Production Optimization using MILP

I try to solve a manufacturing problem. I got K products over T periods of time with given demand d(t,k) for every time period and every product. I got only one machine with a given capacity in each timeperiod t c(t) Setuptimes for each product ts(k), productiontimes tp(k), Setupcosts s(k) and holdingcosts h(k). I want to minimize the costfunction:

    -> Sum[1:T](Sum[1:K](s * y + h(k)*I(t,k)))

where y is a binary variable and I(t,k) is the stored amount of units which we did not used to fullfill demand. Q(t,k) will be the amount of produced products. y,I,Q are to optimize as integers using intlinprog()

I got the functions and all conditions. The only thing where i got problems is, to formulate the condition of

 -> I(t-1,k) + Q(t,k) - I(t,k) = d(t,k)

which has to consider the left stock from the previous period. and i don't know how to refer to the previous persiod.

       for ii = 1:T
            for jj = 1:K
                xtemp = clearer2;
                xtemp(ii,jj) = -1;
                xtemp2 = clearer3;
                xtemp2(ii,jj) = 1;
                xtemp = sparse([clearer12;xtemp(:);xtemp2(:)]'); % Change to sparse row
                Aeq(counter,:) = xtemp; % Fill in row
                if ii > 1  
                    beq(counter) = d(ii,jj); % Problem -> can not subtract I(t-1,k)
                else
                    beq(counter) = d(ii,jj); % first Period no Stock 
                end
                counter = counter + 1;
            end
        end

Answer 1

[Christoph Decker]

Thank you Alan. I figured it out myself today. The problem was to implement the I(t-1,k) to the linear equality condition. I wanted to have the condition

 I(t-1,k) + Q(t,k) - I(t,k) = d(t,k)

Which will set the previous stock I(t-1,k) plus the amount of units produced in this period Q(t,k) minus the left amount of units, which will be stored, equal to the demand of the current period d(t,k) I solved the problem:

 % Demand is satisfied
for ii = 1:T
    for jj = 1:K
        xtemp = clearer2;
        xtemp(ii,jj) = -1; % Every I(t,k) * -1
        xtemp2 = clearer3;
        xtemp2(ii,jj) = 1; % Every Q(t,k)
        if(ii > 1) 
            xtemp(ii - 1,jj) = 1; % Every I(t-1,k)
        else
            xtemp(1,jj) = -1; % First period there is no I(t-1,k)
        end
        xtemp = sparse([clearer12;xtemp(:);xtemp2(:)]'); % Change to sparse row
        Aeq(counter,:) = xtemp'; % Fill in 
        beq(counter) = d(ii,jj); 
        counter = counter + 1;
    end
end

Answer 2

[Alan Weiss]

What do you think the decision variables are, namely, the variables that the solver will change to minimize your cost? Do they include I(t,k) and Q(t,k)? If so, I believe that you can write a set of linear equality constraints to represent your condition.

If not, then what are your decision variables?

Alan Weiss

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