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# lecture13 - Assembly Line Balancing Uncapacitated Lot...

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Assembly Line Balancing Uncapacitated Lot Sizing IE170: Algorithms in Systems Engineering: Lecture 13 Jeff Linderoth Department of Industrial and Systems Engineering Lehigh University February 21, 2007 Jeff Linderoth IE170:Lecture 13 Assembly Line Balancing Uncapacitated Lot Sizing Taking Stock Last Time Assembly Line Balancing (Intro to) Lot Sizing Knapsack Problem This Time: DP + Ilya’s Favorite Algorithm Lot Sizing Greedy Algorithm Jeff Linderoth IE170:Lecture 13 Assembly Line Balancing Uncapacitated Lot Sizing Uncapacitated Lot Sizing Lot sizing is the canonical production planning problem Given a planning horizon T = { 1 , 2 , . . . , T } You must meet given demands d t for t ∈ T You can meet the demand from a combination of production ( x t ) and inventory ( s t - 1 ) Production cost: c ( x t ) = K + cx t if x t > 0 0 if x t = 0 Inventory cost: I ( s t ) = h t s t Jeff Linderoth IE170:Lecture 13 Assembly Line Balancing Uncapacitated Lot Sizing Let’s Solve it with DP What should our stages be? Hint: Typically stages have type “from beginning until now” (like S ij ) or from “now until end” (like in capital budgeting) Stage Let f t ( s ) : be the minimum cost of meeting demands from t, t + 1 , . . . T if s units are in inventory at the beginning of period t Jeff Linderoth IE170:Lecture 13

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Assembly Line Balancing Uncapacitated Lot Sizing Let’s Solve an Example T = 3 d = [2 , 1 , 2] h = [1 , 1 , 0] K = 2 , c = 1 Busy Going Backwards f 3 (0) = 2 + 2(1) = 4 f 3 (1) = 2 + 1(1) = 3 f 3 (2) = 0 Jeff Linderoth IE170:Lecture 13 Assembly Line Balancing Uncapacitated Lot Sizing In General A General Recursive Relationship f t ( s ) = min x 0 , 1 , 2 ,... { c t ( x ) + h t ( s + x - d t ) + f t +1 ( s + x - d t ) } . Let’s do a couple by hand. This gets tedious – so let’s code it up... Jeff Linderoth IE170:Lecture 13 Assembly Line Balancing Uncapacitated Lot Sizing Oh Dear! What if K = 250 , d = [220 , 280 , 360 , 140 , 270] , c t = 2 , h t = 1 This might be a problem, as you need to consider producing every possible amount between 0 and 1270 Instead, as is often the case in dynamic programming, we look
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