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```RELIABILITY ENGINEERING – I SEM 2020-21
BITS Pilani
Pilani Campus
Rajesh P Mishra
Probability Density Function : f(t)
• Probability Density Function is the first derivative of the Failure Density
Function and given by f(t)
• F(t) = 1 – R(t)
• f(t) =
𝑑𝐹(𝑡)
𝑑𝑡
• f(t) ≥ 0
• F(t) =
=−
and
𝑡
𝑓(𝑡)
0
𝑑𝑅(𝑡)
𝑑𝑡
∞
𝑓(𝑡)
0
𝑑𝑡
𝑑𝑡 = 1
and R(t) =
∞
𝑓(𝑡)
𝑡
𝑑𝑡
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Design Life
• A design life is defined to be the time to failure t R that corresponds to a
specified / desired reliability R.
• R (t R) = R
• Failure occurring within some interval of time [a, b] = F(b) – F(a)
= R(a) – R(b)
=
𝑏
𝑓(𝑡)
𝑎
𝑑𝑡
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Mean Time to Failure (MTTF)
MTTF is the first moment of probability density function f(t)
MTTF =
=
∞
𝑡
0
𝑓(𝑡) 𝑑𝑡
∞
𝑅(𝑡)
0
𝑑𝑡
Integrating by parts, we get
MTTF =
∞
𝑅(𝑡)
0
𝑑𝑡
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Median Time to Failure (t med)
• Median Time to Failure is the time with 50% of the failure occurring before
the median time to failure and 50% occurring after the median
• R (t med) = 0.5
• VARIANCE =
∞
0
𝑡 − 𝑀𝑇𝑇𝐹
2
𝑓 𝑡 𝑑𝑡 =
∞
𝑡
0
2 𝑓 𝑡 𝑑𝑡 − 𝑀𝑇𝑇𝐹
2
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Hazard Rate Function
•
Hazard Rate Function is defined as the limit of the failure rate as the interval approaches zero.
•
The hazard function is the instantaneous failure rate
•
ƛ (t) = Lim △t⤑0
ƛ (t)
=
ƛ (t) =
-
𝑅 𝑡 −𝑅(𝑡+ △𝑡)
𝑅 𝑡
△𝑡
𝑑𝑅(𝑡)
𝑑𝑡
1
x 𝑅(𝑡)
𝑓(𝑡)
𝑅(𝑡)
ƛ (t) dt = -
R(t) = exp [
𝑑𝑅(𝑡)
𝑅(𝑡)
𝑡
- 0ƛ
Integrating both sides then
𝑡
ƛ
0
𝑡 𝑑𝑡 = - ln R(t)
𝑡 𝑑𝑡 ]
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CFR and AFR
• Cumulative Failure Rate (CFR) over a period of time t is defined by
L(t) =
𝑡
ƛ
0
𝑡 𝑑𝑡
• Average Failure Rate (AFR) between two time periods t1 and t2 is defined by
AFR (t1,t2) =
If t1 = 0 and t 2 = t
𝑡2
1
ƛ
𝑡2−𝑡1 𝑡1
then AFR =
𝑡 𝑑𝑡 =
ln 𝑅 𝑡1 −ln 𝑅 (𝑡2)
𝑡2 −𝑡1
ln 𝑅 0 −ln 𝑅 (𝑡)
𝑡
=
0 −ln 𝑅(𝑡)
𝑡
=
𝐿(𝑡)
𝑡
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Bath Tub
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SOLUTION 2.1
Given R(t) =
1
0.001𝑡+1
t≥0
Find
R(100) =
1
0.001 𝑥 100+1
R(1000) =
=
1
0.001 𝑥 1000+1
1
1.1
= 0.9091
1
2
= = 0.5
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b.
𝑡
ƛ
0
𝑡 𝑑𝑡 = - ln R(t)
ƛ 𝑡 = -
𝑑 ln(1/0.001𝑡+1)
𝑑𝑡
=
0.001
0.001𝑡+1
It is an increasing function as the value is positive.
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Solution 2.2
ƛ 𝑡 = 0.4 t
a.
where t is in years
R(t) = exp [ -
𝑡
ƛ
0
Hence when t = 1/12 year = 1 month
b.
