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/* Food Manufacture 2, section 12.2 in
* Williams, "Model Building in Mathematical Programming"
*
* Sebastian Nowozin <nowozin@gmail.com>
*/
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;
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