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