I am trying to do a maximization in R that I have done previously in Excel with the solver. The problem is that I don't know how to deal with it (i don't have a good level in R).
let's talk a bit about my data. I have 26 Swiss cantons and the Swiss government (which is the sum of the value of the 26 cantons) with their population and their "wealth". So I have 27 observatios by variable. I'm not sure that the following descriptions are useful but I put them anyway. From this, I calculate some variables with while loops. For each canton [i]:
resource potential = mean(wealth2011 [i],wealth2012 [i],wealth2013 [i])
population mean = mean(population2011 [i],population2012 [i],population2013 [i])
resource potential per capita = 1000*resource potential [i]/population [i]
resource index = 100*resource potential capita [i]/resource potential capita [swiss government]
Here a little example of the kind of loops I used:
RI=0
i = 1
while(i<28){
RI[i]=resource potential capita [i]/resource potential capita [27]*100
i = i+1
}
The resource index (RI) for the Swiss government (i = 27) is 100 because we divide the resource potential capita of the swiss government (when i = 27) by itself and multiply by 100. Hence, all cantons that have a RI>100 are rich cantons and other (IR<100) are poor cantons. Until here, there was no problem. I just explained how I built my dataset.
Now the problem that I face: I have to create the variable weighted difference (wd). It takes the value of:
0 if RI>100 (rich canton)
(100-RI[i])^(1+P)*Pop[i] if RI<100 (poor canton)
I create this variable like this: (sorry for the weakness of the code, I did my best).
wd=-1
i = 1
a = 0
c = 0
tot = 0
while(i<28){
if(i == 27) {
wd[i] = a
} else if (RI[i] < 100) {
wd[i] = (100-RI[i])^(1+P)*Pop[i]
c = wd[i]
a = a+c
} else {
wd[i]= 0
}
i = i+1
}
However, I don't now the value of "p". It is a value between 0 and 1. To find the value of p, I have to do a maximization using the following features:
RI_26 = 65.9, it is the minimum of RI in my data
RI_min = 100-((x*wd [27])/((1+p)*z*100))^(1/p), where x and z are fixed values (x = 8'677, z = 4'075'977'077) and wd [27] the sum of wd for each canton.
We have p in two equation: RI_min and wd. To solve it in Excel, I used the Excel solver with the following features:
p_dot = RI_26/RI_min* p ==> p_dot =[65.9/100-((x* wd [27])/((1+p)*z*100))^(1/p)]*p
RI_26 = RI_min ==>65.9 =100-((x*wd [27])/((1+p)*z*100))^(1/p)
In Excel, p is my variable cell (the only value allowed to change), p_dot is my objective to define and RI_26 = RI_min is my constraint.
So I would like to maximize p and I don't know how to do this in R. My main problem is the presence of p in RI_min and wd. We need to do an iteration to solve it but this is too far from my skills.
Is anyone able to help me with the information I provided?
you should look into the optim function.
Here I will try to give you a really simple explanation since you said you don't have a really good level in R.
Assuming I have a function f(x) that I want to maximize and therefore I want to find the parameter x that gives me the max value of f(x).
First thing to do will be to define the function, in R you can do this with:
myfunction<- function(x) {...}
Having defined the function I can optimize it with the command:
optim(par,myfunction)
where par is the vector of initial parameters of the function, and myfunction is the function that needs to be optimized. Bear in mind that optim performs minimization, however it will maximize if control$fnscale is negative. Another strategy will be to change the function (i.e. changing the sign) to suit the problem.
Hope that this helps,
Marco
From the description you provided, if I'm not mistaken, it looks like that everything you need to do it's just an equation.
In particular you have the following two expressions:
RI_min = 100-((x*y)/((1+p)*z*100))^(1/p)
and, since x,y,z are fixed, the only variable is p.
Moreover, having RI_26 = RI_min this yields to:
65.9 =100-((x*y)/((1+p)*z*100))^(1/p)
Plugging in the values of x,y and z you have provided, this yields to
p=0.526639915936052
I don't understand what exactly you are trying to maximize.
