Minimize a function with two variables
I would like to keep a function with two variables to a minimum.
I first created a function (rba)
that is needed inside a function (kvasum)
that I need to keep to a minimum. The minimized values are part of rba
.
# Data
vpk = data.frame(V1 =c(3650000000, 19233, 2211.2, 479.47, 168.46, 83.447, 52.349, 38.738,
32.34, 29.588), V2 = 1:10)
n = nrow(vpk)
# functions to minimize
# This function returns a vector with 10 values
rba = function(par){
v <- matrix(ncol = 1, nrow = 10)
for (p in 1:10){
k<- ifelse (par[1] < 1-1/p && par[1]>0 && p > par[2] &&
par[2]>0 && par[2]<2, par[2]*p,
ifelse(par[1] < 1-1/par[2] && par[1] > 0 &&
p < par[2] && par[2]>0 && par[2]<2, -1+(par[1]+1/par[2]),
ifelse(par[1] > (1 - 1 / max(p,par[2])) &&
par[2]>0 && par[2]<2, -1+p, "error")))
v[p] <- k
}
return(v)
}
# This function uses the function rba, and returns a value
kvasum = function(par){
sum( (log(vpk$V1)/log(1/n) - rba(par) )^2)
}
# what I would I to do is to find par[1] and par[2] such that kvasum is minimized
m1 = optim(par=c(0.1,0.4),kvasum, lower=0)
I tried to use the optim function, but I cannot get it to work. I am getting a non-numeric argument and trying everything I can think of. Any help is appreciated.
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There are several issues with your overall process that are causing problems.
First of all, as @ user227710 mentions in the comments, you should replace &&
with &
. They have different meanings.
Now for the optimizer
It sounds like you want to set limits on your parameters (i.e. what is known as in-box constraints). To do this, and thus use an argument lower
, you need to use the method L-BFGS-B
. When you use this, you need to provide an argument as well upper
.
Received error, you get it because your operator ifelse
only works when values are between 0 and 1 roughly. Otherwise the variable k gets a value error
(this value is returned if all conditions in the statements are ifelse
FALSE), so you get
Error in log(vpk$V1)/log(1/n) - rba(par) :
non-numeric argument to binary operator
mistake.
So if you set your window limits appropriately (or maybe take a look at your ifelse statement, because you may have coded it wrong), this seems to work fine:
# Data
vpk = data.frame(V1 =c(3650000000, 19233, 2211.2, 479.47, 168.46, 83.447, 52.349, 38.738,
32.34, 29.588), V2 = 1:10)
n = nrow(vpk)
# functions to minimize
# This function returns a vector with 10 values
rba = function(par){
v <- matrix(ncol = 1, nrow = 10)
for (p in 1:10){
k<- ifelse (par[1] < 1-1/p & par[1]>0 & p > par[2] &
par[2]>0 & par[2]<2, par[2]*p,
ifelse(par[1] < 1-1/par[2] & par[1] > 0 &
p < par[2] & par[2]>0 & par[2]<2, -1+(par[1]+1/par[2]),
ifelse(par[1] > (1 - 1 / max(p,par[2])) &
par[2]>0 & par[2]<2, -1+p, "error")))
#I am adding a line here so that you know why the optim failed
if(k=='error') stop('your ifelse function returned an error')
v[p] <- k
}
return(v)
}
# This function uses the function rba, and returns a value
kvasum = function(par){
sum( (log(vpk$V1)/log(1/n) - rba(par) )^2)
}
# what I would I to do is to find par[1] and par[2] such that kvasum is minimized
m1 = optim(par=c(0.1,0.4),kvasum, method='L-BFGS-B', lower= c(0.1,0.1), upper=c(0.9,0.9))
Output:
> m1
$par
[1] 0.1 0.1
$value
[1] 171.5774
$counts
function gradient
2 2
$convergence
[1] 0
$message
[1] "CONVERGENCE: NORM OF PROJECTED GRADIENT <= PGTOL"
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