R help - Dec 2002 - optimize() - Solution is invariant with respect to parame ters - l exical function

Solved the problem myself, thks, for any efforts taken!

-----Original Message-----
From: Pfaff, Bernhard [mailto:Bernhard.Pfaff at drkw.com]
Sent: 19 December 2002 14:58
To: r-help at stat.math.ethz.ch
Subject: [R] optimize() - Solution is invariant with respect to
parameters - l exical function


Dear R-List member,

suppose the following single valued functional form is given: y = f(x | a,
b, c). Now I want to use optimize to solve for the value of x that satisfy
the function.
As an example, pls. take a look at the following:

input <- function(y,a,b,c){
f <- function(x){
  (a + b*x + c*x^2 - y)^2
}
optimize(f,c(1,100))
}
#
# example with x=2
#
input(y=7,a=1,b=1,c=1)
temp1 <- input(y=7,a=1,b=1,c=1)$minimum
temp1
1 + temp1 + temp1^2
#
# example with x=3
#
input(y=14,a=2,b=1,c=1)
temp2 <- input(y=14,a=2,b=1,c=1)$minimum
temp2
2 + temp2 + temp2^2

So far, so good(?). Now, the function that I am interesting in has the
following form:


bsiv.opt <- function(op, s, x, r, tt){
  iv <- function(sig){
   (s*pnorm((log(s/x)+(r+0.5*sig^2)*tt)/sig*sqrt(tt)) -
x*exp(-r*tt)*pnorm((log(s/x)+(r+0.5*sig^2)*(tt))/sig*sqrt(tt)-sig*sqrt(tt))
- op)^2
  }
  optimize(iv,c(1,100))
}
#
# example 1
#
bsiv.opt(4.75, 93.625, 90, 0.0512, 22)
#
# example 2
#
bsiv.opt(4.75, 93.625, 80, 0.0512, 22)

My question is as follows: although, I have altered one parameter value, the
solved minimum value does not change; how come? What have I made wrong?
Any hints and pointers are most welcome.

Rgds,
Bernhard 




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