Markov Reliability and Availability Analysis Firma

Firma convenzione Politecnico di Milano e Veneranda Fabbrica

del Duomo di MilanoAula Magna – Rettorato

Mercoledì 27 maggio 2015

Markov Reliability and Availability AnalysisPart II: Continuous Time Discrete State

Markov Processes

Piero Baraldi

Continuous Time Discrete State Markov Processes

• The stochastic process may be observed at:

• Discrete times

• Continuously

t10 t2 t3 tn Tm

0 Tm

t

t

DISCRETE-TIME DISCRETE-STATE MARKOV PROCESSES

CONTINUOUS-TIME DISCRETE-STATE MARKOV PROCESS

Piero Baraldi

The conceptual model: Continuous-Time

3

• The stochastic process is observed continuously and transitions are assumed to occur continuously in time

0 Tmt

state j = 0 state j = 5 state j = 2

state j = N

Piero Baraldi

The conceptual model: Finite State Space

4

• The random process of system transition between states in time is described by a stochastic process {X(t); t ≥ 0}

• 𝑋𝑋 𝑡𝑡 ≔ system state at time 𝑡𝑡• 𝑋𝑋 3.6 = 5: the system is in state number 5 at time 𝑡𝑡 = 3.6

OBJECTIVE:Computing the probability that the system is in a given state

as a function of time, for all possible states

( )[ ] [ ] NjTtjtXP m ...,,1,0,,0, =∈=

Piero Baraldi

Needed Information

5

What do we need?

Objective:

Transition Probabilities!

𝑃𝑃 𝑋𝑋 𝑡𝑡 = 𝑗𝑗 , 𝑡𝑡 ∈ 0,𝑇𝑇𝑚𝑚 , 𝑗𝑗 = 0,1, … ,𝑁𝑁

Piero Baraldi

The Conceptual Model: Transition Probabilities

6

• Transition probability that the system moves to state j at time 𝑡𝑡 + 𝜈𝜈 giventhat it is in state i at current time t and given the previous system history

i j

t + νt0 Tmt

( ) ( ) ( ) ( )[ ]tuuxuXitXjtXP <≤===+ 0,,|ν(i = 0, 1, …, N, j = 0, 1, …, N)

u

Piero Baraldi

The Conceptual Model: Markov Assumption

7

( ) ( ) ( ) ( )[ ]tuuxuXitXjtXP <≤===+ 0,,|ν

• IN GENERAL STOCHASTIC PROCESSES: the probability of a future state of the system usually depends on its entire life history

( ) ( ) ( ) ( )[ ]tuuxuXitXjtXP <≤===+ 0,,|ν

• IN MARKOV PROCESSES: the probability of a future state of the system only depends on its present state

=

THE PROCESS HAS “NO MEMORY”

( ) ( )[ ]itXjtXP ==+ |ν

(i = 0, 1, …, N, j = 0, 1, …, N)

(i = 0, 1, …, N, j = 0, 1, …, N)

Piero Baraldi

The Conceptual Model: homogeneous Markov process

8

If the transition probability depends on the interval ν and not on the individual times 𝑡𝑡 and 𝑡𝑡 + 𝜈𝜈

• the transition probabilities are stationary• the Markov process is homogeneous in time

( ) ( ) ( )[ ] ( )ννν ijij pitXjtXPttp ===+=+ |,

i j

𝑡𝑡10 Tmt

i j

0 Tmt

𝑡𝑡1 + 𝜈𝜈

𝑡𝑡2 𝑡𝑡2 + 𝜈𝜈

Piero Baraldi

The Conceptual Model

• Homogeneus process without memory

Transition time Exponential distribution

9

Piero Baraldi

10

The conceptual model: Transition Rates

HYPOTHESIS:• The time interval 𝜈𝜈 = 𝑑𝑑𝑡𝑡 is small such that only one event (i.e., one

stochastic transition) can occur within it

( )0

lim 0dt

dtdt

θ→

=( ),dtdtij θα +⋅

=

ijα = transition rate from state i to state j

(Taylor 1𝑠𝑠𝑠𝑠 order expansion)

