[R] Simplex function in R
Gabor Grothendieck
ggrothendieck at gmail.com
Thu Dec 11 13:14:54 CET 2008
Try the pseudo inverse:
m <- rbind(c(1, 1, 1), c(1, 0, 1), c(0, 1, 0))
b <- c(5, 2, 3)
library(MASS)
ginv(m) %*% b
On Thu, Dec 11, 2008 at 2:20 AM, Chris Line <chris.line at tibra.com.au> wrote:
> I have a set of linear equations and would like to find any feasible solution. A simplex solution works in Case 1 below, but not in Case 2. I would be grateful for any help.
>
> Case 1: Find any feasible solution for the set of linear equations:
> a + b + c = 5
> a + b + 0c = 4
> 0a + b + c = 4
>
> Solution - a feasible (and unique) solution is a=1, b=3, c=1.
>
> The following R code returns a feasible solution:
> A3M = matrix(c(1,1,0,1,1,1,1,0,1),nrow=3)
> b3M = matrix(c(5,4,4),ncol=1)
> A1M = matrix(c(1,0,0,0,1,0,0,0,1),nrow=3)
> b1M = matrix(c(10,10,10),ncol=1)
> AM = matrix(c(1,1,1),nrow=1)
> simplex(a = AM, A1 = A1M, b1 = b1M, A2 = NULL, b2 = NULL, A3 = A3M, b3 = b3M, maxi = TRUE)
>
>
> Case 2: Find any feasible solution for the set of linear equations
> a + b + c = 5
> a + 0b + c = 2
> 0a + b + 0c = 3
>
> Solution - one feasible solution of many is a=1, b=3, c=1. There are infinite possible solutions in Case 2.
>
> However, the following R code fails to return any feasible solution:
> A3M = matrix(c(1,1,0,1,0,1,1,1,0),nrow=3)
> b3M = matrix(c(5,2,3),ncol=1)
> A1M = matrix(c(1,0,0,0,1,0,0,0,1),nrow=3)
> b1M = matrix(c(10,10,10),ncol=1)
> AM = matrix(c(1,1,1),nrow=1)
> simplex(a = AM, A1 = A1M, b1 = b1M, A2 = NULL, b2 = NULL, A3 = A3M, b3 = b3M, maxi = TRUE)
>
> The code returns the error:
> Error in A.out[, basic] <- iden(M) : subscript out of bounds
>
> Am I using the Simplex function incorrectly? There may be a better way to approach the problem of finding a feasible solution.
>
> Cheers,
> Chris.
>
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