Identification of the System Reliability

Reliability – lesson 2

Identification of the system reliability

Testing huge number of system products with reports about failures

Reports:

time of failure of systems

number of failures for all systems in short time period (short time with respect to the total test time)

Statistical data processing

approximate detected values

compare with some standard probability distribution and set the parameters of distribution

Selected distribution enable to compute others parameters of system (including all other parameters like time to wait for repair, failure of reserve)

Distribution of continuous variable

Exponential distribution

has one parameter , can be easy express as analytic functions

R(t)=

Q(t)=

f(t)=

(t)=

TS=

It is the most used distribution

MTTF is 1/, the distribution is defined by mean value

Very good approximation of systems in time of normal operation

Do not match the early stage and end-of-life stage

Some systems with flexible failure rate are not convenient for exponential distribution

Rayleight distribution

is defined by one parameter k and the failure rate is increasing linearly

suppose k>=0

R(t)=

Q(t)=

f(t)=

(t)= k t

TS=

Weibull distribution

has two parameters, m and t0

R(t)=

Q(t)=

f(t)=.

(t)=

TS=

Wiebull distribution contains exponential and Rayleight distribution as a special cases.

for m=1 it is exponential distribution for m=2 Rayleight distribution

Wiebull distribution for m<1 is very good approximation of the product the early stage

Wiebull distribution is the second most used distribution (after exponential)

Gama distribution

two parameters m>0, c>0, system with backup

f(t)=

TS=m.c

D=m.c2

if m is a natural number the definition of

so:

f(t)=

where:m…change the shape of function f(t)

c…change the scale factor of time

gama distribution is for m=1 equal to exponential distribution with =1/c

5 Truncated normal distribution:

two parameters > 0

f(t)=

where

reliability theory use only t0 it means that fort<0 the probability of failure is zero
characterization of reliability with truncated normal distribution t>0

…is distribution function of normal distribution n=0 n2=1

probability density of normal distribution is:

for  > 2 is TS very similar to 
truncated normal distribution is suitable for approximation of reliability of product in end-of-life stage (see figure)

Logarithmic – normal distribution

is defined by logarithm of random variable with normal distribution:

random variable x has normal distribution with parameters ,  and x=ln(t), so and t > 0

The normal distribution need not to be truncated

probability density.

where:

with respect to variable t:

and

where constant M0,4343 is for transformation lnlog

Mean time to failure:

can be evaluated

Dispertion:

for small size of  < 0.1 is similar to normal distribution and can be used for time of system back up

previous simple distribution cannot approximate the whole system for the life time of product combination of distributions

Superimposing of two exponential distribution

approximation of reliability in early stage and normal working product

R(t)=

f(t)=

as the probability distribution is probability it need to satisfies

from this we can calculate

start point =1c1+2c2

Mean time to failure =

Typically we have 12 for big t is significantlly smaller than , so for big time t this distribution is transforming into exponential distribution with t)

Combination of exponential and truncated normal distribution

is suitable for approximation of reliability in normal working product and end-of-life stage of product

failure rate is combination: t)=t)+t)

probability of working without failure is

Partially Linear Distribution

so called bathtub curve

stages are approximated by linear functions:

constant value exponential distribution

increasing Reileight distribution

decreasing small problem - new type distribution

Decreasing linear function has for t=0 value 0 and then is linearly going to 0, so the function is defined as (t)=0-K1t. But we know, that (t)0 for all t0. Therefore this definition is valid only in interval . But this function does not satisfies the condition that

if t . The solution is that we define (t)=constant_value for

probability of working without failure is defined as integration of (t). The integration is divided into three intervals (see the bathtub curve)

This partially linear distribution satisfies:

t; 0tt1:(t)=0-K1t

R(t)= =R1(t)

the end of the first interval is

t; t1tt2:(t)=

R(t)=

using the previous equation we receive:

R(t)=R1(t1)=R2(t)

t; t2tt3:(t)=+K2(t)

R(t)= R2(t2)

Remarks:1)the distribution is defined by parameters 0, K1, , t2, K2, t3

2)the end of the interval t3 is possible to shift till .

Discrete distributions

The most used distributions are:

binomial (Bernoulli)
Poisson

Binomial distribution

Used:probability of some event in selection from defined set or in repeating tests. The event is present or not present.

Suppose that we make n test, in m tests the event is present and the probability of this event is p

Probability distribution of binomial distribution is:

m – tests satisfy event

n-m – test don’t satisfy event

number of combinations is:

all combinations are independent the product defines the probability of all components in one time
probability that the event is in all n tests:

probability that the event is not in any test:

Distribution function of random variable M with binomial distribution is:

Remark:for big n the F(m) is similar to normal distribution

Mean value:

Dispersion:

Remark: with substitution to random variable we receive:

Poisson distribution

used:to describe presence of isolated events isolated in time, duration. The number of events in one unite has to be defined

Example: number of failures in time unit, number of failures in distance unit (for a car)

Remark:Poisson distribution is limit case of binomial distribution when

p0 … probability of event

n… number of tests

Probability density of Poisson distribution is:

m=0,1,2,…is number of event occurrence

a…………mean number of event occurrence in one time unit

Mean value:

Dispertion:

The shape of Poisson distribution for different process in time:

suppose the events are independent
the occurrence of event in interval t with probability p that is proportional to size of time interval t
mark  number of event occurrence in time unit t

number of event and … mean value of event in time unit

Remark:

probability that in time t was no event expression for working without failure: