# Re: [R] Simplex(boot) returning invalid answer

From: Duncan Murdoch <murdoch_at_stats.uwo.ca>
Date: Mon 14 Mar 2005 - 07:58:50 EST

On Fri, 11 Mar 2005 09:09:15 -0800, Andrew Stoneman <stoneman@otsys.com> 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))