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1
By
Behzad Akbari
Tarbiat Modares University
Spring 2009
Probability Overview and
Introduction to Reliability Analysis
In the Name of the Most High
These slides are based on the slides of Prof. K.S. Trivedi (Duke University)
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Sample Space
Probability implies random experiments. A random experiment can have many possible
outcomes; each outcome known as a sample point (a.k.a. elementary event) has some probability assigned. This assignment may be based on measured data or guestimates.
Sample Space S : a set of all possible outcomes (elementary events) of a random experiment. Finite (e.g., if statement execution; two outcomes) Countable (e.g., number of times a while statement is
executed; countable number of outcomes) Continuous (e.g., time to failure of a component)
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Events
An event E is a collection of zero or more sample points from S
S and E are sets use of set operations.
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Algebra of events
Sample space is a set and events are the subsets of this (universal) set.
Use set algebra and its laws on p. 9. Mutually exclusive (disjoint) events
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Probability axioms
(see pp. 15-16 for additional relations)
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Probability system
Events, sample space (S), set of events. Subset of events that are measurable. F : Measurable subsets of S
F be closed under countable number of unions and intersections of events in F .
-field: collection of such subsets F . Probablity space (S, F , P)
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Combinatorial problems
Deals with the counting of the number of sample points in the event of interest.
Assume equally likely sample points:
P(E)= number of sample points in E / number in S Example: Next two Blue Devils games
S = {(W1,W2), (W1,L2), (L1,W2), (L1,L2)}
{s1, s2, s3, s4} P(s1) = 0.25= P(s2) = P(s3) = P(s4) E1: at least one win {s1,s2,s3} E2: only one loss {s2, s3} P(E1) = 3/4; P(E2) = 1/2
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Conditional probability
In some experiment, some prior information may be available, e.g., P(e|G): prob. that e occurs, given that ‘G’ has occurred.
In general,
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Mutual Independence
A and B are said to be mutually independent, iff,
Also, then,
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Independent set of events
Set of n events, {A1, A2,..,An} are mutually independent iff, for each
Complements of such events also satisfy,
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Series-Parallel systems
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Series system
Series system: n statistically independent components.
Let, Ri = P(Ei), then series system reliability:
For now reliability is simply a probability, later it will be a function of
time
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Series system (Continued)
This simple PRODUCT LAW OF RELIABILITIES,is applicable to series systems of independentcomponents.
n
iis RR
1
R1 R2 Rn
(2)
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Series system (Continued)
Assuming independent repair, we have product law of availabilities
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Parallel system
System consisting of n independent parallel components.
System fails to function iff all n components fail.
Ei = "component i is functioning properly"
Ep = "parallel system of n components is functioning
properly."
Rp = P(Ep).
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Parallel system (Continued)
Therefore:
"" failedhassystemparallelTheEp
"" failedhavecomponentsnAll____
2
__
1 ... nEEE
)...()(____
2
__
1
__
np EEEPEP
)( ...)()(____
2
__
1 nEPEPEP
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Parallel system (Continued)
• Parallel systems of independent components follow the PRODUCT LAW OF UNRELIABILITIES
R1
Rn
...
...
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Parallel system (Continued)
Assuming independent repair, we have product law of unavailabilities:
n
iip AA
1
)1(1
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Series-Parallel System
Series-parallel system: n-series stages, each with ni
parallel components.
Reliability of series parallel system
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Series-Parallel system (example)
Example: 2 Control and 3 Voice Channels
control
control
voice
voice
voice
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Each control channel has a reliability Rc
Each voice channel has a reliability Rv
System is up if at least one control channel and at least 1
voice channel are up.
