Therefore this is expressed mathematically PT \u00bc P \u00f0 8 2 \u00de 114 8 Markov

# Therefore this is expressed mathematically pt ¼ p ð

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Therefore, this is expressed mathematically: PT ¼ P ð 8 : 2 Þ 114 8 Markov Processes where P is the limiting state probability vector and T is the transition matrix. This may be rewritten as: P ð T ± I Þ ¼ 0 ð 8 : 3 Þ where I is the identity matrix. Substituting the transition matrix T into the above equation gives the following expression. P 1 P 2 P 3 P 4 ½ ² ±ð k 1 þ k 2 Þ k 1 k 2 0 l 1 ±ð l 1 þ k 2 Þ 0 k 2 l 2 0 ±ð l 2 þ k 1 Þ k 1 0 l 2 l 1 ±ð l 1 þ l 2 Þ 2 6 6 6 4 3 7 7 7 5 ¼ 0 0 0 0 ½ ð 8 : 4 Þ By taking the transpose of the equation, the general form of the equation is obtained. ±ð k 1 þ k 2 Þ l 1 l 2 0 k 1 ±ð l 1 þ k 2 Þ 0 l 2 k 2 0 ±ð l 2 þ k 1 Þ l 1 0 k 2 k 1 ±ð l 1 þ l 2 Þ 2 6 6 4 3 7 7 5 P 1 P 2 P 3 P 4 2 6 6 4 3 7 7 5 ¼ 0 0 0 0 2 6 6 4 3 7 7 5 ð 8 : 5 Þ 8.2.4 Step 4: Full Probability Condition The sum of all the individual probabilities is equal to 1. P 1 þ P 2 þ P 3 þ P 4 ½ ¼ 1 This condition is required to be able to solve the above equation as it contains only n - 1 independent equations and there are four state variables involved. Therefore, any row within the above equation can be replaced with this condition, such as the first row. 00 11 01 10 μ 1 μ 2 μ 1 μ 2 λ 1 λ 1 λ 2 λ 2 State 1 State 3 State 2 State 4 Fig. 8.1 Diagram for a two- component repairable system for availability analyses 8.2 Systems Availability Analyses 115 1 1 1 1 k 1 ±ð l 1 þ k 2 Þ 0 l 2 k 2 0 ±ð l 2 þ k 1 Þ l 1 0 k 2 k 1 ±ð l 1 þ l 2 Þ 2 6 6 4 3 7 7 5 P 1 P 2 P 3 P 4 2 6 6 4 3 7 7 5 ¼ 1 0 0 0 2 6 6 4 3 7 7 5 ð 8 : 6 Þ 8.2.5 Step 5: Solving the Markov Matrix Equation Using Linear Algebra As equation now contains four independent equations, it may be solved using linear algebra. In general, this yields the following expressions. P 1 ¼ l 1 l 2 ð l 1 þ k 1 Þð l 2 þ k 2 Þ ð 8 : 7 Þ P 2 ¼ k 1 l 2 ð l 1 þ k 1 Þð l 2 þ k 2 Þ ð 8 : 8 Þ P 3 ¼ l 1 k 2 ð l 1 þ k 1 Þð l 2 þ k 2 Þ ð 8 : 9 Þ P 4 ¼ k 1 k 2 ð l 1 þ k 1 Þð l 2 þ k 2 Þ ð 8 : 10 Þ 8.3 Example of Markov Chains for Reliability Analyses A simple system with one active component named primary system and one standby redundant component named secondary system is connected with another component, i.e., switch [ 7 ]. Figure 8.2 shows this system. When switch fails, it is unable to switch. The failure of the switch matters, if it is required to switch from the primary to the secondary system. If the switch fails after the standby spare is already in use, then the system can continue operation. However, if the switch fails before the primary unit fails, then the secondary system cannot be turned on and the system fails when the primary unit fails. The order in which the primary system and switch fail determines whether the system continues operation.  • • • 