lecture12 - Capital Budgeting Assembly Line Balancing IE170...

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Unformatted text preview: Capital Budgeting Assembly Line Balancing IE170: Algorithms in Systems Engineering: Lecture 12 Jeff Linderoth Department of Industrial and Systems Engineering Lehigh University February 12, 2007 Jeff Linderoth IE170:Lecture 12 Capital Budgeting Assembly Line Balancing Taking Stock Last Time Intro to Dynamic Programming: Capital Budgeting This Time: More DP Assembly Line Balancing Lot Sizing Jeff Linderoth IE170:Lecture 12 Capital Budgeting Assembly Line Balancing Dynamic Programming Not really an algorithm but a technique. Not really “programming” like Java programming Dynamic Programming in a Nutshell 1 Characterize the structure of an optimal solution 2 Recursively define the value of an optimal solution 3 Compute the value of an optimal solution “from the bottum up” 4 Construct optimal solution (if required) Jeff Linderoth IE170:Lecture 12 Capital Budgeting Assembly Line Balancing DP for Assembly Line Scheduling You have been hired to optimize the Yugo Factory in Prattville, AL There are two assembly lines. Each line has n different stations: S 11 , S 12 , . . . , S 1 n and S 21 , S 22 , . . . , S 2 n . Stations S 1 j and S 2 j perform the same function, put take a different amount of time: ( a 1 j and a 2 j ) Once a Yugo is processed at station S ij , it can either 1 Stay on the same line ( i ) with no time penalty 2 Transfer to the other line, but is then delayed by t ij Jeff Linderoth IE170:Lecture 12 Capital Budgeting Assembly Line Balancing Your Mission Problem: Given this setup, what stations should be chosen from each line in order to minimize the time that a car is in the factory? Note: We can’t (efficiently) just check all possibilities? How many are there? Jeff Linderoth IE170:Lecture 12 Capital Budgeting Assembly Line Balancing A Better Way A better way to find an optimal solution is to think about what properties an optimal solution must have. Question What is the fasest way to get through station S 1 j ? If j = 1 : a 11 If j ≥ 2 , then we have two choice for how to get through S 1 j Through S 1 ,j- 1 then to S 1 j Through S 2 ,j- 1 then to S 1 j Jeff Linderoth IE170:Lecture 12 Capital Budgeting Assembly Line Balancing Key Obervation Suppose fastest way through S 1 j is through S 1 ,j- 1 We must have taken a fastest way to get through S 1 ,j- 1 in this fastest solution through S 1 j ....
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This note was uploaded on 08/06/2008 for the course IE 170 taught by Professor Ralphs during the Spring '07 term at Lehigh University .

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lecture12 - Capital Budgeting Assembly Line Balancing IE170...

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