𝑡 𝑑𝑡 ] = exp [ -
𝑡
0.4𝑡
0
𝑑𝑡 ] = 𝑒^ − 0.2𝑡2
R(t) = 0.9986
R (tr) = R = 0.95
exp^ -0.2 t^2 = 0.95
t = 0.5064 years = 6 months
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Solution 2.3
Given
f(t) = 0.01
a. R(t) =
100
0.01 𝑑𝑡
𝑡
b. ƛ (t) =
𝑓(𝑡)
𝑅(𝑡)
c. MTTF =
=
where 0 ≤ t ≤ 100 days
=
0.01 (100 − 𝑡)
0.01
0.01 (100 −𝑡)
100
0.01 (100
0
− 𝑡) dt = 50 days
100
d. Variance = 0 0.01 𝑡2𝑑𝑡 − 𝑀𝑇𝑇𝐹 2 = 833.33 STD DEV = 28.87
e. R(t) = 1 – 0.01t = 0.5
tmed = 50 days
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SOLUTION 2.4
Given
f(t) =
3𝑡2
10 9
0 ≤ t ≤ 1000 hrs.
a. Probability of failure within 100 hr warranty period
F(t) =
b. MTTF =
1000
𝑡
0
c. R(tr) = 0.99 =
𝑡
𝑓(𝑡)
0
𝑑𝑡 =
100
𝑓(𝑡)
0
𝑑𝑡 = 0.001
𝑓(𝑡) dt = 750 hrs.
1000
𝑓(𝑡)
𝑡
dt
= 215.44 hrs.
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EXPONENTIAL RELAIBILITY FUNCTION
• A failure distribution that has a constant failure rate is called exponential
probability distribution
• If the failure rates of all failure modes of a component are constant and
independent, then the overall failure rate of the component is also constant
• Failures due to random or chance events will follow this distribution
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CFR MODEL
R(t) = exp [ -
𝑡
ƛ
0
𝑡 𝑑𝑡 ]
If ƛ is constant then R(t) = e – ƛt where t ≥ 0
F(t) = 1 - R(t) = 1 - e – ƛt
f(t) =
𝑑𝐹(𝑡)
𝑑𝑡
MTTF =
=0 −
∞
𝑅(𝑡)
0
Variance =
∞
0
𝑑𝑅(𝑡)
𝑑𝑡
= ƛ e – ƛt
𝑑𝑡 =
∞
e
0
𝑡 − 𝑀𝑇𝑇𝐹
–
2
ƛt
𝑑𝑡 =
𝑓 𝑡 𝑑𝑡 =
1
= Standard Deviation
ƛ
∞
0
𝑡 −
1 2
ƛ
ƛe –
ƛt𝑑𝑡=
1 = (MTTF)2
ƛ2
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• Reliability of component with CFR has slightly more than one-third chance of
surviving to its mean time to failure:
R (MTTF) = e –MTTF/MTTF = e -1 = 0.368
• Design Life t R
R(t R) = 0.5 = e – ƛ t R
t R= - ln R / ƛ, if R 0.5 then,
Median = - ln 0.5 MTTF = 0.69315 MTTF
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Reliability in Series
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Reliability in Parallel (Redundnacy)
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Problem 3.1 Page 55
A component experiences chance (CFR) failures with an MTTF of 1100 hour.
Find the following:
i. Reliability for a 200-hr mission
ii. The design life for a 0.9 reliability
iii. The median time to failure
iv. The reliability of 200 hr mission if a second, redundant (and independent)
v. The reliability of 200 hr mission if a second, single component is serially
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Solution 3.1
Given MTTF = 1100 hr hence ƛ = 1/1100
i. R(200) = exp – (1/1100* 200) = 0.9555
ii. 0.90 = exp – (1/1100*t) taking ln both side and solve td = 115.9 hours
iii. 0.50 = exp –(1/1100 * t) taking ln both side and solve tmed = 762.46 hrs
iv. R(t) = 1 – (1- exp – ƛt)2 = 1 – ( 1 – exp – 1/1100* 200)2 = 0.9723
v. R(t) = exp - 2ƛt = exp – 2*1/1100*200 = 0.9131
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Characterisitics
• The time to failure of a component is not dependent on how long the
component has been operating
• The probability that the component will operate for next t hours is the same
regardless of whether the component is brand new, has been operating for
several hours
• Using Condition Probability
R (t I To) = R (t + To)/ R(To) =
exp −ƛ (𝑡+𝑇𝑜)
exp( − ƛ𝑇𝑜)
= exp (-ƛt)
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Problem 3.2 Page 55
• A CFR system with ƛ = 0.0004 has been operating for 1000 hrs. What is the
probability that it will fail in the next 100 hr? the next 1000 hr?
