The Constant Failure Rate Model - …academic.udayton.edu/charlesebeling/ENM 565/PDF PPT...
37
C. Ebeling, Intro to Reliability & Maintainability Engineering, 2 nd ed. Waveland Press, Inc. Copyright © 2010 Chapter 3 The Constant Failure Rate Model Exponential Probability Distribution Chapter 3 1 Hurry, come here! Chapter 3 is starting now. You don’t want to miss any of this.
Transcript of The Constant Failure Rate Model - …academic.udayton.edu/charlesebeling/ENM 565/PDF PPT...
C. Ebeling, Intro to Reliability & Maintainability Engineering, 2nd ed. Waveland Press, Inc. Copyright © 2010
Chapter 3The Constant Failure Rate Model
Exponential ProbabilityDistribution
Chapter 3 1
Hurry, come here! Chapter 3 is starting now. You don’t want to miss
any of this.
Presenter
Presentation Notes
One of the most frequently used reliability functions is based upon the exponential probability distribution. The exponential distribution is the first of several theoretical probability distributions that will be discussed. Perhaps because of its relative simplicity and ease of use, it was one of the first distributions to be applied to failure times. Any failure mode having a constant failure rate over time will have an exponential distribution.
Recall the Bathtub Curve
Chapter 2 2
CFR
Presenter
Presentation Notes
A well known plot of a hazard rate function is the bathtub curve. It obviously gets its name from its shape. It suggests that for some items, early or infantile failures occur that decrease (DFR) with age as design and manufacturing problems are recognized and corrected. This is followed by a period of nearly Constant Failure Rates (CFR) in which most failures are a result of external random occurrences. Can you suggest what probability distribution may best describe this period? Finally, as the component ages and wearout begins to dominate, an IFR is observed. It is not clear how many components actually behave in this manner. Certainly some do. All three failure modes may be continuously present; however, the dominate failure mode depends upon the age of the component.
The Exponential Distribution
Chapter 3 3
0t , = (t) ≥λλ
0t ,e =e = R(t) t-dt- t0 ≥∫ λλ '
Then
Let
Presenter
Presentation Notes
We derive the exponential distribution by starting with the assumption that the hazard rate function is constant (CFR) and equal to lambda. Certainly this is the simplest assumption that can be made concerning the functional form of the hazard rate function. Then using the relationship between the reliability function and the hazard rate function presented in Chapter 2, the exponential reliability function is obtained. Check and see if this function has the properties of a reliability function! You do recall what those properties are, don’t you?
The Reliability Function
Chapter 3 4
Presenter
Presentation Notes
The reliability function is plotted for three different values of its parameter lambda. From the graph it can be seen that as lambda gets larger, the reliability as a function of time more quickly approaches zero.
The CDF and PDF
Chapter 3 5
e-1 = F(t) t-λ
e = dtF(t) d=f(t) t-λλ
Presenter
Presentation Notes
The CDF and PDF are obtained using the relationships presented in Chapter 2. Test your understanding of these functions by convincing yourself that they obey the properties of a CDF and PDF.
Probability Density Function (PDF)
Chapter 3 6
Presenter
Presentation Notes
The PDF provides the “shape” of the distribution. For the exponential, the distribution is skewed right with the mode always occurring at t = 0.
The MTTF
Chapter 3 7
.368 = e = e = R(MTTF) 1-MTTF-MTTF
note that
or
0|- t
- t0
1eMTTF = dt = = e-
λλ
λ λ∞ ∞∫
0
1tMTTF te dtλλλ
∞ −= =∫
Presenter
Presentation Notes
Using the definitional form of the mean of a probability distribution, the MTTF is found to be the inverse of the (constant) failure rate. It is only for the exponential distribution that this true. The failure rate is failures per time unit (e.g. .01 failures per day) and the MTTF is time units to failure (100 days to failure). If we set t = MTTF and find R(t = MTTF), it will always be .368 for this distribution. Why? This suggests that only 36.8 percent of all units will survive beyond the MTTF. Finding the mean of the exponential is easier if we integrate the reliability function. As an exercise, try it.
The Standard Deviation
Chapter 3 8
λλ
λσ λ
2t-
2
02 1 = dt e 1-t = ⎟
⎠⎞
⎜⎝⎛
∫∞
and MTTFσλ
= =1
Presenter
Presentation Notes
The variance is found through integration. Interesting, the standard deviation is equal to the mean. What can be concluded from this? Perhaps, the more reliable a component (as measured by its MTTF), the more variability in failure times will be observed.