Reliability:])1(1][)1(1[ 32
vc RRR
Series-Parallel system (Continued)
(3)
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Theorem of Total Probability
Any event A: partitioned into two disjoint events,
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Example
Binary communication channel:
P(R0|T0)
P(R1|T1)
P(R1 |T
0)
P(R 0|T1
)T0
T1 R1
R0 Given: P(R0|T0) = 0.92; P(R1|T1) = 0.95P(T0) = 0.45; P(T1) = 0.55
P(R0) = P(R0|T0) P(T0) + P(R0|T1) P(T1) = 0.92 x 0.45 + 0.08 x 0.55 = 0.4580
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Bridge Reliability using
conditioning/factoring
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Bridge: conditioning
C1
C5
C2
C4
C3
Non-series-parallel block diagram
Factor (condition)on C3
S T
C3 up
C3 downC1
C5
C2
C4
S T
C1
C4
C2
C5S T
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Bridge (Continued)
Component C3 is chosen to factor on (or condition on)
Upper resulting block diagram: C3 is down
Lower resulting block diagram: C3 is up
Series-parallel reliability formulas are applied to both the
resulting block diagrams
Use the theorem of total probability to get the final result
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Bridge (Continued)
RC3down= 1 - (1 - RC1RC2) (1 - RC4RC5)
AC3down= 1 - (1 - AC1AC2) (1 - AC4AC5)
RC3up = (1 - FC1FC4)(1 - FC2FC5)
= [1 - (1-RC1) (1-RC4)] [1 - (1-RC2) (1-RC5)]
AC3up = [1 - (1-AC1) (1-AC4)] [1 - (1-AC2) (1-AC5)]
Rbridge = RC3down . (1-RC3 ) + RC3up RC3
also
Abridge = AC3down . (1-AC3 ) + AC3up AC3
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Fault Tree
Reliability of bridge type systems may be modeled using a fault tree
State vector X={x1, x2, …, xn}
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Fault tree (contd.)
Example:
CPU
DS1
DS3
DS2
NIC1
NIC2
/CPU
/DS1
/DS3
/DS2
/NIC2
/NIC1
SystemFail
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Bernoulli Trial(s)
Random experiment 1/0, T/F, Head/Tail etc. e.g., tossing a coin P(head) = p; P(tail) = q.
Sequence of Bernoulli trials: n independent repetitions. n consecutive execution of an if-then-else statement
Sn: sample space of n Bernoulli trials
For S1:
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Bernoulli Trials (contd.)
Problem: assign probabilities to points in Sn
P(s): Prob. of successive k successes followed by (n-k) failures. What about any k failures out of n ?
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Bernoulli Trials (contd.)
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Non-homogenuous Bernoulli Trials
Non-homogenuous Bernoulli trials Success prob. for ith trial = pi
Example: Ri – reliability of the ith component.
Non-homogeneous case – n-parallel components such that k or more out n are working:
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Homework :
For the following system, write down the expression for system reliability:
Assuming that block i failure probability qi
C
A B
D
C
C
E
D
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Methods for non-series-parallel RBDs
Factoring or conditioning
State enumeration (Boolean truth table)
Min-paths
inclusion/exclusion
SDP (Sum of Disjoint Products)
BDD (Binary Decision Diagram)
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Basic Definitions
tFtXPtR 1
00
dttRdtttfXEMTTF
F(t): : distribution function of system lifetime
f(t): density function of system lifetime
Reliability R(t):
X : time to failure of a system
Mean Time To system Failure
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Availability
This result is valid without making assumptions on the form of the distributions of times to failure & times to repair.
