why while loop repeats so long?

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)

Answer 1

[Torsten]

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

[arian hoseini]

thank u legend