hw2_solution (1)

# hw2_solution (1) - Question1 b AMPL MOD and DATA FILES set...

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Question1- b- AMPL MOD and DATA FILES set dining; set resource; param price{dining}; param avail{resource}; param laborcost{dining}; param matcost{dining}; param cons{resource,dining}; var x{dining}>=0; maximize Tvalue: sum{j in dining}(price[j]-32*laborcost[j]-matcost[j])*x[j]; subject to res_con { i in resource}: sum{j in dining}cons[i,j]*x[j]<=avail[i]; set dining:= 1 2 3; set resource:= mach1 mach2 mach3; param: price laborcost matcost:= 1 2800 15 800 2 3100 22 680 3 2500 11 930; param avail:=mach1 400 mach2 625 mach3 200; param cons: 1 2 3:= mach1 2 3 2 mach2 5 4 1.5 mach3 1.5 1.3 .7;

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Solution 2 iterations, objective 266250 ampl: display x; x [*] := 1 75 2 0 3 125 ; Question2: AMPL MOD and DATA FILES set Nutrition; set Food; param cost {Food}; #price per pound of food param reqNut {Nutrition}; #the required amount of each nutrient param amt {Nutrition, Food}; #the amount of nutrient gained for each pound of each type of food var x {j in Food} >= 0; #pounds of each type of food bought minimize Total_Cost: sum {j in Food} cost[j] * x[j]; #minimize the cost, which is the cost per pounds multiplies by the amount of pounds purchased for each type of food
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## This note was uploaded on 07/08/2011 for the course IE 121 taught by Professor Perevalov during the Spring '08 term at Lehigh University .

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hw2_solution (1) - Question1 b AMPL MOD and DATA FILES set...

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