Supplement for Repairable System Reliability
Reference: Probability and Reliability for
Engineers, Miller, TA340.M648 1985
primarily chapter 15
NIST e-book Engineering Statistics Handbook,
sections labelled
PDF = probability_density_function = f( t)
t
⌠
CDF = cumulative_distribution_function = F( t) = ⎮ f( x ) dx
⌡
0
8.1.2.2 Reliability or Survival function
Reliability_function = probability_unit_survives_beyond_t
R( t) = 1 − F( t)
F( t) = 1 − R( t)
or ...
8.1.2.3 Failure (or Hazard) rate
h ( t) = failure_rate =
f( t)
1 − F( t)
=
f( t)
conditional probability
R( t ) f( t) = R( t) ⋅ h ( t)
therefore ...
d
d
F( t) = − R( t) = f( t) dt
dt
R( t) = 1 − F( t ) now ...
h ( t) =
f( t)
R( t )
=−
d
R( t )
dt
R( t )
d
= − ln( R( t) )
dt
t
integrate from 0 to t
⌠
⎮ h ( x ) dx = −ln( R( t))
⌡
0
t
exponentiate ..
⌠
− ⎮ h( x) dx
⌡
0
R( t) = e
t
f( t) = R( t) ⋅ h ( t) = h ( t) ⋅ e
therefore ...
⌠
− ⎮ h( x) dx
⌡
0
now .. if assume (observe) failure rate h(t) = constant = λ
t
h ( t) := λ
λ
f( t) := h ( t) ⋅ e
⌠
− ⎮ h( x) dx
⌡
0
t
(− λ )⋅
f( t) → λ⋅e
⌠
(− λ ) ⋅ x
(− λ ) ⋅ t⎤
F( t) := ⎮ λ⋅e
dx → ⎡⎣−e
⎦+1
⌡
F( t) := 1 − e
− λ ⋅t
0
have exponential assumption of probability of failure times
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exponential pdf and cdf example
PDF_of_time_to_next_failure = f( t) = λ⋅e
− λ ⋅t
λ :=
1
t := 0 .. 10000
1000
f( t) := λ⋅e
− λ ⋅t
probability density of time to next failure (lambda = 1/1000)
probability
0.001
5 .10
4
0
0
2000
4000
6000
1 .10
8000
4
time to next failure
CDF = F( t ) = 1 − e
− λ ⋅t
(
= CDF_of_waiting_time_to_next_failure
F( t) := 1 − e
)
− λ ⋅t
or .. CDF of "interarrival" time between failures
cumulative probability density of time to next failure (lambda = 1/1000)
probability
1
0.5
0
0
1000
2000
3000
4000
5000
6000
7000
8000
9000
1 .10
4
time to next failure
exponential pdf and cdf example
reset variables
interpret time to failure as a waiting time, it can be shown that this can be represented as a Poisson
process, if a component which fails is immediately replaced with a new one having the same failure rate λ.
Some results from this observation:
mean_waiting_ttime_between_successive_failures =
1
λ
= MTBF
Some results for exponential model
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R( t) := 1 − F( t)
Reliability or Survival function
R( t) → e
Reliability_function = probability_unit_survives_beyond_t
(− λ ) ⋅ t
− 0.05
e.g. if component has a failure rate of 0.05/1000 hours, probability that
it will survive at least 10,000 hrs is given by:
e
⋅10000
1000
= 0.607
n components in series
if a system consists of n components in series, with respective failure rates λ1 , λ2 ... λn
n
n
Rs( t) =
∏e
−
− λ i⋅t
=e
∑
i
λ i⋅t
=1
so it also is an exponential distribution ...
and the MTBF for the system is:
i= 1
MTBF series_system =
1
1
=
n
n
∑
λi
i= 1
⎛ 1 ⎞
⎜ MTBF ⎟
i ⎠
i = 1 ⎝
∑
for a parallel system
... with respective failure rates λ1 , λ2 ... λn
in this case we need to deal with "unreliabilities"
F =1−R
i
is probability component i will fail
i
n
probability_all_will_fail = unreliability = Fp =
∏ Fi
i= 1
n
Rp ( t) = 1 − Fp ( t) = 1 −
and probability of survival =R p (t)
∏ ⎣i
n
⎡F ( t)⎤ = 1 −
i= 1
⎦
∏ ⎡⎣1 − Ri(t)⎤⎦
i= 1
in this case; exponential probability of failure
∏ ⎡⎣Fi(t)⎤⎦ = ∏ (
1 − e
n
Fp ( t) =
n
i= 1
h p( t) =
fp ( t)
Rp( t)
)
− λ i⋅t
this will not show exponential distribution ...
i= 1
=
d
Fp ( t
)
dt
difficult to evaluate, but notice at least it is f(t).
Rp ( t )
Rp (t) difficult to obtain in general, but when all components have same failure rate
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n
Rp( t) = 1 −
− λ ⋅t
− λ ⋅t
∏ (1 − e ) = 1 − (1 − e )
n
∏⎣
⎡1 − R ( t)⎤ = 1 −
⎦
i
i= 1
i = 1
n_choose_k( n , k ) :=
(
1− 1−e
)
− λ ⋅t
n!
= 1 − 1 − n_choose_k( n , 1) ⋅ e
Rp ( t) = n_choose_k( n , 1 ) ⋅ e
− λ ⋅t
binomial coefficient
k!⋅(n − k)!
(
n
n
− λ ⋅t
+ n_choose_k( n , 1) ⋅ e
− n_choose_k( n , 2 ) ⋅ e
− λ ⋅t
− 2λ ⋅ t
)
− ........
+ ......................... ( −1 )
n− 1 − n⋅ λ ⋅ t
⋅
e
binomial theorem from mathworld.wolfram.com/BinomialTheorem.html
it can be shown see reference page 460
after differentiating to find fp (t)
fp ( t) =
d
Rp ( t )
dt
and then calculating the mean (MBTF parallel)
MBTF parallel =
1
α
⋅ ⎛⎜ 1 +
⎝
1
2
+ ............. +
1⎞
⎟
n⎠
e.g. if use two identical components in parallel
4
1 3
MBTF parallel = ⋅
α 2
increase of 50% not double
∑
k=1
1
k
= 2.083
four to double
another example if time permits on board
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