Question

1. The optimal solution of the linear programming problem is at the intersection of constraints 1 and 2.

Please answer the following questions by using graphical sensitivity analysis.

Max 2x₁ + x₂

s.t. 4x₁ +1x₂ ≤8

4x₁ +3x₂ ≤12

  1x₁ +2x₂ ≤6

x₁ , x₂ ≥ 0

A. Over what range can the coefficient of x₁ vary before the current solution is no longer optimal?

B. Over what range can the coefficient of x₂ vary before the current solution is no longer optimal?

C. Compute the dual price for the first constraint.

Answer 1

Solution using Graphical method is following:

A) Optimality range of coefficient of X1 is such that slope of objective function is between the slope of constraint 1 and 2 (because these are binding constraints)

Therefore, range of coefficient of X1 is from __ 1.33 __  (=1*4/3) to __ 4 __  (=1*4/1)

B) Optimality range of coefficient of X2 is such that slope of objective function is between the slope of constraint 1 and 2 (because these are binding constraints)

Therefore, range of coefficient of X2 is from __ 0.50 __  (=2*1/4) to __ 1.5 __  (=2*3/4)

C) If we increase the RHS of first constraint by 1 unit, then the objective function shifts to the right.

The new point of intersection between C1 and C2 is X2 = (12-9)/(3-1) = 1.5 and X1 = (9-1.5)/4 = 1.875

New objective value = 2*1.875+1.5 = 5.25

We see that the increase in objective value = 5.25 - 5 = 0.25

Therefore, dual price of first constraint is 0.25