Also:
MTTRMTTF
MTTFASS
)yearminutes(
60*8760*)1(
perin
Adowntime ss
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Exponential Distribution
0 1 tetF t
0 t tetf 0 t tetR
tR
tfth
/1MTTF
Distribution Function:
Density Function:
Reliability:
Failure Rate:
failure rate is age-independent (constant)
MTTF:
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Reliability Block Diagrams
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Reliability Block Diagrams: RBDs
Combinatorial (non-state space) model type Each component of the system is represented as a block System behavior is represented by connecting the blocks
Blocks that are all required are connected in series Blocks among which only one is required are connected in parallel When at least k of them are required are connected as k-of-n
Failures of individual components are assumed to be independent
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Reliability Block Diagrams (RBDs)(continued) Schematic representation or model Shows reliability structure (logic) of a system Can be used to determine
If the system is operating or failed Given the information whether each block is in operating or failed
state A block can be viewed as a “switch” that is “closed” when the
block is operating and “open” when the block is failed System is operational if a path of “closed switches” is found from
the input to the output of the diagram
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Reliability Block Diagrams (RBDs)(continued) Can be used to calculate
Non-repairable system reliability given Individual block reliabilities Or Individual block failure rates Assuming mutually independent failures events
Repairable system availability and MTTF given Individual block availabilities Or individual block MTTFs and
MTTRs Assuming mutually independent failure events Assuming mutually independent restoration events Availability of each block is modeled as an alternating renewal
process (or a 2-state Markov chain)
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Series system in RBD
Series system of n components.
Components are statistically independent
Define event Ei = "component i functions properly.”
For the series system:
ceindependenby ,
)"properly gfunctionin is system The"
)()...()(
)...(
(
21
21
n
n
EPEPEP
EEEP
P
R1 R2 Rn
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Reliability for Series system
Product law of reliabilities:
where Ri is the reliability of component i
For exponential Distribution:
For weibull Distribution:
n
iis
n
iis tRtRRR
11
)()(or
n
ii
i
t
st
i etRthenetRif 1)()(
n
ii
i
t
st
i etRthenetRif 1
)(
)()(
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Availability for Series System
Assuming independent repair for each component,
where Ai is the (steady state or transient) availability of component i
n
iis
n
i ii
in
iis
tAtA
MTTRMTTF
MTTFAA
1
11
)()(
or ,
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MTTF for Series System
Assuming exponential failure-time distribution with constant failure rate i for each component, then:
1/1
nMTTF
ii
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Parallel system in RBD
A system consisting of n independent components in
parallel.
It will fail to function only if all n components have failed.
Ei = “The component i is functioning”
Ep = "the parallel system of n component is functioning
properly."
R1
Rn
...
...
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Parallel system in RBD(Continued)
"failedhassystemparallelThe"pE
"failedhavecomponentsnAll"____
2
__
1 ... nEEE
)...()(____
2
__
1
__
np EEEPEP )()...()(____
2
__
1 nEPEPEP
Therefore:
)(1)( pp EPEP
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49
Reliability for parallel system
Product law of unreliabilities
where Ri is the reliability of component i
For exponential distribution:
))(1(1)()1(111
n
iip
n
iip tRtRRR or ,
n
i
tp
ti
ii etRthenetR1
)1(1)(,)(
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Availability for parallel system
Assuming independent repair,
where Ai is the (steady state or transient) availability of component i.
n
iip
n
i ii
in
iip
tAtAor
MTTRMTTF
MTTRAA
1
11
))(1(1)(
1)1(1
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Homework :
For a 2-component parallel redundant system with EXP( ) behavior, write down expressions for:
Rp(t) MTTFp
Further assuming EXP(µ) behavior and independent repair, write down expressions for: Ap(t) Ap
downtime
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Homework :
For a 2-component parallel redundant system with EXP( ) and EXP( ) behavior, write down expressions for:
Rp(t) MTTFp
Assuming independent repair at rates µ1 and µ2, write down expressions for: Ap(t) Ap
downtime
1 2
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Series-Parallel system
2 Control and 3 Voice Channels Example
control
control
voice
voice
voice
•System is up as long as 1 control and 1 voice channel are up
•The whole system can be treated as a series system with two blocks, each block being a parallel system
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Each control channel has a reliability Rc(t)
Each voice channel has a reliability Rv(t)
System is up if at least one control channel and at
least 1 voice channel are up.
