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Chapter 8 Continuous Time Markov Chains. Markov Availability Model.

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Presentation on theme: "Chapter 8 Continuous Time Markov Chains. Markov Availability Model."— Presentation transcript:

1 Chapter 8 Continuous Time Markov Chains

2 Markov Availability Model

3 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. UP 1 DN 0

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 P 1 (0)=1

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

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24 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. Markov availability model

25 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.

26 210 210 Non-shared (independent) repair Shared repair Markov availability model (Continued)

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

28 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

29 Steady-state balance equations (Continued) Hence Since We have or

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

31 Steady-state balance equations

32 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.

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40 Markov Reliability Model

41 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). Markov reliability model with repair

42 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. Markov reliability model with repair (Continued) Absorbing state

43 Assume that the initial state of the Markov chain is 2, that is, P 2 (0) = 1, P k (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 Markov reliability model with repair (Continued)

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45 After solving these equations, we get R(t) = P 2 (t) +P 1 (t) Recalling that, we get: Markov reliability model with repair (Continued)

46 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 Markov reliability model with repair (Continued)

47 Markov Reliability Model with Imperfect Coverage

48 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:

49 Now allow for Imperfect coverage c

50 Markov model with imperfect coverage (Continued) Assume that the initial state is 2 so that: Then the system of differential equations are:

51 Markov model with imperfect coverage (Continued) After solving the differential equations we obtain: R(t)=P 2 (t) + P 1 (t) From R(t), we can system MTTF: It should be clear that the system MTTF and system reliability are critically dependent on the coverage factor.

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58 SOURCES OF COVERAGE DATA Measurement Data from an Operational system: Large amount of data needed; Improved Instrumentation Needed Fault/Error Injection Experiments Costly yet badly needed: tools from CMU, Illinois, Toulouse

59 SOURCES OF COVERAGE DATA (Continued) A Fault/Error Handling Submodel Phases of FEHM: Detection, Location, Retry, Reconfig, Reboot Estimate Duration & Prob. of success of each phase IBM(EDFI), HARP(FEHM), Draper(FDIR)

60 Homework 6: Modify the Markov model with imperfect coverage to allow for finite time to detect as well as imperfect detection. You will need to add an extra state, say D. The rate at which detection occurs is . Draw the state diagram and using SHARPE GUI investigate the effects of detection delay on system reliability and mean time to failure.

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