3 4 5 6 Requirements 335 200 140 440 300 200 The setup cost for the construction of the base assembly is \$200, and the holding cost is \$0.30 per assembly per week. a)  b)  c)  d)  e)  f)  What lot sizing do you obtain from the EOQ formula? Determine the lot sizes using the Silver- Meal heuristic. Determine the lot sizes using the least unit cost heuristic. Determine the lot sizes using part period balancing. Compare the holding and setup costs obtained over the six periods using the policies found in parts (a) through (d) with the cost of a lot- for- lot policy. Find the optimal solution using the Wagner- Whitin algorithm. 4QA3 F12 A. Gandomi 34 Average demand = (335 + 200 + 140 + 440 + 300 + 200)/6 = 269.17 EOQ = Week Demand Production Inventory 4QA3 F12 1 335 599 264 2(200 )(269.17) = 599 0.3 2 3 4 5 6 200 140 440 300 200 0 599 0 599 0 64 523 83 382 182 A. Gandomi 35...
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