Translate GAUSS code to R code

hamudenben · 11/1/11

That's a good idea. Unfortunately, I'm a mac-man

- Anyone familiar with Amazon AWS EC2 ?

alain · 11/1/11

hamudenben said:
That's a good idea. Unfortunately, I'm a mac-man
- Anyone familiar with Amazon AWS EC2 ?

the OSX version of R is already optimized.

Yike Lu · 11/1/11

http://stackoverflow.com/questions/5260068/multithreaded-blas-in-python-numpy

Go about halfway down, there are some benchmarks.

Indeed, the Mac BLAS is already much better than the basic one that would come on a Linux system. The most important thing is to convert as many for loops as possible into vector code, that is typically the largest time sink.

For comparison, I was coding up something to fill a matrix for the purpose of Fractional Brownian Motion Monte Carlo. The matrix is 7560x7560. I first did the fill using a double for-loop which took 420 seconds for the whole matrix. With vectorized code, it took only 6 seconds. Note that this is with Python, which is compiled into bytecode so presumably it is somewhat faster than R, even in the for-loops scenario.

I also had a Cholesky decomposition I needed to do. Converting Cholesky from base BLAS to MKL, the time went from 121s to 12s.

IlyaKEightSix · 11/1/11

Several things you can do:

A) Parallelize it. Look at the package doSMP for windows or doMC for linux.
B) Vectorize it. EG if you're doing a matrix inner product, instead of saying sum=0, for(i in 1:ncol(vector)){sum=vector1[,i]*vector2[i,]}, you can instead do t(vector1)%*%vector2

yeah, sorry I couldn't do it because of the GAUSS interpretation issues, but I am absolutely swamped with this quant project for my Chicago mentor =X

Pawel · 11/6/11

I have recently been trying do create a high performance R environment, and I can try to test it by running your program - I have a cluster of 5 PCs with total 8+4*4 cores.
Could you post your R code ?

hamudenben · 11/6/11

I most certainly can. Though some of it is not ALL correct yet.

This is my code for fig. 2 in the afford mentioned article (by George Pennacchi).

I know it's ugly as hell (sorry), but all you really want to do is:
- Set npath <- 10, and run it.
- If it works, go ahead and test it (npath > 1000)

Thanks,
Jens

Code:

## PROGRAM TO CREATE TWO (2) SEQUENCES OF RANDOM NUMBERS WITH A PREDETERMINED LENGTH (n) AND CORRELATION (rho) VIA THE CHOLESKY DECOMPOSITION ##

## RUN BEFOREHAND:
# NONE

# Syntax:
# a.b = a has been applied to b, i.e. t.A is t(A)
# or
# b.a = b has superscript (or similar) a, i.e. b^a or a.tilt = ã
# b_a = b has subscript a
# CAPITAL LETTERS usually SYMBOLISE A MATRIX

##

# Clear all ? Use:
# rm(list=ls())

# More than one plot in output? Use:
# par(mfrow=c(1,1))

# More than one plot in same frame? Use:
# lines(x,y)

##

# "Same" random numbers
set.seed(1987)
# Check? Use: rnorm(10)

##

#Cholesky
# Time
T <- 5 # Years maturity

# Number of paths (for interest rates and assets)
npath <- 10

rho <- -0.2 # Correlation between interest rate and ln(x)-return

#CIR
r0 <- 0.035 # Interest rate @ t=0 => same as r[1]
r.bar <- 0.069 # Equilibrium level for the interest rate
sigma_r <- 0.07 # Standard deviation for the interest rate
kappa <- 0.114 # Mean reversion parameter for the interest rate

#Jumps
mu_Y <- -0.01 # Mean for jump size
sigma_Y <- 0.02 # Standard deviation for jump size
lambda <- c(1,0,1) # Frequency (1 = once per year)

