Help with differential equation Kolmogorov in queuing theory
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Le Duc Long
el 29 de Jun. de 2020
Comentada: Le Duc Long
el 29 de Jun. de 2020
Hi everybody,
I have a problem with differential equation Kolmogorov in queuing theory. Now I need to write a code to solve the equation system (1) with the condition (2) and (3) according to Euler method. I wrote the code but the output didn't seem right (please see images). Any one can see my code and give me some advices. Thanks so much
clear all
tspan = [0:0.01:5];
n=6;
ic=zeros(1,n+1);
for ii=1:n+1
ic(1,1)=1;
end
[t, p] = ode45(@odeFun, tspan, ic);
figure ('name','xac suat theo time')
plot(t, p)
legend({'p7'})
function dpdt = odeFun(t, p)
lambda = 4.8;
mu = 2;
A=[-1 0 0 0 0 0 0;
1 -1 0 0 0 0 0;
0 1 -1 0 0 0 0;
0 0 1 -1 0 0 0;
0 0 0 1 -1 0 0;
0 0 0 0 1 -1 0;
0 0 0 0 0 1 0];
B=[0 1 0 0 0 0 0;
0 -1 0 0 0 0 0;
0 0 -2 2 0 0 0;
0 0 0 -2 2 0 0;
0 0 0 0 -2 2 0;
0 0 0 0 0 -2 2;
0 0 0 0 0 0 -2];
dpdt = (lambda.*A + mu.*B)*p;
end
.
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Respuesta aceptada
Alan Stevens
el 29 de Jun. de 2020
Your code doesn't maintain condition (2) for all times. You should eliminate p6 from equations (1) using condition (2), then use the ode solver to solve for the others. You can subsequently calculate p6 for each timestep.
8 comentarios
Alan Stevens
el 29 de Jun. de 2020
Try this:
tspan = 0:0.01:5;
n=6;
ic=zeros(1,n);
for ii=1:n
ic(1,1)=1;
end
[t, p] = ode45(@odeFun, tspan, ic);
p6 = (1 - sum(p,2));
figure ('name','xac suat theo time')
plot(t, p)
legend('p0','p1','p2','p3','p4','p5')
function dpdt = odeFun(~, p)
lambda = 4.8;
mu = 2;
A=[-1 0 0 0 0 0 ;
1 -1 0 0 0 0 ;
0 1 -1 0 0 0 ;
0 0 1 -1 0 0 ;
0 0 0 1 -1 0 ;
0 0 0 0 1 -1 ];
B=[0 1 0 0 0 0 ;
0 -1 2 0 0 0 ;
0 0 -2 2 0 0 ;
0 0 0 -2 2 0 ;
0 0 0 0 -2 2 ;
-2 -2 -2 -2 -2 -4] ;
v = [0 0 0 0 0 2]';
dpdt = (lambda.*A + mu.*B + mu*v)*p;
end
Notice that the book plots only p0 to p5. It's still not identical to the book! I don't know if this is because the book used a simple Euler method (as you mentioned in your original post) and MATLAB is using a more accurate Runge-Kutta method, or not. I'll leave you to investigate!
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