# Dual-Primal / Simplex

• Aug 7th 2009, 07:54 AM
MAX09
Dual-Primal / Simplex
Hi

I have solved the PRIMAL problem given in the attached image using in the Big M/ Charne's Penalty method. I am pretty sure that the solution obtained is correct because I have verified with the software "TORA".

Now, I would like to know how to read the solution of the DUAL of this problem. Can someone help me with that please?

Thanks,
Max,
• Aug 11th 2009, 05:25 AM
MAX09
A minor error
Hi,

I fell into confusion because I was stupid enough to overlook a small mistake in the input data. This led me into believing that there was a big doubt in my understanding of the dual - primal concept. But I've realized my mistake and have posted the correct output result in the attachment.

As for people who would need a guidance in reading the solution of the dual from the primal problem, here are the steps I usually work out.

We start with the dual constraints corresponding to the starting solution of the primal. In this case, we have the 3 artificial variables A1, A2 and A3 as the starting solution variables of the primal.

The corresponding dual constraints are, x1<= M, x2<=M and x3<=M

Now using the property of primal-dual relationship, we equate the co-efficients of these three variables in the optimal primal table.

x1 -M = -M => x1 =0
x2 -M = 100-M => x2 = 100
x3 -M = 230-M => x3 = 230.

Hence the solution of the dual would be 0,100 and 230.

Hope it helped someone :)
• Aug 12th 2009, 07:42 AM
MAX09
Quote:

Originally Posted by MAX09
Hi,

I fell into confusion because I was stupid enough to overlook a small mistake in the input data. This led me into believing that there was a big doubt in my understanding of the dual - primal concept. But I've realized my mistake and have posted the correct output result in the attachment.

As for people who would need a guidance in reading the solution of the dual from the primal problem, here are the steps I usually work out.

We start with the dual constraints corresponding to the starting solution of the primal. In this case, we have the 3 artificial variables A1, A2 and A3 as the starting solution variables of the primal.

The corresponding dual constraints are, x1<= M, x2<=M and x3<=M

Now using the property of primal-dual relationship, we equate the co-efficients of these three variables in the optimal primal table.

x1 -M = -M => x1 =0
x2 -M = 100-M => x2 = 100
x3 -M = 230-M => x3 = 230.

Hence the solution of the dual would be 0,100 and 230.

Hope it helped someone :)

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