#x
g <- c(0.5,0.5,0.25) # Mean reversion parameter for capital ratio target
x.hat <- 1.1 # Capital ratio target
b0 <- 0.04 # CCB-to-deposit ratio, B/D
p <- 1 # Conversion parameter: par=1, premium>1, discount<1
e.bar <- 0.02 # equity-to-deposit ratio
sigma_x <- 0.02 # Standard deviation for assets
#x0 <- 1.075 # Starting capital-to-deposit ratio
x0.low <- 1+0.065
x0.high <- 1+0.15
x0.nint <- 10 # number of intervals between x0.low and x0.high
B <- 1 # Principal
#c <- 0.05 # Coupon
c.low <- 0.04
c.high <- 0.058
c.nint <- 10 # number of intervals between c.low and c.high
c.fit.matrix <- matrix(0,x0.nint,length(lambda))

#PENN.fn <- function(T, npath, rho, r0, r.bar, sigma_r, kappa, mu_Y, sigma_Y, lambda, g, x.hat, b0, p, e.bar, sigma_x, x0.low, x0.high, x0.nint, B, c.low, c.high, c.nint)
#{
    ##############
    n <- T*250 # 250 days per year
    dt <- T/n # Time-step

    ## Cholesky START ##
    # Building correlation-matrix, RHO
    vect <- c(1,rho,rho,1)
    RHO <- matrix(vect,nrow=2) # Correlation-matrix

    # Cholesky-decomposition of RHO
    chol.RHO <- t(chol(RHO))

    # Two UNcorrelated BMs
    dW_1 <- matrix(1,n,npath) # Making nxnpath-matrix
    dW_2 <- matrix(1,n,npath) # Same
    for(j in 1:npath) # use 'j' when npath
    {
        dW_1[,j] <- rnorm(n)*sqrt(dt)
        dW_2[,j] <- rnorm(n)*sqrt(dt)
    }

    # New correlated process (using Cholesky-decomposition)
    dW_2corr <- matrix(1,n,npath) # Making nxnpath-matrix
    for(j in 1:npath)
    {
        for(i in 1:n)
        {
            dW_2corr[i,j] <- dW_1[i,j]*chol.RHO[2,1]+dW_2[i,j]*chol.RHO[2,2]
        }
    }

    # Check/Output
    # cor.vector <- 1:npath
    # for(j in 1:npath)
    #{
    #    print(cor(dW_1[,j],dW_2corr[,j]))
    #    cor.vector[j] <- cor(dW_1[,j],dW_2corr[,j])
    #}
    #matrix(c("Number of sequences", "Random numbers per sequence", "Number of paths", "Mean correlation", 2, n, npath, mean(cor.vector)), nrow=4)
    ## Cholesky END ##

    ##################

    ## PROGRAM TO SIMULATE A CIR-INTEREST RATE WITH A PREDETERMINED CORRELATION TO ASSETS ##

    ## RUN BEFOREHAND:
    # PENN.CHOL

    ## Interest rate START ##
    # The change in the default-free interest rate from day t to day t+dt (p. 17)

    # Parametres for interest rate modeling (p. 18)
    r <- matrix(r0,n+1,npath) # Making r a "(n+1)xnpath"-matrix, and effectively assigning r0 as first entry in every column

    # Stochastic interest rate, CIR-process
    for(j in 1:npath)
    {
        for(i in 1:n)
        {
            r[i+1,j] <- r[i,j] + kappa*(r.bar-r[i,j])*dt+sigma_r*sqrt(r[i,j])*dW_2corr[i,j]
        }
    }

    ## Interest rate END ##

    ##################

    for(w in 1:length(lambda))
    {
            ## PROGRAM TO SIMULATE A JUMPS IN ASSETS VALUE WITH A PREDETERMINED INTENSITY (lambda) AND SIZE (mu_Y) ##

        ## RUN BEFOREHAND:
        # PENN.CHOL(esky)
        # PENN.CIR.RATE

        ## Jumps START ##
        # Jump-process (p.17)

        # Parametres for jump modeling (p. 18)
        phi <- matrix(rbinom(n%*%npath,1,dt*lambda[w]),n,npath) # Jump or not? Binomial variable
        ln.Y <- matrix(rnorm(n%*%npath,mu_Y,sigma_Y),n,npath) # (Actual) jump size
        ## Jumps END ##

