# Qdm_2-Decision variables

May 5th, 2015
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Decision variables Let x represent the number of pumps produced in month m, m = 1, 2, 3, 4 Let s represent the number of pumps in ending inventory in month m, m = 1, 2, 3, 4

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Qdm 2Assignment 173-320-01Question 1Decision variablesLet x represent the number of pumps produced in month m, m = 1, 2, 3, 4Let s represent the number of pumps in ending inventory in month m, m = 1, 2, 3, 41 = May, 2 = June, 3 = July, 4 = AugustDeviation variablesLet d+,- represent the number of pumps deviated from target in month m, m = 1, 2, 3, 41, 5 = May, 2, 6 = June, 3, 7 = Julya) Month Constraints:May: x1 s1 = 200 , x1 < 500June: s1 + x2 s2 = 600 ,x2 < 400July: s2 + x3 s3 = 600 ,x3 < 800August: s3 + x4 = 600 ,x4 < 500b) Goal equations for fluctuation production from month to month:May to June: x1 d1+ +d1- = x2June to July: x2 d2+ +d2- = x3July to August: x3 d3+ +d3- = x4c) Goal equation for targeted zero ending inventory in months May, June, July:May: s1 d5+ +d5- = 0June: s2 d6+ +d6- = 0July: s3 d7+ +d7- = 0d) Other Constraints:Demand for May: x1 s1 > 200Demand for June: s1 + x2 s2 > 600Demand for July: s2 + x3 s3 > 600Demand for August: s3 + x4 = 600Capacity Constraint for May: x1 < 500Capacity Constraint for June: x2 < 400Capacity Constraint for July: x3 < 800Capacity Constraint for August: x4 < 500Non negativity for decision: xm > 0Non negativity for deviation:dm+,- > 0Ending Inventory for August:s4 = 0Total demand for contract: x1 + s1 + x2 + s2 + x3 + s3 + x4 + s4 = 2000 e) Solution:May Production of pumps (x1) = 500May Ending Inventory (s1) = 300June Production of pumps (x2) =

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