fellow programmers. I'm studying a book on numerical solutions for economics (Judd 1998). I'm trying to reproduce a problem from that same book in R so I can use the optim package to see if I can get similar results.
The problem established by the author is this one: and his results were these.
I have tried to transcribe this problem to R, which resulted in this code chunk:
DisutilityJudd <- function(L){
if(L == 0){
return(0)
}else{
return(0.1)
}
}
AgentUtilityJudd <- function(w, L){
(-exp(-2*w) + 1) - DisutilityJudd(L)
}
reservation.utility.judd <- AgentUtilityJudd(1, 1)
MaxEffortUtility <- function(w1, w2, L = 1){
0.8 * AgentUtilityJudd(w1, L) + 0.2 * AgentUtilityJudd(w2, L)
}
LeastEffortUtility <- function(w1, w2, L = 0){
0.4 * AgentUtilityJudd(w1, L) + 0.6 * AgentUtilityJudd(w2, L)
}
UtilityDifferenceJudd <- function(w1, w2){
MaxEffortUtility(w1, w2) - LeastEffortUtility(w1, w2)
}
PenaltyFunctionJudd <- function(w1, w2, P = 100000){
if(length(w1) == 2){
y <- -1 * (0.8 * (2 - w1[1]) - 0.2 * w1[2] - P *
(pmax(0, -MaxEffortUtility(w1[1], w1[1]) - reservation.utility.judd))^2 -
P * (pmax(0, -UtilityDifferenceJudd(w1[1], w1[1])))^2)
}else{
y <- -1 * (0.8 * (2 - w1) - 0.2 * w2 - P *
(pmax(0, -MaxEffortUtility(w1, w2) - reservation.utility.judd))^2 -
P * (pmax(0, -UtilityDifferenceJudd(w1, w2)))^2)
}
return(y)
}
There were no errors, but the results generated by my code were nowhere near to what I was expecting:
optim(c(1.1, 0.5), PenaltyFunctionJudd)
$par
[1] 1.343909e+49 -2.370681e+51
$value
[1] -4.633849e+50
$counts
function gradient
501 NA
$convergence
[1] 1
$message
NULL
Perhaps there is a problem to my penalty function. I'm assuming that it is due to the pmax function. Could somebody help me identify it? Thank you, I appreciate your attention.
Edit: a typo.
I believe you meant w1[2] in when if(length(w1) == 2) is true.
I have modified your code, without touching how you define the previous function. It is not clear if it the result expected : what does IV(-1) mean, is it the result minus 1 ? a power if 10 ?
PenaltyFunctionJudd <- function(w1, w2, P = 1e5){
if(length(w1) > 1){
w2 <- w1[2]
w1 <- w1[1]
}
# cat("length is 2 \n")
y <- 0.8 * (2 - w1) - 0.2 * w2 - P *
( pmax(0, -MaxEffortUtility(w1, w2) - reservation.utility.judd) )^2 -
P * ( pmax(0, -UtilityDifferenceJudd(w1, w2)) )^2
# cat("pmax1 :", pmax(0, -MaxEffortUtility(w1, w2) - reservation.utility.judd), "\n")
# cat("pmax2 :", pmax(0, -UtilityDifferenceJudd(w1, w2)), "\n")
return(y)
}
optim(c(1.1, 0.5), PenaltyFunctionJudd, control = list(fnscale = -1) )
optim(c(11, 5), PenaltyFunctionJudd, method = "BFGS", control = list(fnscale = -1, maxit = 100) )
You can use cat or print to check your values (here I noticed some Inf and 0 the leaded me to notice code error).
Friendly warning : provided you defined correctly the previous function, there is lot of instability in optimisation (problem badly set ? More penalty needed ?). Indeed when running twice or more the algorithm parameters fluctuate a lot...
Related
I have two equations. They are as follows:
( 1 - 0.25 ^ {1/alpha} ) * lambda = 85
( 1 - 0.75 ^ {1/alpha} ) * lambda = 11
I would like to compute the values of alpha and lambda by solving the above two equations. How do I do this using R?