        ##################

        ##################

        ## PROGRAM TO SIMULATE THE DAILY RISK NEUTRAL PROCESS FOR THE LOG OF THE BANK'S ASSET-TO-DEPOSIT RATIO (p. 17) ##

        ## RUN BEFOREHAND:
        # PENN.CHOLESKY
        # PENN.CIR.RATE
        # PENN.JUMPS

        ## ln(x) START ##
        # Daily risk-neutral process for the log of the bank's asset-to-deposit ratio (p. 17)

        # Parametres for ln(x) modeling (p. 18)
        b <- matrix(b0,n+1,npath) # Making b a (n+1)xnpath matrix and effectively assigning b0 as the first entry of b (of all entries actually)
        x.bar0 <- 1+e.bar+p*b0 # Total capital-to-deposit ratio (conversion threshold)
        x.bar <- matrix(x.bar0,n+1,j)

        h <- matrix(1,n,npath) # Making h a nxnpath-matrix

        # For now:
        k <- exp(mu_Y+0.5*sigma_Y^2)-1 # k = E[Y-1], p. 14 (???)

        # Let the games begin
        # First: The regression part
        c <- seq(c.low, c.high, length=c.nint)
        x0 <- seq(x0.low,x0.high,length=x0.nint)
        reg.matrix <- matrix(0,c.nint, length(x0))

        reg.fit1 <- 1:x0.nint
        reg.fit2 <- 1:x0.nint
        c.fit <- 1:x0.nint

        for(l in 1:x0.nint) # Use l for x0
        {
            for(m in 1:c.nint) # Use m for c
            {
                x <- matrix(x0[l],n+1,npath) # Making x a nxnpath-matrix, and assigning x0 as the value of all entries in x (thus effectively making x0 the first value of x)
                ln.x0 <- matrix(log(x0[l]),n+1,npath) # Same
                ln.x <- ln.x0 # Same
                binom.c <- matrix(1,n+1,npath) # Making binom.c a nxnpath-matrix

                # Calculating: d_1, d_2, h, b, ln.x, x, binom.c
                for(j in 1:npath)
                {
                    for(i in 1:n)
                    {
                        # d_1 & d_2
                        d_1 <- (ln.x[i,j]+mu_Y)/sigma_Y
                        d_2 <- d_1+sigma_Y

                        # h
                        h[i,j] <- lambda[w]*(pnorm(-d_1)-exp(ln.x[i,j])*exp(mu_Y+0.5*sigma_Y^2)*pnorm(-d_2))

                        # b
                        b[i+1,j] <- b[i,j]*exp(-g[w]*(exp(ln.x[i,j])-x.hat)*dt)

                        # ln.x
                        ln.x[i+1,j] <- ln.x[i,j] + ( (r[i,j]-lambda[w]*k) - (r[i,j]+h[i,j]+c[m]*b[i,j])/exp(ln.x[i,j]) - g[w]*(exp(ln.x[i,j])-x.hat) - 0.5*sigma_x^2 )*dt + sigma_x*dW_1[i,j] + ln.Y[i,j]*phi[i,j]

                        # x
                        x[i+1,j] <- exp(ln.x[i+1,j])

                        # x.bar
                        x.bar[i+1,j] <- 1+e.bar+p*b[i+1,j]

                        # binom.c
                        # Creating vector with 1 if payment (i.e. trigger point not reached) and 0 if no payment (i.e. trigger point reached)
                        if(x[i+1,j]>=x.bar[i+1,j] && binom.c[i,j]>0.5)
                        {
                            binom.c[i+1,j] <- 1
                        }else
                        {
                            binom.c[i+1,j] <- 0
                        }
                    }
                }

