why converting to double takes forever?

Question

NooshinY 2018-10-22

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/425413-why-converting-to-double-takes-forever

回答： NooshinY 2018-10-22

Hi I have this function, and I want to just calculate some intergrals. running the code without converting to double takes 1 minutes, but when I add Double(down) it takes for ever and I can't get an answer. Would anyone help me with this?

my code is:

function B=downtime(H12,H22,tau1)
syms u x y tau;
alpha1=2;   
alpha2=3;   
beta1=1.3;  
beta2=1.5;  
aa=0.000025; 
mu1=0.5;
mu2=0.4;
sigma1=0.1;
sigma2=0.15;
muw1=1.2;
muw2=1;
sigmaw1=0.2;
sigmaw2=0.3;
D1=7.5;
D2=7;
a = 0.001841;
b = 0.004071;
m=3;
fu = a*u+b;
y11= normpdf(y,mu1,sigma1);
y12= normpdf(y,mu1*2,sigma1*2^0.5);
y13= normpdf(y,mu1*3,sigma1*3^0.5); 
            g=((x^(alpha1*tau-1))*exp(-x/beta1))/((beta1^(alpha1*tau))*gamma(alpha1*tau));
            G1=int(g,x,0,H12-u);
            G2=int(g,x,0,H12-u-y);
           K1=int(G2*y11,y,0,H12-u);
           K2=int(G2*y12,y,0,H12-u);
           K3=int(G2*y13,y,0,H12-u);
           R=int(G1*fu,u,0,H12)*exp(-aa*tau)+int(K1*fu,u,0,H12)*exp(-aa*tau)*(aa*tau)+int(K2*fu,u,0,H12)*exp(-aa*tau)*(aa*tau)^2/2+int(K3*fu,u,0,H12)*exp(-aa*tau)*(aa*tau)^3/6;
          y21= normpdf(y,mu2,sigma2);
          y22= normpdf(y,mu2*2,sigma2*2^0.5);
          y23= normpdf(y,mu2*3,sigma2*3^0.5);
           g2=((x^(alpha2*tau-1))*exp(-x/beta2))/((beta2^(alpha2*tau))*gamma(alpha2*tau)); 
           G12=int(g2,x,0,H22-u);
           G22=int(g2,x,0,H22-u-y);
           K12=int(G22*y21,y,0,H22-u);
           K22=int(G22*y22,y,0,H22-u);
           K32=int(G22*y23,y,0,H22-u);
           R2=int(G12*fu,u,0,H22)*exp(-aa*tau)+int(K12*fu,u,0,H22)*exp(-aa*tau)*(aa*tau)+int(K22*fu,u,0,H22)*exp(-aa*tau)*(aa*tau)^2/2+int(K32*fu,u,0,H22)*exp(-aa*tau)*(aa*tau)^3/6; 
Rtotal=R*R2;
down=int(1-Rtotal,0,tau1);
down=double(down)
   B=down;