markov reliability and availability analysis firma
TRANSCRIPT
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
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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, =∈=
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Needed Information
5
What do we need?
Objective:
Transition Probabilities!
𝑃𝑃 𝑋𝑋 𝑡𝑡 = 𝑗𝑗 , 𝑡𝑡 ∈ 0,𝑇𝑇𝑚𝑚 , 𝑗𝑗 = 0,1, … ,𝑁𝑁
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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
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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)
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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
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The Conceptual Model
• Homogeneus process without memory
Transition time Exponential distribution
9
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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
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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
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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)
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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)
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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)
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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
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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
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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
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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
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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
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- 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
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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
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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:
⇒
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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
⇒
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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
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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)
Exercise 4: one component/one repairman – Solution to the fundamental equation (3)
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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
Exercise 4: one component/one repairman – Solution to the fundamental equation (4)
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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
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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
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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
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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
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41
Sojourn Time in a state (2)
• 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
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42
Sojourn Time in a state (3)
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
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 ?
operating failed
ABSORBING STATE!
operating
Piero Baraldi
49
System Reliability: Reparaible Systems (1)
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
System Reliability: Reparaible Systems (2)
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
Exercise 6: Solution
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
Exercise 6: Solution
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
2. Solve the reduced problem of for the probabilitiesof being in these (transient) safe states
'A ( ) SitPi ∈,*
Exercise 6: Solution
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
Exercise 7: Solution
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
Exercise 7: Solution
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
2. Solve the reduced problem of for the probabilitiesof being in these (transient) safe states
'A ( ) SitPi ∈,*
Exercise 7: Solution
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=
Exercise 7: Solution
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
Exercise 7: Solution