9.2.2 - ampl: solve MINOS 5.5: ignoring integrality of 3...

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Sheet1 Page 1 data set SITES := 1 2 3 param: buildingcost watercost pollutant1 pollutant2 := 1 100000 20 0.4 0.3 2 6000 30 0.25 0.2 3 40000 40 0.2 0.25 param pollutant1_lim := 80000 param pollutant2_lim := 50000 param M := 400000 model set SITES # pollution control stations param buildingcost {i in SITES} param watercost {i in SITES} param pollutant1 {i in SITES} param pollutant2 {i in SITES} param pollutant1_lim param pollutant2_lim param M var station {i in SITES} binary var water {i in SITES} minimize treatcost: sum {i in SITES} buildingcost[i]*station[i] + sum {i in SITES} watercost[i]*water[i] subject to Pollution1_lim: sum {i in SITES} pollutant1[i]*water[i]>=pollutant1_lim subject to Pollution2_lim: sum {i in SITES} pollutant2[i]*water[i]>=pollutant2_lim subject to if_then {i in SITES}: water[i] - M*station[i] <= 0 subject to nonnegativity {i in SITES}: water[i] >= 0 AMPL sw: ampl ampl: model 9.2.2.mod ampl: data 9.2.2.dat
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Unformatted text preview: ampl: solve MINOS 5.5: ignoring integrality of 3 variables MINOS 5.5: optimal solution found. 0 iterations, objective 4050000 ampl: display station station [*] := 1 0.5 2 3 ampl: display water water [*] := 1 2e+05 2 Sheet1 Page 2 3 Sheet1 Page 3 #the cost of building site i #the cost of treating one ton of water in site i #the amount of pollutant 1 can remove from one ton of water in site i #the amount of pollutant 2 can remove from one ton of water in site i #the minimum amount of pollutant 1 needed to remove #the minimum amount of pollutant 2 needed to remove #the large number M #if site i is chosen = 1 and 0 for otherwise #number of tons of water treated in site i #objective function #minimum amount of pollutant 1 needed to remove #minimum amount of pollutant 2 needed to remove #amount of water removed at site i &gt; 0, station i = 1 #nonegativity Sheet1 Page 4...
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This note was uploaded on 01/25/2010 for the course IEOR 162 taught by Professor Zhang during the Fall '07 term at University of California, Berkeley.

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9.2.2 - ampl: solve MINOS 5.5: ignoring integrality of 3...

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