• Solution:
a. F(t) = 1 - exp (– 0.0004 * 100) = 0.0392
b. F(t) = 1- exp (– 0.0004 * 1000) = 0.3297
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Failure Modes
• Failure occurs due to different physical phenomena or different characteristics of
individual components
• If R(t) is the reliability function for the ith failure mode, assuming independence
among the failure modes, then
• System reliability is the product of these probabilities that none of thee modes occur
before time t
• System hazard rate function is the sum of the individual hazard rate of function of all
the failure modes
• ƛ s = ƛ1 + ƛ2 + ƛ3 +……..
R(t) = e - ƛ s t
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MTTF =
ƛs =
1
ƛ𝑠
1
𝑀𝑇𝑇𝐹1
=
1
ƛ1+ ƛ2+ ƛ3+ ….
1
+
𝑀𝑇𝑇𝐹2
=
1
1
1
+
+⋯
𝑀𝑇𝑇𝐹1 𝑀𝑇𝑇𝐹2
+⋯
If all n components have identical ƛ then ƛs = nƛ and MTTF =
1
𝑛ƛ
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Bath-tub Failure Modes
Failure modes in bath tub includes: Burn-in, Useful Life and Wear-Out
The failure rates of these failure modes can be summed up to find ƛs
Then we can find the reliability of the system for the mission hour
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Problem 3.3 Page 55
• A gearbox has two independent failure modes: a constant failure rate of
0.0003 and a linearly increasing wear out failure given by ƛ (t) = t / 5 x 10 5,
Find the reliability of the gear box for 100 hr of operation.
• Solution:
ƛs = 0.0003 + t / 5 x 10 5
R(t) = exp – (0.0003t + t 2 / 10 x 10 5)
R(100) = exp – (0.0003*100 + (100)2 / 10 x 10 5) = exp -0.04 = 0.9608
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Failure on Demand
• Components fail from switching from idle mode to operating mode like bulbs
or air-conditioners
• If constant failure rate is assumed for each failure mode and a constant
probability of failure on demand is assumed, then the component will have an
effective failure rate on a clock hour basis
• ƛi is the average failure rate when idle
• ƛo is the average failure rate while operating
• p is the probability of failure on demand
• ti is the average length of the idle time period per cycle
• to is the average length of the operating time period per cycle
•
ƛ eff =
𝑡𝑖
𝑡𝑖+𝑡𝑜
ƛi +
𝑡𝑜
𝑡𝑖+𝑡𝑜
ƛo +
𝑝
𝑡𝑖+𝑡𝑜
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Problem 3.18 Page 57
• A 60 watt outdoor lightbulb is advertised as having an average life (i.e.
MTTF) of 1000 operating hours. However, experience has shown that it will
also fail on demand an average of once every 120 cycles. A particular bulb is
turned once each evening for an average of 10 hr. If it is desired to have a
reliability of 90 percent, what is the design life.
ƛ eff =
ƛ eff =
𝑡𝑖
𝑡𝑜
𝑝
ƛi +
ƛo +
𝑡𝑖+𝑡𝑜
𝑡𝑖+𝑡𝑜
𝑡𝑖+𝑡𝑜
14
10
∗0+
(1/1000)
14+10
14+10
0.90 = exp (-0.000764 t)
+
1/120
14+10
= 0.000764
therefore td = 137.9 hrs
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Renewal Process
• A failed component is immediately replaced with a new one, will cause the
system to reach a steady state constant number of failures per unit of time
• For a renewal process in which failed components are replaced as they fail, a
steady state constant failure rate is obtained such that :
ƛ = 𝑛𝑖 1/𝑀𝑇𝑇𝐹𝑖
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Problem 3.7 Page 56
• A system contains 20 identical and critical components that will be replaced
on failure (renewal process). As a result, a constant failure rate for the
system will be observed. If a design life of 10 year with a reliability of 0.99 is
required, what should the system MTTF and median time to failure? If each
component has CFR, what will the component MTTF and median time to
failure be?
• Solution:
0.99 = exp ( - ƛ * 10) therefore ƛs = 0.001/ year
MTTF =
1
0.001
= 1000 years
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Component ƛi = ƛs / n = 0.001/20 = 0.00005 / year
Component MTTF = 1/ ƛi = 1/0.00005 = 20000 years
System Median time to failure :
0.5 = exp (– 0.001 * tmed)
tmed = 693.15 years
Component Median time to failure :
0.5 = exp (- 0.00005 * tmed)
t med = 13863 years
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• Let p be the probability of failure as a result of a load or stress on the system
and if independent loads are applied at constant, fixed interval of time, an
approximate reliability function results
R = 1 – p for single load
Rn = (1-p) n for n loads
= exp ln (1-p) n
= exp n ln (1-p) but as ln (1-p) = - p
= exp (– np)
As n = t / △t , △t being the fixed time between loads, so
R(t) = exp (- p/ △t) t = exp – ƛ t , where ƛ = p/ △t, a constant failure rate.
