On Wed, May 23, 2012 at 10:56:43PM -0700, kylmala wrote:
> Hi,
>
> and thanks for replying this. Yes, you are right that the term
> min(24/bb,26/cc) is actually min(bb/24,cc/26) - my mistake. But still I
> don't get it. If the objective function is
>
> #min(m1,m2,m3)
> f.obj <- c(1, 1, 1)
>
> #and we now that
> # a+aa<=15
> # b+bb+bbb<=35
> # cc+ccc<=40
>
> # so
>
> # 1a + 0b + 0c + 1aa + 0bb + 0cc + 0aaa + 0bbb + 0ccc <=15
> # 0a + 1b + 0c + 0aa + 1bb + 0cc + 0aaa + 1bbb + 0ccc <=35
> # 0a + 0b + 0c + 0aa + 0bb + 1cc + 0aaa + 0bbb + 1ccc <=40
> # and the matrix of numeric constraint coefficients is
>
> C = matrix(nrow=3, data=c(1,0,0,0,1,0,0,0,0,1,1,0,0,0,1,0,1,0,0,0,1))
>
> #Constraits are:
>
> f.rhs <- c(15, 35, 40)
>
> #and
>
> f_dir = c("<=", "<=", "<=")
>
> Because the command lp is form lp ("max", f.obj, f.con, f.dir, f.rhs), how
> can I get those other constraints in
>
> m1 <= a/10
> m1 <= b/11
> m2 <= aa/13
> m2 <= bb/12
> m2 <= cc/10
> m3 <= bb/24
> m3 <= cc/26
Hi.
There are 12 variables, namely a, b, c, aa, bb, cc, aaa, bbb, ccc, m1, m2, m3
and 3 + 7 constraints. So, the matrix should be 10 times 12 and may be created
as follows
C11 <- rbind(
c(1, 0, 0, 1, 0, 0, 0, 0, 0),
c(0, 1, 0, 0, 1, 0, 0, 1, 0),
c(0, 0, 0, 0, 0, 1, 0, 0, 1))
C12 <- matrix(0, nrow=3, ncol=3)
A <- matrix(0, nrow=7, ncol=12)
colnames(A) <- c("a", "b", "c", "aa", "bb", "cc", "aaa", "bbb", "ccc", "m1",
"m2", "m3")
A[1, c( "a", "m1")] <- c(1/10, -1)
A[2, c( "b", "m1")] <- c(1/11, -1)
A[3, c("aa", "m2")] <- c(1/13, -1)
A[4, c("bb", "m2")] <- c(1/12, -1)
A[5, c("cc", "m2")] <- c(1/10, -1)
A[6, c("bb", "m3")] <- c(1/24, -1)
A[7, c("cc", "m3")] <- c(1/26, -1)
f.mat <- rbind(cbind(C11, C12), A)
The variables "c" and "aaa" are not used at all. Is this correct?
The other parameters and the call are
library(lpSolve)
f.obj <- c(0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1)
f.rhs <- c(15, 35, 40, 0, 0, 0, 0, 0, 0, 0)
f.dir = c("<=", "<=", "<=", rep(">=", times=7))
out <- lp("max", f.obj, f.mat, f.dir, f.rhs)
out
Success: the objective function is 2.612179
out$solution
[1] 0.000000 0.000000 0.000000 15.000000 35.000000 37.916667 0.000000
[8] 0.000000 0.000000 0.000000 1.153846 1.458333
The constraint on m1 requires only m1 <= min(a/10,b/11), however, since
we maximize m1 + m2 + m3, the optimum will satisfy m1 = min(a/10,b/11).
Similarly for m2, m3.
Hope this helps.
Petr Savicky.
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