𝑝𝑝𝑖𝑖𝑖𝑖 𝑑𝑑𝑡𝑡 = 𝑃𝑃 𝑋𝑋 𝑡𝑡 + 𝑑𝑑𝑡𝑡 = 𝑗𝑗 𝑋𝑋 𝑡𝑡 = 𝑖𝑖 = 1 − 𝑒𝑒−𝛼𝛼𝑖𝑖𝑖𝑖�𝑑𝑑𝑠𝑠

Piero Baraldi

11

The conceptual model: Transition Rates

HYPOTHESIS:• The time interval 𝜈𝜈 = 𝑑𝑑𝑡𝑡 is small such that only one event (i.e., one

stochastic transition) can occur within it

( )0

lim 0dt

dtdt

θ→

=( ),dtdtij θα +⋅

=

ijα = transition rate from state i to state j

(Taylor 1𝑠𝑠𝑠𝑠 order expansion)

𝑝𝑝𝑖𝑖𝑖𝑖 𝑑𝑑𝑡𝑡 = 𝑃𝑃 𝑋𝑋 𝑡𝑡 + 𝑑𝑑𝑡𝑡 = 𝑗𝑗 𝑋𝑋 𝑡𝑡 = 𝑖𝑖 = 1 − 𝑒𝑒−𝛼𝛼𝑖𝑖𝑖𝑖�𝑑𝑑𝑠𝑠

Piero Baraldi

The conceptual model: The Transition Probability Matrix (1)

12

Discrete-time transition probability matrix

00 01 0

10 11 1

0 1

0 1 ......0...1

... ... ... .........

N

N

N N NN

i j Np p p

A p p p

p p pN

=

( )0

lim 0dt

dtdt

θ→

=( ) ( ),dtdtdtp ijij θα +⋅=

Continuous-time transition probability matrix

0 01 01

10 1 101

1 ...

1 ...

... ... ... ...

N

j Nj

N

j Njj

dt dt dt

A dt dt dt

α α α

α α α

=

=≠

− ⋅ ⋅ ⋅

= ⋅ − ⋅ ⋅

∑

∑

• In analogy with the discrete-time formulation:

𝑝𝑝𝑖𝑖𝑖𝑖 𝑑𝑑𝑡𝑡 = 1 −�𝑖𝑖≠𝑖𝑖

𝑝𝑝𝑖𝑖𝑖𝑖 𝑑𝑑𝑡𝑡 = 1 − 𝑑𝑑𝑡𝑡 ��𝑖𝑖≠𝑖𝑖

𝛼𝛼𝑖𝑖𝑖𝑖 + 𝜃𝜃(𝑑𝑑𝑡𝑡)

Piero Baraldi

13

• In analogy with the discrete-time formulation:

( ) ( )P t dt P t A+ = ⋅

0 01 01

10 1 101

1 ...

1 ...

... ... ... ...

N

j Nj

N

j Njj

dt dt dt

A dt dt dt

α α α

α α α

=

=≠

− ⋅ ⋅ ⋅

= ⋅ − ⋅ ⋅

∑

∑( ) ( ) ( )[ ]⋅tPtPtP N...10

( ) ( ) ( )[ ]dttPdttPdttP N +++ ...10

=

• First-equation:

( ) ( ) ( ) ( )0 0 0 10 1 01

1 ...N

j N Nj

P t dt dt P t P t dt P t dtα α α=

+ = − + ⋅ + +

∑

The conceptual model: the fundamental matrix equation (1)

Piero Baraldi

14

( ) ( ) ( ) ( )0 0 0 10 1 01

1 ...N

j N Nj

P t dt dt P t P t dt P t dtα α α=

+ = − + ⋅ + +

∑

subtract P0(t) on both sides

( ) ( ) ( )00 0 10 1 0

1...

N

j N Nj

dP P t P t P tdt

α α α=

= − ⋅ + ⋅ + + ⋅∑( ) ( )=

−+→ dt

tPdttPdt

00

0lim

divide by dt

let dt → 0

( ) ( ) ( ) ( ) ( ) ( ) ( )dttPdttPdttPtPtPtPdttP NN

N

jj 0110

1000000 ... ααα +++−−=−+ ∑

=

( ) ( ) ( ) ( ) ( )tPtPtPdt

tPdttPNN

N

jj 0110

100

00 ... ααα +++−=−+ ∑

=

The conceptual model: the fundamental matrix equation (2)

Piero Baraldi

15

The conceptual model: The Transition Probability Matrix

System of linear, first-order differential equations in the unknown state probabilities

( ) , 0,1,2,..., , 0jP t j N t= ≥

( )

0 01 01

10 1 101

...

, ...

... ... ... ...

N

j Nj

N

j Njj

d P P t A Adt

α α α

α α α

=

∗ ∗

=≠

−

= ⋅ = −

∑

∑α11

α00

TRANSITION RATEMATRIX

• Extending to the other equations:

* *

It will be indicated as A

Piero Baraldi

Exercise 1

16

Consider a system made by one component which can be in two states: working or failed. Assume constant failure rate 𝜆𝜆 and constant repair rate 𝜇𝜇. You are required to:• Draw the Markov diagram• Find the transition rate matrix, A

Piero Baraldi

17

Discrete states = 0 component working1 component failed

Transition rates = λ rate of failure (i.e., from 0 to 1)= μ rate of repair (i.e., from 1 to 0)

−

−=

µµλλ

A TRANSITION RATE MATRIX

MARKOV DIAGRAM

Exercise 1 (Solution)

Piero Baraldi

Exercise 2

18

Consider a system made by 𝑵𝑵 identical components which can be in two states: working or failed. Assume constant failure rate 𝜆𝜆, that 𝑵𝑵 repairman are availableand that the single component repair rate is constant and equal to 𝜇𝜇. You are required to:• Draw the Markov diagram• Find the transition rate matrix, A

Piero Baraldi

Exercise 2 - Solution

19

-System discrete states:• State 0: none failed, all components function• State 1: one component failed, N-1 function• State 2: two components failed, N-2 function• …• State N: all components failed, none function

Consider a system made by 𝑵𝑵 identical components which can be in two states: working or failed. Assume constant failure rate 𝜆𝜆, that 𝑵𝑵 repairmen are availableand that the component repair rate is constant and equal to 𝜇𝜇. You are requiredto:• Draw the Markov diagram• Find the transition rate matrix

Piero Baraldi

20

all functioning all failed

HYPOTHESES:- one event (failure or repair of one component) can occur in the small ∆t- the events are mutually exclusive

Explanation: Probability of transition 0 1 = = probability { anyone of the N components fails in Δt } == probability { component 1 fails in Δt or component 2 fails in Δt or …} == probability { component 1 fails in Δt } + probability { component 2 fails in Δt } + … == λΔt + λΔt + … = NλΔt

Exercise 2 - Solution

Piero Baraldi

Exercise 3: system with 𝑵𝑵 identical components and 𝟏𝟏repairman available

Consider a system made by 𝑵𝑵 identical components which can be in two states: working or failed. Assume constant failure rate 𝜆𝜆, that 1 repairman is available and that the component repair rate is constant and equal to 𝜇𝜇. You are required to:• Draw the Markov diagram• Find the transition rate matrix

Piero Baraldi

- One repairman is available

- Each component can be in two states: working or failed

- The transition rates are constant = λ rate of failure= μ rate of repair

- The system is made of N identical componentsSYSTEM CHARACTERISTICS:

all functioning all failed

Exercise 3: Solution

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Solution to the Fundamental Equation

23

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Solution to the fundamental equation of the Markov processcontinuous in time

24

System of linear, first-orderdifferential equations in the unknown

state probabilities

( ) , 0,1,2,..., , 0jP t j N t= ≥

0 01 01

10 1 101

...

