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I'm trying to solve a system of two non-linear equations in Matlab in a for loop. I am aware that the for loop may cause the program to run slow but right now it takes around an hour for one instance to be solved. On debugging the code, I realized that this line causes the program to hang: 

[res,status] = mupadmex(statement);

If I terminate the program I get the following: 

Operation terminated by user during mupadengine/evalin (line 97)
In mupadengine/feval (line 150)
            [S,err] = evalin(engine,stmt);
In solve (line 160)   sol = eng.feval('symobj::solvefull',eqns,vars);
In algorithm2 (line 139)
            answer = solve (f , g);

where line 97 refers to the same line of code. Here's the section of my code where I call the solve function: 

syms ro1 t_prime1
assume(t_prime1 > 0)
for i=1:1:length(n1ps)
    n1p = n1ps(1,i); 
    n2p = n2ps(1,i); 
    w1 = (ro1/theta_p)*(1-exp((-1)*theta_p*t_prime1)); 
    w2 = exp((-1)*theta_p*t_prime1);
    w3 = (ro1/theta_p)*(1-exp((-1)*theta_p*T1))-q(1,1);
    w4 = (1-exp((-1)*n1p*theta_p*T1))/(1-exp((-1)*theta_p*T1)); 
    w5 = exp((-1)*theta_p*(T1-t_prime1));
    Q1 = (w1 + w2*w3*w4)*w5 - q(1,1); 

    w6 = ((ro-ro1)/theta_p)*(1-exp((-1)*theta_p*(n1p*T1 + t_prime1 - n2p*T2))); 
    w7 = exp((-1)*theta_p*(n1p*T1 + t_prime1 - n2p*T2));
    w8 = ((ro-ro1)/theta_p)*(1-exp((-1)*theta_p*T2))-q(1,2);
    w9 = (1-exp((-1)*n2p*theta_p*T2))/(1-exp((-1)*theta_p*T2)); 
    w10 = exp((-1)*theta_p*(T2-(n1p*T1 + t_prime1 - n2p*T2)));
    Q2 = (w6 + w7*w8*w9)*w10 - q(1,2);

    Q1_prime = q(1,1)*((exp((tN1 - n1p - 1)*theta_p*T1)-1)/(1-exp((-1)*theta_p*T1))); 
    Q2_prime = q(1,2)*((exp((tN2 - n2p - 1)*theta_p*T2)-1)/(1-exp((-1)*theta_p*T2)));

    f=Q1_prime-Q1;
    g=Q2_prime-Q2;

    answer = solve (f , g);
end
tckmn
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Alireza Kazeminia
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  • What is your question? Your code is not runnable as many things are left undefined. You're not even saving your results: e.g., `answer` isn't a function of `i`. The loop likely has nothing to do with this being slow. Can you run even one iteration with out the loop (just set `i=1`)? Is there a reason you're using symbolic math instead of a numeric method like `fzero` or `fsolve`? – horchler Oct 28 '14 at 19:17
  • Just a general advice; if you want to speed up solving symbolic equations, add assumptions and constraints on the variables such as an interval where to find the solution, it can make a big difference – Amro Oct 28 '14 at 20:43

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