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Solution 3.6 Page 56
• A landing gear system has repetitive stresses placed on it twice a day as a result of
loadings. The probability of a failure during landing is 0.0028. Determine the
reliability of the landing gear system over a 30-day contingency operation. What is
the probability of a failure occurring between days 10 and 20 of the operation?
• Solution: p = 0.0028
△t = ½ day t = 30 days
• i. R(t) = exp (- 0.0028 *2) * 30 = exp – 0.168 = 0.8454
• ii. F (t) = F(b) – F(a) = R(a) – R(b) = exp (-0.0028*2*10) – exp (-0.0028*2*20)
= 0.9455 – 0.8940 = 0.0515
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Two-Parameter Exponential Distribution
Guaranteed Lifetime : to :The minimum time prior to which no failure occurs.
• It is a location parameter that shifts the distribution an amount equal to to to
the right on the time axis.
• f(t) = ƛ exp –ƛ (t – to) where 0 ≤ to ≤ t ≤ ∞
• R(t) = exp –ƛ (t – to)
𝑡
1
• MTTF = t ƛ t exp –ƛ (t – to) dt = to+
ƛ
o
• R (tmed)= exp –ƛ (tmed – to) = 0.5
• tmed = to+
0.69315
ƛ
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• Design life t R = to+
ln 𝑅
−ƛ
• Variance (1/ ƛ 2) and standard deviation (1/ƛ) of the two parameter
exponential distribution are not affected by the location parameter.
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Problem 3.15
• Consider two identical and redundant CFR components having a guaranteed
life of 2 months and a failure rate of 0.15 failure per year. What is the system
reliability for 10,000 hr of continuous operation?
• Solution: to = 2/12 = 0.167
ƛ = 0.15
t = 10000/365*24 = 1.14 years
• R(t) = 1- {1- exp – 0.15 ( 1.14 – 0.167)}2 = 0.982
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Poisson’s Distribution
• If a component having a constant failure rate is immediately repaired or
replaced upon failing, then the number of failures observed over a period of
time t has a Poisson Distribution
• The probability of observing n failures in time t is given by
e – ƛt ƛ𝑡 𝑛
P (t) =
𝑛ǃ
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Problem 3.13
• An electronic circuit board with ƛ (t) = 0.00021 per hour is replaced on failure.
What is the probability that the third failure will occur by 10000 hours?
• Solution:
ƛ = 0.00021
ƛ t = 0.00021* 10000 = 2.1
𝑒 −2.1 2.1 0
p0=
= 0.122
0
ǃ
𝑒 −2.1 2.1 1
P1 =
= 0.257
1
ǃ
𝑒 −2.1 2.1
P2 =
2
ǃ
2
= 0.270
P3 = 1- p0-p1-p2= 0.351
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Redundancy and CFR Model
Two Components in Parallel: The system will fail when both the components fail
F(t) = (1- exp –ƛt) * (1- exp –ƛt) = (1- exp –ƛt) 2
R ( t) = 1 - (1- exp –ƛt) 2 = 2 exp –ƛt – exp -2 ƛt
ƛ(t) =
𝑓(𝑡)
𝑅(𝑡)
MTTF =
=
2ƛ𝑒−ƛ𝑡 − 2ƛ exp − 2ƛ𝑡
2 exp –ƛt – exp −2 ƛt
1.5
ƛ
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MODULE 1: TIME DEPENDENT FAILURE MODELS: Weibull Distribution
Lecture No. 3 Chapter 4 T1
WEIBULL DISTRIBUTION
• Used to model both increasing and decreasing failure rates
• Characterized by the hazard rate function
•
ƛ (t) =
ß
Ɵ
𝑡
Ɵ
ƛ (t) = a t b
ß−1
•
ß is the shape parameter
•
•
Ɵ is the scale parameter and also called the characteristic life
BX Life : The time at which X% of units will fail.
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Weibull Shape Parameter: ß or k
• The Weibull shape parameter, β, is also known as the Weibull slope
• The value of β is equal to the slope of the line in a probability plot
• Different values of the shape parameter can have marked effects on the
behaviour of the distribution
• Some values of the shape parameter will cause the distribution equations to
reduce to those of other distributions. For example, when β = 1, the pdf of the
three-parameter Weibull reduces to that of the two-parameter exponential
distribution.
