Question

Duality Theory :

Consider the following LP problem:

Maximize Z = 2x₁ + x₂ - x₃

subject to

2x₁ + x2+ x₃ ≤ 8

4x₁ +x₂ - x₃ ≤ 10

x₁ ≥ 0, x₂ ≥ 0, x₃ ≥ 0.

(a) Find the dual for this LP

(b) Graphically solve the dual of this LP. And interpret the economic meaning of the optimal solution of the dual.

(c) Use complementary slackness property to solve the max problem (the primal problem).

Clearly answer each part of the question with Excel file thanks

Unfortunately it is excel is required. thanks for asking though!

unfortunately excel is required. thanks for asking though!

Answer 1

Part (a) and (c) has nothing to do with Excel. Part (c) is just arithmetic. Part (a) is writing the dual. So, if the question asks for Excel for these two parts also, please simply paste the answers in the Excel file.

Only the part (b) (the graph) can be drawn in Excel.

(a)

Min W = 8 y1 + 10 y2
s.t.
2 y1 + 4 Y2 >= 2
1 y1 + 1 y2 >= 1
1 y1 - 1 y2 >= -1
y1, y2 >= 0

(b)

Steps to develop the graph is Excel:

Develop the following table in Excel and use the scatter plot option to develop the chart as shown:

The cost line, moving in minimization direction (towards the origin), leaves the feasibility area at a point (1, 0). So, the optimal solution of the dual is as follows:

y1 = 1
y2 = 0

Min. W = 8*1 + 10*0 = 8

The optimal solution of the dual represents the dual prices of the corresponding primal constraints. So, if the RHS of the first primal constraint is increased by one unit, the corresponding increase in the 'Z' value is $1. Similarly, if the RHS of the second primal constraint is increased by one unit, the corresponding increase in the 'Z' value is $0.

(c)

The dual variable y1 = 1; using complementary slackness, the first, primal constraint is an equality

So,

2x1 + x2 + x3 = 8 --------(1)

The third dual constraint has a slack, using complementary slackness, x3 will be zero in primal

So,

x3 = 0

The second constraint of primal will have slack for y2 = 0 i.e. 4x1 + x2 - x3 < 10

Making x3 equal to zero,

2x1 + x2 = 8
4x1 + x2 < 10
Also, Min W = Max Z = 2x1 + x2 = 8

We can set x1 = 0 and x2 = 8 to satisfy them. So, the optimal solution is x1 = x3 = 0 and x2 = 8

and Max Z = Min W = 8