Related
I am doing some projects related to statistics simulation using R based on "Introduction to Scientific Programming and Simulation Using R" and in the Students projects session (chapter 24) i am doing the "The pipe spiders of Brunswick" problem, but i am stuck on one part of an evolutionary algorithm, where you need to perform some data perturbation according to the sentence bellow:
"With probability 0.5 each element of the vector is perturbed, independently
of the others, by an amount normally distributed with mean 0 and standard
deviation 0.1"
What does being "perturbed" really mean here? I dont really know which operation I should be doing with my vector to make this perturbation happen and im not finding any answers to this problem.
Thanks in advance!
# using the most important features, we create a ML model:
m1 <- lm(PREDICTED_VALUE ~ PREDICTER_1 + PREDICTER_2 + PREDICTER_N )
#summary(m1)
#anova(m1)
# after creating the model, we perturb as follows:
#install.packages("perturb") #install the package
library(perturb)
set.seed(1234) # for same results each time you run the code
p1_new <- perturb(m1, pvars=c("PREDICTER_1","PREDICTER_N") , prange = c(1,1),niter=200) # your can change the number of iterations to any value n. Total number of iteration would come to be n+1
p1_new # check the values of p1
summary(p1_new)
Perturbing just means adding a small, noisy shift to a number. Your code might look something like this.
x = sample(10, 10)
ind = rbinom(length(x), 1, 0.5) == 1
x[ind] = x[ind] + rnorm(sum(ind), 0, 0.1)
rbinom gets the elements to be modified with probability 0.5 and rnorm adds the perturbation.
I've got a function, KozakTaper, that returns the diameter of a tree trunk at a given height (DHT). There's no algebraic way to rearrange the original taper equation to return DHT at a given diameter (4 inches, for my purposes)...enter R! (using 3.4.3 on Windows 10)
My approach was to use a for loop to iterate likely values of DHT (25-100% of total tree height, HT), and then use optimize to choose the one that returns a diameter closest to 4". Too bad I get the error message Error in f(arg, ...) : could not find function "f".
Here's a shortened definition of KozakTaper along with my best attempt so far.
KozakTaper=function(Bark,SPP,DHT,DBH,HT,Planted){
if(Bark=='ob' & SPP=='AB'){
a0_tap=1.0693567631
a1_tap=0.9975021951
a2_tap=-0.01282775
b1_tap=0.3921013594
b2_tap=-1.054622304
b3_tap=0.7758393514
b4_tap=4.1034897617
b5_tap=0.1185960455
b6_tap=-1.080697381
b7_tap=0}
else if(Bark=='ob' & SPP=='RS'){
a0_tap=0.8758
a1_tap=0.992
a2_tap=0.0633
b1_tap=0.4128
b2_tap=-0.6877
b3_tap=0.4413
b4_tap=1.1818
b5_tap=0.1131
b6_tap=-0.4356
b7_tap=0.1042}
else{
a0_tap=1.1263776728
a1_tap=0.9485083275
a2_tap=0.0371321602
b1_tap=0.7662525552
b2_tap=-0.028147685
b3_tap=0.2334044323
b4_tap=4.8569609081
b5_tap=0.0753180483
b6_tap=-0.205052535
b7_tap=0}
p = 1.3/HT
z = DHT/HT
Xi = (1 - z^(1/3))/(1 - p^(1/3))
Qi = 1 - z^(1/3)
y = (a0_tap * (DBH^a1_tap) * (HT^a2_tap)) * Xi^(b1_tap * z^4 + b2_tap * (exp(-DBH/HT)) +
b3_tap * Xi^0.1 + b4_tap * (1/DBH) + b5_tap * HT^Qi + b6_tap * Xi + b7_tap*Planted)
return(y=round(y,4))}
HT <- .3048*85 #converting from english to metric (sorry, it's forestry)
for (i in c((HT*.25):(HT+1))) {
d <- KozakTaper(Bark='ob',SPP='RS',DHT=i,DBH=2.54*19,HT=.3048*85,Planted=0)
frame <- na.omit(d)
optimize(f=abs(10.16-d), interval=frame, lower=1, upper=90,
maximum = FALSE,
tol = .Machine$double.eps^0.25)
}
Eventually I would like this code to iterate through a csv and return i for the best d, which will require some rearranging, but I figured I should make it work for one tree first.
