% Coefficients for the cost function (cost per ton of each ore)
f = [27; 25; 32; 22; 20; 24];
 
% Metal content inequalities (if there are specific minimum content requirements)
Aeq = [1 1 1 1 1 1];   % Equality constraint: sum of ore = 1 ton
beq = 1;               % Total of ores should sum to 1 ton
 
% Impurities constraint (less than or equal to 50%)
A_ineq = [40 15 30 50 35 29];  % Impurities
b_ineq = 50;                   % Impurity threshold
 
% Metal content constraints (optional: set minimum required metal content if needed)
A = [-19 -43 -17 -20 0 -12; 
     -15 -10 0 -12 -24 -18; 
     -12 -25 0 0 -10 -16; 
     -14 -7 -53 -18 -31 -25];
 
% Lower bounds and upper bounds for decision variables
lb = zeros(6,1);  % No negative ore amounts
ub = ones(6,1);   % Each ore amount is between 0 and 1 ton
 
% Solve the LP problem using glpk
ctype = "S";  % Equality constraint for total tonnage
vartype = "C"; % Continuous variables
 
% Now run the solver
[x, z, exitflag] = glpk(f, [A_ineq; Aeq], [b_ineq; beq], lb, ub, ctype, vartype, 1);
 
% Check if a valid solution is found
if exitflag == 180
    disp('Optimal amounts of ores:');
    disp(x);
    disp('Minimum cost per ton of alloy:');
    disp(z);
else
    disp('No feasible solution found');
end
 

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