A = {1 .. 2}, B = {3 .. 5},

T = {1 .. maxt}, I = {1 .. maxi};

cost[T,I] << "cost.dat",

hard[I] << "hard.dat";

b[I,T], u[I,T], s[I,T],

Prod[T];

loil{i in I, 1}: b[i,1] - u[i,1] - s[i,1] = -500,

loil{i in I, t in T, t > 1}: s[i,t-1] + b[i,t] - u[i,t] - s[i, t] = 0,

vveg{t in T}: sum{i in A} u[i,t] <= 200,

voil{t in T}: sum{i in B} u[i,t] <= 250,

uhrd{t in T}: sum{i in I} hard[i]*u[i,t] - 6*Prod[t] <= 0,

lhrd{t in T}: sum{i in I} hard[i]*u[i,t] - 3*Prod[t] >= 0,

cont{t in T}: sum{i in I} u[i,t] - Prod[t] = 0,

/* Specify bounds for the variables */

for {i in I, t in T, t= 1000, t<maxt} s[i,t] <= 1000,

for {i in I} s[i,maxt] = 500;

END MODEL

solve Food;

print solution fro Food >> "Food.sol";

quit;

A = {1 .. 2}, B = {3 .. 5},

T = {1 .. maxt}, I = {1 .. maxi};

cost[T,I] << "cost.dat",

hard[I] << "hard.dat";

b[I,T], u[I,T], s[I,T],

Prod[T];

loil{i in I, 1}: b[i,1] - u[i,1] - s[i,1] = -500,

loil{i in I, t in T, t > 1}: s[i,t-1] + b[i,t] - u[i,t] - s[i, t] = 0,

vveg{t in T}: sum{i in A} u[i,t] <= 200,

voil{t in T}: sum{i in B} u[i,t] <= 250,

uhrd{t in T}: sum{i in I} hard[i]*u[i,t] - 6*Prod[t] <= 0,

lhrd{t in T}: sum{i in I} hard[i]*u[i,t] - 3*Prod[t] >= 0,

cont{t in T}: sum{i in I} u[i,t] - Prod[t] = 0,

/* Specify bounds for the variables */

for {i in I, t in T, t<maxt} s[i,t] <= 1000,

for {i in I} s[i,maxt] = 500;

VARIABLES

d[I, T] BINARY;

for {t in T} { ! Formulate second condition

for { i in A } { ! UB on amount of veg. oil refined/month = 200 u[i,t] <= 200*d[i,t], u[i,t] >= 20*d[i,t] },

for { i in B } { ! UB on amount of non-veg. oil refined/month = 250 u[i,t] <= 250*d[i,t], u[i,t] >= 20*d[i,t] },

! Formulate first condition sum{i in I} d[i,t] <= 3,

! Formulate third condition d[1,t] <= d[5,t], d[2,t] <= d[5,t] };

solve FoodB;

print solution for FoodB >> "FoodB.sol";

quit;