Markov Models for Reliability Modeling

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1 Markov Models for Reliability Modeling Prof. Naga Kandasamy ECE Department, Drexel University, Philadelphia, PA 904 Many complex systems cannot be easily modeled in a combinatorial fashion. The corresponding reliability block diagrams can be difficult to construct. Also, the repair process in high-availability computing systems is very difficult to model in a combinatorial fashion. In the face of these difficulties, Markov models are often used for reliability and availability modeling. Two major concepts underlying the Markov models are the system state and state transition. In the case of reliability modeling, the state of a system represents a distinct combination of working and failed modules. As time passes, the system progresses through different states as modules fail and perhaps, are repaired. These changes in states are called state transitions. Reliability Modeling of a TMR System Assume we have a TMR system with three identical computers in a majority voting arrangement with a perfect voter. We can define the state of this system as S = (S, S 2, S 3 ) where S i = if module i is fault free and S i = 0 if module i has failed. So, the TMR system has eight distinct states in which it can operate: (000), (00), (00), (0), (00), (0), (0), and (). Each state represents a unique combination of faulty and fault-free modules in the system. For a TMR system, at leat two modules must operate correctly for the system to operate correctly. So, state (000), (00), (00), and (00) are failed states of the system. The state diagram for the TMR system is shown in Fig.. It can be partitioned into three major cate- Fig. : The Markov model of the TMR system, showing the various system states and state transitions.

2 Fig. 2: The reduced Markov model of the TMR system with the minimum number of states. gories: the perfect state with three functioning modules (), the one-failed states (0), (0), (0), and the system-failed states (00), (00), (00), (000). Each state transition has associated with it a transition probability describing the probability of that transition happening within a specified period of time. Assume that each module in the TMR system obeys the exponential failure law with a failure rate of. So, the probability that a module has failed in the time interval [t, t + ] is e. The exponential can be written in a series expansion as e = + ( ) + ( )2 2! Ignoring the higher-order terms, for a small, we obtain e = + The probability that a module will fail during time period is approximately e = ( ) = If the modules have the same failure probability, we can reduce the Markov model of the TMR system shown in Fig. to that shown in Fig. 2. The probability of the system being in one of the three states is given by p 3 (t + ) = ( 3)p 3 (t) p 2 (t + ) = (3)p 3 (t) + ( 2)p 2 (t) p F (t + ) = (2)p 2 (t) + p F (t) The above equations can be rewritten as p 3 (t + ) p 3 (t) = 3p 3 (t) p 2 (t + ) p 2 (t) = 3p 3 (t) 2p 2 (t) p F (t + ) p F (t) = 2p 2 (t) 2

3 If we take the limit at approaches zero, we obtain the continuous-time Markov model as dp 3 (t) = 3p 3 (t) dp 2 (t) = 3p 3 (t) 2p 2 (t) dp F (t) = 2p 2 (t) The above simultaneous differential equations can be solved using the method of Laplace transforms (see for a Laplace transform table), sp 3 (s) p 3 (0) = 3P 3 (s) sp 2 (s) p 2 (0) = 3P 3 (s) 2P 2 (s) sp F (s) p F (0) = 2P 2 (s) where P 3 (s), P 2 (s), and P F (s) are the Laplace transforms of p 3 (t), p 2 (t), and p F (t), respectively. We assume that the system starts out in perfect shape at time t = 0, and so, p 3 (0) =, and p 2 (0) = p F (0) = 0. The Laplace transforms can be written as P 3 (s) = s P 2 (s) = (s + 2)(s + 3) 6 2 s(s + 2)(s + 3) Using the method of partial fractions, the above equations can be rewritten as P 3 (s) = s P 2 (s) = (s + 2) + 3 (s + 3) s + 3 (s + 2) + 2 (s + 3) Taking the inverse Laplace transforms, we obtain p 3 (t) = e 3t p 2 (t) = 3e 2t 3e 3t p F (t) = 3e 2t + 2e 3t Recall that the reliability of the TMR system, obtained using combinatorial techniques, is 3e 2t 2e 3t. Also, p 3 (t) + p 2 (t) + p F (t) =, as expected. The Laplace transform of dy is sy (s) y(0). 3

4 Fig. 3: Markov model of a simplex or non-redundant system with repair. 2 Markov Models with Repair Let us consider the Markov model of a simple system consisting of one computer and no redundancy. Assume that the computer has a constant failure rate of and a constant repair rate of µ. We can formulate the simple Markov model for this system as shown in Fig. 3. The equations for the Markov model shown in Fig. 3 can be written as p O (t + ) = ( )p O (t) + (µ)p F (t) p F (t + ) = ()p O + ( µ)p F (t) and re-formulated as p O (t + ) p O (t) = p O (t) + µp F (t) p F (t + ) p F (t) = p O (t) µp F (t) Taking the limit as approaches zero, we obtain the following differential equations dp O (t) = p O (t) + µp F (t) dp F = p O (t) µp F (t) Using Laplace transforms and assuming that the initial conditions are p O (0) = and p F (0) = 0, we obtain sp O (s) = P O (s) + µp F (s) s P O (s) µp F (s) 4

5 Solving the simultaneous equations for P O (s) and P F (s), we obtain P O (s) = s + ( + µ) + µ s(s + ( + µ))) s(s + ( + µ)) which can be rewritten using partial fractions as P O (s) = µ +µ s +µ s + +µ s + ( + µ) +µ s + ( + µ) Taking the inverse Laplace transform, we obtain p O (t) = µ + µ + p F (t) = + µ e (+µ)t + µ + µ e (+µ)t As time approaches, p O (t) approaches the constant value of µ +, or the steady-state availability, and p F (t) approaches the constant value of µ +. 5