LINEAR PROGRAMMING ASSIGNMENT Case Solution
Problem 4
We are presented with:
z = 10 + x2 + x4âˆ’ w2 x1 = 1 â€“
x2âˆ’ w2 x3 = 3 âˆ’ 3x2 + 4x4 + w2
w3 = 5 + x2âˆ’ x4âˆ’ w2
a).
The entering variables for the above standard form problem would be x1. The leaving variables using the minimum ratio test would be:
x2âˆ’ w2 x3 +3×2 +4×4+w2 = 3 Â ————— (3/3 = 1)
w3 â€“ x2 +x4 +w2 = 5 ——————-(5/1=5)
The leaving variable with the minimum ratio would be w2.
b).
If Blandâ€™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
- t.
x1âˆ’ x2 + 2x4âˆ’ x5 = 4
2x1 + x4âˆ’ 4x5 â‰¥âˆ’4 x2 + 3x3 + 2x4 + x5 = 2
x1,x2,x3,x4,x5 â‰¥ 0
Rearranging each constraint, so that the right hand side is equal to 0.
x1âˆ’ x2 + 2x4âˆ’ x5 â€“ 4 = 0
2x1 + x4âˆ’ 4x5 â€“s1 âˆ’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
+ Î»1 (x1âˆ’ x2 + 2x4âˆ’ x5 â€“ 4)
+ Î»2 (2x1 + x4âˆ’ 4x5 â€“s1 âˆ’4 x2 + 3x3 + 2x4 + x5 â€“ 2)
This is the formulation of the duality and finally we rewrite the objective function as:
Minimize: 4(a-b) Î»1 + 3(a+b)Î»2-4a-4b
Subject to:
-16/4Î»1- 20/4Î»2 = -20/4
4 Î»1 – 5Î»2 = 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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