1 | /* Food Manufacture 1, section 12.1 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 | |
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25 | /* Actual amount of output oil produced in each month */ |
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26 | var production{m in month} >= 0; |
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27 | var useoil{m in month, o in oils} >= 0; |
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28 | |
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29 | maximize totalprofit: |
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30 | sum{m in month} productprice*production[m] |
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31 | - sum{m in month, o in oils} buyingprices[m,o]*buys[m,o] |
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32 | - sum{m in month, o in oils} storagecost*stock[m,o]; |
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33 | |
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34 | /* Constraints */ |
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35 | |
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36 | /* 1. Starting stock */ |
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37 | s.t. startstock{o in oils}: |
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38 | stock[1,o] = 500; |
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39 | s.t. endstock{o in oils}: |
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40 | stock[6,o] + buys[6,o] - useoil[6,o] >= 500; |
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41 | |
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42 | /* 2. Stock constraints */ |
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43 | s.t. stocklimit{m in month, o in oils}: |
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44 | stock[m,o] <= 1000; |
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45 | |
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46 | s.t. production1{m in month, o in oils}: |
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47 | useoil[m,o] <= stock[m,o] + buys[m,o]; |
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48 | s.t. production2{m1 in month, m2 in month, o in oils : m2 = m1+1}: |
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49 | stock[m2,o] = stock[m1,o] + buys[m1,o] - useoil[m1,o]; |
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50 | |
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51 | s.t. production3a{m in month}: |
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52 | sum{o in oils} oilhardness[o]*useoil[m,o] >= 3*production[m]; |
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53 | s.t. production3b{m in month}: |
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54 | sum{o in oils} oilhardness[o]*useoil[m,o] <= 6*production[m]; |
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55 | |
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56 | s.t. production4{m in month}: |
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57 | production[m] = sum{o in oils} useoil[m,o]; |
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58 | |
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59 | /* 3. Refining constraints */ |
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60 | s.t. refine1{m in month}: |
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61 | useoil[m,"VEG1"]+useoil[m,"VEG2"] <= 200; |
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62 | s.t. refine2{m in month}: |
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63 | useoil[m,"OIL1"]+useoil[m,"OIL2"]+useoil[m,"OIL3"] <= 250; |
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64 | |
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65 | solve; |
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66 | |
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67 | for {m in month} { |
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68 | printf "Month %d\n", m; |
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69 | printf "PRODUCE %4.2f tons, hardness %4.2f\n", production[m], |
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70 | (sum{o in oils} oilhardness[o]*useoil[m,o]) / (sum{o in oils} useoil[m,o]); |
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71 | |
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72 | printf "\tVEG1\tVEG2\tOIL1\tOIL2\tOIL3\n"; |
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73 | printf "STOCK"; |
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74 | printf "%d", m; |
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75 | for {o in oils} { |
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76 | printf "\t%4.2f", stock[m,o]; |
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77 | } |
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78 | printf "\nBUY"; |
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79 | for {o in oils} { |
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80 | printf "\t%4.2f", buys[m,o]; |
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81 | } |
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82 | printf "\nUSE"; |
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83 | printf "%d", m; |
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84 | for {o in oils} { |
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85 | printf "\t%4.2f", useoil[m,o]; |
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86 | } |
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87 | printf "\n"; |
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88 | printf "\n"; |
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89 | } |
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90 | printf "Total profit: %4.2f\n", |
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91 | (sum{m in month} productprice*production[m] |
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92 | - sum{m in month, o in oils} buyingprices[m,o]*buys[m,o] |
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93 | - sum{m in month, o in oils} storagecost*stock[m,o]); |
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94 | printf " turnover: %4.2f\n", |
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95 | sum{m in month} productprice*production[m]; |
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96 | printf " buying costs: %4.2f\n", |
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97 | sum{m in month, o in oils} buyingprices[m,o]*buys[m,o]; |
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98 | printf " storage costs: %4.2f\n", |
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99 | sum{m in month, o in oils} storagecost*stock[m,o]; |
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100 | |
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101 | |
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102 | data; |
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103 | |
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104 | param : oils : oilhardness := |
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105 | VEG1 8.8 |
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106 | VEG2 6.1 |
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107 | OIL1 2.0 |
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108 | OIL2 4.2 |
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109 | OIL3 5.0 ; |
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110 | |
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111 | set month := 1 2 3 4 5 6; |
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112 | |
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113 | param buyingprices |
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114 | |
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115 | : VEG1 VEG2 OIL1 OIL2 OIL3 := |
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116 | |
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117 | 1 110 120 130 110 115 |
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118 | 2 130 130 110 90 115 |
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119 | 3 110 140 130 100 95 |
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120 | 4 120 110 120 120 125 |
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121 | 5 100 120 150 110 105 |
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122 | 6 90 100 140 80 135 ; |
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123 | |
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124 | param productprice := 150; |
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125 | param storagecost := 5; |
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126 | |
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127 | end; |
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