One approach is to translate it into an optimization problem by introducing an loss function:
loss <- function(X) {
L = X[1]
a = X[2]
return(sum(c(
(1 - 0.25^(1/a))*L - 85,
(1 - 0.75^(1/a))*L - 11
)^2))
}
nlm(loss, c(-1,-1))
If the result returned from nlm() has a minimum near zero, then estimate will be a vector containing lambda and alpha. When I tried this, I got an answer that passed the sniff test:
> a = -1.28799
> L = -43.95321
> (1 - 0.25^(1/a))*L
[1] 84.99999
> (1 - 0.75^(1/a))*L
[1] 11.00005
#olooney's answer is best.
Another way to solve these equations is to use uniroot function. We can cancel the lambda values and can use the uniroot to find the value of alpha. Then substitute back to find lambda.
f <- function(x) {
(11/85) - ((1 - (0.75) ^ (1/x)) / (1 - (0.25) ^ (1/x)) )
}
f_alpha <- uniroot(f, lower = -10, upper = -1, extendInt = "yes")
f_lambda <- function(x) {
11 - ((1 - (0.75) ^ (1/f_alpha$root)) * x)
}
lambda = uniroot(f_lambda, lower = -10, upper = -2, extendInt = "yes")$root
sprintf("Alpha equals %f", f_alpha$root)
sprintf("Lambda equals %f", lambda)
results in
[1] "Alpha equals -1.287978"
[1] "Lambda equals -43.952544"
I have quite simple question yet I cannot seem to solve it.
I'm trying to find a parameter with optim() r function.
Here is the case:
library(rootSolve)
d <- read.table(text="indx rate n d
1 0.12 158 14
2 0.095 135 9
3 0.057 123 4
4 0.033 115 5
5 0.019 90 4", header=T)
d$real <- with(d, d/n)
opt <- d[ ,c("rate","real", "n")]
# this is close to the correct solution!
scaler <- apply(opt, 1, function(z) uniroot.all(
function(alpha) z[2] - (1 / (1 + alpha * ( (1 - z[1]) / z[1] )) ), interval = c(0,10)))
# check for solution (not fully correct!)
round(crossprod(scaler * opt$real, opt$n)/sum(opt$n), 3) == round(crossprod(round(opt$rate, 3), opt$n)/sum(opt$n), 3)
# using optim() - completely wrong results
infun <- function(data, alpha){ l <- with(data,
( rate - (1 / ( 1 + alpha[1] * ( (1 - real)/real ))) )); return( -sum( l ) ) }
opt_out <- optim(c(0,0), infun, data=opt, method = "BFGS", hessian = TRUE)
with(opt, (1 / ( 1 + opt_out$par[1] * ( (1 - real)/real ))))
You are trying, with your code, to get an unique alpha for all, but you want to have five values ..
So, you are leaded to make a sum .. but if you have negative and positive values, your sum could go near zero even with individual terms far from 0 ..
Moreover, your infun function is not in accordance with your previous function ..
