Download presentation
Presentation is loading. Please wait.
Published byKurt Woodstock Modified over 9 years ago
1
A. BobbioReggio Emilia, June 17-18, 20031 Dependability & Maintainability Theory and Methods 5. Markov Models Andrea Bobbio Dipartimento di Informatica Università del Piemonte Orientale, “A. Avogadro” 15100 Alessandria (Italy) bobbio@unipmn.itbobbio@unipmn.it - http://www.mfn.unipmn.it/~bobbio/IFOA IFOA, Reggio Emilia, June 17-18, 2003
2
A. BobbioReggio Emilia, June 17-18, 20032 States and labeled state transitions State can keep track of: –Number of functioning resources of each type –States of recovery for each failed resource –Number of tasks of each type waiting at each resource –Allocation of resources to tasks A transition: –Can occur from any state to any other state –Can represent a simple or a compound event State-Space-Based Models
3
A. BobbioReggio Emilia, June 17-18, 20033 Transitions between states represent the change of the system state due to the occurrence of an event Drawn as a directed graph Transition label: –Probability: homogeneous discrete-time Markov chain (DTMC) –Rate: homogeneous continuous-time Markov chain (CTMC) –Time-dependent rate: non-homogeneous CTMC –Distribution function: semi-Markov process (SMP) State-Space-Based Models (Continued)
4
A. BobbioReggio Emilia, June 17-18, 20034 Modeler’s Options Should I Use Markov Models? State-Space-Based Methods + Model Dependencies + Model Fault-Tolerance and Recovery/Repair + Model Contention for Resources + Model Concurrency and Timeliness + Generalize to Markov Reward Models for Modeling Degradable Performance
5
A. BobbioReggio Emilia, June 17-18, 20035 Modeler’s Options Should I Use Markov Models? + Generalize to Markov Regenerative Models for Allowing Generally Distributed Event Times + Generalize to Non-Homogeneous Markov Chains for Allowing Weibull Failure Distributions + Performance, Availability and Performability Modeling Possible - Large (Exponential) State Space
6
A. BobbioReggio Emilia, June 17-18, 20036 In order to fulfil our goals Modeling Performance, Availability and Performability Modeling Complex Systems We Need Automatic Generation and Solution of Large Markov Reward Models
7
A. BobbioReggio Emilia, June 17-18, 20037 Model-based evaluation Choice of the model type is dictated by: –Measures of interest –Level of detailed system behavior to be represented –Ease of model specification and solution –Representation power of the model type –Access to suitable tools or toolkits
8
A. BobbioReggio Emilia, June 17-18, 20038 State space models A transition represents the change of state of a single component x i s s’ Pr {s s’, t} = Pr {Z(t+ t) = s’ | Z(t) = s} Z(t) is the stochastic process Pr {Z(t) = s} is the probability of finding Z(t) in state s at time t.
9
A. BobbioReggio Emilia, June 17-18, 20039 State space models If s s’ represents a failure event: x i s s’ Pr {s s’, t} = = Pr {Z(t+ t) = s’ | Z(t) = s} = i t If s s’ represents a repair event: Pr {s s’, t} = = Pr {Z(t+ t) = s’ | Z(t) = s} = i t
10
A. BobbioReggio Emilia, June 17-18, 200310 Markov Process: definition
11
Transition Probability Matrix initial
12
State Probability Vector
13
Chapman-Kolmogorov Equations
14
Time-homogeneous CTMC
16
The transition rate matrix
17
C-K Equations for CTMC
18
Solution equations
19
Transient analysis Given that the initial state of the Markov chain, then the system of differential Equations is written based on: rate of buildup = rate of flow in - rate of flow out for each state (continuity equation).
20
Steady-state condition If the process reaches a steady state condition, then:
21
Steady-state analysis (balance equation) The steady-state equation can be written as a flow balance equation with a normalization condition on the state probabilities. (rate of buildup) = rate of flow in - rate of flow out rate of flow in = rate of flow out for each state (balance equation).
