Nested if conditions to compute flow

My dataset includes time series data of rainfall and CN for one year. I am trying to determine the flow for two conditions i.e

if Rainfall > 0.2 then  
    Runoff = 1000/CN-10 
otherwise 
    Runoff = (Rainfall - 0.2 * (1000/CN -10)^2)/Rainfall + 0.8 *(1000/CN -10).

I have tried following script but it did not work.

   idx = Rainfall > 0.2;
   Runoff(idx) = 1000/CN - 10;
   Runoff(~idx) = (Rainfall - 0.2 * (1000/CN -10)^2)/Rainfall + 0.8 *(1000/CN -10);

Can any body suggest me script?

Regards, Saleem

[per isakson]

"but it did not work" &nbsp What happened? Error or what? &nbsp Rainfall and Runoff how are they defined?

Answer 1

[Eng. Fredius Magige]

Try this one:

dataset=[]; % copy you three columns of time, rainfall and CN into [], preferable from excel sheet 
% note arrangement of you data very important 1st time, 2nd rainfall and 3rd CN

[datarow datacol]=size(dataset)
hold on
for n=1:datarow
Rainfall=data(n,2)
if (Rainfall > 0.2) 
    Runoff = 1000/(dataset(n,3)-10)
else
    Runoff = (Rainfall - 0.2 * (1000/(dataset(n,3)-10))^2)/Rainfall + 0.8 *(1000/(dataset(n,3)-10)). 
end
hold off

[Walter Roberson]

You would need to assign to Runoff(n) not just to Runoff

[Eng. Fredius Magige]

% The previous just after for, the data is edited be dataset. otherwise be specific, inclusive CN %

[datarow datacol]=size(dataset) hold on for n=1:datarow Rainfall=dataset(n,2) if (Rainfall > 0.2) Runoff = 1000/(dataset(n,3)-10) else Runoff = (Rainfall - 0.2 * (1000/(dataset(n,3)-10))^2)/Rainfall + 0.8 *(1000/(dataset(n,3)-10)). end hold off

Answer 2

[Thorsten]

Does this work for you?

 idx = Rainfall > 0.2;
 Runoff(idx) = 1000/CN(idx) - 10;
 Runoff(~idx) = (Rainfall(~idx) - 0.2 * (1000./CN(~idx)-10).^2)./Rainfall(~idx) + 0.8 *(1000./CN(~idx) -10);

If not, please report exactly what went wrong.

[Walter Roberson]

Runoff(idx) = 1000./CN(idx) - 10;

[Saleem Sarwar]

No, this will not work as I have to compute excess rainfall fulfilling these two conditions 1. if Rainfall - 0.2*S < 0, then excess rainfall will be 0 otherwise excess rainfall = (Rainfall -0.2*s)^2/ Rainfall + 0.8*S) where rainfall and S are daily time series data of years.

[Thorsten]

What about this? And do your really mean -0.2*s instead of -0.2*S in the "otherwise" condition?

 idx = Rainfall - 0.2*S < 0;
 ExcessRainfall(idx) = 0;
 ExcessRainfall(~idx) = (Rainfall(~idx) - 0.2*s(~idx)).^2./Rainfall(~idx) + 0.8*S(~idx);