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Let's say we have the following optimization problem :

Max cx

Subject to $Ax=b$ with $x\geq0$

How can we find the interval of the values for the coefficient $c_{j}$ of the objective where the optimal solution stays always the same ?

Do I have to think of the slope ? I don't get it.

Normally if we imagine the graph, we can say that if we have a line that passes by a point, if we change the slope, the optimal value would stay the same as long as we are between the lines on the consrtaints.

I would appreciate any answer/ guide.

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    Can you please tell us what the objects are? From the wording of your problem and with some guessing, $A$ is an $n\times n$ matrix, $x$ is $n$ vector, $c$ is $n$ vector, cx is $c\cdot x$, $b$ is $n$ vector? Is $A$ invertible? If so, there seems to be at most one admissible $x$? – 2017-02-06
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    Oh your question is a little bit hard lol. Euh for A, not necessarly.. It can be $ n \space x \space n $ or not. – 2017-02-06
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    My question is hard. You don't even know the question you are asking? – 2017-02-06
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    @Jan I don't get it what do you mean by what the objects are, we have an objective function of maximization problem. And I don't know if it the size of the A or even the vector x would make a difference in our theoretical case.. Normally the vector x depends on the matrix A's length – 2017-02-06
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    That is what I am asking. Is $A$ a matrix? What is its size? Is $x$ a vector of the same length? Is $c$ a number (so that $c_{j}$ is a typo) or is $c$ a vector so that your "cx" should in fact be $c\cdot x$? – 2017-02-06
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    Oh sorry now I get it.. Yeah A is a matrix, size is not known, we can say $ 2 \space x \space 3$ with x a vector of 2. or we can say A is $ 5 \space x \space 5 $ with x a vector of 5. c is a vector normally, so if we have x a vector of 3 (x1,x2,x3) we would get (c1,c2,c3).. It can be a point if x has a dimention of 1 – 2017-02-06
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    What makes you think there is anything like the '"values for coefficients $c_{j}$" so that the solution stays the same? Trivially, when you multiply $c$ by a positive constant, the solution stays the same. But nothing more? – 2017-02-06

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You can solve problems like these with the revised simplex method. The solution $x = B^{-1}b$ remains optimal as long as the dual variables are negative: $c_N - c_B B^{-1} N \leq 0$. If you fill in $B$ and $N$ you obtain the condition on $c_j$.

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    Sorry for the late comment, ok So if we have a problem like max 3c1x1 +5c2x2 Subject to x1<=4, x2<=6, 3x1+2x2<=18,x1>=0,x2>=0. How can we determine c1 and c2 so that the optimal solution doesn't change ? – 2017-02-13