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How to solve the markov transition matrix

  1. May 2, 2012 #1
    1. The problem statement, all variables and given/known data
    I design a markov model to study the availability of a node which will not work unless the network work: I make the following matrix. In which (i,j) means i is node up /down and j is network up/down. R1,R2 means the repair rate for node and network. F1,F2 means the failure rate for node and network.
    (0,0) (1,0) (0,1) (1,1)
    (0,0) 0 R1 R2 0
    (1,0) F1 0 0 R2
    (0,1) F2 0 0 R1
    (1,1) 0 F2 F1 0

    2. Relevant equations

    I do the following equations to detemine the P(1,1)
    (R1+R2)P(0,0)= F1 P(1,0) + F2 P(0,1)
    (F1+R2)P(1,0) = R1 P(0,0) +F1 P(1,1)
    (F2+R1)P(0,1) = R2 P(0,0) +F1 P(1,1)
    (F2+F1)P(1,1) = R2 P(1,0) +R1 P(0,1)

    P(0,0) + P(1,0) + P(0,1) + P(1,1) = 1

    I do this symbolically in which it gives me long recursive and I couldn't complete it.

    3. The attempt at a solution
    is my equation right or wrong?
    I need advise in how to solve this equations to get P(1,1).
    I try matlab to calculate this equations as linear system
    it gives me
    [ 0
    0
    0
    0 ]

    Thanks
     
  2. jcsd
  3. May 2, 2012 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    The first 4 equations have a redundancy: given that any three of them hold, the 4th one also holds. So, keep any three of the first four equations, and append the normalization equation P(0,0) + ... + P(1,1) = 1. That 4x4 linear system does have a unique solution.

    It is not too bad, just lengthy. You can get P(0,0) = [F1 P(1,0) + F2 P(0,1)]/(R1+R2) from the first equation. You can get P(1,1) = [R2 P(1,0) +R1 P(0,1)]/(F2+F1) from the fourth equation. Now you can substitute these expressions for P(0,0) and P(1,1) into the second (or third) equation and into the sum(P) = 1 equation to get two equations in the two unknowns P(0,1) and P(1,0). It is messy and time-consuming, but is nevertheless straightforward.

    RGV
     
  4. May 3, 2012 #3
    Hi Ray :)
    Thank you very much, it's helpful.
     
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