Reliability:
]))(1(1][))(1(1[)( 32 tRtRtR vc
Series-Parallel system (Continued)
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Homework :
Specialize formula (3) to the case where:
Derive expressions for system reliability and system mean time to failure.
tv
tc
vc etRandetR )()(
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A Workstations’ File-server 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
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Computer Network
Workstation 1 Workstation 2
File Server
The WFS Example
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RBD for the WFS Example
Workstation 1
Workstation 2
File Server
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Rw(t): workstation reliability
Rf (t): file-server reliability
System reliability R(t) is given by:
Note: applies to any time-to-failure distributions
RBD for the WFS Example (cont.)
tRRWtR ft
1 21
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Assuming exponentially distributed times to failure: failure rate of workstation failure rate of file-server
The system mean time to failure (MTTF) is
given by:
fwfw
dttRMTTF
2
0
12)(
tft eetR w ])1(1[)( 2
RBD for the WFS Example (cont.)
W
f
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0
0.2
0.4
0.6
0.8
1
1.2
0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000
time
Rel
iab
ility
exp
w eib
Comparison Between Exponential and Weibull
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Assume that components are repairable : repair rate of workstation : repair rate of file-server
: availability of workstation
et tff
ff
f
ff
f
fA )()(
et tww
ww
w
ww
wwA )()(
: availability of file-server
Availability Modeling for the WFS Example
w
f
tA f
tAw
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System instantaneous availability A(t) is given by:
The steady-state system availability is:
)()( 2
)22()(lim
ffww
fwwwtAA
tss
Availability Modeling for the WFS Example (cont.)
tAAtA fW t 1 21
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Homework :
For the following system, write down the expression for system availability:
Assuming for each block a failure rate i and independent restoration at rate i
C
A B
D
C
C
E
D
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K-of-N System in RBD
System consisting of n independent components
System is up when k or more components are operational.
Identical K-of-N system: each component has the same failure
and/or repair distribution
Non-identical K-of-N system: each component may have
different failure and/or repair distributions
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Reliability for identical K-of-N
)(1)(
,)](1[)]()[()()(1
tFtR
tRtRtFtR jnjn
kj
njYkofn kn
Reliability of identical k out of n system
is the reliability for each component
k=n, series system n
s tRtR )]([)( k=1, parallel system
np tRtR )](1[1)(
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Steady-state Availability for Identical K-of-N System
Identical K-of-N Repairable System The units operate and are repaired independently All units have the same failure-time and repair-time
distributions Unit failure rate:
Unit repair rate:
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Steady-state availability of identical k-of-n system:
where is the steady-state unit availability
Steady-state Availability for Identical K-of-N System(continued)
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Binomial Random Variable
In fact, the number of units that are up at time t (say Y(t)) is binomially
distributed. This is so because:
where Xi ’s are independent identically distributed Bernoulli random variables.
Another way to say this is we have a sequence of n Bernoulli trials.
)(...)()()( 21 tXtXtXtY n
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Binomial Random Variable (cont.)
Y(t) is binomial with parameters n,p
x
0k
k -nk p)-(1p k)C(n,))(()( xtYPxF
nptYE )]([
k -nk
k p)-(1p k)C(n,ktYPp ))((
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Binomial Random Variable: pmf
pk
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0
0.2
0.4
0.6
0.8
1
1.2
0 1 2 3 4 5 6 7 8 9 10
x
CD
F
Binomial Random Variable: cdf
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Homework
Consider a 2 out of 3 system Write down expressions for its:
Steady-state availability Average cumulative downtime System MTTF and MTTR
Verify your results using SHARPE
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Homework :
The probability of error in the transmission of a bit over a communication channel is p = 10–4.
What is the probability of more than three errors in transmitting a block of 1,000 bits?
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Homework :
Consider a binary communication channel transmitting coded words of n bits each. Assume that the probability of successful transmission of a single bit is p (and the probability of an error is q = 1-p), and the code is capable of correcting up to e (where e > 0) errors. For example, if no coding of parity checking is used, then e = 0. If a single error-correcting Hamming code is used then e = 1. If we assume that the transmission of successive bits is independent, give the probability of successful word transmission.
