Repeating for loop until getting a list of solution

Hi to all

I'm not that good in Matlab and I tried to learn it. I used for loop to create a solution for each value of (n) and repeating the for loop. The for-loop is from 1 to 8760 while n is from 0 to 1 with increament of 0.1.

I want to repeat the for-loop for each (n) and finally make a vecto of [n, solution from for-loop]. Attached flow chart may explain more about what I want.

[Abdullah Al Shereiqi]

More clear flow chart attached

[Sriram Tadavarty]

Can you place the code you tried so far and the issue that came up?

[Rik]

Have a read here and here. It will greatly improve your chances of getting an answer.

It seems you need something like this, but I'm not sure what you are trying to do exactly.

n=0;
for k=1:8760
    n=SomeFunction(k);
    if n==1
        break
    else
        n=n+0.1;
    end
end

[Abdullah Al Shereiqi]

Thanks Saruram and Rik for quick support.

I tried to run the code as per Rik comment. But, the value of (s) reach beyond 1 (more than 1000)!!

s=0;
for apv = 1:length(P_wind)
    if P_PV(apv)==0 ;
        N_PV_1(apv)=0;
    else
        N_PV_1(apv)= ceil (((s*P_ref(apv)))./(P_PV(apv)));
    end
    if s==1;
        break
    else
        s = s+0.01
    end
end

[Rik]

Apart from the suggestion from Guillaume; I forgot to account for floating point rounding errors.

s=0;
for k=1:10
    s=s+0.1;
end
clc
fprintf('%.53f\n',s) %not quite equal to 1

The solution in this case is to use a tolerance:

s==1 %returns false, because s is 0.99999999999999988897769753748434595763683319091796875
abs(s-1)<=2*eps %returns true

[Sriram Tadavarty]

You could even break when s is greater than 1, example, the condition of if s > 1, can be placed, rather than s == 1 to deal with tolerance issues.

Or it can be directly placed a loop for 100 times, in each time 0.01 can be added.

Answer 1

[Guillaume]

If it's a flow chart you've been given, then you are entitled to complain loudly. That flow chart is very misleading. The layout is a more representative of a while loop than a for loop.

This is what you're meant to do:

for n = 0:0.1:1 %represented by 4 boxes on the flow chart! (n = 0, for-loop, n=1?, increment by 0.1)
    %do calculation inside the loop (1 box on the chart)
end
%get a vector of solution (1 box on the chart)

Answer 2

[Abdullah Al Shereiqi]

Thanks to all of you; Saruram , Rik and Guillaume.

It seems working fine after using another for-loop (100 times).

Appreciated your support and guidance