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SYSTEM RELIABILITY
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ENN530 Slide <2> System Reliability
SYSTEM RELIABILITY • A system is formed by a number of
components connected together. The reliability of a system is the probability that a device or system will operate for a given period of time and under given operating conditions.
• There are categories of failure modes, which are studied through two simplest systems: series and parallel.
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ENN530 Slide <3> System Reliability
Series Reliability • A set of components, connected together so
as to form a system, is said to be in series reliability if the failure of any one component causes the failure of the total system (The system can be regarded as being “fault-intolerant”).
• Note: The components themselves need not be physically or topologically in series, what relevant is only the fact that all of them must succeed for the system to succeed.
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ENN530 Slide <4> System Reliability
Reliability Block Diagram • Reliability Block Diagram (RBD) is used to
model the effect of component failures on system performance.
Figure 1: Series RBD
Unit 1 Cause Unit 2 Unit n Effect
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ENN530 Slide <5> System Reliability
Series Reliability
What is reliability of such system?
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ENN530 Slide <6> System Reliability
Series Reliability Function
• Dependent
Where Rs represents system reliability, n is the total number of components in a system. ti is time to failure of i th item.
• Independent
12121312121 ......)(},...{ −== nnns ttttPtttPttPtPtttPR
∏∏==
≡==n
ii
n
iins RtPtttPR
1121 )(},...{
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ENN530 Slide <7> System Reliability
Characteristics of Reliability function of a series system
1. The value of the reliability function of the system, Rs(t), for a series configuration is less than or equal to the minimum value of the individual reliability function of the constituting items, which is:
)}({)(0,...,2,1
tRMintR iis =≤
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ENN530 Slide <8> System Reliability
Characteristics of Reliability function of a series system
2. If λi (t) represent the failure rate of item i, then the system reliability of a series system can be written as:
dxxtRt n
iis )]}([exp{)(
0 1∫ ∑
=
−= λ
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ENN530 Slide <9> System Reliability
Example: A system consists of four items, each of them are necessary to maintain the required function of the system. The time to failure distribution and their corresponding parameter values are given in the following table. Find the reliability of the system for 500 and 700 hours of operation.
Item Time to failure distribution
Parameter values
Item 1 Exponential λ= 0.001 Item 2 Weibull η= 1200 h β= 3.2 Item 3 Normal μ= 800 h σ= 350
Item 4 Weibull η =2000 h β=1.75
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ENN530 Slide <10> System Reliability
Solution: From the information given in the table, the reliability function of various items can be written as:
])2000
(exp[)(
)350
800()(
])1200
(exp[)(
)001.0exp()(
75.14
3
2.32
1
ttR
ttR
ttR
ttR
−=
−Φ=
−=
×−=
])2000
(exp[)(
)350
800()(
])1200
(exp[)(
)001.0exp()(
75.14
3
2.32
1
ttR
ttR
ttR
ttR
−=
−Φ=
−=
×−=
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ENN530 Slide <11> System Reliability
Solution (Cont.): Since the items are connected in series, the reliability function of the system is given by:
])2000
(exp[)350
800(])1200
(exp[)001.0exp()( 75.12.3 tttttRs −×−
Φ×−××−=
Substituting t = 500 and 750 in the above equation, we get:
1759.08355.05568.08003.04723.0)750(4202.09154.08043.09410.06065.0)500(
=×××==×××=
RR
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ENN530 Slide <12> System Reliability
Failure Rate of series configuration with n items
)()(1
ttn
iis ∑
=
= λλ)exp()(
tlExponentia
ii λλ −
))(exp()(
)(
1 ii
iii
i ttWeibull
ββ
ηηηβ
−−1
1))(()( −
=∑= i
i
n
i i
is
tt β
ηηβλ
})(21exp{
21
)(
2
i
i
i
tNormal
σµ
πσ−
− )(/)()(1 i
in
iis
ttftσ
µλ −Φ= ∑
=
Probability density function of i-th item, fi(t)
Failure rate of the system, λs(t)
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ENN530 Slide <13> System Reliability
Mean time to failure (MTTF) of a series system
The mean time to failure, MTTF, of a series configuration, denoted by MTTFs, can be written as:
dttRdtRMTTF i
n
iss )(0
10
∫∫∞
=
∞
Π==
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ENN530 Slide <14> System Reliability
Characteristics of MTTFs of series system
The mean time to failure of a system with series RBD will be less than the mean time to failure of any its constituting items, where MTTFi denote the mean time to failure of the item i.