• The parameter β is a pure number (i.e., it is dimensionless).
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Interpretation
• Weibull distributions with β < 1 have a failure rate that decreases with time,
also known as infantile or early-life failures
• Weibull distributions with β close to or equal to 1 have a fairly constant failure
rate, indicative of useful life or random failures
• Weibull distributions with β > 1 have a failure rate that increases with time,
also known as wear-out failures
• These comprise the three sections of the classic "bathtub curve." A mixed
Weibull distribution with one subpopulation with β < 1, one subpopulation
with β = 1 and one subpopulation with β > 1 would have a failure rate plot
that was identical to the bathtub curve. An example of a bathtub curve is
shown in the following chart
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Weibull Scale Parameter: Ɵ or ƞ
• A change in the scale parameter, Ɵ, has the same effect on the distribution as a
change of the abscissa scale.
• Increasing the value of Ɵ while holding β constant has the effect of stretching out
the pdf. Since the area under a pdf curve is a constant value of one, the "peak" of
the pdf curve will also decrease with the increase of Ɵ, as indicated in the following
figure.
• If Ɵ is increased, while β and γ are kept the same, the distribution gets stretched out
to the right and its height decreases, while maintaining its shape and location.
• If Ɵ is decreased, while β and γ are kept the same, the distribution gets pushed in
towards the left (i.e., towards its beginning or towards 0 or γ), and its height
increases.
• Ɵ has the same unit as T, such as hours, miles, cycles, actuations, etc.
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Effect of varying Scale parameter (Ɵ or ƞ)
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Location Parameter: Ƴ or To
• Frequently, the location parameter is not used, and the value for this parameter can
be set to zero
• When this is the case, the pdf equation reduces to that of the two-parameter Weibull
distribution
• There is also a form of the Weibull distribution known as the one-parameter Weibull
distribution. This in fact takes the same form as the two-parameter Weibull pdf, the
only difference being that the value of β is assumed to be known beforehand
• This assumption means that only the scale parameter needs be estimated, allowing
for analysis of small data sets. It is recommended that the analyst have a very good
and justifiable estimate for β before using the one-parameter Weibull distribution for
analysis
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Effect of ß on Weibull PDF i.e. f (t)
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Effect of ß on CDF i.e. F(t)
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Effect of ß on Weibull Reliability i.e. R (t)
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Effect of ß on Weibull Hazard Rate
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Reliability and Failure Rate
•
R(t) = exp -
• f(t) = •
•
•
•
𝑑 𝑅(𝑡)
𝑑𝑡
=
𝑡ß
0 Ɵ
ß
Ɵ
𝑡
Ɵ
𝑡
Ɵ
ß−1
ß−1
𝑑𝑡 = exp –
exp –
𝑡
Ɵ
𝑡
Ɵ
ß
R(Ɵ) = exp (-1) = 0.368
ß
For ß = 1, ƛ (t) is constant and equal 1/Ɵ
For ß < 1, PDF is similar in shape to the exponential
For ß ≥ 3, PDF is symmetrical like normal distribution
For 1 < ß < 3 PDF is skewed
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MTTF and Variance
• MTTF =
∞
𝑡
0
Let y =
𝑡
Ɵ
