Question

Consider the following LP

max z=3x1+x2

s.t. −2x1 + x2 ≤ 3

x1 + 2x2 ≤ 5

x1,x2 ≥0

(a) Find the dual (or shadow) prices of the binding constraints

(b) Find the dual (or shadow) prices of the binding “dual” constraints.

Answer 1

Solution using graphical method is following:

x1 is plotted along x-axis and x2 is plotted along y-axis

Feasible region is the shaded region bounded by corner points, as highlighted on the graph.

Corner points are: (5, 0), (0, 2.5)

Maximum Objective function value is 15 at corner (5, 0)

Therefore, optimal solution is:

x1 = 5

x2 = 0

Objective function value = 15

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(a)

Optimal lies on curve (x1 + 2x2 ≤ 5)

Therefore, it is the binding constraint.

Shadow price is determined by increasing the Right Hand Side of the binding constraint by 1 unit and again finding the optimal solution and objective value. Shadow price is equal to the difference in objective value for unit change in Right Hand Side (RHS) value of the constraint.

The new graphical solution after increasing the RHS of the constraint by 1 unit, is as follows:

New objective value = 18

Shadow price = 18 - 15

= 3

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(b)

Dual problem is following:

Minimize W = 3y1+5y2

s.t.

-2y1+1y2 >= 3

1y1+2y2 >= 1

y1, y2 >= 0

It solution by graphical method is following:

In the above graph, y1 is plotted along x-axis and y2 is plotted along y-axis

Optimal solution is:

y1 = 0

y2 = 3

Objective value = 15

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The binding constraint is: -2y1+y2>=3

After increasing the RHS of binding constraint by 1 unit, the new optimal solution is determined as below:

New optimal solution is:

y1 = 0

y2 = 4

New objective value = 20

Shadow price = 20 - 15

= 5