[R] Optimize function in R: unable to find maximum of logistic function
Hello Everyone,
I want to perform a 1-D optimization by using the optimize() function. I
want to find the maximum value of a logistic function. The optimize()
function gives the wrong result.
My code:
f= function (k) {
T_s = 20
result = (2- 2/(1+ exp(-2*T_s*k)))
return(result)
}
optimize(f, c(0, 5), tol = 0.1, maximum= TRUE)
The maximum value for the function happens at k=0, and the maximum value is
1. Yet the optimise function, says that the maximum value happens at k=
4.9995, and the maximum value is 0.
Thanks in advance!
Warm Regards,
Shantanu
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Re: [R] Optimize function in R: unable to find maximum of logistic function
Hello,
Thanks a lot for the replies.
I tried to use the optim function in R, and chose the L-BFGS-B method
of optimization. Now the result is very sensitive to the starting values. I
use the same function, but I give the line of code I use for optim.
optim(0.25, f, method= L-BFGS-B, lower = 0, upper=5). Here 0.25 is the
starting value.
Result: $ par = 0, $Value = 1 (this is correct). $par indicates that the
maximum value of the function is at 0, and $ Value says that the maximum
value is 1.
However, I produced a quick table for the results with different starting
values:
Starting Value $par $Value
Result
0.25 0 1
Correct
0.5 0.4999 4.1223e-09
Wrong
11 0
Wrong
3 3 0
Wrong
I wanted to optimise the logistic function, as the logistic function
approximates an indicator function. The indicator function is defined as
follows: I(u), where I(u) = 1, where u = 0, and I(u) = 0, for all other
u. u is strictly positive.
My objective function contains a indicator function.As the indicator
function is non-continuous, the optimization would become non-convex. To
bypass this limitation, I used the logistic function to approximate the
indicator function.
A simple example: f(u) = I(u) + u/6;Maximise f(u) where u is between (0
to 5).
Correct Answer: f(u) =1, where u=0.
Code:
f= function (k) {
T_s = 20
result = (2- 2/(1+ exp(-2*T_s*k))) + (k/6)
return(result)
}
optim(0.15, f, method= L-BFGS-B, lower = 0, upper=5,control =
list(fnscale=-1) )
## Here 0.15 is the starting value. I produce a similar table as above:
Starting Value $par $Value
Result
0.15 0 1
Correct
0.16 5 0.833
Wrong
0.17 5 0.833
Wrong
2 5 0.833
Wrong
Questions:
1. If my objective function contains an Indicator function, is there an
efficient way to implement it?
2. Does R have any quick non-convex optimiser functions?
Best Regards,
Shantanu Mullick
PhD Candidate
ESSEC Business School
On 28 October 2013 22:01, Rolf Turner r.tur...@auckland.ac.nz wrote:
This could be described as a bug, perhaps. Or it could be described as
an indication that numerical optimization is inevitably tricky. Notice that
if you narrow down your search interval from [0,5] to [0,0.5] you get the
right answer:
optimize(f, c(0, 0.5), maximum= TRUE,tol=1e-10)
$maximum
[1] 4.192436e-11
$objective
[1] 1
I guess there's a problem with finding a gradient that is effectively
(numerically) zero when k is equal to 5.
cheers,
Rolf Turner
On 10/29/13 06:00, Shantanu MULLICK wrote:
Hello Everyone,
I want to perform a 1-D optimization by using the optimize() function. I
want to find the maximum value of a logistic function. The optimize()
function gives the wrong result.
My code:
f= function (k) {
T_s = 20
result = (2- 2/(1+ exp(-2*T_s*k)))
return(result)
}
optimize(f, c(0, 5), tol = 0.1, maximum= TRUE)
The maximum value for the function happens at k=0, and the maximum value
is
1. Yet the optimise function, says that the maximum value happens at k=
4.9995, and the maximum value is 0.
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and provide commented, minimal, self-contained, reproducible code.
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