Probability and Random Processes
(15B11MA301)
Lecture-24
(Content Covered: Exponential and Weibull Failure distributions)
Department of Mathematics
Jaypee Institute of Information Technology, Noida
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Contents
• Exponential Failure Distribution
• Solved Examples
• Weibull Failure Distribution
• Solved Examples
• Practice Questions
• References
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Special Failure Distributions
(i) Exponential Distribution
If the time to failure T follows an exponential distribution with parameter 𝜆, then
its probability density function is given by
𝑓 𝑡 = 𝜆𝑒 −𝜆𝑡 , 𝑡 ≥0
The reliability function is given by 𝑅 𝑡 =
Using (1) in (2), we get 𝑅 𝑡 =
∞
‫𝑓 𝑡׬‬
……….(1)
∞
‫𝑓 𝑡׬‬
𝑡 𝑑𝑡 =
𝑡 𝑑𝑡……(2)
∞
‫ 𝑒𝜆 𝑡׬‬−𝜆𝑡 𝑑𝑡
= 𝑒 −𝜆𝑡 …..(3)
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Hazard Function (𝜆 (t)) : We know, 𝜆 𝑡 =
𝑓(𝑡)
𝑅(𝑡)
Using (1) and (3) in hazard function, we get 𝜆 𝑡 =
𝜆𝑒 −𝜆𝑡
𝑒 −𝜆𝑡
=𝜆
It means that, when the failure distribution is an exponential distribution with parameter 𝜆, then the failure
rate at any time is a constant, equal to 𝜆.
Conversely, when hazard rate = 𝜆, then using the definition of 𝑓 𝑡 , we have
𝑓 𝑡 =𝜆 𝑡 𝑒
𝑓 𝑡 = 𝜆𝑒
𝑡
𝑡
− ‫׬‬0 𝑓 𝑡 𝑑𝑡
− ‫׬‬0 𝜆 𝑑𝑡
= 𝜆𝑒 −𝜆𝑡 , 𝑡 ≥ 0
Due to this property, the exponential distribution is also referred as Constant Failure rate distribution.
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▪ Mean Time to failure (MTTF) = E(T) =
1
𝜆
It means that the time to failure of a component is not dependent on how long the
component has been functioning. This property is known as the memoryless property of
the constant failure rate distribution.
▪ Variance = Var(T) =
1
𝜆2
▪ Conditional Reliability
𝑅(𝑇𝑜 +𝑡)
𝑅(𝑡/𝑇𝑜 ) = 𝑅(𝑇
𝑜)
=
𝑒 −𝜆(𝑇𝑜+𝑡)
−𝜆𝑡
=𝑒
.
𝑒 −𝜆𝑇𝑜
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Example: A manufacturer determines that, on the average, a television set is used 1.8 hours per day. A one
year warranty is offered on the picture tube having a MTTF of 2000 hours. If the distribution is exponential,
what percentage of the tubes will fail during the warranty period ?
Solution: Since the distribution of the time to failure of the picture tube is exponential,
𝑅(𝑡) = 𝑒 −𝜆𝑡 , 𝜆 being the failure rate.
Given, MTTF = 2000 hours
∞
i.e. ‫׬‬0 𝑒 −𝜆𝑡 𝑑𝑡 = 2000
i.e.
1
𝜆
= 2000 or 𝜆 = 0.0005 / hour
𝑃(𝑇 ≤ 1 𝑦𝑒𝑎𝑟 ) = 𝑃(𝑇 ≤ 365 × 1.8 ℎ𝑜𝑢𝑟𝑠) (since, TV is operated for 1.8 hours/day)
= 1 − 𝑃 (𝑇 > 657) = 1 − 𝑅(657)
= 1 − 𝑒 −0.0005 𝑋 657 = 0.28
i.e. 28% of the tubes will fail during the warranty period.
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Example: Given that 𝑅 (𝑡) = 𝑒 −
(a)
(b)
(c)
(d)
0.001𝑡
,𝑡≥0
Compute the reliability for a 50 hours mission.
Show that the hazard rate is decreasing.
Given a 10- hour wear-in period, compute the reliability for a 50-hour mission.
What is the design life for a reliability of 0.95, given a 10-hour wear-in period?
Solution: (a) Given 𝑅 (𝑡) = 𝑒 −
0.001𝑡
𝑅(50) = 𝑒 −
(b) We know, 𝜆(𝑡) =
=
,𝑡≥0
0.001 𝑋 50
= 0.9512
−𝑅 ′ (𝑡)
𝑅(𝑡)
− 𝑒 − 0.001 𝑡 ×− 0.001
𝑒 − 0.001 𝑡
×𝑡
1
2
−
1
2
−
= 0.001 × 𝑡 , which is a decreasing function of t.
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(c) 𝑅(𝑡/𝑇𝑜 ) =
𝑅(𝑇𝑜 +𝑡)
𝑅(𝑇𝑜 )
Therefore, 𝑅(50/10) =
=
𝑅(60)
𝑅(10)
𝑒 − 0.001 ×60
𝑒 − 0.001 ×10
= 0.8651
(d) Given, 𝑅(𝑡𝐷 /10) = 0.95
𝑅(𝑡 𝐷 +10)
𝑅(10)
Therefore, 𝑒 −
= 0.95
0.001×(𝑡𝐷 +10)
= 0.95 × 𝑒 −
0.001×10
0.001 × (𝑡𝐷 + 10) = 0.15129
𝑡𝐷 = 12.89 ℎ𝑜𝑢𝑟𝑠
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(ii) Weibull Distribution
We know that, the pdf of the Weibull Distribution is defined as
𝛽
𝑓 𝑡 = 𝛼𝛽𝑡 𝛽−1 𝑒 −𝛼𝑡 , t ≥ 0
On using, 𝛼 =
1
,
𝜃𝛽
…….(4)
in above pdf, an alternative form of Weibull’s pdf can be written as:
𝑓 𝑡 =
𝛽
𝛽 𝑡 𝛽−1 − 𝑡
𝑒 𝜃
𝜃 𝜃
, t ≥ 0 , 𝜃 > 0, 𝛽 > 0
……(5)
Where, 𝛽 : Shape parameter
𝜃: Characteristic life or scale parameter of Weibull’s distribution given by (5).
