using summation and modulus?

i think this is clearer

and that Fi(Pi) have to substituted here.

Answer 1

[Walter Roberson]

I am having a difficult time reading that. It is a low quality image and my eyes are having trouble.

It looks to me as if it might read

F[i](P[G,i]) = a[i]*P[G,i]^2 + b[i]*P[G,i] + c[i] + abs( conjugate(e[i]) * sin( conjugate(f[i]) * (P[G,i min] - conjugate(P[G,i])) )

where I have represented subscript by [].

Please confirm that I have the operations right. For example, why did they use implicit mulitplication for a[i]*P[G,i]^2 but explicit multiplication inside the abs() ? And the conjugate bars are displaced from where I would expect them to be. What is "PG,i min" intended to mean?

[JL555]

sir why are u using abs for that modulus? PG,imin is minimum power which is given. and sir why "conjugate"..what does it mean?

[Stephen23]

They look like abs signs to me too. If they are modulo signs then it is a very non-standard way of showing it. In fact wikipedia lists thirteen mathematical uses of the vertical bar |, but modulo is not one of them.

If it is modulo, then where is the divisor/modulus specified?

[JL555]

And why the conjugate?

[Walter Roberson]

I used conjugate because the previous image had bars over some positions. The bars were positioned a bit oddly, such as being above subscripts, but conjugate was the only meaning I could think of for them.

modulus applied to real or complex number is another name for absolute value.

There are other meanings of modulus in other special forms of mathematics or physics, but they use different symbols.

[Walter Roberson]

Your revised image appears to be

F[i](P[i]) = a[i] + b[i]*P[i] + c[i]*P[i]^2 + abs( e[i] * sin( f[i](P[i, min] - P[i])) )

except it is not clear whether the f[i](P[i,min] - P[i]) is intended to represent multiplication or a function call. If it is a function call then is the F[i] on the left side the same as the f[i] on the right side?

One way of expressing the above in MATLAB code would be

    F_i = @(i) a(i) + b(i).*P(i) + c(i).*P(i).^2 + abs( e(i) .* sin( f(i)*(Pmin(i) - P(i))) );

This would take as input i, the index, and calculate a value. This assumes that the f[i] part is multiplication rather than function call. It also assumes that P(i) is a scalar, and that P[i,min] can be rewritten as a vector of values indexed by i.

[JL555]

no no F[i] & f[i] are two different variables. They are not functions. f[i](P[i,min] - P[i]) this is intended to be multiplied

[Walter Roberson]

Then the anonymous function should be what you are looking for, provided that all of the subscripted quantities mentioned are intended to be scalars and that P subscript i,min is something you can calculate ahead of time and then index by i.

[JL555]

but im always getting an error with this equation.. error doesn't change if i use * or.* any suggestion?

[Walter Roberson]

a = rand(1,5);b = rand(1,5); c = rand(1,5); e = rand(1,5); f = rand(1,5); Pmin = rand(1,5) * 10; P = rand(1,5)*10;
F_i = @(i) a(i) + b(i).*P(i) + c(i).*P(i).^2 + abs( e(i) .* sin( f(i)*(Pmin(i) - P(i))) );
Fvalues = arrayfun(F_i, 1:length(a))

[JL555]

This is the whole programming. The eqn can be found in the first script and the data in the second.Please take a look http://in.mathworks.com/matlabcentral/answers/241425-undefined-function-fun-for-input-argument-of-type-double

[Walter Roberson]

You must have updated the source since that time. I will need the updated source.

[JL555]

ok ..give your mail i'll send you.

[Walter Roberson]

You can post it here.

[JL555]

    function [Ft,Pi,Pl]=FPAeld(x)
    global objfun data pmin pmax DRi URi  pi0 Pd B 
    n=length(data(:,1));
    [m, n1]=size(x);
    B11=B(1,1);
    P=x(1:m,2:n);
    B1n=B(1,2:n);
    Bnn=B(2:n,2:n);
    A=B11;
    BB1=2*B1n*P';
    B1=(BB1-1)';
    C1=(P*Bnn*P');
    C1=diag(C1);
    Pl=Pd-(sum(P))'+C1;
    x1=roots([A B1 Pl]);
    % x=.5*(-B1-sqrt(B1^2-4*A*C))/A
    Pi=abs(min(x1));
    if Pi>data(1,5);
       Pi=data(1,5);
    else
    end
    if Pi<data(1,4);
       Pi=data(1,4);
    else 
    end
        Pi=[Pi' P];
        ai=data(:,1);
        bi=data(:,2);
        ci=data(:,3);
        ei=data(:,6);
        fi=data(:,7);
        pmin=(data(:,4))';
        for i=1:n
        T(i)=ai.*Pi.^2+bi.*Pi+ci+abs(ei.*(fi.*(pmin-Pi)));
        end

data

clear;
clc;
tic;
global objfun data pmax pmin B Pd pi0 URi DRi 
objfun=@FPAeld;
data=[150  2    0.0016    50  200 50 0.063
    25  2.5   0.01         20   80 40 0.098
    0    1    0.0625        15  50 0 0 
    0   3.25  0.00834       10  35 0 0
    0   3     0.025        10  30 0 0
    0   3     0.025         12  40 0 0];
B=[0.0224 0.0103 0.0016 -0.0053 0.0009 -0.0013;0.0103 0.0158 0.0010 -0.0074 0.0007 0.0024;0.0016 0.0010 0.0474 -0.0687 -0.0060 -0.0350;-0.0053 -.0074 -.0687 .3464 0.0105 0.0534;0.0009 0.0007 -0.0060 0.0105 0.0119 .0007;-0.0013 0.0024 -0.0350 0.0534 0.0007 0.2353];
Pd=500;
pi0=[150;35;39;20;18;20];
URi=[60;28;10;10;10;15];
DRi=[80;10;20;05;05;06];
pmax=data(:,5);
pmin=data(:,4)

i'm getting dimension this error can u help please.

