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Machine Repair Model
Summary: (Blank Abstract)
One example of the use of a discrete time Markov chain to solve a problem in computer systems performance evaluation comes from the machine repair model. The machine repair model presented below is actually only one possible model from a family of models that can be used to look at the reliability and performance of a computer system consisting of one or more parts that can fail and be repaired.
The particular machine repair model we use in this example has a single computer that may fail, and a single repairman who services it when it does fail. The machine can be in any of five states:
| state | |
|---|---|
| idle | not broken, but with no work to do |
| busy | actively processing a job |
| waiting | in the midst of processing a job but waiting for an I/O operation to complete |
| broken | some part of the system has failed, and the repairman has not arrived |
| in repair | some part of the system has failed, and the repairman is working to fix the problem |
We can represent the system as a discrete time Markov chain with the following state transition probabilities (rows represent present states, columns represent next states):
| idle | busy | waiting | broken | in repair | |
|---|---|---|---|---|---|
| idle | 0.05 | 0.93 | 0.02 | ||
| busy | 0.10 | 0.43 | 0.43 | 0.04 | |
| waiting | 0.60 | 0.35 | 0.05 | ||
| broken | 0.80 | 0.20 | |||
| in repair | 0.75 | 0.25 |
Blanks indicate that it is not possible to make a transition directly between the two states.
Ordering states in the state probability vector as they appear
above, we have the following single step transition probability
matrix.
As shown in the module on solving discrete-time Markov chains, this
chain can be solved using Matlab in several ways. One way to
find the steady state probability vector in Matlab is to raise
the matrix to a power such that the rows are all the same, to
the precision you want.
A second way of solving this chain is to use the eigenvector approach.
[v,d] = eig(P')
v(:,2) = [-0.1458 -0.7832 -0.5181 -0.3007 -0.0802]
To find the actual steady state probability vector, we need to normalize the vector so that its elements sum to one:
sum(v(:,2)) = -1.8280
pi = v(:,2)/sum(v(:,2))
Finally, we can find the solution directly as the solution of a
set of linear equations. Begin by subtracting the identity
matrix from the single-step probability transition matrix. J5
is the identity matrix of rank 5.
pi = [0 0 0 0 1]/P1
With these steady state probabilities, we can answer several questions of possible interest. For example, the fraction of time that the system is broken is the probability that the model is in either of the states broken or in repair, or 20.84% of the time.
We could improve the fraction of time that the system is not
broken by either increasing its reliability (reducing the
probability that the model makes a transition from state idle,
busy, or waiting to state broken) or by reducing the time it
takes the repairman to begin work on the system when it is
broken. Suppose we take the latter approach. If we reduce the
probability that the model makes the transition back to state
broken given that it is in state broken from 0.8 to 0.2, we find
that
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This work is licensed by Bart Sinclair under a Creative Commons Attribution License (CC-BY 1.0).
Last edited by Elizabeth Gregory on Jun 9, 2005 10:41 am GMT-5.