1.
2. For complex repairable systems, MTTFs represents the mean time to first failure
}{0,...2,1 iis MTTFMinMTTF
=≤
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ENN530 Slide <15> System Reliability
Example: A system consists of three items connected in series. The time-to-failure distribution and their corresponding parameter values are given at the following table. Find the mean time to failure of the system. Compare the value of MTTFs with mean time to failure of individual items.
Item Distribution Parameter Values Item 1 Weibull η1= 10, β1=2.5
Item 2 Exponential λ= 0.2
Item 3 Weibull 2= 20, β2= 3
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ENN530 Slide <16> System Reliability
Solution: Mean time to failure of the system is given by:
dtttt ))(exp()exp()(exp( 21
20 1
ββ
ηλ
η−×−×−= ∫
∞
dttRMTTF ii
s )(0
3
1∫Π∞
=
=
hrdtttt 48.3))20
(exp()2.0exp()10
(exp( 35.2
0
=−×−×−= ∫∞
Compare with MTTF for Item1: 8.87 hr, Item2: 5 hr, Item3:17.86 hr. Note that the formula can only be evaluated using numerical integration.
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ENN530 Slide <17> System Reliability
• Consider a series system of n components, each has a constant failure rate
• since:
Constant failure rate in series reliability
∏=
=n
iis RR
1
])...(exp[)(),exp(
21 ttRthentR
and
ns
ii
λλλλ
+++−=−=
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ENN530 Slide <18> System Reliability
Constant failure rate in series reliability
• The system’s failure rate is: • MTTF:
∑=
λ=λn
iis
1
∑=
λ=
λ=θ n
ii
ss
1
11
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ENN530 Slide <19> System Reliability
Example • The pump system in Slide 5: Assume failure
rate of the motor is 4.0X10-6/hr, the gear box is 3.2X10-6/hr, and the pump is 9.8X10-6/hr.
• What is system failure rate?
• At what probability the system can operate for at least 1000 hours?
663
1101710)8.92.34( −−
=
×=×++== ∑i
is λλ
983.0)( == − ts
setR λ
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ENN530 Slide <20> System Reliability
• A set of components are involved in a system. The system fails only when all components have failed.
Figure 2: Parallel RBD • Such a system is regarded as being “fault-tolerant”.
Parallel Reliability
Unit 1
Unit 2
Unit n
Effect Cause
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ENN530 Slide <21> System Reliability
Pump system
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ENN530 Slide <22> System Reliability
Parallel Reliability Function
}...{ 21 ns tttPR +++=
∏=
==n
iins tFtFtFtFF
121 )()()...()(
∏∏==
−−=−=−=n
ii
n
iiss tRtFFR
11
)}(1{1)(11
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ENN530 Slide <23> System Reliability
• A limit switch has a probability of failure of 17.8% over a one year period. If two switches are installed in parallel, what is the probability that both fail in a one year period.
• Fs= F1 xF2 = .178 x .178 = .032 = 3.2% • Thus the system is much more reliable with two
switches.
• With three switches, Fs=0.1783 = 0.56%.
