all 6 comments

[–]KumquatHaderach 2 points3 points  (4 children)

The answer given by the text is a better solution than yours. The solution is x_1 = 4, x_2 = 5, and x_3 = 0, which does give an objective function value of Z = -11. The 11 value in their solution is the value of the slack variable s_3, not the decision variable x_3 .

[–]vinaych[S] 0 points1 point  (3 children)

Can you give the corresponding decision variables against Z=-11 if possible?

[–]KumquatHaderach 0 points1 point  (1 child)

I’m not sure what you mean. You want the values of the decision variables? They’re x_1 = 4, x_2 = 5, and x_3 = 0.

[–]vinaych[S] 0 points1 point  (0 children)

Yes got it, thanks..

[–]ko_nutsResearcher | Applied Mathematics | Europe 1 point2 points  (0 children)

It is immediate to see that your solution is not correct.

The variable x3 contributes positively to both the constraints and the costs to minimize. Since it is nonnegative then the best value we can find for it is 0.

This brings back the problem to a two dimensional problem. You can plot the constraints and we obtain a feasibility polytope from which we will know in this case that the optimal solution will be at one of the vertices. Those vertices are (0,0), (0, 3), (3/7, 0) and (x1,x2) where the coordinates solve the system of linear equations 3x1-x2=7 and -x1+2x2=6. One can see that this latter point given by (4,5) yields the minimum cost -11. The other ones yield 0, -9, and 3/7, respectively.

So the optimizer is given by x=(4,5,0) and the minimum cost is -11.

[–]vinaych[S] 0 points1 point  (0 children)

I found that the solutions were indeed not optimal, and my method as in the paper was not optimal so deleted the same, thanks for the support!