...

... ... ... ...

N

j Nj

N

j Njj

A

α α α

α α α

=

=≠

−

= −

∑

∑( )d P P t A

dt= ⋅

( )0P C=

USE LAPLACE TRANSFORM

where

Piero Baraldi

25

• Laplace Transform:

• First derivative:

• Apply the Laplace operator to

( ) ( ) ( )0

stj j jP s L P t e P t dt

∞ − = = ∫ Nj ,...,1,0, =( ) ( ) ( )0 , 0,1,...,j

j j

dP tL s P s P j N

dt

= ⋅ − =

( )d P P t Adt

= ⋅

( ) ( )[ ]AtPLdt

tPdL ⋅=

Linearity( ) ( )sP s C P s A− = ⋅ First derivative

( ) 1P s C s I A

− = ⋅ ⋅ −

= inverse transform of( )tP ( )sP~

Solution to the fundamental equation of of the Markov processcontinuous in time: the Lapace Transform Method

Piero Baraldi

26

( ) 0=dt

tPd ( ) ( ) 0=⋅Π=⋅= AAtPdt

tPd

0=⋅Π A

01

N

jj=Π =∑

Solution to the Fundamental Equation: Steady State Probabilities

• At steady state

• Solve the (linear) system:

⇒

Piero Baraldi

27

( ) 0=dt

tPd ( ) ( ) 0=⋅Π=⋅= AAtPdt

tPd

0=⋅Π A

01

N

jj=Π =∑

0

0,1,2,...,jj N

ii

Dj N

D=

Π = =

∑

jD = determinant of the square matrix obtained fromby deleting the j-th row and column

A

Solution to the Fundamental Equation: Steady State Probabilities

• At steady state

• Solve the (linear) system:

• It can be shown that

⇒

Piero Baraldi

Exercise 4: one component/one repairman – Solution to the fundamental equation (1) 28

Consider a system made by one component which can be in two states: working (‘0’) orFailed (‘1’). Assume constant failure rate 𝜆𝜆 and constant repair rate 𝜇𝜇 an that the componentis working at t = 0: C = [1 0]

- You are required to find the component steady state and instantaneous availability

Piero Baraldi

29

Exercise 4: one component/one repairman – Solution to the fundamental equation (2)

• Solve

• Compute

( ) ( ) 1P s C sI A

−= ⋅ −

( ) 1sI A

−−

( )

( )[ ]

+

+++

=

+

+−

=

+−−+

=−−

−

λµλµ

µλ

λµλµ

µµλλ

ss

sss

ss

AIs

ss

AIs

2

11

1

det1

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30

( ) ( ) ( )

++++

+=

µλλ

µλµ

ssssssP~• Anti-Transform

• It is known that axeas

L −− =

+11

( ) ( )axeaass

L −− −=

+

1111and

( ) ( )( ) t tP t e eλ µ λ µµ λ λ λλ µ λ µ λ µ λ µ

− + ⋅ − + ⋅ = + ⋅ − ⋅ + + + +

STATE PROBABILITY VECTOR

( )tP0( )tP1

system instantaneous availability(probability of being in operational state

0 at time t)

system instantaneous unavailability(probability of being in failed state 1

at time t)


Piero Baraldi

31

• TIME-DEPENDENT STATE PROBABILITIES

• STEADY STATE PROBABILITIES

( )MTBFMTTR

MTBFtPt +

=+

=+

==Π∞→ λµ

λλµ

µ11

1lim 00

( )MTBFMTTR

MTTRtPt +

=+

=+

==Π∞→ λµ

µλµ

λ11

1lim 11

= average fraction of time the system is functioning

= average fraction of time the system is down (i.e., under repair)

(system instantaneous availability)

(system instantaneous unavailability)

