/* Food Manufacture 2, section 12.2 in * Williams, "Model Building in Mathematical Programming" * * Sebastian Nowozin */ set oils; set month; /* Buying prices of the raw oils in the next six month. */ param buyingprices{month,oils}; /* Actual amount bought in each month. */ var buys{month,oils} >= 0; /* Stock for each oil. */ var stock{month,oils} >= 0; /* Price of the produced product */ param productprice >= 0; param storagecost; param oilhardness{oils} >= 0; param M >= 0; /* Actual amount of output oil produced in each month */ var production{m in month} >= 0; var useoil{m in month, o in oils} >= 0, <= M; var useoilb{m in month, o in oils}, binary; maximize totalprofit: sum{m in month} productprice*production[m] - sum{m in month, o in oils} buyingprices[m,o]*buys[m,o] - sum{m in month, o in oils} storagecost*stock[m,o]; /* Constraints */ /* 1. Starting stock */ s.t. startstock{o in oils}: stock[1,o] = 500; s.t. endstock{o in oils}: stock[6,o] + buys[6,o] - useoil[6,o] >= 500; /* 2. Stock constraints */ s.t. stocklimit{m in month, o in oils}: stock[m,o] <= 1000; s.t. production1{m in month, o in oils}: useoil[m,o] <= stock[m,o] + buys[m,o]; s.t. production2{m1 in month, m2 in month, o in oils : m2 = m1+1}: stock[m2,o] = stock[m1,o] + buys[m1,o] - useoil[m1,o]; s.t. production3a{m in month}: sum{o in oils} oilhardness[o]*useoil[m,o] >= 3*production[m]; s.t. production3b{m in month}: sum{o in oils} oilhardness[o]*useoil[m,o] <= 6*production[m]; s.t. production4{m in month}: production[m] = sum{o in oils} useoil[m,o]; /* 3. Refining constraints */ s.t. refine1{m in month}: useoil[m,"VEG1"]+useoil[m,"VEG2"] <= 200; s.t. refine2{m in month}: useoil[m,"OIL1"]+useoil[m,"OIL2"]+useoil[m,"OIL3"] <= 250; /* 4. Additional conditions: * i) The food may never be made up of more than three oils every month */ s.t. useoilb_calc{m in month, o in oils}: M*useoilb[m,o] >= useoil[m,o]; s.t. useoilb_limit{m in month}: sum{o in oils} useoilb[m,o] <= 3; /* ii) If an oil is used in a month, at least 20 tons must be used. */ s.t. useminimum{m in month, o in oils}: 20*useoilb[m,o] <= useoil[m,o]; /* iii) If either of VEG1 or VEG2 is used in a month, OIL2 must also be used */ s.t. use_oil2a{m in month}: useoilb[m,"VEG1"] <= useoilb[m,"OIL3"]; s.t. use_oil2b{m in month}: useoilb[m,"VEG2"] <= useoilb[m,"OIL3"]; solve; for {m in month} { printf "Month %d\n", m; printf "PRODUCE %4.2f tons, hardness %4.2f\n", production[m], (sum{o in oils} oilhardness[o]*useoil[m,o]) / (sum{o in oils} useoil[m,o]); printf "\tVEG1\tVEG2\tOIL1\tOIL2\tOIL3\n"; printf "STOCK"; printf "%d", m; for {o in oils} { printf "\t%4.2f", stock[m,o]; } printf "\nBUY"; for {o in oils} { printf "\t%4.2f", buys[m,o]; } printf "\nUSE"; printf "%d", m; for {o in oils} { printf "\t%4.2f", useoil[m,o]; } printf "\n"; printf "\n"; } printf "Total profit: %4.2f\n", (sum{m in month} productprice*production[m] - sum{m in month, o in oils} buyingprices[m,o]*buys[m,o] - sum{m in month, o in oils} storagecost*stock[m,o]); printf " turnover: %4.2f\n", sum{m in month} productprice*production[m]; printf " buying costs: %4.2f\n", sum{m in month, o in oils} buyingprices[m,o]*buys[m,o]; printf " storage costs: %4.2f\n", sum{m in month, o in oils} storagecost*stock[m,o]; data; param : oils : oilhardness := VEG1 8.8 VEG2 6.1 OIL1 2.0 OIL2 4.2 OIL3 5.0 ; set month := 1 2 3 4 5 6; param buyingprices : VEG1 VEG2 OIL1 OIL2 OIL3 := 1 110 120 130 110 115 2 130 130 110 90 115 3 110 140 130 100 95 4 120 110 120 120 125 5 100 120 150 110 105 6 90 100 140 80 135 ; param productprice := 150; param storagecost := 5; param M := 1000; end;