From: Andrew Stoneman <stoneman_at_otsys.com>

Date: Sat 12 Mar 2005 - 04:09:15 EST

> ubVector

[1] 2 1 -1

R-help@stat.math.ethz.ch mailing list

https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html Received on Mon Mar 14 10:34:06 2005

Date: Sat 12 Mar 2005 - 04:09:15 EST

* > ubMatrix <- matrix(c(1,1,-1,0,-1,-1), 3, 2)
*

> ubVector <- c(2,1,-1)

> objective <- c(0,1)

[1,] 1 0 [2,] 1 -1 [3,] -1 -1

> ubVector

[1] 2 1 -1

> smplx <- simplex( a = objective, A1 = ubMatrix, b1 = ubVector)

> smplx$solved

[1] 1

> smplx$soln

x1 x2

0 0

> ubMatrix %*% smplx$soln <= ubVector

[,1]

[1,] TRUE [2,] TRUE [3,] FALSE

The correct answer to the problem, which also has a value of 0, but satisfies the constraints is [1, 0] :

> ubMatrix %*% c(1,0) <= ubVector

[,1]

[1,] TRUE [2,] TRUE [3,] TRUE

Any ideas what's going on here? Am I missing something obvious? Any advice on how I might work around this problem would be greatly appreciated.

Andrew

R-help@stat.math.ethz.ch mailing list

https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html Received on Mon Mar 14 10:34:06 2005

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