DTMC Markov Chain - How to get the stationary vector - markov-chains

Question

For a Discrete Time Markov Chain problem, i have the following:
1) Transition matrix:
0.6 0.4 0.0 0.0
0.0 0.4 0.6 0.0
0.0 0.0 0.8 0.2
1.0 0.0 0.0 0.0
2) Initial probability vector:
1.0 0.0 0.0 0.0
So, i wrote the following SciLab code to get to the stationary vector:
P = [0.6, 0.4, 0, 0; 0, 0.4, 0.6, 0; 0, 0, 0.8, 0.2; 1,0,0,0]
PI = [1,0,0,0]
R=PI*P
count=0;
for i = 1 : 35 // stationary vector is obtained at iteration 33, but i went futher to be sure
R=R*P;
count=count+1
disp("count = "+string(count))
end
PI // shows initial probability vector
P // shows transition matrix
R // shows the resulting stationary vector
After iteration number 33, the following resulting stationary vector is obtained:
0.2459016 0.1639344 0.4918033 0.0983607
What manual calculations do i have to perform in order to get to the stationary vector above without having to multiply the transition matrix 33 times then multiply the result by the initial vector?
I was told that the calculations are quite simple but i just could not realize what to do even after reading some books.
Of course explanations are welcome, but above all things i would like to have the exact answer for this specific case.

Answer 1

You can solve DTMC on Octave by using this short code:
P = [
0.6, 0.4, 0, 0;
0, 0.4, 0.6, 0;
0, 0, 0.8, 0.2;
1, 0, 0, 0
]
pis = [P' - eye(size(P)); ones(1, length(P))] \ [zeros(length(P), 1); 1]
Or with SAGE with this code:
P = matrix(RR, 4, [
[0.6, 0.4, 0, 0],
[ 0, 0.4, 0.6, 0],
[ 0, 0, 0.8, 0.2],
[ 1, 0, 0, 0]
])
I = matrix(4, 4, 1); # I; I.parent()
s0, s1, s2, s3 = var('s0, s1, s2, s3')
eqs = vector((s0, s1, s2, s3)) * (P-I); eqs[0]; eqs[1]; eqs[2]; eqs[3]
pis = solve([
eqs[0] == 0,
eqs[1] == 0,
eqs[2] == 0,
eqs[3] == 0,
s0+s1+s2+s3==1], s0, s1, s2, s3)
On both, the result of the steady state probabilities vector is:
pis =
0.245902
0.163934
0.491803
0.098361
I hope it helps.
WBR,
Albert.