                # Creates vector(s) with payments
                # Finds the lump sum (lump.c) that will be paid in the event of trigger/no trigger
                payments <- matrix(c(rep(c[m]*dt, n-1),B), n, npath)*binom.c[1:n,] # OBS: dt multiplied here!!
                for(j in 1:npath)
                {
                    for(i in 2:n)
                    {
                        if(payments[i,j] == 0 && p*b[sum(binom.c[,j])+1,j] <= x[sum(binom.c[,j])+1,j]-1 )
                        {
                            payments[i,j] <- p*B
                            break
                        }else if(payments[i,j] == 0 && 0 < x[sum(binom.c[,j])+1,j]-1 && x[sum(binom.c[,j])+1,j]-1 < p*b[sum(binom.c[,j])+1,j])
                        {
                            payments[i,j] <- (x[sum(binom.c[,j])+1,j]-1)*B/b[sum(binom.c[,j])+1,j]
                            break
                        }
                        else
                        {
                            payments[i,j] <- payments[i,j]
                        }
                    }
                }

                # Plotting 1+x
                #plot(1:(n+1), x[,1], type="l", ylim=c(min(x),max(x)), ylab="1+x", #xlab="t")
                #for(j in 2:npath) lines(1:(n+1), x[,j], col=j)
                #legend("topleft",expression(ln(x_t)),lty=1,col="black")
                ## ln(x) END ##

                ##################

                ## PROGRAM TO VALUATE THE CONTINGENT CAPITAL FOR A GIVEN FIXED-COUPON RATE (p. 15) ##

                ## RUN BEFOREHAND:
                # PENN.CHOLESKY
                # PENN.CIR.RATE
                # PENN.JUMPS
                # PENN.LNX

                ## Fixed-coupon START ##

                # S exp(-S r ds) * v(t) dt
                vec.disc.v <- rep(0, npath) #disc = discounting
                for(j in 1:npath)
                {
                    disc.v <- 0
                    int.r <- 0

                    for(i in 1:n)
                    {
                        int.r <- int.r + r[i,j]*dt
                        disc.v <- disc.v + exp(-int.r)*payments[i,j]

                    }

                    vec.disc.v[j] <- disc.v
                }

                V0 <- mean(vec.disc.v)

                reg.matrix[m,l] <- V0
                ## Fixed-coupon END ##

            }

            reg.fit1[l] <- lm(reg.matrix[,l]~1+c)$coef[1]
            reg.fit2[l] <- lm(reg.matrix[,l]~1+c)$coef[2]

            c.fit[l] <- (B-reg.fit1[l])/reg.fit2[l]
        }
        c.fit.matrix[,w] <- c.fit
    }

#}

#PENN.fn(5,100,-0.2,0.035,0.069,0.07,0.114,-0.01,0.02,1,0.5,1.1,0.04,1,0.02,0.02,1.065,1.15,5,1,0.04,0.06,5)

# Graphics
plot((x0-1), c.fit.matrix[,1], type="l", col=2, ylim=c(c.low,c.high), xlab="Capital to Deposits", ylab="Coupon Rate")
for(w in 2:length(lambda)) lines((x0-1),c.fit.matrix[,w], col=w+1)
lines((x0-1),rep(0.0423,length(x0)), col=1, lty=2)
legend("topright", c("A","B","C","D"), col=c(1,2,3,4), lty=c(2,1,1,1))

hamudenben · 11/6/11

The program is sort of a fusion (with a special extension in order to get the wanted graph) of the following programs:

Code:

## PROGRAM TO CREATE TWO (2) SEQUENCES OF RANDOM NUMBERS WITH A PREDETERMINED LENGTH (n) AND CORRELATION (rho) VIA THE CHOLESKY DECOMPOSITION ##

## RUN BEFOREHAND:
# NONE

# Syntax:
# a.b = a has been applied to b, i.e. t.A is t(A)
# or
# b.a = b has superscript (or similar) a, i.e. b^a or a.tilt = ã
# b_a = b has subscript a
# CAPITAL LETTERS usually SYMBOLISE A MATRIX