What you can do is something like that :
infun <- function(alpha){ l <- with(cbind(d, alpha), ( real - (1 / ( 1 +
alpha * ( (1 - rate)/rate ))) )); return( sum(abs(l)) ) }
param <- c(5,5,5,5,5)
opt_out <- optim(par = scaler, infun, method = "BFGS", hessian = TRUE)
And in order to check the result you should have written :
with( cbind(opt,opt_out$par), real -1 / ( 1 + opt_out$par * ( (1 - rate)/rate )))
To get the true solution, you can do (after a very litle mathematics on a paper) :
sol <- -((opt[,2]-1)/(opt[,2]))*(opt[,1]/(1-opt[,1]))
and test it :
with( cbind(opt,sol), real -1 / ( 1 + opt_out$par * ( (1 - rate)/rate )))
I have a small MATLAB script mainly doing derivatives using symbolic toolbox that I want to rewrite into R. I chose Ryacas package because I found rSymPy too tricky to install... Here is my R code
# install.packages('Ryacas')
library(Ryacas)
z <- Sym("z")
psi=c()
psi[1]=z^2*exp(-z)/(1-exp(-z))
psi[2]=z^2*exp(-z)/(1-exp(-z))*log(z)
psi[3]=z^2*exp(-z)/(1-exp(-z))*log(z)^2
f=matrix(NA,4,4)
f[1,1]=z^2*exp(-z)/(1-exp(-z))
for(i in 2:4){
f[i,1]=deriv.Sym(psi[i-1],z)
j=2
while(j<=i){
f[i,j]=deriv.Sym(expression(f[i,j-1]/f[j-1,j-1]),z)
j=j+1
}
}
It does not report any error. However, the output shows that R isn't actually doing symbolic computation but returns characters. So I cannot evaluate the result. I tried
> i=2
> deriv.Sym(psi[i-1],z)
expression(((1 - exp(-z)) * (2 * (z * exp(-z)) - z^2 * exp(-z)) -
z^2 * exp(-z)^2)/(1 - exp(-z))^2)
> f[i,1]
[1] "( D( z , 1 ) ( ( ( z ^ 2 ) * ( Exp ( ( - z ) ) ) ) / ( 1 - ( Exp ( ( - z ) ) ) ) ) )"
It seems that deriv.Sym(psi[i-1],z) does the symbolic derivative and get the correct result. But if the result is assigned to a variable, it becomes character class. I feel confused about expression(), yacas(), Sym() and character. Anyone can point out my mistake or help me clarify these concept? Thank you so much.
Below corresponding MATLAB code for reference. The MATLAB code works just fine.
syms c;
psi(1)=c^2*exp(-c)/(1-exp(-c));
psi(2)=c^2*exp(-c)/(1-exp(-c))*log(c);
psi(3)=c^2*exp(-c)/(1-exp(-c))*log(c)^2;
f(1,1)=c^2*exp(-c)/(1-exp(-c));
for i=2:4
f(i,1)=diff(psi(i-1),c);
j=2;
while j<=i
f(i,j)=diff(f(i,j-1)/f(j-1,j-1),c);
j=j+1;
end
end
g11=matlabFunction(f(1,1));
fplot(g11,[0,10])
figure
g22=matlabFunction(f(2,2));
fplot(g22,[0,10])
figure
g33=matlabFunction(f(3,3));
fplot(g33,[0,10])
figure
g44=matlabFunction(f(4,4));
fplot(g44,[0,10])
There are several problems with the R code in the question:
it is attempting to assign an S3 object to elements of a logical matrix:
typeof(NA)
## [1] "logical"
so R has converted it to character (since Sym objects are internally
character) which is as far as it can go. f needs to be defined as a list
with 2 dimensions so that it can hold such objects:
f <- matrix(list(), 4, 4)
since f is a list with 2 dimensions all references to elements of f should use double square brackets as in:
f[[1, 1]] <- z^2 * exp(-z) / (1 - exp(-z))
similarly psi should be initialized as:
psi <- list()
and then referenced as:
psi[[1]] <- z^2 * exp(-z) / (1 - exp(-z))
to evaluate f[[i, 1]] use Eval:
Eval(f[[i, 1]], list(z = 1))
## [1] 0.2432798
This also works but overwrites the Sym object z:
z <- 1
Eval(f[[i, 1]])
in general code should be calling the generic deriv and not by directly going to the specific method deriv.Sym
The revised code is in the section at the end which makes these changes as well as some stylistic improvements.
Suggest you review the vignette that comes with Ryacas. From the R console enter:
vignette("Ryacas")
Also review the Ryacas demos:
demo(package = "Ryacas")
Revised code
# install.packages('Ryacas')
library(Ryacas)
z <- Sym("z")
psi <- list()
psi[[1]] <- z^2 * exp(-z) / (1 - exp(-z))
psi[[2]] <- z^2 * exp(-z) / (1 - exp(-z)) * log(z)
psi[[3]] <- z^2 * exp(-z) / (1 - exp(-z)) * log(z)^2
f <- matrix(list(), 4, 4)
f[[1,1]] <- z^2 * exp(-z) / (1 - exp(-z))
for(i in 2:4) {
f[[i, 1]] <- deriv(psi[[i-1]], z)
j <- 2
while(j <= i) {
f[[i, j]] <- deriv(f[[i, j-1]] / f[[j-1, j-1]], z)
j <- j + 1
}
}
i <- 2
deriv(psi[[i-1]], z)
f[[i, 1]]
Eval(f[[i, 1]], list(z = 1))
I'm trying to modeling a prey-prey-predator system using differential equations based on the LV model. For the sake of the precision, i need to use the runge-kutta4 method.
