Rhs sensitivity analysis example

Sensitivity Analysis via an example
Consider the following LP
min –x1 – 2 x2
s.t. x1 + x2 = 0
Optimal solution: (x1,x2,x3,x4) = (3,3,0,0) and optimal objective value: z = ‐ 9
The Dual Problem (of the standard form):
max 6 y1 + 3 y2
s.t. y1

Note that we can write the dual problem (since the optimal value is the maximum of the objective
values of the extreme points) as:
z = max { (6+∆)*(‐2) + 3*(0); (6+∆)*(‐1)+3*(‐1)} = max {‐12 ‐ 2∆; ‐9‐∆}
Note that primal and dual problem have the same objective value.
The objective z(∆) as a function of ∆ is:
z(∆)
∆
1. Non‐increasing : Since increasing rhs enlarges the feasible region
of the primal problem and objective cannot get worse
and we are minimizing
2. Convex: The slopes are non‐decreasing (becomes less negative), since the marginal
benefit of having an extra unit of component A can only stay the same or get
worse as we have more and more of it.
3. Piece‐wise linear: Since it is given as the maximum of a number of linear functions.
What is the marginal value (also called shadow price) of the rhs of constraint 1? Currently ∆ = 0, the
slope of z(∆) at ∆=0 is ‐1 which is precisely the current value of optimal dual solution of constraint 1.
How long is this marginal value valid? As long as the current dual solution (y1=‐1,y2=‐1) remains optimal.
Note that when ∆

We can get the above information directly from Excel solver (and also other programs by querying “dual
solutions” and “ranges”).
Here is the Excel model:
x1
x2
0 0 (negative) profit = = ‐1*C3‐2*D3
Required
Component A =1*C3+1*D3 6
Component B = 1*D3 3
Available
The target cell is (negative) profit and the changeable cells are x1 and x2. The constraints are Required