Good that you tried it now. Lets see the problem again:
You want to find minimum value for the expression, where P belongs to an interval or a point dependent on P0. Thus you need to minimize twice and then find the minima between those.
irst lets define your function F, which I posted before but I'll post again.
function Fp = F(P)
R = 3;
beta = 8;
Fp = 1 - exp(-((2^R-1)./P).^(beta/2));
Now second, as P and P0 both can vary, we need to find a minima while optimizing the values of both P and P0. Here P belong to 0 to Pb and P0 belong to 0 to 12. fminbnd cannot solve this, however fmincon can. First we'll write the function that we need to minimize. The function myFunc1 is that function. It takes a 2 variable vector where the first element is P and second element is P0.
function Y = myFunc1(P)
N = 2;
Pa = 9;
k = floor((P(2)*N/Pa));
Y = F(P(1))+k*F(N*P(2)-P(2)*k);
There is another place where P can belong to that P = ((N*Po)/(k+1)). This value of P can lie inside and outside of [0 Pb]. So we'll separately solve this for one variable P0 (as P is dependent on P0 here). The function myFunc2 does that (written below):
function Y = myFunc2(P0)
N = 2;
Pa = 9;
k = floor((P0*N/Pa));
P = N*P0/(k+1);
Y = F(P)+k*F(N*P0-P0*k);
Now lets start minimization:
First for the region where P belongs to (0 Pb)
[X1,Fval1]= fmincon(@myFunc1,rand(1,2),[],[],,[],[],[0 0],[7 12]);
Second for the case where P = ((N*Po)/(k+1))
[X2,Fval2] = fminbnd(@myFunc2,0,12);
In the end, you objective is Pi which is the minimum between both optimizations. Thus Pi wil lbe
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