22
State Classification
23
A. BobbioReggio Emilia, June 17-18, 200323 2-component system
24
A. BobbioReggio Emilia, June 17-18, 200324 2-component system
25
A. BobbioReggio Emilia, June 17-18, 200325 2-component system
26
A. BobbioReggio Emilia, June 17-18, 200326 2-component series system A1A1A2 2-component parallel system A1A1 A2
27
A. BobbioReggio Emilia, June 17-18, 200327 2-component stand-by system A B
28
Markov Models Repairable systems - Availability
29
A. BobbioReggio Emilia, June 17-18, 200329 Repairable system: Availability
30
A. BobbioReggio Emilia, June 17-18, 200330 Repairable system: 2 identical components
31
A. BobbioReggio Emilia, June 17-18, 200331 Repairable system: 2 identical components
32
A. BobbioReggio Emilia, June 17-18, 200332 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. 2-component Markov availability model
33
A. BobbioReggio Emilia, June 17-18, 200333 Markov availability model 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.
34
A. BobbioReggio Emilia, June 17-18, 200334 210 210 Non-shared (independent) repair Shared repair Markov availability model
35
A. BobbioReggio Emilia, June 17-18, 200335 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
36
A. BobbioReggio Emilia, June 17-18, 200336 Steady-state balance equations For any state: Rate of flow in = Rate of flow out Considering the shared case i : steady state probability that system is in state i
37
A. BobbioReggio Emilia, June 17-18, 200337 Steady-state balance equations Hence Since We have Or
38
A. BobbioReggio Emilia, June 17-18, 200338 Steady-state balance equations (Continued) Steady-state Unavailability: For the Shared Case = 0 = 1 - A shared Similarly, for the Non-Shared Case, Steady-state Unavailability = 1 - A non-shared Downtime in minutes per year = (1 - A)* 8760*60
39
A. BobbioReggio Emilia, June 17-18, 200339 Steady-state balance equations
40
A. BobbioReggio Emilia, June 17-18, 200340 Absorbing states MTTF
41
A. BobbioReggio Emilia, June 17-18, 200341 Absorbing states - MTTF
42
Markov Reliability Model with Imperfect Coverage
43
A. BobbioReggio Emilia, June 17-18, 200343 Markov model with imperfect coverage Next consider a modification of the 2-component parallel system 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:
44
A. BobbioReggio Emilia, June 17-18, 200344 Now allow for Imperfect coverage c
45
A. BobbioReggio Emilia, June 17-18, 200345 Markov model with imperfect coverage Assume that the initial state is 2 so that: Then the system of differential equations are:
46
A. BobbioReggio Emilia, June 17-18, 200346 Markov model with imperfect coverage After solving the differential equations we obtain: R(t)=P 2 (t) + P 1 (t) From R(t), we can obtain system MTTF: It should be clear that the system MTTF and system reliability are critically dependent on the coverage factor.
47
A. BobbioReggio Emilia, June 17-18, 200347 Source of fault coverage data Measurement data from an operational system Large amount of data needed Improved instrumentation needed Fault-injection experiments Expensive but badly needed Tools from CMU,Illinois, LAAS (Toulouse) A fault/error handling submodel (FEHM) Phases: detection, location, retry, reconfig, reboot Estimate duration and probability of success of each phase
48
A. BobbioReggio Emilia, June 17-18, 200348 Redundant System with Finite Detection Switchover Time 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 investigate the effects of detection delay on system reliability and mean time to failure.
49
A. BobbioReggio Emilia, June 17-18, 200349 Redundant System with Finite Detection Switchover Time Assumptions: Two units have the same MTTF and MTTR; Single shared repair person; Average detection/switchover time t sw =1/ ; We need to use a Markov model.