MTTF =
∞
𝑡
0
∞
ß
𝑓 𝑡 𝑑𝑡 =
dy =
ß
Ɵ
∞ß
0 Ɵ
𝑡
Ɵ
ß−1
𝑡
Ɵ
ß−1
exp –
𝑡
Ɵ
ß
t dt
dt
1
𝑒 − 𝑦 𝑑𝑦 = Ɵ
∞
𝑦ß
0
𝑒 − 𝑦 𝑑𝑦 = Ɵ Γ (1 + 1/ß)
−
Γ(x) = 0 𝑦 𝑥 1 𝑒 − 𝑦 𝑑𝑦
Variance = σ2 = Ɵ2 [Γ (1 + 2/ß) – {Γ (1 + 1/ß)}2]
For x > 0, Γ(x) = (x-1)ǃ
or Γ(x) = (x-1) Γ(x-1)
Γ1/2 = √ᴫ
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Design Life, Median and Mode
Design Life
R(t) = exp –
𝑡
Ɵ
ß
=R
Design Life t R = Ɵ (- ln R) 1/ß
Median Life : R = 0.5 therefore t med = Ɵ (- ln 0.5) 1/ß= Ɵ (0.69315) 1/ß
Mode: when ß > 1, t mode = Ɵ ( 1- 1/ß) 1/ß
when ß ≤ 1
t mode = 0
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Problem 1 Page 79
• For a system having a Weibull failure distribution with a shape parameter of
1.4 and a scale parameter of 550 days. Find the following:
i. R(100 days)
ii. The B1 Life
iii. MTTF
iv. The standard deviation
v. tmed
vi. tmode
vii. Design life for a reliability of 0.90
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Solution
i. R(t) = exp –
𝑡
Ɵ
ß
ii. R = 0.99 = exp – (
= exp – (
𝑡 1.4
)
550
100 1.4
)
550
= 0.912
t = 20.58 days
iii. MTTF = Ɵ Γ (1 + 1/ß) = 550 Γ(1 + 1/1.4) = 500.81 days
iv. σ2 = Ɵ2 [Γ (1 + 2/ß) – {Γ (1 + 1/ß)}2]
Standard Deviation σ = 363.95 days
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v. tmed
R = 0.5
0.5 = exp – (t/550) 1.4
tmed = 423.32 days
vi. tmode = Ɵ ( 1- 1/ß) 1/ß tmode = 550 ( 1 – 1/1.4) ^1/1.4 tmode = 224.77 days
vii. 0.90 = exp – (t/550) 1.4
t = 110.22 days
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Burn-In screening for Weibull
If To is Burn-in Period, then conditional reliability
R(T| To) = exp – [(t + To)/Ɵ] ß
exp - (To/Ɵ) ß
= exp [ -
𝑡+𝑇𝑜
(
Ɵ
)
ß
+
𝑇0 ß
( ) ]
Ɵ
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Problem 4.2
• A turbine blade has demonstrated a Weibull failure pattern with a decreasing
failure characteristics by a shape parameter of 0.6 and a scale parameter of
800 hr.
a. Compute the reliability for a 100 – hr mission
b. If there is a 200 – hr burn-in of the blades, what is the reliability for a 100-hr
mission
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Solution 4.2
Given: ß = 0.6
a. R(t) = exp –
= exp – (
Ɵ = 800
𝑡
Ɵ
ß
100 0.6
) =
800
b. R(t) = exp [- (
t = 100
𝑡+𝑇𝑜
Ɵ
0.75
𝑇0
Ɵ
) ß + ( ) ß ] = exp [- (
100+200
800
) 0.6 + (
200 0.6
) ]
800
= 0.887
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Failure Modes
• For a system having n serially related components or having n independent failure
modes, each having an independent shape parameter ß and scale parameter Ɵi:
• Failure rate is summation from i = 1 to i = n
ƛ(t) = ∑
ß
Ɵ𝑖
𝑡 ß-1
( )
Ɵ𝑖
1
Ɵ𝑖
= ß t ß-1∑ ( ) ß
1
Ɵ𝑖
Characteristic Life = [ ∑ ( ) ß ] -1/ß
• If the failure modes are all Weibull but with differing shape parameter, then the
system failure distribution will not be Weibull.
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Identical Weibull Components
• If a system of a serially related and independent components have identical
hazard rate function
•
ƛ(t) = ∑
=
ß
Ɵ𝑖
𝑛ß
(t
Ɵß
𝑡
Ɵ𝑖
( ) ß-1
) ß-1
R(t) = exp {– n
𝑡 ß
( ) }
Ɵ
This follows Weibull distribution with shape parameter ß and scale parameter
Ɵ
𝑛 ^1/ß
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Problem 4.7
• What is the maximum number of identical and independent Weibull
components having a scale parameter of 10,000 operating hours and a
shape parameter of 1.3 that can be put in series if a reliability of 0.95 at 100
operating hours is desired? What is the resulting system MTTF?
