How to modify a value in code.

This has had me stumped all day. Any help is appreciated. I'm trying to do the following with my code. I want to generate a randomly values of ""p", that must be in limits of a and b and total sum of p in a row i.e p(:,1)+p(:,2)+p(:,3) must be equal to Demand plus PL. The value of PL is to be computed by the values of p by the formula.

clear all ; clc

Demand=300;

N=3;

a(1,1)=100; b(1,1)=600; %bounds on variable 1

a(1,2)=100; b(1,2)=400; %bounds on variable 2

a(1,3)=50; b(1,3)=200; %bounds on variable 3

x=a+(b-a).*rand(1,3);

T=x(:,1)+x(:,2)+x(:,3); %total

z=[x(:,1)./T(:,1) x(:,2)./T(:,1) x(:,3)./T(:,1)]; % Equlity constraint

p= z.*Demand;

Total= p(:,1)+p(:,2)+p(:,3); % Total must be equal to Demand + PL

f1=561+ 7.92.*p(:,1)+0.00156.*(p(:,1).^2); f2=310+ 7.85.*p(:,2)+0.00194.*(p(:,2).^2); f3=78+ 7.97.*p(:,3)+0.00482.*(p(:,3).^2); TC=f1+f2+f3;

PL= 10+p.^(1/4);

--> At the end I just found that sum(p) is equal to Demand as I made it in (p= z.*Demand) and it is not equal to PL+Demand that's the requirement. Please guide me in this regard.

[Thorsten]

Why do you compute TC?

[Roger Stafford]

Thorsten asks a good question which you should pay attention to. What does the computation of 'TC' have to do with 'PL'? Your line

   PL = 10+p.^(1/4);

produces a three-element vector, rather than the needed scalar, and 'TC' is never used. You need to clear up this point before a sensible answer to your question can be given. Please give us a precise formula for finding a scalar 'PL' that makes sense.

[Zahid Iqbal Rana]

TC is the total cost that's I need to compute, PL is the power Loss and Demand is the load demand. I need to compute TC for load demand + PL.

Answer 1

[Sean de Wolski]

http://www.mathworks.com/matlabcentral/fileexchange/9700-random-vectors-with-fixed-sum

[Sean de Wolski]

Why is it not helpful? It does exactly what you want. Did you even read the help?

This generates m random n-element column vectors of values, [x1;x2;...;xn], each with a fixed sum, s, and subject to a restriction a<=xi<=b

[Zahid Iqbal Rana]

Kindly tell me how to use this function for my code.

[John D'Errico]

Sean - randfixedsum assumes the points lie in a cube, not a hyper rectangle. The question here has lower and upper limits that differ for each variable. Note that the restriction for randfixed sum is

a <= x(i) <= b

as opposed to

a(i) <= x(i) <= b(i)

The latter is what is needed to solve this question.