To find the minimum of a function which are constrained problems

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Ashly Kurian
Ashly Kurian 2014-1-26
评论： Amit 2014-1-30
Pi = arg min F(P) + k* F( NPo − Pk)
P∈₱
with ₱ = [0, Pb) ∪ ( (N*Po) / (k+1) )
Pb=7;
F(P) = 1 − exp(−( (2^R – 1) / P ) ^ ( β/2) )
R =3;
β=8;
N=2;
k=floor((Po*N)/Pa);
Pa=9;
Po varies from 0 to 12
find the minimum value for Pi .....
pls suggest a code for this
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Amit
Amit 2014-1-28
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)
% P is [P P0]
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]);
% where [0 0] is lower bounds of [P P0] and [7 12] upper bounds
% Fval is the minimum value found in this region.
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
Pi = min(Fval1,Fval2);
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Ashly Kurian
Ashly Kurian 2014-1-29
when I run myFunc1 to find minimum following warning occurs
Warning: To use the default trust-region-reflective algorithm you must supply the gradient in the objective function and set the GradObj option to 'on'. FMINCON will use the active-set algorithm instead. For information on applicable algorithms, see Choosing the Algorithm in the documentation. > In fmincon at 520 Warning: Your current settings will run a different algorithm (interior-point) in a future release. > In fmincon at 525
Local minimum found that satisfies the constraints.
Optimization completed because the objective function is non-decreasing in feasible directions, to within the default value of the function tolerance, and constraints are satisfied to within the default value of the constraint tolerance.
criteria details
No active inequalities.
Amit
Amit 2014-1-30
The warning is self explanatory. Fmincon's default algorithm requires gradient which is not there. The gradient is not a requirement though as fmincon has other solvers that it will choose in that.
The warning is just to tell you that automatically different algorithm is selected. See afterwards, it even says 'local minima found' and blah.

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更多回答（1 个）

Amit
Amit 2014-1-26
Step 1: Make you function
function Y = myFunc(P,P0)
N = 2;
Pa = 9;
k = floor((P0*N/Pa));
Y = F(P)+k*F(N*P0-P*k);
function Fp = F(P)
R = 3;
beta = 8;
Fp = 1 - exp(-((2^R-1)./P).^(beta/2));
Step 2: Minimize it within the bounds:
P0 = 9;
[Pi, FVal] = fminbnd(@(x) myFunc(x,P0),0,7);
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Ashly Kurian
Ashly Kurian 2014-1-28
编辑：Ashly Kurian 2014-1-28
Is this code correct?
function Y = myFunc(P)
N = 2;
Pa = 9;
for P0 = 1:1:10
k = floor((P0*N/Pa));
Y = F(P)+k*F(N*P0-P*k)
end
function Fp = F(P)
R = 3;
beta = 8;
Fp = 1 - exp(-((2^R-1)./P).^(beta/2));
Step 2: Minimize it within the bounds:
[Pi, FVal] = fminbnd(@(x) myFunc(x),0,7.1)

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