( ) ( )tetP µλ

λµλ

λµµ +−

++

+=0

( ) ( )tetP µλ

λµλ

λµλ +−

++

+=1


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Quantity of Interest

32

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Frequency of departure from a state

33

• Unconditional probability of arriving in state j in the next 𝑑𝑑𝑡𝑡 departing from state i at time t:

• Frequency of departure from state i to state j:

= depij ij iv α= ⋅Π

• Total frequency of departure from state i to any other state j:

( ) ( ) ( )0

N

i ij i ii ijj i

v t P t P t=≠

= ⋅ = − ⋅∑α α i ii iv = − ⋅Πα=

(at steady state)

(at steady state)

𝑃𝑃[𝑋𝑋 𝑡𝑡 + 𝑑𝑑𝑡𝑡 = 𝑗𝑗,𝑋𝑋 𝑡𝑡 = 𝑖𝑖]

𝑃𝑃[𝑋𝑋 𝑡𝑡 + 𝑑𝑑𝑡𝑡 = 𝑗𝑗,𝑋𝑋 𝑡𝑡 = 𝑖𝑖]=𝑃𝑃 𝑋𝑋 𝑡𝑡 + 𝑑𝑑𝑡𝑡 = 𝑗𝑗 𝑋𝑋 𝑡𝑡 = 𝑖𝑖 � 𝑃𝑃 𝑋𝑋 𝑡𝑡 = 𝑖𝑖 = 𝑝𝑝𝑖𝑖𝑖𝑖 𝑑𝑑𝑡𝑡 𝑃𝑃𝑖𝑖(𝑡𝑡)

𝜈𝜈𝑖𝑖𝑖𝑖𝑑𝑑𝑑𝑑𝑑𝑑 𝑡𝑡 = lim

𝑑𝑑𝑠𝑠→0

𝑝𝑝𝑖𝑖𝑖𝑖 𝑑𝑑𝑡𝑡 𝑃𝑃𝑖𝑖 𝑡𝑡𝑑𝑑𝑡𝑡

= 𝛼𝛼𝑖𝑖𝑖𝑖𝑃𝑃𝑖𝑖(𝑡𝑡)

dep dep

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Frequency of arrival to a state

34

• In analogy, the total frequency of arrivals to state i from any other state k:

( ) ( )0

0

Narri ki k

kk i

Narri ki k

kk i

v t P tα

ν α

=≠

=≠

= ⋅

= ⋅Π

∑

∑ (at steady state)

AT STEADY STATE:frequency of departures from state i = frequency of arrivals to state i

Piero Baraldi

System Failure Intensity

35

• Rate at which system failures occur • SYSTEM FAILURE INTENSITY Wf:

( ) ( )f i i Fi S

W t P t λ →∈

= ⋅∑

S = set of success states of the system

= probability of the system being in the functioning state i at time t( )tPiλi F = conditional (transition) probability of leaving success state i towards

failure states

F = set of failure states of the system

Piero Baraldi

System Failure Intensity

36

• Rate at which system failures occur • SYSTEM FAILURE INTENSITY Wf:

( ) ( )f i i Fi S

W t P t λ →∈

= ⋅∑

S = set of success states of the system

= probability of the system being in the functioning state i at time t( )tPiλi F = conditional (transition) probability of leaving success state i towards

failure states

F = set of failure states of the system

• Steady state- Expected number of system failure per unit of time

Piero Baraldi

37

System Repair Intensity

• Rate at which system repairs occur • Expected number of system repairs per unit of time

• SYSTEM REPAIR INTENSIT 𝑾𝑾𝒓𝒓:

F = set of failure states of the system= probability of the system being in the failure state j at time t( )tPj

μj S = conditional (transition) probability of leaving failure state j towardssuccess states

( ) ( )r j j Sj F

W t P t µ →∈

= ⋅∑S = set of success states of the system

Piero Baraldi

Exercise 5: one component/one repairman

38

Consider a system made by one component which can be in two states: working (‘0’) orFailed (‘1’). Assume constant failure rate 𝜆𝜆 and constant repair rate 𝜇𝜇 an that the componentis working at t = 0: C = [1 0]

