On Fri, 11 Mar 2005 09:09:15 -0800, Andrew Stoneman <[EMAIL PROTECTED]> wrote :
>In trying to use simplex() from the boot package, I have run into a >situation that doesn't seem like it should be possible. It is claiming >that it has solved the LP, but returns a vector of all zeros, which >does not satisfy the constraints I passed in. A small example: > > > ubMatrix <- matrix(c(1,1,-1,0,-1,-1), 3, 2) > > ubVector <- c(2,1,-1) > > objective <- c(0,1) > > > ubMatrix > [,1] [,2] >[1,] 1 0 >[2,] 1 -1 >[3,] -1 -1 > > ubVector >[1] 2 1 -1 > > > smplx <- simplex( a = objective, A1 = ubMatrix, b1 = ubVector) You missed this in the help page: > b1: A vector of length 'm1' giving the right hand side of the > "<=" constraints. This argument is required if 'A1' is given > and ignored otherwise. All values in 'b1' must be > non-negative. The reason for this requirement is that the origin should be a feasible solution; that's where the algorithm starts. It's been a while since I looked at this stuff so I forget if there's an easier transformation, but one way to solve the problem you're interested in (-x-y <= -1) is to multiply through by -1 giving (x + y >= 1), i.e. ubMatrix <- matrix(c(1,1,-1,0), 2, 2) ubVector <- c(2,1) lbMatrix <- matrix(c(1,1), 1, 2) lbVector <- 1 objective <- c(0,1) simplex(a = objective, A1 = ubMatrix, b1 = ubVector, A2 = lbMatrix, b2 = lbVector) which gives the answer you were looking for. I suppose you might suggest to the maintainer to add stopifnot(all(c(b1, b2, b3) >= 0)) to the beginning of the function rather than giving a bad answer for bad input. Duncan Murdoch ______________________________________________ [email protected] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