But given the equations, some of the populations become quickly negative.
So I tried to use the events/root system of ODE but it seems that rk4 and rootfun are not compatibles...
eventFunc <- function(t, y, p){
if (y["N1"] < 0) { y["N1"] = 0 }
if (y["N2"] < 0) { y["N2"] = 0 }
if (y["P"] < 0) { y["P"] = 0 }
return(y)
}
rootFunction <- function(t, y, p){
if (y["P"] < 0) {y["P"] = 0}
if (y["N1"] < 0) {y["N1"] = 0}
if (y["N2"] < 0) {y["N2"] = 0}
return(y)
}
out <- ode(func=Model_T2.2,
method="rk4",
y=state,
parms=parameters,
times=times,
events = list(func = eventFunc,
root = TRUE),
rootfun = rootFunction
)
This code give me the followin error :
Error in checkevents(events, times, Ynames, dllname) :
either 'events$time' should be given and contain the times of the events, if 'events$func' is specified and no root function or your solver does not support root functions
Is there any solution to use rk4 and forbid the functions to go under 0?
Thanks in advance.
For those who might ask, here is what works :
if(!require(ggplot2)) {
install.packages("ggplot2"); require(ggplot2)}
if(!require(deSolve)) {
install.packages("deSolve"); require(deSolve)}
Model_T2.2 <- function(t, state, par){
with(as.list(c(state, par)), {
response1 <- (a1 * N1)/(1+(a1*h1*N1)+(a2*h2*N2))
response2 <- (a2 * N2)/(1+(a1*h1*N1)+(a2*h2*N2))
dN1 = r1*N1 * (1 - ((N1 + A12 * N2)/K1)) - response1 * P
dN2 = r2*N2 * (1 - ((N1 + A21 * N2)/K2)) - response2 * P
dP = ((E1 * response1) + (E2 * response2)) * P - Mp
return(list(c(dN1, dN2, dP)))
})
}
parameters<-c(
r1=1.42, r2=0.9,
A12=0.6, A21=0.5,
K1=50, K2=50,
a1=0.77, a2=0.77,
b1 = 1, b2=1,
h1=1.04, h2=1.04,
o1=0, o2=0,
Mp=0.22,
E1=0.36, E2=0.36
)
## inital states
state<-c(
P=10,
N1=30,
N2=30
)
times <- seq(0, 30, by=0.5)
out <- ode(func=Model_T2.2,
method="rk4",
y=state,
parms=parameters,
times=times,
events = list(func = eventFunc,
root = TRUE),
rootfun = rootFunction
)
md <- melt(as.data.frame(out), id.vars=1, measure.vars = c("N1", "N2", "P"))
pl <- ggplot(md, aes(x=time, y=value, colour=variable))
pl <- pl + geom_line() + geom_point() + scale_color_discrete(name="Population")
pl
And the result in a graph :
Evolution of prey1, prey2 and predator populations
As you can see, the population of predators become negative which is clearly impossible in the real world.
Edit : missing variables, sorry about that.
This is a problem you will have with all explicit solvers like rk4. Reducing the time step will help, up to a point. Better use a solver with an implicit method, lsoda seems universally available in one form or another.
Another way to explicitly force positive values is to parametrize them as exponentials. Set N1=exp(U1), N2=exp(U2) then the ODE function code translates to (as dN = exp(U)*dU = N*dU)
N1 <- exp(U1)
N2 <- exp(U2)
response1 <- (a1)/(1+(a1*h1*N1)+(a2*h2*N2))
response2 <- (a2)/(1+(a1*h1*N1)+(a2*h2*N2))
dU1 = r1 * (1 - ((N1 + A12 * N2)/K1)) - response1 * P
dU2 = r2 * (1 - ((N1 + A21 * N2)/K2)) - response2 * P
dP = ((E1 * response1*N1) + (E2 * response2*N2)) * P - Mp
For the output you have then of course to reconstruct N1, N2 from the solutions U1, U2.