- You are required to find the failure and repair intensities

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Exercise 5: one component/one repairman – Failure and repair intensities 39

( ) ( )f i i Fi S

W t P t λ →∈

= ⋅∑

( ) ( )r j j Sj F

W t P t µ →∈

= ⋅∑functioning failed

S = set of success states of the system = {0}F = set of failure states of the system = {1}λi F = λμj S = μ

( ) ( ) ( )tf etPtW µλ

λµλ

λµλµλ +−

++

+==

2

0

( ) ( ) ( )tr etPtW µλ

λµµλ

λµµλµ +−

++

+== 1

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Sojourn Time in a state (1)

40

• Time of occupance of state i (sojourn time) = 𝑻𝑻𝒊𝒊

• Markov property and time homogeneity imply that if at time t the process is in state i,the time remaining in state i is independent of time already spent in state 𝑖𝑖

P 𝑇𝑇𝑖𝑖 > 𝑡𝑡 + 𝑠𝑠 𝑇𝑇𝑖𝑖 > 𝑡𝑡 = 𝑃𝑃 𝑋𝑋 𝑡𝑡 + 𝑢𝑢 = 𝑖𝑖, 0 ≤ 𝑢𝑢 ≤ 𝑠𝑠 𝑋𝑋 𝜏𝜏 = 𝑖𝑖, 0 ≤ 𝜏𝜏 ≤ 𝑡𝑡) =

= P(𝑋𝑋 𝑡𝑡 + 𝑢𝑢 = 𝑖𝑖, 0 ≤ 𝑢𝑢 ≤ 𝑠𝑠|𝑋𝑋 𝑡𝑡 = 𝑖𝑖) (by Markov property)

= P(𝑋𝑋 𝑢𝑢 = 𝑖𝑖, 0 ≤ 𝑢𝑢 ≤ 𝑠𝑠|𝑋𝑋 0 = 𝑖𝑖) (by homogeneity)

= ℙ(𝑇𝑇𝑖𝑖 > 𝑠𝑠) Memoryless Property

• The only distribution satisfying the memoryless property is the Exponential distribution

𝑇𝑇𝑖𝑖~𝐸𝐸𝐸𝐸𝑝𝑝

Piero Baraldi

41


• System departure rate from state i (at steady state): −𝛼𝛼𝑖𝑖𝑖𝑖

• Expected sojourn time 𝒍𝒍𝒊𝒊: average time of occupancy of state 𝑖𝑖

𝑇𝑇𝑖𝑖~𝐸𝐸𝐸𝐸𝑝𝑝(−𝛼𝛼𝑖𝑖𝑖𝑖)

𝑙𝑙𝑖𝑖 = 𝔼𝔼 𝑇𝑇𝑖𝑖 =1

−𝛼𝛼𝑖𝑖𝑖𝑖

Piero Baraldi

42


i ii iν = −α ⋅Π• Total frequency of departure at steady state:

lii ii iiΠ

ν = −α ⋅Π =

• Average time of occupancy of state:1liii

=−α

iii l⋅=Π ν

The mean proportion of time Πi that the system spends in state i is equal to the total frequency of arrivals to state i multiplied by the mean

duration of one visit in state i

dep

dep

𝑣𝑣𝑖𝑖𝑑𝑑𝑑𝑑𝑑𝑑 = 𝑣𝑣𝑖𝑖𝑎𝑎𝑎𝑎𝑎𝑎

arr

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System Availability

43

S = set of success states of the systemF = set of failure states of the system

• System instantaneous availability at time t= sum of the probabilities of being in a success state at time t

( ) ( ) ( ) ( )1 1i ji S j F

p t P t q t P t∈ ∈

= = − = −∑ ∑

( ) ( ) ( )1i j

i S j Fp s P s P s

s∈ ∈

= = −∑ ∑

In the Laplace domain

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System Reliability

44

• TWO CASES:

