Problem: A Company can use 3 different procedures to produce a product, for the production of every product are necessary 3 machines as below:

The numbers relate the hours necessary. every machine is avaible for 50 hours. The profict of the product depends of the procedure that has been used:


Proc 1=7

Proc 2=9

Proc 3=5

this is how I will proceed:


$Xi$= product with procedure i=1,2,3;

max $7x_1+9x_2+5x_3$ subject to:

2$ x_1+ x_2+3 x_3 ≤50$

4 $x_1+2 x_2+3 x_3 ≤50$

3 $x_1+4 x_2+2 x_3 ≤50$


MINIMIZE hours of usage of machine 2 with the obligation than profit must be at least 100

$X_i,_j$ where i=machine and j=proc

min $X_2,_1+X_2,_2+X_2,_3$

subject to




$7(2X_1,_1+4X_2,_1+3X_3,_1)+9(2X_1,_2+2X_2,_2+4X_3,_2)+ 5(3X_1,_3+4X_2,_3+2X_3,_3)≥100$

Is my doing correct for the resolution of the problem?

  • $\begingroup$ Looks good to me. However, I think you can express the second model using the same variables as the first. In particular I don't think you need 9 variables. After all, the number of hours machine 2 is used is still a function of the amount of product. $\endgroup$ – Michael Grant Apr 25 '13 at 15:20
  • $\begingroup$ Actually, I take that back. Your second model is actually not correct. The profit is a function of the amount of product used, not the total number of hours spent by all machines. But that's exactly how you expressed your objective. $\endgroup$ – Michael Grant Apr 25 '13 at 15:28

Here's my proposed improvement to the second model: $$\begin{array}{ll} \text{minimize} & 4 x_1 + 2 x_2 + 3 x_3 \\ \text{subject to} & 2 x_1 + 1 x_2 + 3 x_3 \leq 50 \\ & 4 x_1 + 2 x_2 + 3 x_3 \leq 50 \\ & 3 x_1 + 4 x_2 + 2 x_3 \leq 50 \\ & 7 x_1 + 9 x_2 + 5 x_3 \geq 100 \end{array}$$ Now, you might consider it redundant to constrain machine 2's hours to 50 when we're also minimizing its usage. But what if it is not possible to achieve 100 in profit without exceeding 50 hours on machine 2? If you know for sure that you can hit 100 profit without exceeding the 50-hour limit for machine 2, you can delete that constraint. But just to be safe, I recommend leaving it in.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.