##

# Clear all ? Use:
# rm(list=ls())

# More than one plot in output? Use:
# par(mfrow=c(1,1))

# More than one plot in same frame? Use:
# lines(x,y)

##

# Time
T <- 5 # Years maturity
n <- T*250 # 250 days per year
dt <- T/n # Time-step

# Number of paths (interest rates and assets)
npath <- 1000

# "Same" random numbers
set.seed(1987)
# Check? Use: rnorm(10)

## Cholesky START ##
# Building correlation-matrix, RHO
rho <- -0.2 # Correlation between interest rate and ln(x)-return
vect <- c(1,rho,rho,1)
RHO <- matrix(vect,nrow=2) # Correlation-matrix

# Cholesky-decomposition of RHO
chol.RHO <- t(chol(RHO))

# Two UNcorrelated BMs
dW_1 <- matrix(1,n,npath) # Making nxnpath-matrix
dW_2 <- matrix(1,n,npath) # Same
for(j in 1:npath) # use 'j' when npath
{
    dW_1[,j] <- rnorm(n)*sqrt(dt)
    dW_2[,j] <- rnorm(n)*sqrt(dt)
}

# New correlated process (using Cholesky-decomposition)
dW_2corr <- matrix(1,n,npath) # Making nxnpath-matrix
for(j in 1:npath)
{
    for(i in 1:n)
    {
        dW_2corr[i,j] <- dW_1[i,j]*chol.RHO[2,1]+dW_2[i,j]*chol.RHO[2,2]
    }
}

# Check/Output
cor.vector <- 1:npath
for(j in 1:npath)
{
    print(cor(dW_1[,j],dW_2corr[,j]))
    cor.vector[j] <- cor(dW_1[,j],dW_2corr[,j])
}
matrix(c("Number of sequences", "Random numbers per sequence", "Number of paths", "Mean correlation", 2, n, npath, mean(cor.vector)), nrow=4)
## Cholesky END ##

and

Code:

## PROGRAM TO SIMULATE A CIR-INTEREST RATE WITH A PREDETERMINED CORRELATION TO ASSETS ##

## RUN BEFOREHAND:
# PENN.CHOL

## Interest rate START ##
# The change in the default-free interest rate from day t to day t+dt (p. 17)

# Parametres for interest rate modeling (p. 18)
r0 <- 0.035 # Interest rate @ t=0 => same as r[1]
    r <- matrix(r0,n+1,npath) # Making r a "(n+1)xnpath"-matrix, and effectively assigning r0 as first entry in every column
r.bar <- 0.069 # Equilibrium level for the interest rate
sigma_r <- 0.07 # Standard deviation for the interest rate
kappa <- 0.114 # Mean reversion parameter for the interest rate

# Stochastic interest rate, CIR-process
for(j in 1:npath)
{
    for(i in 1:n)
    {
        r[i+1,j] <- r[i,j] + kappa*(r.bar-r[i,j])*dt+sigma_r*sqrt(r[i,j])*dW_2corr[i,j]
    }
}

# Plotting r
plot(1:(n+1),r[,1], type="l", xlab="t", ylab="r", col=1)
for(j in 2:npath) lines(1:(n+1),r[,j], col=j)
#legend("bottomright","r",lty=1,col="black")

# Check/Output
matrix(c("Number of sequences", "Random numbers per sequence", "Mean reversion level", "Mean for CIR-process", 1, n, r.bar, mean(r)), nrow=4)
## Interest rate END ##

and

Code:

## PROGRAM TO SIMULATE A JUMPS IN ASSETS VALUE WITH A PREDETERMINED INTENSITY (lambda) AND SIZE (mu_Y) ##

## RUN BEFOREHAND:
# PENN.CHOL(esky)
# PENN.CIR.RATE

## Jumps START ##
# Jump-process (p.17)

# Parametres for jump modeling (p. 18)
mu_Y <- -0.01 # Mean for jump size
sigma_Y <- 0.02 # Standard deviation for jump size
lambda <- 1 # Frequency (1 = once per year)
phi <- matrix(rbinom(n%*%npath,1,dt*lambda),n,npath) # Jump or not? Binomial variable
ln.Y <- matrix(rnorm(n%*%npath,mu_Y,sigma_Y),n,npath) # (Actual) jump size