The Median Time to Failure
Chapter 3 9
λλ.69315 = .5 1- = tmed ln
= .69315 MTTF
R t e t( ) .= =−λ 5
Presenter
Presentation Notes
The median is easily derived and is always 69.315 percent of the mean. Of course only 50 percent of all components will survive beyond the median.
The Design Life
Chapter 3 10
R = e = )tR( t-R
Rλ
For a given reliability R, let
lnthen R1= - Rtλ
Presenter
Presentation Notes
Solving for the design life is similar to solving for the median. In fact, it is just a generalization of the median isn’t it? What happens if R = .5 in the above formula?>
Example #1 - CFR model
An electrical transmission line has been tested and found to have a CFR of .001 defects per foot. Find:a. R(100)b. R(1000)c. MTTFd. Standard Deviatione. tmed
f. 90 percent design life
Chapter 3 11
Presenter
Presentation Notes
It’s about time – an exercise for you to try. This should not be difficult. The answers follow.
Example #1 - solution
Chapter 3 12
R eR eMTTF ft
MTTF ftt MTTF ftt MTTF ft
med
( ) .( ) .
/. ..
. . .ln(. ) . .
. ( )
. ( )
.
100 90481000 368
1 001 10001000
69315 6931590 1054
001 100
001 1000
90
= =
= == =
= == == − =
−
−
σ
Presenter
Presentation Notes
You should have gotten the correct answers. If not, go back to the beginning, do not pass go, and do not collect any money.
Memoryless Property
Chapter 3 13
ee =
)TR()T+R(t = )T|R(t
T-
)T+(t-
0
00
0
0
λ
λ
R(t) = e = e
e e = t-T-
T-t-
0
0λ
λ
λλ ⋅
Presenter
Presentation Notes
What about conditional reliability you ask! Here is an interesting result. The conditional reliability equals the unconditional reliability. This is just another manifestation of the “memoryless” property of the exponential distribution. This implies that if we have two components, one 2 years old (T0 = 2); another 10 years old (T0 = 10), the reliability that they will survive another 5 years (t = 5) is the same for both of them and is equal to R(5). The components do not age! Can you suggest some situations where this would be true?
Failure Modes
R(t) = n
= R (t)∏
ii
1
Chapter 3 14
Ri(t) is the reliability function for the ith failure mode, then, assuming independence among the failure modes,the system reliability, R(t) is found from
P(A1 ∩A2 … ∩An) = P(A1) P(A2) … P(An)
Presenter
Presentation Notes
A failure mode is simply one manner in which a component can fail. Components may have several failure modes each having their own unique failure distribution. For example, a motor may fail due to a power surge or it may fail because of a short in the wiring. A complex system can have the failure of each of its components defined as a failure mode. The overall reliability of a component having several failure modes is simply the product of the reliability of each failure mode (assuming independence – no common cause).
More on Failure Modes
i- (t )d tR (t) = e 0
tiz ′ ′λ
Chapter 3 15
Let
R(t) = n
= - (t )dt0
t
e∏ z ′
i
i '
1
λ
Then
λ λ(t) = (t)=1
n
ii∑where
01
exp ( ') 'nt
ii
= t dtλ=
⎡ ⎤−⎢ ⎥⎣ ⎦∑∫
t0- (t )d t= e λ ′ ′∫
Presenter
Presentation Notes
Using one of our favorite formulae from Chapter 2, we reach the relatively simple result that the system or overall failure rate is the sum of all the individual failure rates. What could be easier than addition? You follow the derivation, right?
Failure Modes and CFR
Chapter 3 16
λλλ i
n
=1i
==(t) ∑
If a system consists of n independent,serially related components each withCFR, then
andt0- d t - tR(t) = = e eλ λ′∫
Presenter
Presentation Notes
Things get simpler if the failure rate or hazard rate function is constant for all failure modes. In this case the system failure rate is also constant (CFR) which then generates an exponential distribution. Serial related components or failure modes will be discussed some more in Chapter 5. However, for now, it simply means that if any component fails or failure mode occurs, the system will fail.
Example - Failure Modes
An engine tune-up kit consists of 3 parts each having CFRs (in failures per mile) of .000034, .000017, and .0000086.
Find the MTTF, median time to failure, standard deviation, and reliability of the tune-up kit at 10,000 miles.
Chapter 3 17
Presenter
Presentation Notes
Your turn!
