[9] | 1 | /* Food Manufacture 2, section 12.2 in |
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| 2 | * Williams, "Model Building in Mathematical Programming" |
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| 3 | * |
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| 4 | * Sebastian Nowozin <nowozin@gmail.com> |
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| 5 | */ |
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| 6 | |
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| 7 | set oils; |
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| 8 | set month; |
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| 9 | |
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| 10 | /* Buying prices of the raw oils in the next six month. */ |
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| 11 | param buyingprices{month,oils}; |
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| 12 | |
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| 13 | /* Actual amount bought in each month. */ |
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| 14 | var buys{month,oils} >= 0; |
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| 15 | |
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| 16 | /* Stock for each oil. */ |
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| 17 | var stock{month,oils} >= 0; |
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| 18 | |
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| 19 | /* Price of the produced product */ |
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| 20 | param productprice >= 0; |
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| 21 | param storagecost; |
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| 22 | |
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| 23 | param oilhardness{oils} >= 0; |
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| 24 | param M >= 0; |
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| 25 | |
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| 26 | /* Actual amount of output oil produced in each month */ |
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| 27 | var production{m in month} >= 0; |
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| 28 | var useoil{m in month, o in oils} >= 0, <= M; |
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| 29 | var useoilb{m in month, o in oils}, binary; |
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| 30 | |
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| 31 | maximize totalprofit: |
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| 32 | sum{m in month} productprice*production[m] |
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| 33 | - sum{m in month, o in oils} buyingprices[m,o]*buys[m,o] |
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| 34 | - sum{m in month, o in oils} storagecost*stock[m,o]; |
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| 35 | |
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| 36 | /* Constraints */ |
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| 37 | |
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| 38 | /* 1. Starting stock */ |
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| 39 | s.t. startstock{o in oils}: |
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| 40 | stock[1,o] = 500; |
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| 41 | s.t. endstock{o in oils}: |
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| 42 | stock[6,o] + buys[6,o] - useoil[6,o] >= 500; |
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| 43 | |
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| 44 | /* 2. Stock constraints */ |
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| 45 | s.t. stocklimit{m in month, o in oils}: |
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| 46 | stock[m,o] <= 1000; |
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| 47 | |
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| 48 | s.t. production1{m in month, o in oils}: |
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| 49 | useoil[m,o] <= stock[m,o] + buys[m,o]; |
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| 50 | s.t. production2{m1 in month, m2 in month, o in oils : m2 = m1+1}: |
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| 51 | stock[m2,o] = stock[m1,o] + buys[m1,o] - useoil[m1,o]; |
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| 52 | |
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| 53 | s.t. production3a{m in month}: |
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| 54 | sum{o in oils} oilhardness[o]*useoil[m,o] >= 3*production[m]; |
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| 55 | s.t. production3b{m in month}: |
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| 56 | sum{o in oils} oilhardness[o]*useoil[m,o] <= 6*production[m]; |
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| 57 | |
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| 58 | s.t. production4{m in month}: |
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| 59 | production[m] = sum{o in oils} useoil[m,o]; |
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| 60 | |
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| 61 | /* 3. Refining constraints */ |
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| 62 | s.t. refine1{m in month}: |
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| 63 | useoil[m,"VEG1"]+useoil[m,"VEG2"] <= 200; |
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| 64 | s.t. refine2{m in month}: |
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| 65 | useoil[m,"OIL1"]+useoil[m,"OIL2"]+useoil[m,"OIL3"] <= 250; |
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| 66 | |
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| 67 | /* 4. Additional conditions: |
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| 68 | * i) The food may never be made up of more than three oils every month |
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| 69 | */ |
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| 70 | s.t. useoilb_calc{m in month, o in oils}: |
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| 71 | M*useoilb[m,o] >= useoil[m,o]; |
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| 72 | s.t. useoilb_limit{m in month}: |
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| 73 | sum{o in oils} useoilb[m,o] <= 3; |
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| 74 | |