When I print d I get multiple values, so it is iterating through i, but it gets held up at the optimize function.
Defining frame was my most recent tactic, because d returns one NaN at the end, but it may not be the best input for interval. I've tried interval=c((HT*.25):(HT+1)), defining KozakTaper within the for loop, and defining f prior to the optimize, but I get the same error. Suggestions for what part I should target (or other approaches) are appreciated!
-KB
Forestry Research Fellow, Appalachian Mountain Club.
MS, University of Maine
**Edit with a follow-up question:
I'm now trying to run this script for each row of a csv, "Input." The row contains the values for KozakTaper, and I've called them with this:
Input=read.csv...
Input$Opt=0
o <- optimize(f = function(x) abs(10.16 - KozakTaper(Bark='ob',
SPP='Input$Species',
DHT=x,
DBH=(2.54*Input$DBH),
HT=(.3048*Input$Ht),
Planted=0)),
lower=Input$Ht*.25, upper=Input$Ht+1,
maximum = FALSE, tol = .Machine$double.eps^0.25)
Input$Opt <- o$minimum
Input$Mht <- Input$Opt/.3048. # converting back to English
Input$Ht and Input$DBH are numeric; Input$Species is factor.
However, I get the error invalid function value in 'optimize'. I get it whether I define "o" or just run optimize. Oddly, when I don't call values from the row but instead use the code from the answer, it tells me object 'HT' not found. I have the awful feeling this is due to some obvious/careless error on my part, but I'm not finding posts about this error with optimize. If you notice what I've done wrong, your explanation will be appreciated!
I'm not an expert on optimize, but I see three issues: 1) your call to KozakTaper does not iterate through the range you specify in the loop. 2) KozakTaper returns a a single number not a vector. 3) You haven't given optimize a function but an expression.
So what is happening is that you are not giving optimize anything to iterate over.
All you should need is this:
optimize(f = function(x) abs(10.16 - KozakTaper(Bark='ob',
SPP='RS',
DHT=x,
DBH=2.54*19,
HT=.3048*85,
Planted=0)),
lower=HT*.25, upper=HT+1,
maximum = FALSE, tol = .Machine$double.eps^0.25)
$minimum
[1] 22.67713 ##Hopefully this is the right answer
$objective
[1] 0
Optimize will now substitute x in from lower to higher, trying to minimize the difference
Here's the relevant code from the vignette, altered slightly to fit it on the page here, and make it easy to reproduce. Code for visualizations omitted. Comments are from vignette author.
(Full vignette: https://cran.r-project.org/web/packages/pbo/vignettes/pbo.html)
library(pbo)
#First, we assemble the trials into an NxT matrix where each column
#represents a trial and each trial has the same length T. This example
#is random data so the backtest should be overfit.`
set.seed(765)
n <- 100
t <- 2400
m <- data.frame(matrix(rnorm(n*t),nrow=t,ncol=n,
dimnames=list(1:t,1:n)), check.names=FALSE)
sr_base <- 0
mu_base <- sr_base/(252.0)
sigma_base <- 1.00/(252.0)**0.5
for ( i in 1:n ) {
m[,i] = m[,i] * sigma_base / sd(m[,i]) # re-scale
m[,i] = m[,i] + mu_base - mean(m[,i]) # re-center
}
#We can use any performance evaluation function that can work with the
#reassembled sub-matrices during the cross validation iterations.