Parallel Systems - Example
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ENN530 Slide <24> System Reliability
Characteristics of a parallel system
1. The system reliability, Rs(t), is more than reliability of the any of the consisting items. That is,
)}({)(,...,1
tRMaxtR inis =≥
)}({)(,...,1
tRMaxtR inis =≥
• If λi(t) represents the failure rate of item i, then the reliability function of a parallel configuration can be written as:
)])(exp(1[1)(0
1dtttR
t
i
n
is ∫−−Π−==
λ
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ENN530 Slide <25> System Reliability
Example:
A fly-by-wire aircraft has four flight control system electronics (FCSE) connected in parallel. The time-to-failure of FCSE can be represented by Weibull distribution with scale parameter η = 2800 and β = 2.8. Find the reliability of flight control system for 1000 hours of operation.
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ENN530 Slide <26> System Reliability
Solution: Reliability function for a parallel system with four identical items is given by:
44
1)](1[1)](1[1)( tRtRtR iis −−=−Π−=
=
9455.0))28001000(exp())(exp()( 8.2 =−=−= β
ηttR
999991.0]9455.01[1)1000( 4 =−−=sR
Where R(t) is the reliability function of each item. For t=1000, R(t) is given by:
Thus, the reliability of flight control system for 1000 hours of operation is given by:
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ENN530 Slide <27> System Reliability
Failure Rate of a parallel system
n
i
i
n
ji
n
jiij
s
tR
tFtft
1
1 ,1
)](1[1
)}()({)(
=
=≠=
−Π−
Π×=
∏
∑λ ∏
∑
=
=≠=
−−
Π×=
n
ii
n
ji
n
jiij
s
tR
tFtft
1
1 ,1
)](1[1
)}()({)(λ
Where, fi(t) is the probability density function of item i
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ENN530 Slide <28> System Reliability
Example:
A fly-by-wire aircraft has four flight control system electronics (FCSE) connected in parallel. The time-to-failure of FCSE can be represented by Weibull distribution with scale parameter η = 2800 and β = 2.8. Find the failure rate of the system at time t = 100.
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ENN530 Slide <29> System Reliability
Solution: Since all the four items are identical, the failure rate of the system can be written as:
4
3
)]([1)]([)(4)(
tFtFtfts −
××=λ
))(exp()()( 1 ββ
ηηηβ tttf −= − ))(exp(1)( β
ηttF −−=
Substituting t = 100, we get λs(t) = 8.0 x 10-8
Where:
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ENN530 Slide <30> System Reliability
MTTFs of a parallel system
The mean time to failure system can be written as:
dttRdtRMTTF i
n
iss )]}(1[1{0
10
−Π−== ∫∫∞
=
∞
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ENN530 Slide <31> System Reliability
Constant failure rate in parallel reliability
• For each component, failure rate is a constant. • If the system has two units, we assume that
the failure of the two units are independent. • Assume time-to-failure distribution of
component i is exponential with mean λi. Then MTTFs is given by:
dttdtRMTTF i
n
ii
n
is )]}exp(1[1{0
10
1λ−−Π−=Π= ∫∫
∞
=
∞
=
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ENN530 Slide <32> System Reliability
For particular values of n, we can simplify the equation to derive the expression for the MTTFs as following:
Assume n = 2 (Two components):
2121
111λλλλ +
−+=sMTTF
Assume n = 3 (Three components) :
321323121321
1111111λλλλλλλλλλλλ ++
++
−+
−+
−++=sMTTF
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ENN530 Slide <33> System Reliability
REDUNDANCY • Systems in which some components may fail
without causing system failure are “fault tolerant”. To achieve this, some measure of hardware redundancy must be introduced. There are two main types of redundancy: active and passive (or standby) redundancy.
• In Active redundancy, two or more components are placed in parallel reliability. Load is normally shared between the components. If one component fails, then the remaining units carry the full load.
• In Passive (Standby) redundancy, one component carries full load. If this component fails, another standby component (not necessarily identical) is switched on to take the place.
