Transcript Reliability-1

EML 4550: Engineering Design Methods
Probability and Statistics
in Engineering Design:
Reliability
Class Notes
Hyman: Chapter 5
EML4550 2007
1
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
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Mathematical formulation
 Probability of system failing by time t is the integral of the failure probability
density function
t
F (t )   f ( )d
f(t) can be considered as the instantaneous failure
rate
0
dF (t )
f (t ) 
dt
1.0
f(t)
F(t)
t
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R(t)=1-F(t)
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
h(t ) 

dN f (t ) / dt
N s (t )

50
 0.005
10,000
d N o F (t ) / dt d [ F (t )] / dt d [1  R(t )] / dt  dR / dt



N o R (t )
R (t )
R (t )
R
 Nf(t) = number of objects that have failed by time t
 Ns(t) = number of objects that have survived by time t
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Reliability as a function of the per unit failure rate
dR / dt
dR
h (t )  
 h(t )dt 
R
R
Integrate with respect to time  R  e 
 h ( t ) dt
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Relatively constant for an extended period of time
Constant per-unit failure rate
h(t)  
R ( t )  e  t
 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
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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.
m
1
1

MTBF i 1 MTTFi
 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?
1
1 1 5
   ,
MTBF 2 3 6
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MTBF  1.2 (years)
Mean time to failure (MTTF)
 MTTF = Expected, on average, value of the time when failure occurs

MTTF 
 tf (t )dt


dR
For t  0  MTTF    t
dt
dt
0
For a constant hazard rate 
MTTF 
1

.
For constant per unit failure rates, the MTTF is the inverse of
the per - unit failure rate
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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)
R (t )  e
 t
e
105 100
 0.999
N s  N o R(t )  (10000)( 0.999)  9990
N f  N O  N S  10 failures expected
MTTF 
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
 100,000( days )
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).

number of test failed
m


total time tested
total time for failed samples  others
m
m
t
i 1
i
 ( n  m )T
Example : determine the failure rate for a part. Total 20 samples are tested for
20 days nonstop. Four parts failed during the period after 6, 13, 17, 18 days,
respective ly. Estimate the average failure rate.

4
 0.0107(1 / day )
(6  13  17  18)  (20 - 4)(20)
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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?
R (t )  1  F (t )  1  Probabilit y of the item will fail
Use z - normal variable : z 
y-

700  800
 2.5, Pr( z  2.5)  Pr( y  700)  0.0062
40
R (t  700)  1  0.0062  .9938  Out of 10,000 fresheners , only 62 will fail
(a ) z 
850  800
 1.25, Pr( z  1.25)  1  Pr( z  1.25)  1  0.1057  0.8943
40
R (t  850)  1  0.8943  0.1057  only 1057 fresheners will survive after 850 h.
(b) z 
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