using summation and modulus?

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JL555 2015-9-10

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https://ww2.mathworks.cn/matlabcentral/answers/241944-using-summation-and-modulus

评论： Walter Roberson 2015-9-17

i think this is clearer

and that Fi(Pi) have to substituted here.

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Answer 1

Walter Roberson 2015-9-10

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https://ww2.mathworks.cn/matlabcentral/answers/241944-using-summation-and-modulus#answer_191943

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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?

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JL555 2015-9-15

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    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 2015-9-15

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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.

Walter Roberson 2015-9-16

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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 2015-9-17

编辑：JL555 2015-9-17

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 2015-9-17

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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 .*

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using summation and modulus?

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using summation and modulus?

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