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EML4550 2007 1
EML 4550: Engineering Design Methods
Probability and Statistics
in Engineering Design:Reliability
Class NotesHyman: Chapter 5
EML4550 -- 2007
Reliability
Reliability Probability that an item will perform its stated function without failure
under stated conditions of use for a stated measure of the variable (time, distance, batch, etc.)
In the context of reliability we will use the term ‘time’ for the variable (but it may be miles traveled, number of takeoffs and landings, etc.)
Reliability is based only on the ‘available’ time interval and can be expressed as a function of time,
R(t)=Ns(t)/No=(Total # survived til time t)/(Total #)
The unreliability is the opposite of R(t) and can be defined as the probability of failure,
F(t)=Nf(t)/No =(Total # failed til time t)/(Total #)
R(t)+F(t)=1
EML4550 -- 2007
Mathematical formulation
Probability of system failing by time t is the integral of the failure probability density function
dt
tdFtf
dftFt
)()(
)()(0
f(t)
t
F(t)
1.0R(t)=1-F(t)
f(t) can be considered as the instantaneous failure rate
EML4550 -- 2007
Per-unit failure rate
Suppose we have a population of 10,000 transistors under test/operation going out at a rate of 50 failures/hour dNf/dt
If we had 1,000 transistors under the same conditions (possibly coming from a subset of the 10,000 above), we would expect 5 failures/hour
It is convenient to define a “per-unit failure rate” or “hazard rate” of, in this case
R
dtdR
tR
dttRd
tR
dttFd
tRN
dttFNd
tN
dttdNth
o
o
s
f
/
)(
/)](1[
)(
/)]([
)(
/)(
005.0000,10
50
)(
/)()(
Nf(t) = number of objects that have failed by time t
Ns(t) = number of objects that have survived by time t
EML4550 -- 2007
Reliability as a function of the per unit failure rate
dttheR
R
dRdtth
R
dtdRth
)( timerespect to withIntegrate
)(/
)(
Relatively constant for an extended period of time
EML4550 -- 2007
Constant per-unit failure rate
te)t(R
)t(h
For systems with constant per-unit failure rate the reliability decreases exponentially with time
Example: 2000 items tested for 500 hours. Per-unit failure rate is 0.002 per hour, how many will survive after 500 hours?
R(500)=exp-(0.002*500)=exp(-1)=0.37, 0.37*2000=740 will survive after 500 hours
EML4550 -- 2007
Mean Time Between Failures (MTBF) Mean Time To Failure (MTTF)
Mean Time To Failure (MTTF): The mean of the survival time for all components. This is usually applied to parts that are not repairable, such as light bulbs, pens, etc.. This can also be applied to a system with many components
Mean Time Between Failures (MTBF) is the mean of all intervals between failures provided a large enough sample is taken; usually applied to systems that are to be repaired, such as compressor unit in a power plant, etc.. It is also useful for a system with multiple components (m) and each of which is immediately replaced on failure.
Example: a system has 2 components; one with a MTTF of 2 years, the other has a MTTF of 3 years. What is the MTBF?
m
i iMTTFMTBF 1
11
(years) 2.1,6
5
3
1
2
11 MTBF
MTBF
EML4550 -- 2007
Mean time to failure (MTTF)
MTTF = Expected, on average, value of the time when failure occurs
rate failureunit -per the
of inverse theis MTTF therates, failureunit per constant For
.1
MTTF
rate hazardconstant a For
0For )(0
dtdt
dRtMTTFtdtttfMTTF
EML4550 -- 2007
Example: Reliability & MTTF
A manufacturing process has an established per unit failure rate of 10-5 failures per day. (a) Determine the reliability for a period of 100 days?
(b) If the process produces 10,000 parts all together, how many failures we should expect within 100 days?
(c) What is the Mean Time To Failure (MTTF)
)(000,1001
MTTF
expected failures 10
9990)999.0)(10000()(
999.0)( 10010 5
days
NNN
tRNN
eetR
SOf
os
t
EML4550 -- 2007
Average per unit failure rate
The average hazard rate, , can sometimes be estimated by performing test on a specific number of samples (n), recording the total number of failures (m), total time span (T), and individual times to failures (ti).
)/1(0107.04)(20)-(2018)1713(6
4
rate. failure average the Estimately.respective
days, 18 17, 13, 6,after period theduring failed parts Four nonstop. days 20
for testedare samples 20 Total part.a for rate failure thedetermine :Example
)(otherssamples failedfor timetotal tested timetotal
failed test ofnumber
1
day
Tmnt
mmm
ii
EML4550 -- 2007
Example: Normal Failure Analysis
It is known that an air freshener has a mean lifetime of 800 h with a standard deviation of 40 h (normally distributed). Determine what is the reliability of a freshener at 700 h and 850 h?
h. 850after survive willfresheners 1057only 1057.08943.01)850(
8943.01057.01)25.1Pr(1)25.1Pr(,25.140
800850z(b)
fail will62only ,fresheners 10,000 ofOut 9938.0062.01)700(
0062.0)700Pr()5.2Pr(,5.240
800700)(
-yz : variablenormal-z Use
fail willitem theofy Probabilit1)(1)(
tR
zz
tR
yzza
tFtR