1) Non-Reparaible Systems No repairs allowed

2) Reparaible Systems Repairs allowed

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45

• No repairs allowed ⇒

• In the Laplace Domain:

• Mean Time to Failure (MTTF):

Reliability = Availability ( ) ( ) ( )tqtptR −=≡ 1

( ) ( ) ( )1i j

i S j FR s P s P s

s∈ ∈

= = −∑ ∑

( ) ( ) ( ) ( ) ( )0000

~10~0~

=∈∈=

∞−

∞

−===

== ∑∑∫∫

sFjj

Sii

s

st sPs

PRdtetRdttRMTTF

System Reliability: Non-Reparaible Systems

Piero Baraldi

System Reliability

46

• TWO CASES:

1) Non-reparaible systems No repairs allowed

2) Reparaible systems Repairs allowed

Piero Baraldi

47

System Reliability: Reparaible Systems (1)

operating

operating

failed

{ }2=F{ }1,0=S

Can I trasnsform the Markov Diagram in such a way that 𝑅𝑅 𝑡𝑡 = 𝑃𝑃0 𝑡𝑡 + 𝑃𝑃1 𝑡𝑡 ?

Piero Baraldi

48


operating

operating

failed

{ }2=F{ }1,0=S

Can I trasnsform the Markov Diagram in such a way that 𝑅𝑅 𝑡𝑡 = 𝑃𝑃0 𝑡𝑡 + 𝑃𝑃1 𝑡𝑡 ?

operating failed

ABSORBING STATE!

operating

Piero Baraldi

49


1. Transform the failed states 𝑗𝑗 ∈ 𝐹𝐹 into absorbing states (the system cannot be repairedit is not possible to escape from a failed state)

operating

operating

failed { }2=F{ }1,0=S

−+

−=

µµλλµµ

λλ

220

022A

The new matrix contains the transition rates for transitions onlyamong the success states

'ASi∈

(the “reduced” system is virtually functioning continuously with no interruptions)

𝐴𝐴 =−2𝜆𝜆 2𝜆𝜆 0𝜇𝜇 𝜇𝜇 + 𝜆𝜆 𝜆𝜆0 0 0

𝐴𝐴𝐴 = −2𝜆𝜆 2𝜆𝜆𝜇𝜇 𝜇𝜇 + 𝜆𝜆

Piero Baraldi

50


2. Solve the reduced problem of for the probabilitiesof being in these (transient) safe states

'A ( ) SitPi ∈,*

Reliability

( ) ( ) '**

AtPdt

tPd⋅=

( ) ( )ii S

R t P t∗

∈

=∑

Mean Time To Failure (MTTF)

( ) ( ) ( )0

0 0ii S

MTTF R t dt P R∞

∗

∈

= = =∑∫

NOTICE: in the reduced problem we have only transient states ⇒ ( ) 0** =∞=Π ii P

Piero Baraldi

Exercise 6

Consider a system made by 2 identical components in series. Eachcomponent can be in two states: working or failed. Assume constantfailure rate 𝜆𝜆, that 2 repairman are available and that the single component repair rate is constant and equal to 𝜇𝜇. You are required to:• find the system reliability • find the system MTTF

51

Piero Baraldi

52

systemoperating

systemfailed

systemfailed

SERIES LOGIC

System discrete states:State 0: system is operating (both components functioning)State 1: system is failed (only one of the two components functioning)State 2: system is failed (both components failed)

( )

−+−

−=

µµλλµµ

λλ

220

022A


Piero Baraldi

53

( )2 2 0

' 20 2 2

A Aλ λµ λ µ λ λ

µ µ

− = − + ⇒ = − −

1. Exclude all the failed states 𝑗𝑗 ∈ 𝐹𝐹 from the transition rate matrix

• System reliability R(t) :=probability of the system being in safe states 0 or 1 continuously from t = 0

