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Chapter 8

Chapter 8. Continuous Time Markov Chains. Markov Availability Model. UP 1. DN 0. 2-State Markov Availability Model. 1) Steady-state balance equations for each state: Rate of flow IN = rate of flow OUT State1: State0:

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Chapter 8

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  1. Chapter 8 Continuous Time Markov Chains

  2. Markov Availability Model

  3. UP 1 DN 0 2-State Markov Availability Model 1) Steady-state balance equations for each state: • Rate of flow IN = rate of flow OUT • State1: • State0: 2 unknowns, 2 equations, but there is only one independent equation.

  4. 2-State Markov Availability Model(Continued) 1) Need an additional equation: Downtime in minutes per year = * 8760*60

  5. 2-State Markov Availability Model(Continued) 2) Transient Availability for each state: • Rate of buildup = rate of flow IN - rate of flow OUT This equation can be solved to obtain assuming P1(0)=1

  6. 2-State Markov Availability Model(Continued) 3) 4) Steady State Availability:

  7. Markov availability model • Assume we have a two-component parallel redundant system with repair rate . • Assume that the failure rate of both the components is . • When both the components have failed, the system is considered to have failed.

  8. Markov availability model(Continued) • Let the number of properly functioning components be the state of the system. The state space is {0,1,2} where 0 is the system down state. • We wish to examine effects of shared vs. non-shared repair.

  9. Markov availability model(Continued) 2 1 0 Non-shared (independent) repair 2 1 0 Shared repair

  10. Markov availability model(Continued) • Note: Non-shared case can be modeled & solved using a RBD or a FTREE but shared case needs the use of Markov chains.

  11. Steady-state balance equations • For any state: Rate of flow in = Rate of flow out Consider the shared case i: steady state probability that system is in state i

  12. Steady-state balance equations(Continued) • Hence Since We have or

  13. Steady-state balance equations(Continued) • Steady-state unavailability = 0= 1 - Ashared Similarly for non-shared case, steady-state unavailability = 1 - Anon-shared • Downtime in minutes per year = (1 - A)* 8760*60

  14. Steady-state balance equations

  15. Homework 5: • Return to the 2 control and 3 voice channels example and assume that the control channel failure rate is c, voice channel failure rate is v. • Repair rates are c and v, respectively. Assuming a single shared repair facility and control channel having preemptive repair priority over voice channels, draw the state diagram of a Markov availability model. Using SHARPE GUI, solve the Markov chain for steady-state and instantaneous availability.

  16. Markov Reliability Model

  17. Markov reliability model with repair • Consider the 2-component parallel system but disallow repair from system down state • Note that state 0 is now an absorbing state. The state diagram is given in the following figure. • This reliability model with repair cannot be modeled using a reliability block diagram or a fault tree. We need to resort to Markov chains. (This is a form of dependency since in order to repair a component you need to know the status of the other component).

  18. Markov reliability model with repair (Continued) • Markov chain has an absorbing state. In the steady-state, system will be in state 0 with probability 1. Hence transient analysis is of interest. States 1 and 2 are transient states. Absorbing state

  19. Markov reliability model with repair (Continued) Assume that the initial state of the Markov chain is 2, that is, P2(0) = 1, Pk (0) = 0 for k = 0, 1. Then the system of differential Equations is written based on: rate of buildup = rate of flow in - rate of flow out for each state

  20. Markov reliability model with repair (Continued)

  21. Markov reliability model with repair (Continued) After solving these equations, we get R(t) = P2(t) +P1(t) Recalling that , we get:

  22. Markov reliability model with repair (Continued) Note that the MTTF of the two component parallel redundant system, in the absence of a repair facility (i.e.,  = 0), would have been equal to the first term, 3 / ( 2* ), in the above expression. Therefore, the effect of a repair facility is to increase the mean life by  / (2*2), or by a factor

  23. Markov Reliability Model with Imperfect Coverage

  24. Markov model with imperfect coverage Next consider a modification of the above example proposed by Arnold as a model of duplex processors of an electronic switching system. We assume that not all faults are recoverable and that c is the coverage factor which denotes the conditional probability that the system recovers given that a fault has occurred. The state diagram is now given by the following picture:

  25. Now allow for Imperfect coverage c

  26. Markov modelwith imperfect coverage (Continued) Assume that the initial state is 2 so that: Then the system of differential equations are:

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