# [R] simplex

arun smartpink111 at yahoo.com
Wed Nov 13 14:19:00 CET 2013

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Hi Sven,
May be this helps:

If you look into ?simplex.object

"

solved This indicates whether the problem was solved.  A value of
‘-1’ indicates that no feasible solution could be found.  A
value of ‘0’ that the maximum number of iterations was
reached without termination of the second stage.  This may
indicate an unbounded function or simply that more iterations
are needed. A value of ‘1’ indicates that an optimal solution
has been found.
"

s1 <- simplex(a=price * -1, A1=A, b1=b, A2=A, b2=b2, maxi=T) # This cant find a solution
s1\$solved
# -1

A.K.

Dear All,

I do have a problem with a linear optimisation I am trying to achieve:

library(boot)
price <- c(26.93, 26.23, 25.64, 25.97, 26.12, 26.18, 26.49, 27, 27.32, 27.93, 27.72, 27.23)
A <- matrix(0, nrow=length(price)+1, ncol=length(price))
A[1,] <- 1
for(i in 1:length(price)) A[i+1,i] <- 1
b <- c(864000.01, 288000, 288000, 288000, 288000, 288000, 288000, 288000, 288000, 288000, 288000, 288000, 288000)
b2 <- c(864000, 216000, 216000, 216000, 216000, 216000, 216000, 216000, 216000, 216000, 216000, 216000, 216000)

simplex(a=price, A1=A, b1=b) # This works
simplex(a=price, A1=A, b1=b, maxi=T) # This is the maxing function and works as well
simplex(a=price * -1, A1=A, b1=b, A2=A, b2=b2, maxi=T) # This cant find a solution

The result I would expect is that it picks the 4
highest(note I multiplied "price" with -1) numbers and multiplies them
with constraint in b2.

For example this works:

library(boot)
price <- c(1, 2, 3 , 4)
A <- matrix(0, nrow=length(price)+1, ncol=length(price))
A[1,] <- 1
for(i in 1:length(price)) A[i+1,i] <- 1
b <- c(8.01, 2.5, 2.5, 2.5, 2.5)
b2 <- c(8, 2, 2, 2, 2)

simplex(a=price, A1=A, b1=b)
simplex(a=price, A1=A, b1=b, maxi=T)
simplex(a=price * -1, A1=A, b1=b, A2=A, b2=b2, maxi=T)

I cant see a logical difference between the two, why would it not find a solution for the first problem?

Thank you.

Sven

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