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Why at the last step, when we have found optimal basic feasible solution which implies also optimal corner point feasible solution, the right-hand values of equations of system (augmented form of constraints) actually coincide with final values of basic variables? For example let

$Z = 3x_1+5x_2\to max$

$x_1 \le 4$

$2x_2 \le 12$

$3x_1+2x_2 \le 18$

At the last step of solving we'll get $Z = 36 - \frac{3}{2}x_4 - x_5$. So as all coefficients are negative then it is the most optimal solution. And from here we can see that $x_4$ and $x_5$ are non-basic variables at this step. While $x_1, x_2, x_3$ are basic. And equations will be in the form:

$x_3 + \frac{1}{3}x_4 - \frac{1}{3}x_5 = 2$

$x_2 + \frac{1}{2}x_4 = 6 $

$x_1 - \frac{1}{3}x_4 + \frac{1}{3}x_5 = 2$

And the final solution for the problem is $(2; 6; 2; 0; 0)$. You can see that all values in solution coincide with values of equations. Can you please explain me, why is it happpening? What is underlying logic here?

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    @lan, which gradient? Where do you obtain this? In the question I was calculating all values by hand and obtained $(2; 6; 2; 0; 0)$, and suddenly noticed that values coincide with right-hand values of last equations. For me it is unclear what happened there. P.S. I also had the question about basis, you can see in my profile. Maybe I can't understand it fully because of lack of understanding the purpose of basis? – 2017-02-26
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    Oh, then your question is much more simple: it's just because $x_4$ and $x_5$ turned out to be zero. So those terms in the three equations vanish, and all that is left is $x_3=2,x_2=6,x_1=2$. So the question is then "why are $x_4,x_5$ zero?"; I think the answer is because the solution is at the vertex corresponding to the second and third inequalities. – 2017-02-26
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    @lan, oh, thank you! Is it always happening? So that all basic variables are expressed through non-basic variables at the end? And can you please tell me where you got $3x_1+2x_2=18$? – 2017-02-26
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    The $3x_1+2x_2=18$ is just the equality version of the third inequality. It corresponds to the fact that the solution is on that "face" of the triangle. Anyway, what's happening is that the non-basic variables get set to zero. These correspond to the "active" equality constraints, which in this case are $2x_2=12$ and $3x_1+2x_2=18$, which is why $x_4$ and $x_5$ turned out to be zero. The basic variables are then those that you actually have to solve for. – 2017-02-26
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    @lan, can you please say where you learnt all of this? You mention about "face" of the triangle, and I can't even suspect from where I can read about this. – 2017-02-26
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    I actually don't know much about linear programming; I had to google a bit to find out what "basic variable" meant, for example. But I do understand Lagrange multipliers, and that's where linear programming gets its start. For instance here, the gradient of $Z$ is $\begin{bmatrix} 3 \\ 5 \end{bmatrix}$ which does not vanish. So the solution must be on the boundary, which is comprised of the three "faces" $x_1=4,2x_2=12,3x_1+2x_2=18$. (I use the word "face" because this in 3D these are the faces of a polytope. They are of course sides of a polygon in 2D.) – 2017-02-26
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    (Cont.) If you perform Lagrange multipliers on each of these faces, you again find that the gradient of the Lagrangian never vanishes, because $\begin{bmatrix} 3 \\ 5 \end{bmatrix}$ is not parallel to $\begin{bmatrix} 1 \\ 0 \end{bmatrix},\begin{bmatrix} 0 \\ 2 \end{bmatrix}$ or $\begin{bmatrix} 3 \\ 2 \end{bmatrix}$. So you restrict to the boundary of the faces *themselves*, and now the boundary is just the three vertices, so the problem is to find the best value of $Z$ among the three vertices. – 2017-02-26
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    (Cont.) Linear programming and similar techniques can be thought of as a shortcut way to do what I just explained. There's more to it (because there's more structure to these problems than there is in general nonlinear optimization) but this is still the idea. – 2017-02-26

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