# Plotting jumps
plot(1:n,ln.Y[,1]*phi[,1], type="lines", xlab="t", ylab="Jump size", col=1)
for(j in 2:npath) lines(1:n,ln.Y[,j]*phi[,j], col=j)

# Check/Output
matrix(c("lambda check","jump size check",sum(phi)/(T*npath), mean(ln.Y)),2,2)
## Jumps END ##

and finally

Code:

## PROGRAM TO SIMULATE THE DAILY RISK NEUTRAL PROCESS FOR THE LOG OF THE BANK'S ASSET-TO-DEPOSIT RATIO (p. 17) ##

## RUN BEFOREHAND:
# PENN.CHOLESKY
# PENN.CIR.RATE
# PENN.JUMPS

## ln(x) START ##
# Daily risk-neutral process for the log of the bank's asset-to-deposit ratio (p. 17)

# Parametres for ln(x) modeling (p. 18)
B <- 1 #The principal
c <- 0.05
g <- 0.5 # Mean reversion parameter for capital ratio target
x.hat <- 1.1 # Capital ratio target
b0 <- 0.04 # CCB-to-deposit ratio, B/D
    b <- matrix(b0,n+1,npath) # Making b a (n+1)xnpath matrix and effectively assigning b0 as the first entry of b (of all entries actually)
p <- 1 # Conversion parameter: par=1, premium>1, discount<1
e.bar <- 0.02 # equity-to-deposit ratio
x.bar <- 1+e.bar+p*b0 # Total capital-to-deposit ratio (conversion threshold)
sigma_x <- 0.02 # Standard deviation for assets

h <- matrix(1,n,npath) # Making h a nxnpath-matrix

# For now:
k <- 1-exp(-mu_Y+0.5*sigma_Y^2) # k = E[Y-1], p. 14 (???)

# Let the games begin
x0 <- 1.075 # Starting capital-to-deposit ratio
x <- matrix(x0,n+1,npath) # Making x a nxnpath-matrix, and assigning x0 as the value of all entries in x (thus effectively making x0 the first value of x)
ln.x0 <- matrix(log(x0),n+1,npath) # Same
ln.x <- ln.x0 # Same
binom.c <- matrix(1,n+1,npath) # Making binom.c a nxnpath-matrix

# Calculating: d_1, d_2, h, b, ln.x, x, binom.c
for(j in 1:npath)
{
    for(i in 1:n)
    {
        # d_1 & d_2
        d_1 <- (ln.x[i,j]+mu_Y)/sigma_Y
        d_2 <- d_1+sigma_Y

        # h
        h[i,j] <- lambda*(pnorm(-d_1)-exp(ln.x[i,j])*exp(mu_Y+0.5*sigma_Y^2)*pnorm(-d_2))

        # b
        b[i+1,j] <- b[i,j]*exp(-g*(exp(ln.x[i,j])-x.hat)*dt)

        # ln.x
        ln.x[i+1,j] <- ln.x[i,j] + ( (r[i,j]-lambda*k) - (r[i,j]+h[i,j]+c*b[i,j])/exp(ln.x[i,j]) - g*(exp(ln.x[i,j])-x.hat) - 0.5*sigma_x^2 )*dt + sigma_x*dW_1[i,j] + ln.Y[i,j]*phi[i,j]

        # x
        x[i+1,j] <- exp(ln.x[i,j])

        # binom.c
        # Creating vector with 1 if payment (i.e. trigger point not reached) and 0 if no payment (i.e. trigger point reached)
        if(x[i,j]>=(1+b[i,j]) && binom.c[i,j]>0.5)
        {
            binom.c[i+1,j] <- 1
        }else
        {
            binom.c[i+1,j] <- 0
        }
    }
}