Example - solution
Chapter 3 18
λσ λλ
s
s
med s
MTTF mit miR e
= + + == = =
= == =−
. . . ./ , . .
. , .( , ) .. ( , )
000034 000017 0000086 00005961 16 7785
69315 1163010 000 5510000596 10 000
Presenter
Presentation Notes
Ahhhhh, the answers.
Parts Count Approach
An integrated circuit board consists of the following components each having a CFR.
Component a-Failure Rate(10-5) b- Quantity (a) x (b)Diodes, silicon .00041 10 .0041Resistors .014 25 .3500Capacitors .0015 12 .0180Transformer .0020 2 .0040Relays .0065 6 .0390Inductive devices .0004 12 .0048
total .4199 x 10-5
Chapter 3 19
Therefore Rsys(t) = e- .000004199 t and MTTF = 1/.4199 x 105
Presenter
Presentation Notes
A military specification on the prediction of electronic components defines a parts count approach to finding system reliability. It is based upon the assumption that all components are CFR and that the failure rates can be summed to get the system failure rate. Electronic components are more likely to experience CFR than mechanical components. Why?
All Failure Modes are CFR
MTTF = 1 = 11
MTTF
where MTTF = 1i=1
n
ii=1
n
i
ii
= 1λ
λ
λ
∑ ∑
Chapter 3 20
λ λλ
= n MTTF = 1n 1
1and
If all components have identical failure rates, then:
Presenter
Presentation Notes
Things can get simpler (I know, it is hard to believe) if all failure modes or components have the same identical constant failure rate. Give some examples where this would be the case. In general for CFR components, the system MTTF is equal to the inverse of the sum of the component MTTF inverses. Of course, all we are doing here is adding the failure rates.
3.5 Poisson Process
If a component having a constant failure rate λ is
immediately repaired or replaced upon failing, the number of failures observed over a time period t has a Poisson distribution. The probability of observing n failures in time tis given by the Poisson probability mass function pn(t):
Chapter 3 21
( ) ( )
2
0, 1, 2, !
[ ]
nt
n
e tp t n
nE n t
λ λ
σ λ
−
= =
= =
…
The Poisson and Exponential
Chapter 3 22
( ) ( ) ( )0
0 0!
tte t
p t e R tλ
λλ−−= = =
The Gamma DistributionLetTi = time between failure i – 1 and failure i having an exponential distribution with parameter λ.
Yk = the time of the kth failure. The sum of k independent exponential random variables has a gamma distribution with parameters λ and k.
If k is an integer (i.e. the Erlang distribution),
Chapter 3 23
1
k
k ii
Y T=
= ∑
( )1
( ) for , , 0k k t
Yt ef t k t
k
λλ λ− −
= ≥Γ
The Gamma DistributionThe CDF is the probability that the kth failure will occur by
time t.
The mean value for Yk is k/λ, and the variance is k/λ2. The mode is (k – 1)/λ.
Chapter 3 24
{ } ( ) ( )1
0
Pr 1!k
ikt
k Yi
tY t F t e
iλ λ−
−
=
≤ = = − ∑
( ) { } { } ( ) ( )( )
11Pr Pr
; the Poisson!
n nn n n Y Ynt
P t Y t Y t F t F t
e tn
λ λ++
−
= ≤ − ≤ = −
=
Related Failure Distributions
Chapter 3 25
Random Variable Distribution Parameter(s)T, time to failure Exponential λYk, time of the kth failure Gamma (Erlang) λ, kN, number of failures in time t Poisson λ
EXAMPLE 3.9
A specially designed welding machine has a nonrepairable motor
with a constant failure rate of 0.05 failures per year. The company has purchased two spare motors. If the design life of the welding machine is 10 yr, what is the probability that the two spares will be adequate?
Chapter 3 26
λt = 0.05(10) = 0.5.
( )0.52
0.52
0
0.5 0.2510 1 0.5 0.9856! 2
n
n
eR en
−−
=
⎛ ⎞= = + + =⎜ ⎟⎝ ⎠∑
EXAMPLE 3.9
Let Y3 be the time of the third failure. Y3 has a gamma distribution with k = 3 and λ = 0.05.
The expected, or mean, time to obtain 3 failures is 3/0.05 = 60 yr.
