Linear programming in R using lpsolve - r

Question

I'm trying to solve a linear programming problem in R using lpsolve package.
Here is the problem:
Here is the sample in R for reproducible example:
library("lpSolve")
a <- matrix(c(1,2,5,
1/2,1,3,
1/5,1/3,1),nrow=3,byrow=T)
#
f.obj <- c(1,0,0,0)
f.con <- matrix (c(
1,1,-a[1,2],0, #Contraint 1 for a12
1,-1,a[1,2],0, #Contraint 2 for a12
1,1,0,-a[1,3], #Contraint 1 for a13
1,-1,0,a[1,3], #Contraint 2 for a13
1,0,1,-a[2,3], #Contraint 1 for a23
1,0,-1,a[2,3], #Contraint 2 for a23
0,1,1,1, #Contraint 3
0,1,0,0, #Constraint 4
0,0,1,0, #Constraint 4
0,0,0,1 #Constraint 4
), nrow=10, byrow=TRUE)
f.dir <- c(rep("<=",6), "=",rep(">",3))
f.rhs <- c(rep(1,6),1,rep(0,3))
g <- lp ("max", f.obj, f.con, f.dir, f.rhs)
g$solution
I'm able to solve this manually for a small problem, what if I had a 7 X 7 or a n x n matrix a. How would I specify the constraint 1 and 2, especially I'm struggling to define the constraint as it relates to a[i,j]?
a = matrix(
c(1,4,9,6,6,5,5,
1/4,1,7,5,5,3,4,
1/9,1/7,1,1/5,1/5,1/7,1/5,
1/6,1/5,5,1,1,1/3,1/3,
1/6,1/5,5,1,1,1/3,1/3,
1/5,1/3,7,3,3,1,2,
1/5,1/4,5,3,3,1/2,1
),nrow = 7,byrow =T)
the solution to the above matrix is 0.986 0.501 0.160 0.043 0.060 0.060 0.1 0.075 Any help would be greatly appreciated.

Answer 1

Have updated to incorporate revised constraint 4 and have made some minor code improvements.
Assuming the constraint matrix in the question is correct, this uses combn to iterate over all i < j setting the appropriate elements. Note that x[1] is the value of i and x[2] is the value of j in f. make_cons returns the constraint matrix in the same order as shown in the question but the rbind line in make_cons could be simplified to rbind(cons1, cons2, cons3, cons4) if it were OK to use such order.
make_cons <- function(a) {
n <- nrow(a)
f <- function(x) replace(numeric(n), x, c(1, -a[x[1], x[2]]))
cons1 <- cbind(1, t(combn(1:n, 2, f)))
cons2 <- cbind(1, -cons1[, -1])
cons3 <- c(0, rep(1, n))
cons4 <- cbind(0, diag(n))
rbind(t(matrix(rbind(t(cons1), t(cons2)), ncol(cons1))), cons3, cons4)
}
# test
# a and f.con from question
a <- matrix(c(1, 0.5, 0.2, 2, 1, 0.333333333333333, 5, 3, 1), 3)
f.con <- matrix(c(1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 1, -1, 1, -1, 0, 0,
1, 1, 0, 0, -2, 2, 0, 0, 1, -1, 1, 0, 1, 0, 0, 0, -5, 5, -3,
3, 1, 0, 0, 1), 10)
all.equal(f.con, make_cons(a), check.attributes = FALSE)
## [1] TRUE

Answer 2

here is one possibility which uses for loops.
As I mentioned in the commenst, I think that you got condition (4) wrong. Here is my suggestion.
My idea is to first set up a matrix for constraints (4), then for constraint (3)
and then add constraints (2) and (1) in a loop. Note that, in the beginning, I do not consider the column corresponding to \mu. I will add this column in the end.
n<- nrow(a)
f.cons<- diag(n)
f.cons<- rbind(f.cons, rep(1,n))
This sets up a matrix corresponding to constraints (4) (first n rows) and constraint (3). Now I add rows to this matrix, using loops and the command rbind.
for(i in 1:(n-1)){
for(j in (i+1): n){
x<- rep(0, n)
x[i]<- 1 #x corresponds to (1)
x[j]<- -a[i,j]
y<- -x #y corresponds to (2)
f.cons<- rbind(f.cons, rbind(x, y))
}
}
So far, I have ignored the first column, which corresponds to \mu.
I add it with these two simple lines:
f.cons<- cbind(rep(1, nrow(f.cons)), f.cons)
f.cons[1:(n+1), 1]=0
Note that in my matrix f.cond the first n+1 lines correspond to constraints (3) and (4)!