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Solution 4.7
Given : Ɵ = 10,000
R(t) = exp
0.95 = exp
ƛ(t) =
𝑛ß
(t
Ɵß
𝑡 ß
{– n ( ) }
Ɵ
100 1.3
–n(
)
10000
ß = 1.3
R(t) = 0.95
t = 100
Solving we get n = 21
) ß-1
ƛ (t) = 0.0006857
MTTF = 1/ƛ (t) = 1458 hours
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The Three Parameter Weibull
Minimum life : to such that T > to The parameter to is called the location parameter
The distribution assumes that no failure will take place prior to time to
R(t) = exp [ - (
ƛ (t) =
𝑡 −𝑡0
Ɵ
) ß]
t ≥ to
ß 𝑡 −𝑡0 ß-1
(
)
Ɵ
Ɵ
1
ß
MTTF = to + Ɵ Γ(1 + )
tmed = to + Ɵ (0.69315) 1/ß
Td = to + Ɵ (-ln R) 1/ß
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Problem 4.4
• For a three parameter Weibull distribution with ß = 1.54 Ɵ = 8500 hr
to = 50 hr, find the following:
a. R(150)
b. MTTF
c. tmed
d. The standard deviation
e. The design life for a reliability of 0.98
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Solution 4.4
Given t = 150 to = 50 ß = 1.54
a. R(t)
Ɵ = 8500
𝑡 −𝑡0 ß
= exp [ - (
) ] t ≥ to
Ɵ
150−50 1.54
= exp [ - (
)
= 0.9989
8500
MTTF = 50 + 8500 Γ ( 1 + 1/1.54) = 50 + 8500 x 0.9 = 7700 hours
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t med = to + Ɵ (0.69315) 1/ß= 50 + 8500 (0.69315) 1/1.54= 6750 hours
Variance σ 2 = (8500)2 [ Γ(1 + 2/1.54) – { Γ( 1 + 1/1.54)}2 ]
Standard Deviation = 5070 hours
Design Life
td = to + Ɵ (- ln R) 1/ß
= 50 + 8500 ( - ln 0.98 ) 1/1.54=725 hours
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Redundancy with Weibull Failures
Let us assume two identical and independent components used to form
redundant system
Rs = 1 – [ (1 – R(t) ] 2
If R(t) = exp – (t / Ɵ ) ß
Rs = 2 exp – (t / Ɵ ) ß - exp – 2 (t / Ɵ ) ß
MTTF = 2 Ɵ Γ( 1 + 1/ß ) ƛs (t) =
ß
Ɵ
(
𝑡
Ɵ
Ɵ
2 ^ 1/ß
Γ( 1 + 1/ß ) = Ɵ Γ (1 + 1/ß) (2 – 2 -1/ß)
2−2 𝑒 −(t
/Ɵ)ß
) ß-1
2− 𝑒 −(t / Ɵ ) ß
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Problem 4.5
• For the components defined in Exercise 4.1. determine the reliability for 100
hr if there are two redundant components.
• Given : t = 100
Ɵ = 550 ß = 1.4
• Rs = 2 exp – (t / Ɵ ) ß - exp – 2 (t / Ɵ ) ß
= 2 exp – (100/550) 1.4 – exp –2 (100/550) 1.4
= 0.992
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BITS Pilani
Pilani Campus
Reliability of Systems
Chapter – 5
RL 2.1-2.5
Introduction
• A system consist of number of different components
• Components within a system may be related to one another in two primary
ways
-Series configuration or
-Parallel configuration
• In series configuration all components must function for the system to
function.
• However in a parallel or redundant configuration at least one component
must function for the system to function.
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Serial Configuration
• In a series configuration, all assets are considered
critical
Reliability block diagram for components in series
•
•
•
•
•
E1=Event that component 1 does not fail
E2=Event that component 2 does not fail
Then P(E1)=R1 and P(E2)=R2 R1,R2=Reliabilities
Rs=P(E1∩E2)=P(E1) P(E2)= R1 R2
Therefore generelizing system reliability asRs= (R1) x (R2) x (R3) x …………….x (Rn)
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BITS Pilani, Pilani Campus
Serial Configuration
𝑛
𝑅𝑠 =
𝑅𝑖 (𝑡)
𝑖=1
If component has constant failure rate 𝜆𝑖
𝑛
𝑅𝑠 =
𝑒𝑥𝑝(−𝜆𝑖 𝑡)
𝑖=1
𝑛
𝑅𝑠 = exp −
𝜆𝑖 𝑡 = exp −𝜆𝑠t
𝑖=1
MTTF=1/𝜆𝑠 = 1/
𝑛
𝑖=1 𝜆𝑖
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Example
• Consider a four component system of which the components are
independent and identically distributed with CFR. If Rs(100)=0.95 is the
specified reliability , Find individual component MTTF.
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Example
• A system is comprised of four serially related components each having a
Weibull time to failure distribution with parameters as shown in table.
Determine the R(10).
