Chapter 8

- Continuous Time Markov Chains

Definition

- A discrete-state continuous-time stochastic

process is called a Markov

chain if - for t0 lt t1 lt t2 lt . lt tn lt t , the

conditional pmf satisfies the relation - A CTMC is characterized by state changes that can

occur at any arbitrary time - Index space is continuous.
- The state space is discrete valued.

Continuous Time Markov Chain (CTMC)

- A CTMC can be completely described by
- Initial state probability vector for X(t0)
- Transition probabilities.
- Also,

Homogenous CTMCs

- is a time-homogenous CTMC iff
- Or, the conditional pmf satisfies
- A CTMC is said to be irreducible if every state

can be reached from every other state, with a

non-zero probability. - A state is said to be absorbing if no other state

can be reached from it with non-zero probability.

CTMC Chapman-Kolmogorov Equation

- It can also be written as
- In the matrix form, (Matrix Q is called the

infinitesimal generator matrix (or

simply Generator Matrix)

CTMC Steady-state Solution

- Steady state solution of CTMC
- Irreducible CTMCs having ve steady-state pj

values are called recurrent non-null. - Performance measures may be computed by assigning

reward rates to states and computing expected

steady state reward rates - Accumulated reward (over an interval of time)

Continuous Time Birth-Death Process

- The CTMC and i0,1,2, forms a

B-D process, if ?i, i0,1,2,.. and µi,

i1,2,.. exists, and ?i Birth rate (gt 0) and

µi Death rate (gt 0)

Continuous Time Birth-Death Process (contd.)

In Steady-state,

Steady State Equations

These are called balance eqs. Re-arranging

above,

0

M/M/1 Queue

- Arrivals follow Poisson distribution, i.e.,

inter-arrival times are all i.i.d, EXP(?). - Inter-departure times are i.i.d, EXP(µ).
- N(t) birth-death proc., ?k? µkµ.
- Define, ??/µ (traffic intensity, in Erlangs)

Poisson arrival Process with rate ?

M/M/1 queue (contd.)

- From the balance flow equations, we get
- ? lt 1 (for reasons of stability).
- Expected of customers,

M/M/1 queue (contd.)

- This measure can be viewed as a weighted average,

. - By choosing suitable weights to the states of a

CTMC, we can get most measures of interest and

the resulting model is known as the MRM(Markov

Reward Model). - Other measures
- Average queue length (En)
- Average (expected) response time
- Average (expected) wait time etc.

M/M/1 queue Littles formula

- Let the random variable R denote the response

time - (defined as the time elapsed from the instant

of job arrival until its completion) - Littles law states
- ER EN/?
- Here
- Response time (R) wait time (W) service time

(S) - EW ER ES 1/µ(1-?) - 1/ µ .

Response time distribution (tagged job approach)

- Assuming FCFS and steady-state conditions
- If there are already n jobs in the system, the

next job (N1)st will experience a response time

R SS1S2..SN - S service time for the (N1)st job S1

residual service time for job currently

undergoing service (1). - Because of the memory-less property, these times

are EXP( ). - Hence, for some Nn, the LST of R is,
- Therefore,

M/M/m queue

- m-servers service the queue.

µ

Poisson arrivals (?)

M/M/m Queue Solution

M/M/m Queue performance measures

- Average queue length EN rk k

M/M/m Queue performance measures

- Server utilization rv M - number of busy

servers. For number of customers 0 lt k lt m,

the number of busy servers k. Beyond that the

number of busy servers m. - A customer may have to join the queue.

Poisson stream behavior

- M/M/m input/output both form Poisson streams.
- m2 case
- Case 1 Two independent queues
- Case 2 M/M/2 case

Two separate Poisson streams

? 2 separate M/M/1 queues

Two separate Poisson streams

Combined Poisson steams

Comparative performance

- Case 1 For each M/M/1 queue,
- Case 2 Common queue M/M/2

M/M/1/n Queue

- Finite queue size, finite buffer space ? finite

state space.

- Steady State Solution

M/M/1/n Queue Performance Measures

- Mean queue length (expected of jobs in the

system). - rk k,
- Loss probability
- rn 1, rk 0, k0,1,..,n-1
- Throughput
- rk m , k1,2, ..,n r0 0 (or, rk l ,

k0,1,2, ..,n-1 rn 0)

M/M/1/n Response time distribution

- Response time distribution Job may be rejected

(or accepted) - Unconditional
- Conditional (conditioned on the job being

accepted) - Reward assignment for the kth state, response

time experienced by the tagged task is sum of

k-service times, each of which is EXP(µ), i.e.,

k-stage Erlang. - Unconditional
- Conditional

Special cases of Birth-Death Process

- Pure birth processes
- Poisson process
- Software Reliability Growth Model NHPP
- Number of software failures occurring in (0, t

is N(t), and N(t) is Poisson with, ?(t) abe-bt

and m(t) EN(t) a(1- e-bt) - Instantaneous failure intensity, ?(t)

ba-m(t) - Transient solution may be found using Laplace

transforms - Pure death processes
- No-repairs

Markov Availability Model

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.

2-State Markov Availability Model(Continued)

- Need an additional equation

Downtime in minutes per year

876060

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

2-State Markov Availability Model(Continued)

- 3)
- 4) Steady State Availability

Using SHARPE to Solve the models

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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.

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.

Markov availability model (Continued)

2

1

0

Non-shared (independent) repair

2

1

0

Shared repair

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.

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

Steady-state balance equations (Continued)

- Hence
- Since
- We have
- or

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) 876060

Steady-state balance equations

Homework

- 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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Markov Reliability Model

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).

Markov reliability model with repair (Continued)

Absorbing state

- 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)

- 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

Markov reliability model with repair

(Continued)

Markov reliability model with repair

(Continued)

- After solving these equations, we get
- R(t) P2(t) P1(t)
- Recalling that

, we get

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

Markov Reliability Model with Imperfect Coverage

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

Now allow for Imperfect coverage

c

Markov modelwith imperfect coverage (Continued)

- Assume that the initial state is 2 so that
- Then the system of differential equations are

Markov model with imperfect coverage (Continued)

- After solving the differential equations we

obtain - R(t)P2(t) P1(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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2-component Availability model with detection

delay

- 2-component availability model
- Steady state availability Ass 1-p0
- Failures detection stage takes random time,

EXP(d) - Down states are 0 and 1D ? Ass 1- p0- p1D
- Therefore, steady state unavailability U(d) is

given by

2-component availability model with finite

coverage

- Coverage factor c (probability that the fault

is covered) - 1C state is a re-boot (down) state.

2-components availability model delayfinite

coverage

- Model has detection delaycoverage factor
- Down states are 0, 1C and 1D.

Preventive Maintenance example

- Prolonged usage of a component may lead to

increased failure rate (i.e. IFR situation) - Hence, life time may be modeled as HypoEXP()

distribution, say 2-stage Hypo. - Component is inspected randomly. Time between

inspections is a random, following EXP(?i).

Inspection completion time is EXP(µi). - What does inspection do?
- First stage of life no action
- Second stage of life repair
- That is, preventive maintenance
- State ltstage, faultygt

Performance Models

- Example 2-servers with different service times.
- State ltn1, n2gt
- Performance Average no. of jobs in the system,

En1n2 - Reward rate rn1, n2 n1n2
- Except for the lt0,0gt, in all other states, viz.,

ltk,0gt and ltk,1gt, there are k jobs in the system.

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

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)

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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