#Following the original paper we can use the Sharpe ratio as
sharpe <- function(x,rf=0.03/252) {
sr <- apply(x,2,function(col) {
er = col - rf
return(mean(er)/sd(er))
})
return(sr)
}
#Now that we have the trials matrix we can pass it to the pbo function
#for analysis.
my_pbo <- pbo(m,s=8,f=sharpe,threshold=0)
summary(my_pbo)
Here's the portion i'm curious about:
sr_base <- 0
mu_base <- sr_base/(252.0)
sigma_base <- 1.00/(252.0)**0.5
for ( i in 1:n ) {
m[,i] = m[,i] * sigma_base / sd(m[,i]) # re-scale
m[,i] = m[,i] + mu_base - mean(m[,i]) # re-center
}
Why is the data transformed within the for loop, and does this kind of re-scaling and re-centering need to be done with real returns? Or is this just something the author is doing to make his simulated returns look more like the real thing?
Googling and searching through stackoverflow turned up some articles and posts regarding scaling volatility to the square root of time, but this doesn't look quite like what I've seen. Usually they involve multiplying some short term (i.e. daily) measure of volatility by the root of time, but this isn't quite that. Also, the documentation for the package doesn't include this chunk of re-scaling and re-centering code. Documentation: https://cran.r-project.org/web/packages/pbo/pbo.pdf
So:
Why is the data transformed in this way/what is result of this
transformation?
Is it only necessary for this simulated data, or do I need to
similarly transform real returns?
I posted this question on the r-help mailing list and got the following answer:
"Hi Joe,
The centering and re-scaling is done for the purposes of his example, and
also to be consistent with his definition of the sharpe function.
In particular, note that the sharpe function has the rf (riskfree)
parameter with a default value of .03/252 i.e. an ANNUAL 3% rate converted
to a DAILY rate, expressed in decimal.
That means that the other argument to this function, x, should be DAILY
returns, expressed in decimal.
Suppose he wanted to create random data from a distribution of returns with
ANNUAL mean MU_A and ANNUAL std deviation SIGMA_A, both stated in decimal.
The equivalent DAILY returns would have mean MU_D = MU_A / 252 and standard
deviation SIGMA_D = SIGMA_A/SQRT(252).
He calls MU_D by the name mu_base and SIGMA_D by the name sigma_base.
His loop now converts the random numbers in his matrix so that each column
has mean MU_D and std deviation SIGMA_D.
HTH,
Eric"
I followed up with this:
"If I'm understanding correctly, if I’m wanting to use actual returns from backtests rather than simulated returns, I would need to make sure my risk-adjusted return measure, sharpe ratio in this case, matches up in scale with my returns (i.e. daily returns with daily sharpe, monthly with monthly, etc). And I wouldn’t need to transform returns like the simulated returns are in the vignette, as the real returns are going to have whatever properties they have (meaning they will have whatever average and std dev they happen to have). Is that correct?"
I was told this was correct.
I have a function in R that I wish to maximise subject to some simple constraints in optim or constrOptim, but I'm struggling to get my head around ci and uito fit my constraints.
My function is:
negexpKPI <- function(alpha,beta,spend){
-sum(alpha*(1-exp(-spend/beta)))
}
where alpha and beta are fixed vectors, and spend is a vector of spends c(sp1,sp2,...,sp6) which I want to vary in order to maximise the output of negexpKPI. I want to constrain spend in three different ways:
1) Min and max for each sp1,sp2,...,sp6, i.e
0 < sp1 < 10000000
5000 < sp2 < 10000000
...
2) A total sum:
sum(spend)=90000000
3) A sum for some individual components:
sum(sp1,sp2)=5000000
Any help please? Open to any other methods that would work but would prefer base R if possible.
According to ?constrOptim:
The feasible region is defined by ‘ui %*% theta - ci >= 0’. The
starting value must be in the interior of the feasible region, but
the minimum may be on the boundary.
So it is just a matter of rewriting your constraints in matrix format. Note, an identity constraint is just two inequality constraints.
Now we can define in R:
## define by column
ui = matrix(c(1,-1,0,0,1,-1,1,-1,
0,0,1,-1,1,-1,1,-1,
0,0,0,0,0,0,1,-1,
0,0,0,0,0,0,1,-1,
0,0,0,0,0,0,1,-1,
0,0,0,0,0,0,1,-1), ncol = 6)
ci = c(0, -1000000, 5000, -1000000, 5000000, 90000000, -90000000)
Additional Note
I think there is something wrong here. sp1 + sp2 = 5000000, but both sp1 and sp2 can not be greater than 1000000. So there is no feasible region! Please fix your question first.
