# [R] constrOptim and problem with derivative

Berend Hasselman bhh at xs4all.nl
Tue Dec 20 16:22:52 CET 2011

```Berend Hasselman wrote
>
>
> Michael Griffiths wrote
>>
>> Dear List,
>>
>> I am using constrOptim to solve the following
>>
>> fr1 <- function(x) {
>>     b0 <- x[1]
>>     b1 <- x[2]
>>     ((1/(1+exp(-b0+b1))+(1/(1+exp(-b0)))+(1/(1+exp(-b0-b1)))))/3
>> }
>>
>> As you can see, my objective function is
>> ((1/(1+exp(-b0+b1))+(1/(1+exp(-b0)))+(1/(1+exp(-b0-b1)))))/3 and I would
>> like to solve for both b0 and b1.
>>
>> If I were to use optim then I would derive the gradient of the function
>> (grr) as follows:
>>
>> fr2 <-
>> expression(((1/(1+exp(-b0+b1))+(1/(1+exp(-b0)))+(1/(1+exp(-b0-b1)))))/3)
>> grr <- deriv(fr2,c("b0","b1"), func=TRUE)
>>
>> and then simply use optim via
>>
>> optim(c(-5.2,0.22), fr1, grr)
>>
>> My problem is that I wish to place constraints (b0>=-0.2 and b1>= 0.1)
>> upon
>> the values of b0 and b1. I can set the constraints matrix and boundary
>> values to
>>
>> ui=rbind(c(1,0),c(0,1)) and ci=c(-0.2,0.1), however, when I come to run
>> constrOptim function via
>>
>>
>> constrOptim(c(-0.1,0.2), fr1, grr, ui=rbind(c(1,0),c(0,1)),
>> ci=c(-0.2,0.1))
>>
>> I get the following error message:
>>
>> "Error in .expr1 + b1 : 'b1' is missing"
>>
>> So, it seems to me that I am doing something incorrectly in my
>> specification of grr in constrOptim.
>>
>
> grr is a function with two arguments. Do this
>
> grr
>
> and then you will see.
> But the gradient function passed to constrOptim wants a function with a
> vector argument.
>
> So if you do
>
>     b0 <- x[1]
>     b1 <- x[2]
>     grr(b0,b1)
> }
>
>

This is  incorrect.
The gradient function should return a vector. It was returning a scalar with
attributes.

# Correct
b0 <- x[1]
b1 <- x[2]
g <- grr(b0,b1)
}

and this looks better

ci=c(-0.2,0.1))

I'm puzzled why constrOptim or optim didn't issue an error message in the
original case.

Berend

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```