R help - Mar 2005 - Simplex(boot) returning invalid answer

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)

 > 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

On Fri, 11 Mar 2005 09:09:15 -0800, Andrew Stoneman
<stoneman at 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))

to the beginning of the function rather than giving a bad answer for
bad input.

Duncan Murdoch