LINEAR PROGRAMMING ASSIGNMENT Case Solution & Answer

LINEAR PROGRAMMING ASSIGNMENT Case Solution

Problem 4

We are presented with:

z = 10 + x2 + x4− w2 x1 = 1 –

x2− w2 x3 = 3 − 3x2 + 4x4 + w2

w3 = 5 + x2− x4− w2

a).

The entering variables for the above standard form problem would be x1. The leaving variables using the minimum ratio test would be:

x2− w2 x3 +3×2 +4×4+w2 = 3  ————— (3/3 = 1)

w3 – x2 +x4 +w2 = 5 ——————-(5/1=5)

The leaving variable with the minimum ratio would be w2.

b).

If Bland’s rule is used, then x1 and w2 would be chosen again.

Problem 5

The answer for this part is as follows:

Solution
Â		(a)	(b)	( c)
i	Feasible basic solution	Yes	Yes	Yes
ii	Optimal basic solution	No	No	No
iii	unbounded problem	Unknown	Unknown	Unknown
iv	infeasible problem	No	No	No

Problem 6

The initial LP model is :

Min x1 + 3x2 + x3 + 4x4

1. t.

x1− x2 + 2x4− x5 = 4

2x1 + x4− 4x5 ≥−4 x2 + 3x3 + 2x4 + x5 = 2

x1,x2,x3,x4,x5 ≥ 0

Rearranging each constraint, so that the right hand side is equal to 0.

x1− x2 + 2x4− x5 – 4 = 0

2x1 + x4− 4x5 –s1 −4 x2 + 3x3 + 2x4 + x5 – 2= 0

Then we define the non-negative dual variable for each inequality constraint and unrestricted dual variable for each equality constraint,

Min x1 + 3x2 + x3 + 4x4

+ λ1 (x1− x2 + 2x4− x5 – 4)

+ λ2 (2x1 + x4− 4x5 –s1 −4 x2 + 3x3 + 2x4 + x5 – 2)

This is the formulation of the duality and finally we rewrite the objective function as:

Minimize: 4(a-b) λ1 + 3(a+b)λ2-4a-4b

Subject to:

-16/4λ1- 20/4λ2 = -20/4

4 λ1 – 5λ2 = 5

The above formulation is optimal for this problem.

Problem 7

a)

For the values of a= 1 and b= 1 we will have a unique optimal solution.

b)

For the values of a=1 and b= 0 we will have many optimal solutions.

c)

For the values of a= 0 and b= 0 we will have no feasible solution.

d)

For the values of a=0 and b= 1, the solution would be unbounded…..

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