How to optimize an reduce the computation time

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Ruslan
Ruslan 2024-11-28
评论： Walter Roberson 2024-11-28
while Fg >= 0.001
if Fg > 0.15
Psat = Psat + 1;
elseif Fg > 0.02
Psat = Psat + 0.1;
elseif Fg > 0.009
Psat = Psat + 0.01;
else
Psat = Psat + 0.001;
end
Pri = Psat./PcMPa;
k = exp(5.37.*(1+w).*(1-Tri.^(-1)))./Pri;
syms fg
fun = sum(x.*(k-1)./(1+fg*(k-1)));
Fgmass = double(solve(fun==0,fg));
Fg = Fgmass(Fgmass<0.99 & Fgmass>0);
end
Hello. I am a complete beginner. How can i reduce computation time for this process? Now it takes around a minute. I need to do 2-20 thousand of such computations. Thank you in advance.
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Walter Roberson
Walter Roberson 2024-11-28
k = exp(5.37.*(1+w).*(1-Tri.^(-1)))./Pri;
In that calculation, w and Tri are calculated before the while loop. The entire expression exp(5.37.*(1+w).*(1-Tri.^(-1))) is effectively constant . So pre-calculate it,
wexp = exp(5.37.*(1+w).*(1-Tri.^(-1)));
before the loop, and then inside the loop
k = wexp ./ Pri;
Walter Roberson
Walter Roberson 2024-11-28
syms fg
fun = sum(x.*(k-1)./(1+fg*(k-1)));
Fgmass = double(solve(fun==0,fg));
Fg = Fgmass(Fgmass<0.99 & Fgmass>0);
Are you expecting multiple solutions for the solve() ? If not then it would be better to use vpasolve() instead of solve()

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回答（1 个）

Torsten
Torsten 2024-11-28
编辑：Torsten 2024-11-28
Ran in:
I don't understand the sense of your while-loop.
If you want to define Psat as a function of fg, the problem becomes more complicated.
Try to use "fzero" or "fsolve" instead of symbolic computations like done below if you want to make the code perform faster.
x = [0.0035, 0.0314, 0.5426, 0.0857, 0.0572, 0.0076, 0.0245, 0.0075, 0.012, 0.0153, 0.026, 0.0302, 0.021, 0.0174, 0.0136, 0.011, 0.0111, 0.0095, 0.0086, 0.0068, 0.006, 0.0056, 0.0051, 0.0408];
PcMPa = [3.399, 7.382, 4.604, 4.880, 4.249, 3.648, 3.797, 3.381, 3.369, 3.012, 2.736, 2.486, 2.289, 2.096, 1.979, 1.824, 1.7, 1.524, 1.486, 1.4, 1.325, 1.294, 1.173, 1.08];
TcK= [126.3, 304.2, 190.6, 305.4, 369.8, 408.2, 425.2, 460.4, 469.7, 507.4, 540.3, 568.8, 594.6, 617.7,639, 658, 675, 693, 708, 723, 734, 748, 755, 768];
w =[0.045, 0.231, 0.0115, 0.0908, 0.1454, 0.1756, 0.1928, 0.2273, 0.251, 0.2957, 0.3506, 0.3978, 0.4437, 0.4902, 0.535, 0.575, 0.619, 0.659, 0.706, 0.742, 0.77, 0.79, 0.827, 0.907];
TK = 200;
R = 8.314;
PPsat = 0.01:0.01:3.38;
Tri = TK./TcK;
fg0 = 0.7;
for i = 1:numel(PPsat)
Psat = PPsat(i);
Pri = Psat./PcMPa;
k = exp(5.37.*(1+w).*(1-Tri.^(-1)))./Pri;
%syms fg
%sol = vpa(solve(fun==0,fg));
%sol = sol(abs(imag(sol))<1e-6);
%sol = sol(sol>0 & sol<1);
%Fg(i) = double(sol);
fun = @(fg)sum(x.*(k-1)./(1+fg*(k-1)));
Fg(i) = fsolve(fun,fg0,optimset('Display','none'));
fg0 = Fg(i);
end
plot(PPsat,Fg)
xlabel('Psat')
ylabel('fg')
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Stephen23
Stephen23 2024-11-28
Do not reinvent the wheel.
As Torsten wrote, just use FZERO or FSOLVE or similar.
Avoid symbolic mathematics (if computation time is important).
Torsten
Torsten 2024-11-28
编辑：Torsten 2024-11-28
Ran in:
My while loop is trying to find value of Psat at which fg as close to zero as possible
Then insert fg = 0 (or something close to 0) and solve for Psat:
x = [0.0035, 0.0314, 0.5426, 0.0857, 0.0572, 0.0076, 0.0245, 0.0075, 0.012, 0.0153, 0.026, 0.0302, 0.021, 0.0174, 0.0136, 0.011, 0.0111, 0.0095, 0.0086, 0.0068, 0.006, 0.0056, 0.0051, 0.0408];
PcMPa = [3.399, 7.382, 4.604, 4.880, 4.249, 3.648, 3.797, 3.381, 3.369, 3.012, 2.736, 2.486, 2.289, 2.096, 1.979, 1.824, 1.7, 1.524, 1.486, 1.4, 1.325, 1.294, 1.173, 1.08];
TcK= [126.3, 304.2, 190.6, 305.4, 369.8, 408.2, 425.2, 460.4, 469.7, 507.4, 540.3, 568.8, 594.6, 617.7,639, 658, 675, 693, 708, 723, 734, 748, 755, 768];
w =[0.045, 0.231, 0.0115, 0.0908, 0.1454, 0.1756, 0.1928, 0.2273, 0.251, 0.2957, 0.3506, 0.3978, 0.4437, 0.4902, 0.535, 0.575, 0.619, 0.659, 0.706, 0.742, 0.77, 0.79, 0.827, 0.907];
TK = 200;
R = 8.314;
Tri = TK./TcK;
Pri = @(Psat)Psat./PcMPa;
k = @(Psat)exp(5.37.*(1+w).*(1-Tri.^(-1)))./Pri(Psat);
fg = 0;
fun = @(Psat)sum(x.*(k(Psat)-1)./(1+fg*(k(Psat)-1)));
Psat = fsolve(fun,5)
Equation solved. fsolve completed because the vector of function values is near zero as measured by the value of the function tolerance, and the problem appears regular as measured by the gradient.
Psat = 3.3467

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