PARTITION OF THE TRANSITION MATRIX


Piero Baraldi

54

'd P P Adt

∗∗= ⋅

• Easy to solve in the time domain:

( )0P C∗ ∗=

which simplifies to

*0

*0 2 P

dtdP

⋅−= λ

( ) 10*0 =P

20 ( ) tP t e λ∗ −=( ) =tR


'A ( ) SitPi ∈,*


Piero Baraldi

Exercise 7

Consider a system made by 2 identical components in parallel. Eachcomponent can be in two states: working or failed. Assume constantfailure rate 𝜆𝜆, that 2 repairman are available and that the single component repair rate is constant and equal to 𝜇𝜇. You are required to:• find the system reliability • find the system MTTF

55

Piero Baraldi

56

systemoperating

systemfailed

systemoperating

PARALLEL LOGIC

System discrete states:State 0: system is operating (both components functioning)State 1: system is operating (only one of the two components functioning)State 2: system is failed (both components failed)

( )

−+−

−=

µµλλµµ

λλ

220

022A


Piero Baraldi

57

1. Exclude all the failed states 𝑗𝑗 ∈ 𝐹𝐹 from the transition rate matrix

• System reliability R(t) :=probability of the system being in safe states 0 or 1 continuously from t = 0

( ) ( )

2 2 02 2

'0 2 2

A Aλ λ

λ λµ µ λ λ

µ µ λµ µ

−−

= − + ⇒ = − + −

PARTITION OF THE TRANSITION MATRIX


Piero Baraldi

58

( ) ( )2 2d P P t

dtλ λµ λ µ

∗∗ −

= ⋅ − + ( ) ( )0 1 0P∗ =

In the time domain:

In the Laplace domain:

( ) ⇒⋅= '**

AtPdtPd

( ) ( ) ( ) 11 0 'P s sI A

−∗= ⋅ −( ) ( ) ( ) ⇒−⋅= −1**

'0~~ AIsPsP


'A ( ) SitPi ∈,*


Piero Baraldi

59( ) ( ) ( ) 1

1 0 'P s sI A−∗

= ⋅ −

( )

+−

−=

µλµλλ 22

'with A

++−

−+=−

λµµλλ

ss

AIs22

'

( ) ( )( )

+

++−+++

=− −

λµλλµ

λµλµλ 22

221' 1

ss

ssAIs

( ) ( )( )1

0 1

21'2

ssI A

ss sλ µ λµ λω ω

− + + − = +− −

2 2

0,13 6

2λ µ λ λµ µ

ω− − ± + +

=where

Exercise 7: Solution59

Piero Baraldi

60

• It is known that

( ) ( ) ( )( ) ( ) ( )( ) ( )λλµωωλµ

λλµωω

212

2011'~

1010

1**++

−−=

+

++−−

=−⋅= − ssss

sss

AIsCsP

010 1

0 1

( )tte eR t

ωωω ωω ω

⋅⋅⋅ − ⋅=

−

( ) ( ) ( )sPsPsR 10~~~ +=SYSTEM RELIABILITY In the Laplace domain

In the time domainSYSTEM RELIABILITY

𝐿𝐿−11

𝑠𝑠 − 𝑎𝑎=


Piero Baraldi

61

( ) ( ) 1P s C s I A

−∗∗ ′= ⋅ ⋅ −

( ) 1' TMTTF C A w−∗= ⋅ − ⋅ [ ]1 1 1 ... 1 Tw =with

( )

( ) ( )

( )

1

2

2 2

2

2 2 11 0

1

2 111 02 12 2

11 3212 2

32 2

MTTFλ λµ µ λ

µ λ λµ λλ λ µ λµ

λ µµ λ λλ λ

µλ λ

−− = ⋅ ⋅ − +

+ = ⋅ ⋅ ⋅ = + −

+= + ⋅ = =

= +

• Starting from :

• MEAN TIME TO FAILURE ( ) ( )0 0ii

MTTF R P∗= =∑ ( )∑=

=1

0

* 0~i

iP


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