Steady-state Availability and State Probabilities
As time progresses, the system availability and the individual state probabilities reach stable values. This means that the change in these probabilities are negligible or zero. Theoretically, this happens at an infinite time. However, in most calculations, this can be observed at a reasonably large system time. This condition is known as the steady-state condition or long-run behaviour of the system. The probabilities (or availability) can be found by substituting infinity (∞) for the time t. For example, consider the system shown in Figure 8-6. The availability of the system is:
Hence, by substituting ∞ for t, the steady-state availability can be obtained:
To apply this procedure, the analytical availability expression must first be found. As demonstrated earlier, this process is cumbersome. The remainder of this topic shows how steady-state solutions can be more easily found.
 | In order to have a non-trivial solution (non-zero availability), there should be no absorbing state in the system. In other words, the Markov process should be an Ergodic Markov process. |
According to the above discussion, at steady-state conditions, the change in state probabilities are zero. This means P`I(t)=0. Further, if there is no absorbing state in the system, the steady-state values are independent of the initial state of the system.
Therefore, equation (8.10) becomes:
Similarly, equation (8.13) is equivalent to:
B P(∞)=0
For example, equation (8.8) is:
These are a set of linear equations. Therefore, steady-state probabilities can be obtained by solving these equations. However, both of the equations are the same. Hence, effectively, there is only one equation with two variables (P1 and P2). This is because these two equations are not independent. In fact, if there are n states in the system, n-1 independent equations can be found. But, at any time, the sum of the probabilities of all states is equivalent to 1. This information can be used to make the number of unknowns and equations the same:
Any one of the above equations can be replaced with equation (8.48).
For the example:
Solving this equation gives:
These steady-state probability results can be used to find both steady-state frequencies and rewards.