double for loop in Binomial Tree in R - r

Question

I want set up a model for interest rate in binomial tree. The interest rate is path dependent. I want return interest rate (discount factor and payoff) at every step in all scenarios(2^N). The reason I want to return every single interest rate is that I want use the interest rate is compute discount factor. I know how to do this in a complex way. Here I want to use a double loop (or something simpler) to get the results.
w is for "0" or "1" dummy variable matrix representing all scenarios.
r is interest rate. if there is a head(1), then r1=r0+u=r0+0.005; if there is a tail(0), then r1=r0-d.
D is discount factor. D1=1/(1+r0), D2=D1/(1+r1)...
P is payoff.
In this case, period N is 10. therefore, I can compute step by step. However,if N gets larger and larger, I cannot use my method. I want a simple way to compute this. Thank you.
#Real Price
N <- 10
r0 <- 0.06
K <- 0.05
u <- 0.005
d <- 0.004
q <- 0.5
w <- expand.grid(rep(list(0:1),N))
r <- D <- P <- matrix(0,0,nrow=2^N,ncol=N)
for(i in 1:dim(w)[1])
{
r[i,1] <- r0 + u*w[i,1] - d*(1-w[i,1])
r[i,2] <- r[i,1] + u*w[i,2] - d*(1-w[i,2])
r[i,3] <- r[i,2]+u*w[i,3]-d*(1-w[i,3])
r[i,4] <- r[i,3]+u*w[i,4]-d*(1-w[i,4])
r[i,5] <- r[i,4]+u*w[i,5]-d*(1-w[i,5])
r[i,6] <- r[i,5]+u*w[i,6]-d*(1-w[i,6])
r[i,7] <- r[i,6]+u*w[i,7]-d*(1-w[i,7])
r[i,8] <- r[i,7]+u*w[i,8]-d*(1-w[i,8])
r[i,9] <- r[i,8]+u*w[i,9]-d*(1-w[i,9])
r[i,10] <- r[i,9]*+u*w[i,10]-d*(1-w[i,10])
D[i,1] <- 1/(1+r0)
D[i,2] <- D[i,1]/(1+r[i,1])
D[i,3] <- D[i,2]/(1+r[i,2])
D[i,4] <- D[i,3]/(1+r[i,3])
D[i,5] <- D[i,4]/(1+r[i,4])
D[i,6] <- D[i,5]/(1+r[i,5])
D[i,7] <- D[i,6]/(1+r[i,6])
D[i,8] <- D[i,7]/(1+r[i,7])
D[i,9] <- D[i,8]/(1+r[i,8])
D[i,10] <- D[i,9]/(1+r[i,9])
P[i,1] <- D[i,1]*pmax(K-r0,0)*(0.5^N)
P[i,2] <- D[i,2]*pmax(K-r[i,1],0)*(0.5^N)
P[i,3] <- D[i,3]*pmax(K-r[i,2],0)*(0.5^N)
P[i,4] <- D[i,4]*pmax(K-r[i,3],0)*(0.5^N)
P[i,5] <- D[i,5]*pmax(K-r[i,4],0)*(0.5^N)
P[i,6] <- D[i,6]*pmax(K-r[i,5],0)*(0.5^N)
P[i,7] <- D[i,7]*pmax(K-r[i,6],0)*(0.5^N)
P[i,8] <- D[i,8]*pmax(K-r[i,7],0)*(0.5^N)
P[i,9] <- D[i,9]*pmax(K-r[i,8],0)*(0.5^N)
P[i,10] <- D[i,10]*pmax(K-r[i,9],0)*(0.5^N)
}
true.price <- sum(P)
#> true.price
# > true.price
# [1] 0.00292045

Answer 1

You can just use a nested loop, looping over 2:(ncol(w)) within the i loop:
for(i in 1:nrow(w)) {
r[i, 1] <- r0 + u*w[i, 1] - d*(1-w[i, 1])
D[i, 1] <- 1/(1+r0)
P[i, 1] <- D[i, 1]*pmax(K-r0, 0)*(0.5^N)
for (j in 2:(ncol(w))) {
r[i,j] <- r[i, j-1] + u*w[i, j] - d*(1-w[i, j])
D[i,j] <- D[i, j-1]/(1+r[i, j-1])
P[i,j] <- D[i, j]*pmax(K-r[i, j-1], 0)*(0.5^N)
}
}
true.price <- sum(P)