r - Minimize a function with two variables -

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i minimize function 2 variables.

first have made function (rba), needed inside function (kvasum) need minimize. values minimized on 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 minimize   # function returns vector 10 values  rba = function(par){   v <- matrix(ncol = 1, nrow = 10)   (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) }  # function uses function rba, , returns value  kvasum = function(par){   sum( (log(vpk$v1)/log(1/n) - rba(par) )^2) }  # find par[1] , par[2] such kvasum minimized  m1 = optim(par=c(0.1,0.4),kvasum, lower=0)

i've tried use optim function, can't work. non-numeric argument, , have tried can think of. appreciated.

there few problems overall process cause trouble.

first of @user227710 mentions in comments should replace && &. these have different meanings.

now optimiser

it looks want set limits parameters (i.e. known box constraints). in order , use lower argument, need use l-bfgs-b method. when use need specify upper argument well.

the error receive, receive because ifelse statement works when values between 0 , 1 roughly. otherwise, k variable gets value error (which value returned if conditions in ifelse statements false) why 

error in log(vpk$v1)/log(1/n) - rba(par) :    non-numeric argument binary operator

error.

therefore, if specify box constraints accordingly (or maybe have @ ifelse statement because might have coded wrong) seems work perfect:

# 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 minimize   # function returns vector 10 values  rba = function(par){   v <- matrix(ncol = 1, nrow = 10)   (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 adding line here know why optim failed     if(k=='error') stop('your ifelse function returned error')     v[p] <- k    }   return(v) }  # function uses function rba, , returns value  kvasum = function(par){   sum( (log(vpk$v1)/log(1/n) - rba(par) )^2) }  # find par[1] , par[2] such kvasum 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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