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• If the time to failure T follows Weibull’s distribution (5) , then the reliability function is
∞
𝑅 𝑡 = ‫𝑡𝑑 𝑡 𝑓 𝑡׬‬
𝑅 𝑡
𝛽
∞ 𝛽 𝑡 𝛽−1 − 𝑡
= ‫𝜃 𝜃 𝑡׬‬
𝑒 𝜃 𝑑𝑡
∞ −𝑥
𝑡 𝛽
= ‫𝑥𝑑 𝑒 𝑡׬‬, on putting 𝜃
−𝑥
=𝑥
=𝑒
=𝑒
−
𝑡 𝛽
𝜃
• Hazard Function: 𝜆 𝑡 =
=
=
……(6)
𝑓(𝑡)
𝑅(𝑡)
𝑡 𝛽
𝛽−1
𝛽 𝑡
−
𝑒 𝜃
𝜃 𝜃
𝑡 𝛽
−
𝑒 𝜃
𝛽 𝑡 𝛽−1
𝜃 𝜃
……(7)
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• Mean Time to failure (MTTF) = E(T)
𝑡 𝛽
𝛽−1
∞
𝛽 𝑡
−
=‫׬‬0 𝑡 (𝜃 𝜃
𝑒 𝜃 )𝑑𝑡
=θ
▪ Variance = Var(T) = θ2
▪ Conditional Reliability
2
𝛽
+1 −
1
𝛽
𝑅(𝑇𝑜 +𝑡)
𝑅(𝑡/𝑇𝑜 ) =
𝑅(𝑇𝑜 )
1
𝛽
+1
2
+1
=
𝑇𝑜 +𝑡 𝛽
−
𝜃
𝑒
𝑇𝑜 𝛽
−
𝑒 𝜃
= 𝑒𝑥𝑝 −
𝑇𝑜 +𝑡 𝛽
𝜃
+
𝑇𝑜 𝛽
𝜃
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Example For a system having a Weibull failure distribution with a shape parameter of 1.4 and a
scale parameter of 550 days, find
(a)
(b)
(c)
(d)
(e)
R(100 days)
the B1 life
MTTTF
The standard deviation
The design life for a reliability of 0.90
Solution: We know, the pdf of Weibull distribution is given by
𝑓(𝑡) =
𝛽
𝜃
𝛽
𝑡 𝛽−1 − 𝑡
𝑒 𝜃
𝜃
, t ≥ 0 , 𝜃 > 0, 𝛽 > 0
Given 𝛽 = 1.4 and 𝜃 = 550 days
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(a) We know, 𝑅(𝑡) = 𝑒
−
𝑡 𝛽
𝜃
−
100 1.4
550
𝑅(100) = 𝑒
= 0.9122
(b) B1 Life: The lifetime corresponding to a reliability of 0.99 is called the B1 Life.
Let 𝑡𝑅 be the B1 life of the system, Then
𝑅(𝑡𝑅 ) = 0.99
𝑅(𝑡𝑅 ) = 𝑒
𝑡 𝑅 1.4
550
−
𝑡 𝑅 1.4
550
= 0.99
= 0.01005
Therefore, 𝑡𝑅 = 550 (0.01005)(1/1.4)
= 20.6 days
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(c) MTTF = 𝜃
1
𝛽
= 550
+1
1
1.4
+1
= 550 (1.71) = 550 (0.91057) ( using the gamma tables)
= 500.8 days
(d) (S.D.)2 = 𝜃 2
2
𝛽
= (550)2
2
= (550)
+1 −
2
1.4
1
𝛽
2
+1
+1 −
2.43 − 1.71
1
1.4
2
+1
2
= 363.96 days
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(e) Let 𝑡𝐷 be the required design life for R =0.90, Then
𝑅(𝑡𝐷 ) = 0.90
𝑅(𝑡𝐷 ) = 𝑒
𝑡 𝐷 1.4
550
−
𝑡 𝐷 1.4
550
= 0.90
= 0.010536
Therefore, 𝑡𝐷 = 550 (0.010536)(1/1.4)
= 110.2 days
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Practice Question
1. A turbine blade has demonstrated a Weibull failure pattern with a decreasing failure rate
characterised by a shape parameter of 0.6 and a scale parameter of 800 hours.
(a) Compute the reliability for a 100-hour mission.
(b) If there is a 200-hour burn-in of the blades, what is the reliability for a 100- hour
mission.
(Ans. 0.75, 0.89)
2. The one-month reliability on an indicator lamp is 0.95 with the failure rate specified as
constant. What is the probability that more than two spare bulbs will be needed during the
first year of operation (ignore replacement time).
(Ans. 0.0246)
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References
1. Veerarajan, T., Probability, Statistics and Random Processes, 3rd Ed. Tata McGraw-
Hill, 2008.
2. Ghahramani, S., Fundamentals of Probability with Stochastic Processes, Pearson,
2005.
3. Papoulis, A. and Pillai, S.U., Probability, Random Variables and Stochastic
Processes, Tata McGraw-Hill, 2002.
4. Miller, S., Childers, D., Probability and Random Processes, Academic Press, 2012.
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