How to modify a value in code.

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Zahid Iqbal Rana
Zahid Iqbal Rana 2014-12-17
编辑： John D'Errico 2014-12-28
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.
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Roger Stafford
Roger Stafford 2014-12-17
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
Zahid Iqbal Rana 2014-12-23
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.

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回答（1 个）

Sean de Wolski
Sean de Wolski 2014-12-17
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Zahid Iqbal Rana
Zahid Iqbal Rana 2014-12-23
Kindly tell me how to use this function for my code.
John D'Errico
John D'Errico 2014-12-28
编辑：John D'Errico 2014-12-28
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.

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