50
A. BobbioReggio Emilia, June 17-18, 200350 Redundant System with Finite Detection Switchover Time 1 1D2 0
51
A. BobbioReggio Emilia, June 17-18, 200351 Redundant System with Finite Detection Switchover Time After solving the Markov model, we obtain steady-state probabilities:
52
A. BobbioReggio Emilia, June 17-18, 200352 Closed-form
53
A. BobbioReggio Emilia, June 17-18, 200353 WFS Example
54
A. BobbioReggio Emilia, June 17-18, 200354 A Workstations-Fileserver Example Computing system consisting of: –A file-server –Two workstations –Computing network connecting them System operational as long as: –One of the Workstations and –The file-server are operational Computer network is assumed to be fault-free
55
A. BobbioReggio Emilia, June 17-18, 200355 The WFS Example
56
A. BobbioReggio Emilia, June 17-18, 200356 Assuming exponentially distributed times to failure – w : failure rate of workstation – f : failure rate of file-server Assume that components are repairable – w : repair rate of workstation – f : repair rate of file-server File-server has priority for repair over workstations (such repair priority cannot be captured by non-state- space models) Markov Chain for WFS Example
57
A. BobbioReggio Emilia, June 17-18, 200357 Markov Availability Model for WFS 0,0 2,11,1 1,02,0 0,1 f 2 w w ww ww w ff ff ff f f Since all states are reachable from every other states, the CTMC is irreducible. Furthermore, all states are positive recurrent.
58
A. BobbioReggio Emilia, June 17-18, 200358 In the figure, the label (i,j) of each state is interpreted as follows: i represents the number of workstations that are still functioning j is 1 or 0 depending on whether the file-server is up or down respectively. Markov Availability Model for WFS (Continued)
59
A. BobbioReggio Emilia, June 17-18, 200359 For the example problem, with the states ordered as (2,1), (2,0), (1,1), (1,0), (0,1), (0,0) the Q matrix is given by: Markov Availability Model for WFS (Continued) Q =
60
A. BobbioReggio Emilia, June 17-18, 200360 Markov Model (steady-state) : Steady-state probability vector These are called steady-state balance equations rate of flow in = rate of flow out after solving for obtain Steady-state availability
61
A. BobbioReggio Emilia, June 17-18, 200361 We compute the availability of the system: System is available as long as it is in states (2,1) and (1,1). Instantaneous availability of the system: Markov Availability Model
62
A. BobbioReggio Emilia, June 17-18, 200362 Markov Availability Model (Continued)
63
A. BobbioReggio Emilia, June 17-18, 200363 Assume that the computer system does not recover if both workstations fail, or if the file-server fails Markov Reliability Model with Repair
64
A. BobbioReggio Emilia, June 17-18, 200364 Markov Reliability Model with Repair States (0,1), (1,0) and (2,0) become absorbing states while (2,1) and (1,1) are transient states. Note: we have made a simplification that, once the CTMC reaches a system failure state, we do not allow any more transitions.
65
A. BobbioReggio Emilia, June 17-18, 200365 Markov Model with Absorbing States If we solve for P 2,1 (t) and P 1,1 (t) then R(t)=P 2,1 (t) + P 1,1 (t) For a Markov chain with absorbing states: A: the set of absorbing states B = - A: the set of remaining states z i,j : Mean time spent in state i,j until absorption
66
A. BobbioReggio Emilia, June 17-18, 200366 Markov Model with Absorbing States (Continued) Mean time to absorption MTTA is given as: Q B derived from Q by restricting it to only states in B
67
A. BobbioReggio Emilia, June 17-18, 200367 Markov Reliability Model with Repair (Continued) [ ]
68
A. BobbioReggio Emilia, June 17-18, 200368 Mean time to failure is 19992 hours. Markov Reliability Model with Repair (Continued)
69
A. BobbioReggio Emilia, June 17-18, 200369 Assume that neither workstations nor file- server is repairable Markov Reliability Model without Repair
70
A. BobbioReggio Emilia, June 17-18, 200370 Markov Reliability Model without Repair (Continued) States (0,1), (1,0) and (2,0) become absorbing states
71
A. BobbioReggio Emilia, June 17-18, 200371 Mean time to failure is 9333 hours. Markov Reliability Model without Repair (Continued) [ ]
Similar presentations
© 2025 SlidePlayer.com. Inc.
All rights reserved.