why while loop repeats so long?

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arian hoseini
arian hoseini on 30 Nov 2022
Commented: arian hoseini on 1 Dec 2022
i have this code that works correctly but i wrote another one that looks like loop repeats for so long...
clc
clear all
format longG %changes the output display format in the Command Window to the format specified by style.
alpha = [510; 310; 78; 120; 350; 490; 53; 150; 170; 149];
B = [7.20; 7.85; 7.98; 7.10; 7.30; 6.90; 8.01; 7.30; 7.42; 7.16];
C = [0.00142; 0.00194; 0.00482; 0.00321; 0.00210; 0.00178; 0.00212; 0.00350; 0.00267; 0.00390];
D = [0.0000001; 0.0000002; 0.0000003; 0.0000001; 0.0000002; 0.0000003; 0.0000001; 0.0000002; 0.0000003; 0.0000001];
Pmin =[150; 100; 50; 50; 100; 100; 100; 100; 100; 100];
Pmax =[600; 400; 200; 200; 350; 500; 300; 300; 300; 300];
n = 10;
Pr = 2500;
L = 8.6;
%-----------------------------program starts here--------------------------
ep=1;
count=0
while ep>=0.95
count = count+1;
for i=1:n
c=(B(i)-L);b=(2*C(i));a=(3*D(i));
delta=(b^2-(4*a*c));
P(i,:)=((-b+sqrt(delta))/(2*a));
if P(i)< Pmin(i)
P(i)=Pmin(i);
elseif P(i)> Pmax(i)
P(i)=Pmax(i);
end
Pt=sum(P);
ep=Pr-Pt;
M=sum(ones(length(D(i)),1)/(2*(D(i))));
end
DL=ep/M;
if ep>0
L=L+DL;
elseif ep<0
L=L-DL;
end
end
disp(L); %show the final Lambda
vpa(P,6)
this one is not working correctly
clc
clear all
format longG
a = [510; 310; 78; 120; 350; 490; 53; 150; 170; 149];
b = [7.20; 7.85; 7.98; 7.10; 7.30; 6.90; 8.01; 7.30; 7.42; 7.16];
c = [0.00142; 0.00194; 0.00482; 0.00321; 0.00210; 0.00178; 0.00212; 0.00350; 0.00267; 0.00390];
d = [0.0000001; 0.0000002; 0.0000003; 0.0000001; 0.0000002; 0.0000003; 0.0000001; 0.0000002; 0.0000003; 0.0000001];
Pmin =[150; 100; 50; 50; 100; 100; 100; 100; 100; 100];
Pmax =[600; 400; 200; 200; 350; 500; 300; 300; 300; 300];
n = 10;
Pr = 2500;
L = 8.6;
syms p;
F=sym('F',[1 10]);
X=sym('X',[1 10]);
ep=1;
count=0;
while ep>=0.9
count=count+1;
for i=1:10
F(i)=a(i)+b(i).*p+c(i).*p.^2+d(i).*p.^3;
D=diff(F(i),p)-L==0;
S=solve(D,p);
X(i)=max(S);
if X(i)<pmin(i)
X(i)=pmin(i);
elseif X(i)>pmax(i)
X(i)=pmax(i);
end
pt=sum(X);
ep=2500-(pt)
M=sum(ones(length(d(i)),1)/(2*(d(i))));
end
DL=ep/M;
if ep>0
L=L+DL;
elseif ep<0
L=L-DL;
end
end
disp(L)
vpa(X)

Accepted Answer

Torsten
Torsten on 30 Nov 2022
If you output "count" for the numerical solution, it's about 14000.
Now if you measure the time for one symbolic iteration in the while loop (approximately 1.6 seconds) and multiply this value by 14000 - I guess you will have to wait quite a long time for a result:
1.6*14000/3600
ans = 6.2222
6.2 hours. Wow! So stick to the numerical solution.
Your code would be much faster if you solved the nonlinear equation for L using "fzero", e.g.
L0 = 6.0;
L = fzero(@fun,L0)
L = 8.6749
function res = fun(L)
B = [7.20; 7.85; 7.98; 7.10; 7.30; 6.90; 8.01; 7.30; 7.42; 7.16];
C = [0.00142; 0.00194; 0.00482; 0.00321; 0.00210; 0.00178; 0.00212; 0.00350; 0.00267; 0.00390];
D = [0.0000001; 0.0000002; 0.0000003; 0.0000001; 0.0000002; 0.0000003; 0.0000001; 0.0000002; 0.0000003; 0.0000001];
Pmin =[150; 100; 50; 50; 100; 100; 100; 100; 100; 100];
Pmax =[600; 400; 200; 200; 350; 500; 300; 300; 300; 300];
Pr = 2500;
c=B-L;
b=2*C;
a=3*D;
delta=b.^2-(4*a.*c);
P=(-b+sqrt(delta))./(2*a);
P(P<Pmin) = Pmin(P<Pmin);
P(P>Pmax) = Pmax(P>Pmax);
Pt=sum(P);
res=Pr-Pt;
end

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