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| 75 | /* ii) If an oil is used in a month, at least 20 tons must be used. |
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| 76 | */ |
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| 77 | s.t. useminimum{m in month, o in oils}: |
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| 78 | 20*useoilb[m,o] <= useoil[m,o]; |
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| 79 | |
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| 80 | /* iii) If either of VEG1 or VEG2 is used in a month, OIL2 must also be used |
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| 81 | */ |
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| 82 | s.t. use_oil2a{m in month}: |
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| 83 | useoilb[m,"VEG1"] <= useoilb[m,"OIL3"]; |
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| 84 | s.t. use_oil2b{m in month}: |
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| 85 | useoilb[m,"VEG2"] <= useoilb[m,"OIL3"]; |
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| 86 | |
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| 87 | solve; |
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| 88 | |
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| 89 | for {m in month} { |
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| 90 | printf "Month %d\n", m; |
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| 91 | printf "PRODUCE %4.2f tons, hardness %4.2f\n", production[m], |
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| 92 | (sum{o in oils} oilhardness[o]*useoil[m,o]) / (sum{o in oils} useoil[m,o]); |
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| 93 | |
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| 94 | printf "\tVEG1\tVEG2\tOIL1\tOIL2\tOIL3\n"; |
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| 95 | printf "STOCK"; |
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| 96 | printf "%d", m; |
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| 97 | for {o in oils} { |
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| 98 | printf "\t%4.2f", stock[m,o]; |
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| 99 | } |
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| 100 | printf "\nBUY"; |
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| 101 | for {o in oils} { |
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| 102 | printf "\t%4.2f", buys[m,o]; |
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| 103 | } |
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| 104 | printf "\nUSE"; |
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| 105 | printf "%d", m; |
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| 106 | for {o in oils} { |
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| 107 | printf "\t%4.2f", useoil[m,o]; |
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| 108 | } |
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| 109 | printf "\n"; |
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| 110 | printf "\n"; |
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| 111 | } |
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| 112 | printf "Total profit: %4.2f\n", |
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| 113 | (sum{m in month} productprice*production[m] |
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| 114 | - sum{m in month, o in oils} buyingprices[m,o]*buys[m,o] |
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| 115 | - sum{m in month, o in oils} storagecost*stock[m,o]); |
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| 116 | printf " turnover: %4.2f\n", |
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| 117 | sum{m in month} productprice*production[m]; |
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| 118 | printf " buying costs: %4.2f\n", |
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| 119 | sum{m in month, o in oils} buyingprices[m,o]*buys[m,o]; |
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| 120 | printf " storage costs: %4.2f\n", |
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| 121 | sum{m in month, o in oils} storagecost*stock[m,o]; |
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| 122 | |
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| 123 | |
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| 124 | data; |
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| 125 | |
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| 126 | param : oils : oilhardness := |
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| 127 | VEG1 8.8 |
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| 128 | VEG2 6.1 |
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| 129 | OIL1 2.0 |
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| 130 | OIL2 4.2 |
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| 131 | OIL3 5.0 ; |
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| 132 | |
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| 133 | set month := 1 2 3 4 5 6; |
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| 134 | |
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| 135 | param buyingprices |
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| 136 | |
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| 137 | : VEG1 VEG2 OIL1 OIL2 OIL3 := |
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| 138 | |
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| 139 | 1 110 120 130 110 115 |
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| 140 | 2 130 130 110 90 115 |
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| 141 | 3 110 140 130 100 95 |
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| 142 | 4 120 110 120 120 125 |
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| 143 | 5 100 120 150 110 105 |
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| 144 | 6 90 100 140 80 135 ; |
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| 145 | |
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| 146 | param productprice := 150; |
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| 147 | param storagecost := 5; |
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| 148 | param M := 1000; |
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| 149 | |
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| 150 | end; |
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