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