Sorry, I was using sample data that I hadn't fully checked; the true optimisation is for 40 sp values with 92 constraints which would if I'd replicated here in full would have made the problem more difficult to explain. I've added a few extra zeroes to make it feasible now.
I'm trying to fit the information from the G function of my data to the following mathematical mode: y = A / ((1 + (B^2)*(x^2))^((C+1)/2)) . The shape of this graph can be seen here:
http://www.wolframalpha.com/input/?i=y+%3D+1%2F+%28%281+%2B+%282%5E2%29*%28x%5E2%29%29%5E%28%282%2B1%29%2F2%29%29
Here's a basic example of what I've been doing:
data(simdat)
library(spatstat)
simdat.Gest <- Gest(simdat) #Gest is a function within spatstat (explained below)
Gvalues <- simdat.Gest$rs
Rvalues <- simdat.Gest$r
GvsR_dataframe <- data.frame(R = Rvalues, G = rev(Gvalues))
themodel <- nls(rev(Gvalues) ~ (1 / (1 + (B^2)*(R^2))^((C+1)/2)), data = GvsR_dataframe, start = list(B=0.1, C=0.1), trace = FALSE)
"Gest" is a function found within the 'spatstat' library. It is the G function, or the nearest-neighbour function, which displays the distance between particles on the independent axis, versus the probability of finding a nearest neighbour particle on the dependent axis. Thus, it begins at y=0 and hits a saturation point at y=1.
If you plot simdat.Gest, you'll notice that the curve is 's' shaped, meaning that it starts at y = 0 and ends up at y = 1. For this reason, I reveresed the vector Gvalues, which are the dependent variables. Thus, the information is in the correct orientation to be fitted the above model.
You may also notice that I've automatically set A = 1. This is because G(r) always saturates at 1, so I didn't bother keeping it in the formula.
My problem is that I keep getting errors. For the above example, I get this error:
Error in nls(rev(Gvalues) ~ (1/(1 + (B^2) * (R^2))^((C + 1)/2)), data = GvsR_dataframe, :
singular gradient
I've also been getting this error:
Error in nls(Gvalues1 ~ (1/(1 + (B^2) * (x^2))^((C + 1)/2)), data = G_r_dataframe, :
step factor 0.000488281 reduced below 'minFactor' of 0.000976562
I haven't a clue as to where the first error is coming from. The second, however, I believe was occurring because I did not pick suitable starting values for B and C.
I was hoping that someone could help me figure out where the first error was coming from. Also, what is the most effective way to pick starting values to avoid the second error?
Thanks!
As noted your problem is most likely the starting values. There are two strategies you could use:
Use brute force to find starting values. See package nls2 for a function to do this.
Try to get a sensible guess for starting values.
Depending on your values it could be possible to linearize the model.
G = (1 / (1 + (B^2)*(R^2))^((C+1)/2))
ln(G)=-(C+1)/2*ln(B^2*R^2+1)
If B^2*R^2 is large, this becomes approx. ln(G) = -(C+1)*(ln(B)+ln(R)), which is linear.
If B^2*R^2 is close to 1, it is approx. ln(G) = -(C+1)/2*ln(2), which is constant.
(Please check for errors, it was late last night due to the soccer game.)
Edit after additional information has been provided:
The data looks like it follows a cumulative distribution function. If it quacks like a duck, it most likely is a duck. And in fact ?Gest states that a CDF is estimated.
library(spatstat)
data(simdat)
simdat.Gest <- Gest(simdat)
Gvalues <- simdat.Gest$rs
Rvalues <- simdat.Gest$r
plot(Gvalues~Rvalues)
#let's try the normal CDF
fit <- nls(Gvalues~pnorm(Rvalues,mean,sd),start=list(mean=0.4,sd=0.2))
summary(fit)
lines(Rvalues,predict(fit))
#Looks not bad. There might be a better model, but not the one provided in the question.