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Homework :
Assume that the probability of successful transmission of a single bit over a binary communication channel is p. We desire to transmit a four-bit word over the channel. To increase the probability of successful word transmission, we may use 7-bit Hamming code (4 data bits + 3 check bits). Such a code is known to be able to correct single-bit errors. Derive the probabilities of successful word transmission under the two schemes, and derive the condition under which the use of Hamming code will improve performance.
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Reliability for Non-identical K-of-N System
n
kq CS SNj Siijnk
q
rrR )1(,
},,...2,1{ ,...1|},...,{Let 2121 nNniiiiiiC mmm
The reliability for nonidentical k-of-n system is:
That is,
rtR
R
RrRrR
rt
n
nknnknnk
when ,0
1
)1(
,
,0
1,11,,
where ri is the reliability for component i
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Steady-state Availability for Non-identical K-of-N System
n
kq CS SNj Siijnk
q
aaA )1(,
Assuming constant failure rate i and repair rate i for each component i, similar to system reliability, the steady state availability for non-identical k-of-n system is:
That is,
rtA
A
AaAaA
rt
n
nknnknnk
when ,0
1
)1(
,
,0
1,11,,
where is the availability for component iii
ii
ua
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1
5
2
4
3S T
Non-series-parallel RBD-Bridge with Five Components
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Truth Table for the Bridge
1111111111111111
1111111100000000
1111000011110000
1100110011001100
1111111110101000
1 2 3 4 System ProbabilityComponent
AA 21
1010101010101010
5
}54321
__AAAAA
54321
__AAAAA
54321
_AAAAA
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Truth Table for the Bridge
0000000000000000
1111111100000000
1111000011110000
1100110011001100
1100100010001000
1 2 3 4 System ProbabilityComponent
1010101010101010
5
4321
_AAAA}
54321
__AAAAA
54321
__AAAAA
54321
___AAAAA
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Bridge Availability
From the truth table:
5432154321
54321432154321
543215432121
_____
_____
___
AAAAAAAAAA
AAAAAAAAAAAAAA
AAAAAAAAAAAAAbridge
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Bridge: Conditioning
1
5
2
4
3
Non-series-parallel block diagram
Factor (condition)on C3
S T
C3 up
C3 down1
5
2
4
S T
1
4
2
5S T
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Bridge (cont.)
Component 3 is chosen to factor on (or condition on)
Upper resulting block diagram: 3 is down Lower resulting block diagram: 3 is up Series-parallel reliability formulas applied to both
resulting block diagrams Results combined using the theorem of total
probability
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Bridge (cont.)
A3down= 1 - (1 - A1A2) (1 - A4A5) A3up = [1 - (1-A1) (1-A4)] [1 - (1-A2) (1-A5)]
Abridge = A3down . (1-A3 ) + A3up A3
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Homework :
Specialize the bridge reliability formula to the case where:
Ri(t) =
Find Rbridge(t) and MTTF for the bridge
Specialize the bridge availability formula assuming that failure rate of component i is
i and the restoration rate is i
Verify your results using SHARPE
tie
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BTS Sector/Transmitter Example
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BTS Sector/Transmitter Example
3 RF carriers (transceiver + PA) on two antennas
Need at least two functional transmitter paths in order to meet
demand (available)
Failure of 2:1 Combiner or Duplexer 1 disables Path 1 and Path 2
Transceiver 1 Power Amp 1
Transceiver 2 Power Amp 2
2:1 Combiner Duplexer 1
Pass-Thru Duplexer 2Transceiver 3 Power Amp 3
Path 1
Path 2
Path 3
(XCVR 1)
(XCVR 2)
(XCVR 3)
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Measures
Steady state System unavailability
System Downtime
Methodology
Fault tree with repeat events (later)
Reliability Block Diagram
Factoring
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We use Factoring
If any one of 2:1 Combiner or Duplexer 1 fails, then the system is down.
