Originally posted by: StudentLucas

I've simplified my problem:

there are a 2d-array with article/price information and a 3d array with article/day/demand information

The exercise is to get the best sales (combination of price and demand). But it doesnt work (there is "no solution"). If I change "==" to "<=" it works but with a wrong solution, the same is if I would use "=>".

Hier is my code, could please somebody check it? Thanks in advance!

{string} Fuel = {"Gasolin", "Diesel"};
{string} Day = {"Monday", "Saturday"};

 // Index for price/demand category
 // 1: regular price, 2: discount price for members
{int} I = {1, 2};

int price[Fuel][I] = [
        [11, 8],
        [14, 10]
];

int demand[Fuel][Day][I] = [
        [[100, 118],[221, 284]],
        [[122, 132],[319, 405]]
];

dvar int+ sales[Fuel][Day];

maximize sum(f in Fuel, d in Day) sales[f][d];

subject to{

        forall(f in Fuel, d in Day, i in I){
                sales[f][d] == price[f][i] * demand[f][d][i];
        }    
}

Hi,

{string} Fuel = {"Gasolin", "Diesel"};
{string} Day = {"Monday", "Saturday"};

 // Index for price/demand category
 // 1: regular price, 2: discount price for members
{int} I = {1, 2};

int price[Fuel][I] = [
        [11, 8],
        [14, 10]
];

int demand[Fuel][Day][I] = [
        [[100, 118],[221, 284]],
        [[122, 132],[319, 405]]
];

dvar int+ sales[Fuel][Day];
dvar boolean thisistheprice[Fuel][Day][I];

maximize sum(f in Fuel, d in Day) sales[f][d];

subject to{

// for a given day and fuel, only one price

forall(f in Fuel, d in Day) sum(i in I) thisistheprice[f][d][i]==1;

        forall(f in Fuel, d in Day){
                sales[f][d] == sum(i in I) price[f][i] * thisistheprice[f][d][i]*demand[f][d][i];
        }    
}

will work better.

regards

https://www.linkedin.com/pulse/ist-mathematische-optimierung-und-wie-kann-sie-helfen-alex-fleischer/