Step back

[Weng-Kit Tan]

How do you translate this into vba code?

for i = (N-1) downto 0 do
for j = 0 to i do
C[j] = disc * ( p*C[j+1] + (1-p)*C[j] )

1.
Particularly the part about "downto", so instead of running i from 0
to (N-1), this is a step-back running from (N-1) to 0.

2.
How do you make the Function remember each individual value C[j] from
j = 0 to i

 

[Frank Kabel]

Hi
try
Dim N as integer
Dim i as integer
Dim j as integer
Dim C()

N=10
Redim C(N)
for i = (N-1) to 0 step -1
for j = 0 to i
C(j)=disc * ( p*C[j+1] + (1-p)*C[j] )
next j
next i


Whatever disc and p is