To find the minimum of a function which are constrained problems

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

[Ashly Kurian]

the minimum should be found for both parts of the eqn....

[Amit]

So you're trying to find P such that F(P) + k*F(N*P0-Pk) is minimum?

[Ashly Kurian]

s...

[Amit]

what is s...

[Ashly Kurian]

sorry...yes

Answer 1

[Amit]

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);

[Ashly Kurian]

Thanks a lot...

[Ashly Kurian]

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]

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.

Answer 2

[Amit]

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);

[Ashly Kurian]

Po is not 9...it varies from 0 to 12...what is x

[Amit]

X is a simple variable for the function. You said P0 varies from 0 to 12. From that statement, I thought that for a scenario, P0 is constant.

Please state your question clearly. That includes the objective of the problem. Also, MATLAB has a very good help. Try seeing what different function do and how can you use it.

[Ashly Kurian]

also P∈₱ where ₱ = [0, Pb) ∪ ( (N*Po) / (k+1) )

[Amit]

Is P0 integer or a real number?

[Ashly Kurian]

Considering the equation the first part F(P) .. the function is given as follows

F(P) = 1 − exp(−((2^R – 1)/P)^(β/2))

Here R and β are constant and P belongs to ₱ where ₱ = [0, Pb)∪ ((N*Po)/(k+1))…(that U is referred as union)

Pb=7

k=floor((Po*N)/Pa);

 Pa=9; N=2

Po varies from 0 to 12

Considering the second part k* F( (N*Po) − (P*k))……….

F ( NPo − Pk) is also a function as F(P) = 1 − exp(−((2^R – 1)/P)^(β/2)) but instead of P have to substitute NPo − Pk so the function will become as follows

F(NPo − Pk) = 1 − exp(−((2^R – 1)/( NPo − Pk))^(β/2))

Here again R and β are constant again Po varies from 1 to 12 and P is same as mentioned before….

Min value of P should be found from both the parts of the equation and added and should be stored in a variable Pi

[Ashly Kurian]

Po is a real no

[Amit]

The way I'll do this problem is like this. I'll make 3 function files.

One for F(P) as I have done in the answer.

Second where input is [P,P0] and output will be Y. You can optimize this using fmincon for the scenario where P belongs to [0,Pb).

Third, for the case where P = N*P0/(k+1). This function will take only 1 input as P0. N*P0/(k+1) is out of [0,Pb) only when P0 >= 10.5. Thus, the P0 bounds in this case will be [10.5,12]. I can optimize this using fminbnd (as this is a single variable function).

Now I can take the minimum of both solution, which will be the value for pi.

Read MATLAB documentation for these function and try it out. If you can't succeed in doing this, I'll help you. But I need to see your effort and the code you tried.

[Ashly Kurian]

three different m files?....that previous ans u posted ...i got error

[Amit]

what error you got?

Try reading this: This might help you in understanding what I meant by 3 function files. http://www.mathworks.com/help/matlab/ref/function.html

[Ashly Kurian]

Error using myFunc (line 4)

Not enough input arguments.

[Amit]

That means, you're not entering the right amount of input for the function. Did you see the function link I posted here.

[Ashly Kurian]

How to give input to the above mentioned code.....

[Ashly Kurian]

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)

[Amit]

See the answer.