# Help with differential equation Kolmogorov in queuing theory

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Le Duc Long on 29 Jun 2020
Commented: Le Duc Long on 29 Jun 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
.

Alan Stevens on 29 Jun 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.

Le Duc Long on 29 Jun 2020
Hi Alan Stevens
If replace p6=1-(p0+p1+p2+p3+p4+p5) for dp5/dt I will get the non-homogeneous system of differential equations?
Alan Stevens on 29 Jun 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!
Le Duc Long on 29 Jun 2020
Thanks for your help. I do not know why the condition p0(t)+p1(t)+...+pn(t) is diffrent 1. The condition (2) is true with any time value. Can you explain more? Best regards!