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- p := 3
- sum (i in 1..10) x(i,i) = p !podmienka supermarketov je p
- sum (i in 1..10) a(i)*x(i,i) <= 40000 !podmienka rozpoctu pre zakupenie pozemkov
- forall (i in 1..10) sum (j in 1..10) x(i,j) = 1 !ludia z kazdej obce mozu chodit maximalne do jedneho supermarketu
- forall (i in 1..10) x(i,i) >= sum (j in 1..10) x(j,i) - x(i,i)*10 !ludia z kazdej obce mozu chodit len do obce, kde uz je postaveny
- !supermarket (supermarkety su na x(i,i) takze ak je x(i,i) rovne 0,
- !tak cely i-ty stlpec je rovny 0)
- maximise (sum (i in 1..10, j in 1..10) x(i,j)*c(i,j)) !maximalizovat pocet zakaznikov
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