Reliability Engineering Lec Notes #4

7/30/2019 Reliability Engineering Lec Notes #4

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Discrete Probability Distributions Commonly Used in Reliability Engineering

expected value E(r) m mean n

r=0 r p(r)

variance Var(r) 2 n

r=0 (m r)2 p(r)

The quantity is called the standard deviation.

Binomial Distribution

Suppose an event has two possible outcomes and these outcomes can occur in a statisti-

cally independent manner. The problem is to find the probability that rdesired outcomes

occur in n ev ents, irrespective of the order of occurrence.

Example 1

A pump is demanded on the average 3 times per week. The probability of pump failure

per demand is 0.1. If the pump fails, it is replaced with a new one in negligible time.

What is the probability that the pump will fail exactly twice in one week?

Solution

If "g" stands for "pump operates on demand" and, "f" stands for "pump fails on demand"

then the possible outcomes of the demands in one week and their probabilities are:

Demand#1 Demand#2 Demand#3 Probability

g g g (1. 0. 1)3

g g f (1. 0. 1)2(0. 1)

g f g (1. 0. 1)2(0. 1)

g f f (1. 0. 1)(0. 1)2

f g g (1. 0. 1)2(0. 1)

f g f (1. 0. 1)(0. 1)2

f f g (1. 0. 1)(0. 1)2

f f f (0. 1)3

Then the desired probability is


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3 (1. 0. 1)(0. 1)2 = 0. 027

In general, if probability of the desired outcome is x then, the probability that exactly r

desired outcomes occur in n trials is

p(r) =

n

r

xr (1 x)nr =

n!

r!(n r)!xr(1 x)nr

For the Binomial Distribution

m = n x

2 = n x(1 x)

Example 2

What is the probability that the pump of Example 1 will fail at mosttwice per week?

Solution

Note that

P(

r)

=

r

s=0

n!

s!(n s)!xs(1

x)ns

Then

P( 2) =2

s=0 3!s!(3 s)! (0. 1)s(1 0. 1)3s

= (1 0. 1)3 +3!

1!(3 1)!(0. 1)(1 0. 1)2 +

3!

2!(3 2)!(0. 1)2(1 0. 1)

= (1 0. 1)3 + 3(0. 1)(1 0. 1)2 + 3(0. 1)2(1 0. 1) = 0. 999

This result can be confirmed from the table above.


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Poisson Distribution

Now suppose that in the binomial distribution n is large and x


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Continuous Probability Distributions Commonly Used in Reliability Engineering

expected value E(t) m mean

0

dt t p(t)

variance Var(t) 2

0

dt(m t)2 p(t)Erlangian and Exponential Distributions

If(t) is constant, then

Average number of events within [0, t] = = t

and from the Poisson distribution we get

p(r, t) =(t)r

r!et

which gives the probability of exactly rfailures within [0, t]. The probability ofror less

failures within [0, t] is found from

P( r, t) =

r

i=0(t)i

i! et

.

If the system fails for the rth time within dtat t, it must failed r 1 times before prior tot. Then the probability, f(r, t), that the system fails for the rth time within dtat t is

f(r, t) = p(r 1, t) =(t)r1

(r 1)!et, > 0, r> 1

which is called the Erlangian Distribution. The Cfd for the Erlangian distribution is the

probability that r

th

failure will occur within [0, t] and is given by

F(r, t) =

t

0

dt(t)r1

(r 1)!et = 1 et

r1

i=0 (t)

i

i!.

An important case is when the system fails for the first time within dt at t (i.e. r= 1)which gives the exponential distribution


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f(t) = et.

For the exponential distribution

m =

1

2 =1

2

Example 4

Consider the welding machine of the previous example. What is the probability

P(3, 9 t 10) that the company will need a third spare motor during the 10th year ofoperation?

Solution

The failure rate is = 0.05/year. Then

P(3, 9 t 10) =

10

9

dt f(r, t) =10

9

dt0. 05(0. 05t)2

(2)!e0.05t= 0. 0035

Gamma Distribution

Gamma distribution corresponds to the situation when ris not necessarily an integer

in the Erlangian Distribution. In this case,

f(t) =(t)r1

(r)et, > 0, r> 0

where

Gamma Function (x) =

0dy yx1 ey x n (n = 1, 2, . . . )

The Cfd is

F(t) =1

(r)(r, t)

where


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Incomplete Gamma Function (r, x) =x

0

dy yr1 ey

Note that ifris an integer

1

(r)

x

0

dy yr1 ey = 1 exr1

i=0 x

i

i!

and we get the same Cfd we did for the Erlangian distribution. Gamma distribution with

integer r can be applied to situations where the time between failures of a system has

exponential distribution such as:

system consisting ofr units each with constant failure rate and switching on when

the r 1th unit fails,

system subjected to random shocks and failing after the rth shock.

For the pdf given by Gamma Distribution

m =r

2 =

r

2

0 0.5 1 1.5 2 2.5 30

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

lambda*t

f(t)/lambda

r=1.0

r=0.5

r=1.5

r=3.0

r=6.0

Gamma Failure Probability Distribution Function


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0 0.5 1 1.5 2 2.5 30

0.5

1

1.5

2

2.5

lambda*t

lambda(t)/lambda

r=0.5

r=1.0

r=1.5

r=2.0

r=3.0

Failure Rate for Gamma Failure Probability Distribution Function

Example 5

The failure density of a component during its "burn-in" period satisfies the gamma distri-

bution with r= 0. 5 and = 0. 033/day. How long would it take before the failure rate iswithin 10% of its asymptotic value?

