I am trying to solve a function where the values are implicit given within two equations. I have to use an iterative method.
Values that are already given are V_DCL, barrier, us_r, time.
My code for the function is:
Test <- function(V_DCL, barrier, us_r, time, vola_A) {
V_A <- V_DCL/(pnorm(d_1, 0, 1)) + (barrier*exp(us_r*time)*pnorm(d_2, 0, 1))/(pnorm(d_1, 0, 1))
d_1 <- (log(V_A/barrier)+(us_r + 0.5*vola_A^2)*t)/(vola_A*sqrt(time))
d_2 <- d_1 - vola_A*sqrt(time)
outputs <- list(V_A, d_1, d_2, vola)
return(outputs)
}
I need to get a value for V_A and vola_A.
This function is similar to the black scholes formula. But I do not have a value for the value of the asset but for the liabilities, so I rearranged it.
So far I know that I have to make an initial guess for vola_A which need to get changed until all equations fit.
I already looked into the base repeat() function and in the package simecol. But I did not figure out how to apply it on my code.
Can you give me some ideas? Thank you.
Edit (additional information):
The given data for us_r is 0.05, time stands for time horizon and I will work with 1 period for now. The given value for barrier is 2.683782e+13 and for V_DCL it is 4.732741e+11.
Related
I would like to solve a differential equation in R (with deSolve?) for which I do not have the initial condition, but only the final condition of the state variable. How can this be done?
The typical code is: ode(times, y, parameters, function ...) where y is the initial condition and function defines the differential equation.
Are your equations time reversible, that is, can you change your differential equations so they run backward in time? Most typically this will just mean reversing the sign of the gradient. For example, for a simple exponential growth model with rate r (gradient of x = r*x) then flipping the sign makes the gradient -r*x and generates exponential decay rather than exponential growth.
If so, all you have to do is use your final condition(s) as your initial condition(s), change the signs of the gradients, and you're done.
As suggested by #LutzLehmann, there's an even easier answer: ode can handle negative time steps, so just enter your time vector as (t_end, 0). Here's an example, using f'(x) = r*x (i.e. exponential growth). If f(1) = 3, r=1, and we want the value at t=0, analytically we would say:
x(T) = x(0) * exp(r*T)
x(0) = x(T) * exp(-r*T)
= 3 * exp(-1*1)
= 1.103638
Now let's try it in R:
library(deSolve)
g <- function(t, y, parms) { list(parms*y) }
res <- ode(3, times = c(1, 0), func = g, parms = 1)
print(res)
## time 1
## 1 1 3.000000
## 2 0 1.103639
I initially misread your question as stating that you knew both the initial and final conditions. This type of problem is called a boundary value problem and requires a separate class of numerical algorithms from standard (more elementary) initial-value problems.
library(sos)
findFn("{boundary value problem}")
tells us that there are several R packages on CRAN (bvpSolve looks the most promising) for solving these kinds of problems.
Given a differential equation
y'(t) = F(t,y(t))
over the interval [t0,tf] where y(tf)=yf is given as initial condition, one can transform this into the standard form by considering
x(s) = y(tf - s)
==> x'(s) = - y'(tf-s) = - F( tf-s, y(tf-s) )
x'(s) = - F( tf-s, x(s) )
now with
x(0) = x0 = yf.
This should be easy to code using wrapper functions and in the end some list reversal to get from x to y.
Some ODE solvers also allow negative step sizes, so that one can simply give the times for the construction of y in the descending order tf to t0 without using some intermediary x.
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
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.
I am using the ODE function In R in order to solve this equation:
library(deSolve)
FluidH <- function(t,state,parameters) {
with(as.list(c(state,parameters)),
dh <- Qin/A - ((5073.3*h^2+6430.1*h)/(60*A))
list(c(dh))
})
}
parameters <- c(Qin =10, A=6200)
state<- c(h=0.35)
time <- seq(0,2000,by=1)
out <- ode(y= state, func = FluidH, parms = parameters, times = time)
I might be missing something with math, but when I try to calculate h by myself by assigning the initial state I don't get the same numbers as the output of the function!
for example to calculate h at time 1 : h=h0+ dh*dt -> h= 0.35 + 10/6200 - ((5073.3*h^2+6430.1*h)/(60*6200))=0.3438924348
and the output of ode gives 0.343973044412394
Can anyone tell what am I missing?
You computed the Euler step with step size dt=1. The solver uses a higher order method with (usually) a smaller step size that is adapted to meet the default error tolerances of 1e-6 for relative and absolute error. The step-size 1 that you give only determines where the numerical solution is sampled for the output, internally the solver may use many more or sometimes even less steps (interpolating the output values).