Thanks to J_F, I am now able to run my L-V model.
The radau (not randau as you mentionned) function indeed accept root function and events ans implicitly implements the runge-kutta method.
Thanks again, hope this will help someone in the future.
I'm writing a function for Gaussian mixture models with spherical covariance structures--ie $\Sigma_k = \sigma_k^2 I$. This particular function is similar to the mclust package with identifier VII.
http://en.wikipedia.org/wiki/Mixture_model
Anyways, the problem I'm having is running into infinite values for the weight matrix. Definition: Let W be an n x m matrix where n = 1, ..., n (number of obs) and m = 1, ..., m (number of mixtues). Each element of W (ie w_ij) can essentially be defined as a specific form of:
w_im = \frac{a / b * exp(c)}{\sum_i=1^m [a_i / b_i * exp(c_i)]}
Computing this numerically is giving me infinite values. So I'm trying to use the log-identity log(x+y) = log(x) + log(1 + y/x). But the issue is that it's not as simple as log(x+y) but rather log(\sum_i=1^m [a_i / b_i * exp(c_i)]).
Here's some code define:
n_im = a / b * exp(c) ;
d_.m = \sum_i=1^m [a_i / b_i * exp(c_i)] ; and
c_mat[i,j] as the value of the exponent for the [i,j]th term.
n_mat[, i] <- log(a[i]) - log(b[i]) - c[,i] # numerator of w_im
internal_vec1[i] <- (a[i] * b[1])/ (a[1] * b[i]) # an internal for the step below
c_mat2 <- cbind(rep(1, n), c_mat[,1] - c_mat[,-1]) # since e^a / e^b = e^(a-b)
for (i in 1:n) {
d_vec[i] <- n_mat[i,1] + log(sum(internal_vec1 * exp(c_mat2[i,)))
} ## still getting infinite values
I'm trying to define the problem as briefly as possible. the entire function is obviously much larger than this. But, since the problem I'm running into is specifically dealing with infinite (and 1/infinity) values, I'm hoping this snippet is sufficient. Anyone with a coding trick here?
Here is the solution!! (I've spent way too damn long on this)
**The first function log_plus() solves the simple problem where you want log(\sum_{i=1)^n x_i)
**The second function log_plus2() solves the more complicated problem described above where you want log(\sum_{i=1}^n [a_i / b_i * exp(c_i)])
log_plus <- function(xvec) {
m <- length(xvec)
x <- log(xvec[1])
for (j in 2:m) {
sum_j <- sum(xvec[1:j-1])
x <- x + log(1 + xvec[j]/sum_j)
}
return(x)
}
log_plus2 <- function(a, b, c) {
# assumes intended input of form sum(a/b * e^c)
if ((length(a) != length(b)) || (length(a) != length(c))) {
stop("Input equal length vectors")
}
if (!(all(c > 0) || all(c < 0))) {
stop("All values of c must be either > 0 or < 0.")
}
m <- length(a)
# initilialize log sum
x <- log(a[1]) - log(b[1]) + c[1]
# aggregate / loop log sum
for (j in 2:m) {
# build denominator
b2 <- b[1:j-1]
for (i in 1:j-1) {
d1 <- 0
c2 <- c[1:i]
if (all(c2 > 0)) {
c_min <- min(c2[1:j-1])
c2 <- c2 - c_min
} else if (all(c2 < 0)) {
c_min <- max(c2[1:j-1])
c2 <- c2 - c_min
}
d1 <- d1 + a[i] * prod(b2[-i]) * exp(c2[i])
}
den <- b[j] * (d1)
num <- a[j] * prod(b[1:j-1]) * exp(c[j] - c_min)
x <- x + log(1 + num / den)
}
return(x)
}