# [R] None-linear equality constrained optimisation problems

Berend Hasselman bhh at xs4all.nl
Mon Dec 19 12:27:10 CET 2011

```Michael Griffiths wrote
>
> Dear R users,
>
> I have a problem. I would like to solve the following:
> I have
>
> pL = 1/(1+e^(-b0+b1))
> pM = 1/(1+e^(-b0))
> pH = 1/(1+e^(-b0-b1))
>
> My target function is
>
> TF= mean(pL,pM,pH) which must equal 0.5%
>
> My non-linear constraint is
>
> nl.Const = 1-(pM/pH), which must equal 20%, and would like the values of
> both b0 and b1 where these conditions are met.
>
> I have searched widely for an answer, and did think that Rdonlp2 would
> suffice, only to find it no longer supported. I have solved this using
> Excel's solver function, however, because of the non-linear constraint I
> am
> having problems finding the solution in R.
>
> Can the community suggest another method by which this might be solved?
>

>From your description I gather that this is actually a problem of solving a
system of non-linear equations.
So you could for example use package nleqslv. As follows.

library(nleqslv)

f <- function(x) {
b0 <- x
b1 <- x

pL <- 1/(1+exp(-b0+b1))
pM <- 1/(1+exp(-b0))
pH <- 1/(1+exp(-b0-b1))

CR <-  (1 - pM/pH) - 0.2
TF <-  mean(pL,pM,pH) - .005## which must equal 0.5%

c(CR,TF)
}

bstart <- c(.5,.5)
nleqslv(bstart ,f, control=list(trace=0))

with output (excerpt)

\$x
 -5.0685870  0.2247176
\$fvec
 1.337569e-09 8.879125e-10

bstart <- c(.85,.85)
nleqslv(bstart ,f, control=list(trace=0))

with output (excerpt)

\$x
 1.384720 6.678025
\$fvec
  1.402461e-11 -3.595026e-11

So your solution for b0 and b1  depends heavily on the starting values.

BTW: your non-linear constraint is not really non-linear.
1-pM/pH=.2 is equivalent to .8 * pH = pM which is linear.

Berend

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