# Creates vector(s) with payments
# Finds the lump sum (lump.c) that will be paid in the event of trigger/no trigger
payments <- matrix(c(rep(c*dt, n-1),B), n, npath)*binom.c[1:n,] # OBS: dt multiplied here!!
for(j in 1:npath)
{
    for(i in 2:n)
    {
        if(payments[i,j] == 0 && p*b[sum(binom.c[,j])+1,j] <= x[sum(binom.c[,j])+1]-1 )
        {
            payments[i,j] <- p*B
            break
        }else if(payments[i,j] == 0 && 0 < x[sum(binom.c[,j])+1]-1 && x[sum(binom.c[,j])+1]-1 < p*b[sum(binom.c[,j])+1,j])
        {
            payments[i,j] <- (x[sum(binom.c[,j])+1]-1)*B/b[sum(binom.c[,j])+1,j]
            print(j)
            break
        }
        else
        {
            payments[i,j] <- payments[i,j]
        }
    }
}

# Plotting 1+x
plot(1:(n+1), x[,1], type="l", ylim=c(min(x),max(x)), ylab="1+x", xlab="t")
for(j in 2:npath) lines(1:(n+1), x[,j], col=j)
legend("topleft",expression(ln(x_t)),lty=1,col="black")
## ln(x) END ##

dim(payments)
plot(1:dim(payments)[2],payments[7250,])

Final <- payments[dim(payments)[1],]
(1-sum(Final)/length(Final))*0.1
(1-sum(Final)/length(Final))*0.2

hamudenben · 11/8/11

Pawel said:
I have recently been trying do create a high performance R environment, and I can try to test it by running your program - I have a cluster of 5 PCs with total 8+4*4 cores.
Could you post your R code ?

If you manage to run the code please PM me your title and position, so I can mention you in my thesis.

Pawel · 11/8/11

Sorry, for not responding for so long ...,
I did manage to run it on 8 cores just to test the speed up and it seems 8 times faster

- it occured my cluster is not fully functional yet so I could not run it on 24 cores ...

Anyway, it seems to me (as someone wrote) that you could gain some extra speed by replacing element by element calculations with matrix calculations (you should get rid of element access [] whereever possible).

I used "foreach" package and "doMC", I changed the loop of l:

Code:

c.fit.matrix[,w] <- foreach(l = 1:x0.nint, .combine=rbind) %dopar% # Use l for x0

and so the last line of this loop is:

Code:

         #c.fit.matrix[,w] <- c.fit
         c.fit

hamudenben · 11/8/11

1337! And no worries. If I optimized the code could you run it for me? Ofc, I'll pay you for any expenses that may be relate to this.

Andy Nguyen · 11/8/11

I'd like to see the graph of any speedup and all, if just for some nerdy curiosity.

hamudenben · 11/8/11

Well - If my code is just vaguely correct, you could check the appendices in the original article

hamudenben · 11/8/11

hamudenben said:
Well - If my code is just vaguely correct, you could check the appendices in the original article

However, I'm pretty sure it doesn't impress Asian-Father-meme.

Yike Lu · 11/8/11

Oh also, you can use

Code:

library(compile)

compiledversion=cmpfun(myfunction)

which compiles it into bytecode, should get around a 2x or 3x improvement for very little effort.

hamudenben · 11/15/11

Oh snap, do you know how to deal with this:

Code:

> install.packages("compile")
Warning message:
In getDependencies(pkgs, dependencies, available, lib) :
  package 'compile' is not available (for R version 2.13.2)

hamudenben · 11/15/11

Okay, back to the main problem.

In hope that someone might be able to help me, I will go ahead, and post my strategy for how to create figure 9. - 12. I'm supposed to find the change in value of equity for a change in value of X. And X is jump risk (lambda), volatility for jump size (sigma_y) etc. for the different figures 9. - 12.
Since equity + debt by definition is equal to the total value of the firm then changes in debt value must equal minus the (same) changes in equity value (because risk-neutral valuation is used and thus no extra gain can be made - solely value transfer).
This means that I should be able to do the following:
- Run my program for Figure 2 (posted above).
- Obtain value of debt and coupons.
- Change, say, lambda, and run program again.
- BUT: Use (old) coupons.
- Obtain new value of debt.
- Use that Change in debt = - Change in equity
Can anyone follow my logic? I think it makes sense, I just can't seem to implement it. If anyone can see how this could be implemented, I would be very grateful, if you would let me know.
Best regards,
Jens

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