Solution

Recall that

(t) =dF(t)/dt

1 F(t).

If failure density satisfies the gamma distribution then

F(t) =1

(r)(r, t)

which gives


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dF(t)

dt= f(t) =

(t)r1

(r)et

and

(t) =(t)r1

(r) (r, t)et.

10 20 30 40 50 600.02

0.04

0.06

0.08

0.1

0.12

0.14

days

lambda(t)forr=0

.5lambda=

0.0

33/day(perday)

lambda=0.033/day

Sincetlim (t) = , we need to find tvalue which satisfies

abs[(t) ]/ = 0. 1 => g(t) =(t)r1

(r) (r, t)et = 1. 1

=> g(t) =(0. 033t)0.5

(0. 5) (0. 5, 0. 033t)e0.033t = 1. 1

110 115 120 125 130 135 140 145 1501.085

1.09

1.095

1.1

1.105

1.11

1.115

Days

g(t)(perday)

The answer is 127.6 days.


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Lognormal Distribution

The lognormal distribution is again a two parameter distribution where the logarithm

of the stochastic parameter follows a normal or gaussian distribution. If the failure den-

sity satisfies the lognormal distribution

f(t) =1

2texp

ln(t/)2

22

, , > 0

F(t) =1

2

1 erf|z| ift<

1 + erfz otherwisez =

ln(t/)

2

The lognormal distribution is often used in risk analyses because there is usually large

uncertainty on the data. The mean and variance for the lognormal distribution are

m = exp(2/2) 2 = 2 exp{2[exp(2) 1]}

Weibull Distribution

The Weibull distribution is applicable to a large number of situations, including compo-

nent aging. The three parameter form of the distribution (i.e. with time delay ) is

f(t) =

t

1exp

t

, > 0, > 0 , 0 t

and

F(t) = 1 exp{ [(t )/]} , (t) =

t

1

,

m = + (1 + 1/) , 2 = 2{(1 + 2/) [(1 + 1/)]2}.


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0 0.5 1 1.5 2 2.5 30

0.2

0.4

0.6

0.8

1

1.2

(t tau)/beta

f(t)*beta

alpha=3.0

alpha=2.0 (Rayleigh Distribution)

alpha=1.5

alpha=1.2

alpha=1.0 (Exponential Distribution)

Weibull Failure Probability Distribution Function

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

0.5

1

1.5

2

2.5

3

(t tau)/beta

lambda(t)*beta

alpha=3.0

alpha=2.0

alpha=1.5

alpha=1.0

Failure Rate for Weibull Failure Probability Distribution Function

Example 6

The maintenance records for a component indicates that the MTTF for the component is

3 years and the variance on the time to failure is 2 months. Assuming that the time to

failure for the component satisfies the Weibull distribution with no delay, find the time

interval for a new component during which the component will operate with a minimum

probability of 90%.

Solution

From the given data,

(1 + 1/) = 3 , 2{(1 + 2/) [(1 + 1/)]2} = 1/6 = 0. 17


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=> [3/ (1 + 1/)]2{(1 + 2/) [(1 + 1/)]2} = 0. 17

=> g() = (1 + 2/) 1. 019 [(1 + 1/)]2 = 0 => = 0. 987

=> = 3/(1 + 1/) = 3/2. 05 = 1. 464 years

Then the reliability function for the component is

R(t) = 1 F(t) = e (t/1.464)0.987

.

We want to find tfor which

R(t) = e 0.687t0.987

0. 9 ln(1/0. 9) 0. 687 t0.987 => t 0. 15 years

Mixed Distributions

f(t) =I

i=1 ki fi(t) ; 0 ki 1 ,

I

i=1 ki = 1

Composite Distributions

(t) =

I

i=1 H(t Ti1)[i(t) i1(t)] , H(t) =

1 ift> 0

0 otherwise

Convoluted Distributions

If:

a) a device can operate with any one of n units whose failure densities are

fi(t) (i = 1, 2, . . . , n),

b) ith unit comes on line when all the i 1 units have failed, and,

c) switching is perfect,

then

f(t) =

t

0

dtn1 fn(t tn1)tn1

0

dtn2 fn1(tn1 tn2) . . .t2

0

dt1 f2(t2 t1)f1(t1)


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Example 7

A flow delivery system has two pumps. Under normal operation, Pump#1 is on and

Pump#2 is off. If Pump#1 fails-off, then Pump#2 is turned on with perfect switch-

ing. Given that the reliability functions for the pumps are, respectively,

R1(t) = e1t R2(t) = e

2t,

find the reliability function for the flow delivery system.

Solution

First recall that

f(t) = (t)R(t) => f1(t) = 1e1t , f2(t) = 1e

2t.

Then

f(t) =

t

0

dt1 f2(t t1)f1(t1)

=> f(t) = 12

t

0

dt1 e2(t t1)e1t1 = 122 1e1t e2t

.

=> R(t) = 1 F(t) = 1

t

0

dt f(t) = 11 2 e2t 2

1 2e1t

For identical pumps, i.e. 1 = 2

f(t) = (1)2

t

0

dt1 e

1(t t1)e1t1 = t(1)2e1t

=> R(t) = 1 F(t) = 1

t

0

dt f(t) = (1 + 1t)e1t

Note that we could have obtained the same result by letting 2 approach 1:


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21lim R(t) =

21lim

1e2t 2e

1t

1 2

=0lim

1e(1 )t (1 )e

1t

=0lim [t1e

(1 )t + e1t] = (1 + 1t)e1t

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Reliability Engineering Lec Notes #4