Component
Scale parameter
Shape parameter
1
100
1.20
2
150
0.87
3
510
1.80
4
720
1.00
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Parallel Configuration
• In parallel system all units must fail to for system to fail
• Parallel units are represented by block diagram
• System reliability is found by taking 1 minus probability that all n components
fail
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Parallel Configuration
• Rs(t) = 1 - [(1 - R1) (1 - R2) (1 - R3) …………. (1 - Rn)]
𝑛
𝑅𝑠 𝑡 = 1 −
1 − 𝑅𝑖 (𝑡)
𝑖=1
• For a redundant system consisting of all CFR
components
𝑛
1 − 𝑒 −𝜆𝑖𝑡
𝑅𝑠 𝑡 = 1 −
𝑖=1
EXAMPLE- For two component parallel system and having CFR
derive expression for MTTF
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Combined Series Parallel Systems
• Generally systems contain assets in both series and
parallel relationships. Consider, the example below
• To compute the system reliability, the network may be
broken into series or parallel subsystems
• The system reliability may be calculated on the basis
of the relationship among the subsystems
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BITS Pilani, Pilani Campus
Combined Series Parallel Systems
•
•
•
•
•
In the network the subsystems have the following reliabilities
RA= [1 – (1 - R1) (1 - R2)]
RB= (RA) x (R3) RC= (R4) x (R5)
RS= [1 – (1 - RB) (1 - RC)] (R6)
If R1= R2= 0.90, R3= R6= 0.98, and R4= R5= 0.99, then
RB= [1 – (0.10)2] (0.98) = 0.9702
RC= (0.99)2= 0.9801 And
RS= [1 – (1 – 0.9702) (1 – 0.9801)] (0.98) = 0.9794
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BITS Pilani, Pilani Campus
Example
• Given the system, calculate its useful life reliability for a 100-hr mission. All
given failure rates are in failures per million hours of operation.
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Example
• Determine the reliability of the given system.
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Example
• Find the reliability of the following system which is shown in the figure if the
reliability of components are RA =0.96, RB = 0.92, RC = 0.94, RD = 0.89, RE =
0.95, RF = 0.88, RG = 0.90.
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BITS Pilani, Pilani Campus
Solution
• The reliability of the subsystem with components A and B is R1
=[1- (1-0.96) (1-0.92)]= 0.9968
• The reliability of the subsystem with components E, F and G is R2
= [1- (1-(0.95)(0.88))(1-0.90)]= 0.9836
• System Reliability = Rs = 0.9968 X 0.94 X 0.89 X 0.9836 = 0.8202
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BITS Pilani, Pilani Campus
High – level Vs Low-level Redundancy
• System redundancy may be obtained in two ways
i. Each component may have parallel component
ii. Entire system may placed in parallel with another
identical system
High level redundancy
Low level redundancy
Let A and C are identical B and D are identical components
If both the component has same reliability R then
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BITS Pilani, Pilani Campus
High – level vs Low-level Redundancy
• 𝑅𝑙𝑜𝑤 = 1 − 1 − 𝑅 2 = 2 𝑅 − 𝑅2 2
• 𝑅ℎ𝑖𝑔ℎ = 1 − 1 − 𝑅2 2 = 2𝑅2 − 𝑅4
• By Comparing two reliabilities, it can be shown that
reliability of low level redundancy is greater than
reliability of high level redundancy
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Example
• A radio set consists of three major components: a power supply, a receiver,
and an amplifier, having reliabilities of 0.8,0.9 and 0.85 respectively. Compute
system reliabilities for both high level and low level redundancy for systems
with two parallel components.
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Solution
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k-out of –n Redundancy
• A requirement exist for k out of n identical and independent
components to function for the system to function.
• If each component is viewed as independent trial with R as a
constant probability of success, then probability of exactly x
components operating is𝑛
𝑃 𝑥 =
𝑅 𝑥 1 − 𝑅 𝑛−𝑥
𝑥
𝑛
𝑛!
=
𝑥! (𝑛 − 𝑥)!
𝑥
𝑛
𝑅 𝑠 =
𝑃(𝑥)
𝑥=𝑘
Example- A vehicle require more than its 4 engines to operate to work. If
each engine has a reliability of 0.97, determine the reliability of working.
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Complex configuration
• Certain systems cannot be decomposed into series and
parallel systems
• However such networks can be analyzed by using either
decomposition or enumeration
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Complex configuration
• Decomposition• Two sub networks are created : one in which component
E is assumed to be functioning and one in which E has
failed.
• Reliability of each sub network is determined separately
• Then the system reliability is given by-
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Common Mode Failures
• Assumption of independence of failure among n components may
be violated. For example some components may share common
power source, environmental conditions may affect etc
• A common mode failure can be depicted in series with those
common components sharing failure mode
• Rs=[1-(1-R1)(1-R2)(1-R3)]R’
• In order to redundancy network to have an effect, common mode
failure must have high reliability
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Example
• Two parallel, identical and independent components have CFR. If it is desired
that Rs(1000) = 0.95, Find the MTTF. Assume a common-mode CFR of
0.00001 in addition to the components’ independent failure rates.
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THANK YOU
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