Sensitivity analysis in linear programming Could someone please explain in detailed steps how to apply a sensitivity analysis to such problem:
$$maximize \ \ 2x_1 + 3x_2 \\
s.t. \ \ 4x_1+3x_2≤600 \\ 2x_1+2x_2≤320 \\ 3x_1+7x_2≤840 \\ x_i≥0$$
The goal is it to determine the boundaries of $x_2$.
 A: I am not particularly keen to do your homework for you "in detail". But here is a starting point. The red line is a contour of $2x+3y$. The blue line is $4x+3y=600$; the orange line is $x+y=160$; the green line is $3x+7y=840$. The other lines are $x\ge0,y\ge0$.
The first question to settle is obviously where the allowed area is. It is in fact the area "below" the green, orange and blue lines but in the first quadrant. So the allowed area is the irregular pentagonal area shown in the plot.
Ignoring sensitivities we would maximise $2x+3y$ at the intersection of the green and orange lines, because we want the red line to be as far "upwards" as possible (whilst keeping its slope fixed).
You say nothing in the question about what kind of sensitivity analysis you want. There are a whole range of things you can change. You can "change the RHS", meaning that you shift the constraint lines, whilst keeping their slopes unchanged. Here you can see that the blue line is a "non-binding" constraint - shifting it will not change the optimisation for unless you shift it further to the left than the intersection of the green and orange lines. On the other hand, the green and orange constraints are "binding" - any shift however small will shift the optimisation.
Another sensitivity is to "change the objective function coefficients". In other words you change the coefficients 2, 3 in $2x+3y$. That has the effect of changing the slope of the red line and hence the optimisation.
Finally, of course, you can change the coefficients in the constraints, thus changing the slopes of the blue, green, orange lines. Again small changes to the blue line will make no difference, but small changes to the orange and green lines will.

A: To conduct a sensitivity analysis the final tableau is needed. 
Basis   x1     x2     s1      s2      s3     RHS
x3      0       0      1     -2.375   0.25   50
x1      1       0      0      0.875  -­0.25   70
x2      0       1      0     -­0.375   0.25   90
z       0       0      0      0.625   0.25   410

The variables $x_1,x_2$ and $x_3$ are in the basis. Now it can be analyzed under which condition $x_2$ remains in the basic.
Lower bound for the parameter of $x_2$:
We take the reciprocal of every positive value of the non-basic variables in the $x_2$-row at multiply them by the corresponding objective function values. Then we take the minimum of it. In this case we have only one fraction:
$min\left( \frac{0.25}{0.25} \right)=\frac{0.25}{0.25}=1$
If there would be no positive values the lower bound would be $-\infty$.
Upper bound for the parameter of $x_2$:
We take the reciprocal of every negative value of the non-basic variables in the $x_2$-row at multiply them by the corresponding objective function values. Then we take the maximum of it. In this case we have only one fraction as well:
$max\left( \frac{0.625}{-0.375} \right)=\frac{0.625}{-0.375}=-\frac53$
If there would be no negative values the upper bound would be $\infty$.
To get the final lower/upper bound we take the value of the parameter of the initial tableau (objective function) and substract the identified values above. 
Lower bound $3-1=2$
Upper bound $3-(-\frac53)=3+\frac53=4\frac23$
For further information see here, especially page 10 and 11. If you have any question about it feel free to ask.