Error using .* Matrix dimensions must agree.

Error in FPAeld (line 34)
    T(i)=ai.*Pi.^2+bi.*Pi+ci+abs(ei.*(fi.*(pmin-Pi)));

[Walter Roberson]

What are you passing into objfun ?

[Walter Roberson]

Your statement

Pi=[Pi' P];

is starting with Pi being a scalar (that is possibly complex.) P is an array which is something by 5, where the 5 is one fewer than the number of rows in data. [Pi' P] is horizontal concatenation. That concatenation is going to fail if P has two or more rows. Therefore the fact that you got further on to crash with the inner dimensions error shows that P must be a row vector, 1 by 5. P is extracted from x, the input argument, so the input argument must be a row vector. horizontal concatenation between a scalar and a row vector succeeds and will return a row vector. So in order to have gotten past this line, Pi would have to be a row vector.

In the statement

      T(i)=ai.*Pi.^2+bi.*Pi+ci+abs(ei.*(fi.*(pmin-Pi)));

all the other items are either scalars or are column vectors. But Pi is a row vector. So pmin - Pi is a row vector, and fi .* that row vector is trying to use .* to multiply a column vector by a row vector. That is going to fail.

You seem to be expecting a scalar as the output from the expression, even though you are working with column vectors. That suggests to me that you do not want .* multiplication, that instead you want * multiplication, so that the column vector * a row vector gives you a scalar result. If that is what you want you need to be transposing both sides or else have the row vector on the left side of the *, such as

Pi.^2 * ai + Pi * bi + something

the "something" would have to be carefully considered to give a scalar output.

[JL555]

function [Ft,Pi,Pl]=FPAeld(x)
global objfun data pmin pmax Pd B 
n=length(data(:,1));
[m, n1]=size(x);
B11=B(1,1);
P=x(1:m,2:n);
B1n=B(1,2:n);
Bnn=B(2:n,2:n);
A=B11;
BB1=2*B1n*P';
B1=(BB1-1)';
C1=(P*Bnn*P');
C1=diag(C1);
Pl=Pd-(sum(P))'+C1;
x1=roots([A B1 Pl]);
% x=.5*(-B1-sqrt(B1^2-4*A*C))/A
Pi=abs(min(x1));
if Pi>data(1,5);
   Pi=data(1,5);
else
end
if Pi<data(1,4);
   Pi=data(1,4);
else 
end
for i=1:n
    Pi=[Pi Pi(i)];
    Ft=data(i,1).*Pi(i).^2+data(i,2).*Pi(i)+data(i,3)+abs(data(i,6).*(data(i,7).*(data(i,4)-Pi)));

these are my changes

error: Attempted to access Pi(3); index out of bounds because numel(Pi)=1.

[JL555]

and i need Pi to be 1x6

[Walter Roberson]

Your Pi starts as a scalar. Your i starts as 1. The first thing you do inside the loop is Pi=[Pi Pi(i)]; which the first time through is going to be Pi=[Pi Pi(1)] which is going to extend Pi with a copy of the first element. Then the next iteration of i you again have Pi=[Pi Pi(i)] which is going to be Pi=[Pi Pi(2)] and we already know that Pi(2) is a copy of the original Pi(1) so your Pi is now going to be three elements long, all copies of the same value. Are you sure you want to do that?

Your second line of the loop has

      Ft = data(i,1).*Pi(i).^2+data(i,2).*Pi(i)+data(i,3) + abs(data(i,6).*(data(i,7).*(data(i,4)-Pi)));

which indexes Pi(i) in each location except the last one. Your Pi is growing, so each iteration the result of Ft is going to be one value longer.

And then your code supposedly ends without an "end" statement. I would judge that the error message you are encountering is on some code after what you posted here.

You should always post the complete error message with traceback, so that we do not need to guess about where the problem is.

[JL555]

thanks for responding..anyway "end" is not the error and yes the last one should be Pi(i). but i'm getting this error

Attempted to access P(1); index out of bounds because numel(P)=0.

Error in FPAeld (line 28)
    Pi=[Pi P(i)];

[JL555]

I just want something that in the end the result for Pi should be a 1x6 matrix. Thats the problem i dont know to code that part.

[Walter Roberson]

Go back to the code you posted above

Change the lines

        ai=data(:,1);
        bi=data(:,2);
        ci=data(:,3);
        ei=data(:,6);
        fi=data(:,7);

to

        ai = data(:,1) .';
        bi = data(:,2) .';
        ci = data(:,3) .';
        ei = data(:,6) .';
        fi = data(:,7) .';

then they will all agree in size, 1 x 6, with your 1 x 6 Pi, and you will not get the error about matrix dimensions not agreeing for .*