Finding x-axis crossing point for cumulative cash flow

Hi,

I had a small set of data that has years (1-25) and cumulative cash flow (-200,000 to 10,000). I want to find where the line crosses the x-axis and therefore showing the payback period. However, I'm stuck :-(. I can't using interp1 as that returns the y value for a value of x. Fzero also doesn't appear to be an option as I don't know the function. Any help guys? I'm 99% sure I sound really stupid in this question, so apologies..

Thanks again..

[Oleg Komarov]

Post all 25 cash flows here (formatting with the {} button)

[David]

[-£116,549.62 -£109,094.93 -£101,763.22 -£94,558.85 -£87,485.51 -£80,546.27 -£73,743.62 -£67,079.54 -£60,555.51 -£54,172.58 -£47,931.42 -£41,832.31 -£35,875.20 -£30,059.75 -£34,816.97 -£29,282.74 -£23,887.62 -£18,630.31 -£13,509.36 -£8,523.15 -£3,669.93 £1,052.19 £5,645.19 £10,111.16 £14,452.25]

Answer 1

[Sean de Wolski]

What about fzero with interp1 as the engine of the function?

EDIT per comment Not really; N does not need to be large.

Example:

x_pts = 1:7;  %sample x values
y_pts = -5:2:7; %sample y values
f = @(x)interp1(x_pts,y_pts,x,'linear','extrap'); %must be able to extrapolate
fzero(f,6)
%find where the interpolation yields zero.

[David]

Ah, okay. So if I use something like y=interp1(year,cumulative,N), where N is a large enough array to interpolate across. I have, however, got a few problems with the fzero function, as in that I'm unsure of how to turn the above y function into an actual MATLAB function to use in the fzero line. I'm pretty new at this game, hence the slowness.

[Andrei Bobrov]

data = [-116549.62 -109094.93 -101763.22 -94558.85 -87485.51 -80546.27 -73743.62 -67079.54 -60555.51 -54172.58 -47931.42 -41832.31 -35875.20 -30059.75 -34816.97 -29282.74 -23887.62 -18630.31 -13509.36 -8523.15 -3669.93 1052.19 5645.19 10111.16 14452.25];

>>f = @(x)interp1(0:length(data)-1,data,x,'linear','extrap')

f =

@(x)interp1(0:length(data)-1,data,x,'linear','extrap')

>> fzero(f,15)

ans =

20.7771784706869

[Oleg Komarov]

f = @(x)interp1(1:11,data(15:end),x,'linear','extrap')

plot(1:25,data,'-b',fzero(f,0) + 14,0,'or')

fzero + 14 = 21.7772