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ENN530 Slide <34> System Reliability
Pump system
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ENN530 Slide <35> System Reliability
Active Redundancy
• System failure function:
a
b
[ ][ ][ ] [ ] [ ][ ])()()()(1
)(1)(1)()()(
tRtRtRtRtRtR
tFtFtF
baba
ba
bas
+−−=−−=
=
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ENN530 Slide <36> System Reliability
Active Redundancy • Reliability function:
• If a and b are constant failure rate components, what is Reliability function?
• Is the resulted system a constant failure rate system?
)()()()()(1)(
tRtRtRtRtFtR
baba
ss
−+=−=
No
ttts
baba eeetR )()( λλλλ +−−− −+=
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ENN530 Slide <37> System Reliability
“m-out-of-n” Good: • In some cases, m out of n identical units must
stay “up” for the system to be operational. • Rs(m,n,p) – System reliability when there are
“n” number of components in a system, the system is considered being functioning properly if m or more units (m, m+1, m+3,.…, n-1,n) are functioning. Let p be the probability that a unit is functioning. The probability of exactly m functioning units (out of n):
( ) mnm ppmn
pnmp −−
= )1(,,
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ENN530 Slide <38> System Reliability
Therefore, the system reliability:
( ) ∑=
=n
mrs pnrppnmR ),,(,,
∑=
−−
=
n
mr
rnr pprn
)1(
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ENN530 Slide <39> System Reliability
Example:
A system has 4 components, what is the reliability of the system being operational with at least 3 units being operational:
∑=
−−
=
4
3
4)1(34
r
rr pp
43
04343
34
)1(44
)1(34
pp
pppp
−=
−
+−
= −
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ENN530 Slide <40> System Reliability
Standby Redundancy
1
2
ττττ
dtRftRRt
s )()()( 20
11 −⋅+= ∫=
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ENN530 Slide <41> System Reliability
Standby Redundancy • R1(t) – the reliability of unit 1 at time t
• R2(t - τ) − the reliability of the unit 2 after it started to operate at time τ
• F1(t) – the pdf for the time to failure of unit 1(τ)
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ENN530 Slide <42> System Reliability
Constant Failure Rate Special case 1: Time to failure distribution of all units follows an exponential distribution
tetR 1)(1λ−= tetR 2)(2
λ−=
τλ
τλ
τ
λλλλ
τλτλ
τ
λ
deee
deeetR
tttt
tt
tsb
∫
∫
=
−−−−
−−−
=
−
+=
+=
0
)(1
)(
01
2121
211)(
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ENN530 Slide <43> System Reliability
)(
)1(
121
212
1
21
1
)(
21
1
ttt
tt
t
eee
eee
λλλ
λλλ
λ
λλλ
λλλ
−−−
−−−
−
−−
+=
−−
+=
We assume that switch is perfect, no standby failure.
Constant Failure Rate (Cont.)
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ENN530 Slide <44> System Reliability
Series – Parallel system
)](1(1[ ,11 ji
m
i
n
js xRR −Π−Π===
R(xi,j) – Reliability of component i in sub-system j n = no. of sub-system, j = 1,2,…..,n; i=1,…m.
m
.
.
.
.
.
.
.
.
…..
m
.
.
.
.
.
.
.
.
m
.
.
.
.
.
.
.
.
1 2 n
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ENN530 Slide <45> System Reliability
Special case: all units are identical. R(xi,j) = p Rs = [ 1 – (1-p)m]n
Example: n = 3, m = 2
Rs = [ 1 – (1-p)2]3
If p = 0.85, Rs = 0.934007
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ENN530 Slide <46> System Reliability
Increasing System Reliability • During design stage
– Simplifying the system – Use better quality components – Improve environments, e.g. use cooling fans,
reduce vibration, etc. – Use Fault-Tolerance in the system
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ENN530 Slide <47> System Reliability
Increasing System Reliability • During development and production stage
– Design improvement – Burn-in components or system – Improve quality control during manufacturing – Use of preventive maintenance – use of corrective maintenance
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Any Questions?