If 2:1 Combiner and Duplexer 1 are up, then the system availability is given by the RBD
XCVR2
XCVR3
XCVR1
Pass-Thru Duplexer2
2|3
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Hence the overall system availability is captured by the RBD
XCVR2
XCVR3 Pass-Thru Dup2
XCVR1
2:1Com Dup12|3
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Methods for Non-series-parallel RBDs
Factoring or Conditioning (done)
Boolean Truth Table (done)
Min-paths
Inclusion/exclusion
SDP (Sum of Disjoint Products)
BDD (Binary Decision Diagram)
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Homework :
Solve for the bridge reliability Using minpaths followed by
Inclusion/Exclusion
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Fault Trees
Combinatorial (non-state-space) model type
Components are represented as nodes
Components or subsystems in series are connected to OR gates
Components or subsystems in parallel are connected to AND
gates
Components or subsystems in kofn (RBD) are connected as (n-
k+1)ofn gate
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Fault Trees (Continued)
Failure of a component or subsystem causes the
corresponding input to the gate to become TRUE
Whenever the output of the topmost gate becomes TRUE,
the system is considered failed
Extensions to fault-trees include a variety of different gates
NOT, EXOR, Priority AND, cold spare gate, functional
dependency gate, sequence enforcing gate
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Fault tree (Continued)
Major characteristics: Theoretical complexity: exponential in number of
components.
Find all minimal cut-sets & then use sum of disjoint products to compute reliability.
Use Factoring or the BDD approach
Can solve fault trees with 100’s of components
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An Fault Tree Example
or
c1
and and
c2 v1 v2 v3
•Structure Function:
32121 vvvcc
2 Control and 3 Voice Channels Example
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An Fault Tree Example (cont.)
Reliability of the system:
)33)(2(
]))(1(1][))(1(1[)(322
32
ttttt
vc
vvvcc eeeee
tRtRtR
,)()( Assume tv
tc
vc etRandetR
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Fault-Tree For The WFS Example
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Reliability expressions are the same as for the RBD
Structure function
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Availability Modeling Using Fault-Tree
Assume that components are repairable
w: repair rate of workstation
f: repair rate of file-server
Aw(t): availability of workstation
etA tffff
f
ff
ff
)()(
etA twwww
w
ww
ww
)()(
Af(t): availability of file-server
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System instantaneous availability A(t) is given by:
A(t) = [1 - (1 - Aw(t))2] Af(t)
Availability Modeling Using Fault-Tree (Continued)
The steady-state system availability is:
)()( 2
)22()(lim
ffww
fwwwtAA
tss
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Summary -Non-State Space Modeling
Non-state-space techniques like RBDs and FTs are easy to represent and assuming statistical independence solve for system reliability, system availability and system MTTF
Each component can have attached to it A probability of failure A failure rate A distribution of time to failure Steady-state and instantaneous unavailability
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2 Proc 3 Mem Fault Tree
and
or
and
or
and
failure
p1 p2m1 m3 m2 m3
A fault tree example
specialized for dependability analysis
represent all sequences of individual component failures that cause system failure in a tree-like structure
top event: system failure
gates: AND, OR, (NOT), K-of-N
Input of a gate:
-- component
(1 for failure, 0 for operational)
-- output of another gate
Basic component and repeated component
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Fault Tree (Cont.)
For fault tree without repeated nodes We can map a fault tree into a RBD
Use algorithm for RBD to compute MTTF in fault tree
For fault tree with repeated nodes Factoring algorithm BDD algorithm SDP algorithm
Fault Tree RBDAND gate parallel systemOR gate serial system
k-of-n gate (n-k+1)-of-n system
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Factoring Algorithm for Fault Tree
Basic idea:
and
or
and
or
and
failure
p1 p2m1 m3 m2 m3
and
oror
failure
p1 p2m1 m2
failure
and
p1 p2
M3 has failed
M3 has not failed