The probability that the third failure will occur within 10 yr is
Chapter 3 27
( ) ( )3
20.05 10 0.05 10
10 1 1 0.05 10 0.01442!YF e− ×
⎛ ⎞×= − + × + =⎜ ⎟
⎝ ⎠
Note: 0.0144 = 1 – 0.9856 since the probability of two or fewer failures in 10 yr is complementary to the event that the third failure occurs within 10 yr.
Redundancy2 identical components with CFR
R(t) = 1-(1- e )2- tλ
Chapter 3 28
= 1-(1-2e +e )- t -2 tλ λ
= 2e - e- t -2 tλ λ
Presenter
Presentation Notes
Components in a system may be related to one another in a serial fashion as discussed earlier or in parallel or redundant fashion. Redundancy will also be discussed further in Chapter 5. If two components are redundant, both must fail for the system to fail. R(t) is found by taking 1 – the probability that both components fail by time t. Why can we do this?
Redundancy - Hazard Rate Function
λ λ λλ λ
λ λ(t) = f(t)
R(t) = 2 e - 2 e
2 e - e
- t -2 t
- t -2 t
Chapter 3 29
= (1 - e )(1-.5 e )
- t
- t
λ λ
λ
Not a CFR process!
Presenter
Presentation Notes
For a system consisting of two redundant and CFR components, the hazard rate function is no longer CFR. Observing the value of the hazard rate function as t get larger, it can be seen that the failure rate approaches a constant that is equal to common failure rate of the two components. What a neat result!
Chapter 3 30
Presenter
Presentation Notes
Here is that result shown graphically for two different lambda values.
Redundancy, CFR & MTTF
Chapter 3 31
= 2 - 12
= 32
= 1.5λ λ λ λ
- t -2 t0 0MTTF = R(t)dt = (2 - )dte eλ λ∞ ∞∫ ∫
A most useful result:0 0
1atat ee dt
a a
∞∞ −− = =
−∫
Presenter
Presentation Notes
The MTTF of a system composed of two identical and CFR components is found by integrating the system reliability function. The MTTF of this redundant system is 1.5 times the MTTF of each of components.
Two-parameter Exponential
Chapter 3 32
∞≤ <tt<0 ,e = dtR(t) d- =f(t) 0
)t-(t- 0λλ
t t ,e = R(t) 0)t-(t- 0 ≥λ
( )0
0
- t t0t
1MTTF = dt = +t e tλλλ
∞ −∫
Presenter
Presentation Notes
Occasionally, a situation is encountered in which no failures can occur until some time t0 has elapsed. This time is referred to as the minimum life. For the exponential distribution that minimum life becomes a second parameter, a location parameter. Its effect to shift the distribution by the amount t0 on t-axis. This causes the MTTF and median time to failure to shift as well; but not the variance. Why?
More 2-parameter Exponential
Chapter 3 33
R t e
t t t
t t R
medt t
med
R
med o( ) .ln . .
ln
( )= =
= +−
= +
= +−
− −λ
λ λ
λ
505 0 69315
0 0
0
Note that the variance does not change and the mode occurs at t0.
Presenter
Presentation Notes
Some more on the two-parameter exponential.
Example CFR Model
A certain type of surge protector is observed to fail at the constant rate of .0005 failures per day. Find:
a. The MTTF, median, and the standard deviation.
b. Find the reliability during the first year; the second year; the second year given it has survived the first year.
c. The 90% design life.
Chapter 3 34
Presenter
Presentation Notes
The inevitable student exercise.
Example - solution
a. MTTF = 1/.0005 = 2000 days (5.5 yr.)Std Dev = 2000 daystmed = 2000 x .69315 = 1386.3 days (3.8
yr.)b. R(365) = e- 365 x .0005 = .833
R(730) = e-730 x .0005 = .694R(365|365) = R(365) = .833
c. t.90 = - 2000 ln .90 = 210 days (7 mo.)
Chapter 3 35
Presenter
Presentation Notes
Check your answers. Contact your instructor if things are not going well.
Bonus Round - Return Period
“An asteroid 2/3 of a mile across smashing into the Earth at a high speed which would cause a blast with the power of thousands of hydrogen bombs could happen within the next 300,000 years.”
- Dave Morrison (NASA)
Chapter 3 36
Let P = the return period, the mean length of time between a failure event. Then
R(t) = e- t / P = e- t /300,000 yrs
and R(70 yr.) = e- 70 / 300,000 = .9997667
The End
Chapter 3 37
Our next assignment will be in Chapter 4. First, however, you will want work the problems identified for chapter 3.
Presenter
Presentation Notes
We look forward to the next Chapter.
