Estimation of distribution parameters
 data processing from reliability tests
to set:
 probability distribution model (estimation of parameters of the models)
 mean time to failure, dispersion
Number of tested products
–problems with data likelihood, need to compare
different methods
–statistic parameter estimation
–good result do net depend on selected methods
<10
>10 <100
>100


method of the biggest likelihood
method of the least square
Motivation
 Suppose that measured values has normal distribution with mean value a and dispersion 2
 the average from n values is x with dispersion 2/n
 we select  > 0 and looking for probability that the average is from interval x  a   , a  
From normal distribution 




  P a    X  a    0 

–precision of estimation

n

Laplace function
x
Φ0  x     0  y dy
where:
x
probability density of normal distr.
Φ0  x  
or
1 1
 Φx 
2 2
distribution function of normal distr.
Our task:
 data set with normal distribution
 dispersion 2
 make selection of n values with average X
Goal: What is the mean value a ?


  P X    a  X    0 
n


! random borders
a is not random variable !!


Reverse goal from motivation task



interval X   , X   is called confidence interval, the probability  is called the size of
confidence interval
Remark: confidential interval can be limited from one size, it means.:
Pa  ad   1
Pa  ah    2

with previous equations we receive equation for , 1 and 2 :
  Pa  ad   Pa  ah   1  1   2   1   2  1
f(x)
2
ah
x
f(x)

x
ad
ah
Confidence interval limited by one side (above) or by two sides
Remark: for unknown distribution can be used raw
estimation with Tchebyshev inequality. It can
(down)
2
be prove, that for random variable X with mean value E(X) and dispersion D ( X )   this
equation is true for all k>0:
P  X  E  X    .k  
1
k2
for x  Ex      is 3 time worst than boundaries for normal distribution
for x  Ex   3   is 40 time worst than boundaries for normal distribution
Graph methods: exists graphic methods for estimation of distribution – depending of failure rate
Reliability of complex systems


decomposition of complex systems
(it is easier to detect the reliability of subsystems)
analysis of structure and mutual connections:
 usually it is assumed that the failures of subsystems are independent
 dependent failures in subsystems with interaction to another subsystems or with
interaction in common environment
Remark: if nothing else is specified we suppose this limitation to simplification the evaluation:
 every system (and all its subsystems) are in state of working or in state of failure
(there is no third state)
 probability of working without failure is constant in time
 systems are without repair
Serial system:


all subsystems works  the whole system is working
failure of at least one subsystem leads to failure of the whole system
Input
a1
a2
an
Input
a1
a2
an
Output
Output
Flow chart and graph representation of serial system
Direction of signals (in general it is oriented multigraph)
Suppose:
n…is a number of subsystems
X i …state failure of i-th subsystem
X i …working state of i-th subsystem
 
the corresponding probabilities are P X i , P X i
Probability of working without failure (intersection of working states of all subsystems) for
dependent subsystems
R  Px1   Px2 | x1   Px3 | x1 x2     Pxn | x1 x2 ...xn1 
If the subsystems are independent:
n
R   P  xi 
i 1
Probability of failure   subsystem that has the failure:
Q  P x1  x2  ...  xn
(union of failure states)
 failure states are not mutually excluded
the equation can be changed:


 
n
n


Q   P xi   P xi x j  ...   1
i 1
i 1
j 1
i j
n 1


P x1 x2 ...xn : together 2n-1 items
sum of probabilities of all combinations
3…(n-1)
Conclusion:
 It is better to calculate probability of working without failure and the probability of
failure calculate from Q = 1 – R.
 Serial system has always worst reliability then is the reliability of the worst
subsystem
 If the system is composed form n identical subsystems with the same probability of
working without failure p and probability of failure is q 
n
probability of working without failure is : R  p n  1  q 
Probability of failure: Q  1  p n  1  1  q 
n
Parallel system
Input
a1
Output
a1
a2
a2
an
…
Input
Output
an
Flow chart and graph representation of parallel system

parallel system is working without failure <=> at least one subsystem is working without
failure (so called redundant connection => increase reliability of system)
 probability of working without failure: (union of state working for subsystems)
R = P(x1+x2+x3+ … +xn)
By analogy with serial system the equation is changed:
R  P( x1 )  P( x2 )  ...  P( xn )  [ P( x1 , x2 )  P( x1 , x3 )  ...
...  P( xi , x j )]  ...  (1) n1 P( x1 x2 ... xn )
i j

failure of parallel system is if there is failure of all subsystems:
intersection of state failure of subsystems Q  P( x1 x2 x3 ...xn )
for dependent subsystems:
Q  P( x1 )  P( x2 x1 )  P( x3 x1 x2 )...P( xn x1 x2 ...xn1 )
for independent subsystems:
n
Q   P ( xi )
i 1
Conclusion:
 probability of working without failure of parallel system is always higher (better) then the
probability of working without failure of the best subsystem
 for parallel system of identical independent subsystems with the same probability of failure
q:
probability of failure:
Q  q n  (1  p) n
probability of working without failure:
R  (1  q n )  1  (1  p) n
Combined systems:

solution of complex systems
(1) transformation to combination of serial and parallel systems – FAST
(2) partially transformation with (1) followed by specific computation
Another types of system
systems of type "m from n"
 n system components
 for correct work you need m working components
Example: rope created from n fibers. To desired carrying capacity is necessary at least m<n not
broken fibers
Example 2: (system "m from n") system of water-stations with common distribution network
h
 model of this system has   parallel connections input → output
m
 every connection has m-components in serial system (one combination /
selection from n components)
 System is working without failure <=> at least one connection is working
 If we suppose identical independent components => binomial distribution
it means that probability of working without failure:
Probability of working p. components
R
n
 f (k , n, p)
k m
extreme case:
for m = 1 (at least one from n) ≡ parallel system
for m = n (n from n) ≡ serial system
Solving the system reliability using index:
Principle:
(1) create index of all possible cases of events
(3) index there are only isolated events
correct and adverse
Σ probabilities of all Σ probabilities of adverse
correct events (state of events
correct working)
(state of failure)
Example:
a
Input
d
Output
b
c
e
Flow chart of example system


number of subsystems = 5 => 2n = 25 = 32 possible events
allocation to group with fixed number of working subsystems:
Number of failures in system
0
1
2
4
5
A1 =
A2 =
A3 =
A4 =
A5 =
A6 =
A7 =
A8 =
A9 =
…
abcde
a bcde
ab cde
abc de
abcd e
abcde
abcde
abc de
a bcd e
System is working without
failure
yes
yes
yes
yes
yes
yes
yes
yes
yes
…
A27= abcde
A28= abcde
A29= abcde
A29= abcd e
no
no
no
no
A31= abcde
A32= abcde
no
no
Sum:
Failure: 13 combination
Working without failure: 19 combination
5
no failure   events, one failure
0
Probability of failure is:
5
  events, 5 failures
1 
5
  events
5
Q  P( A16  A17  A18  A20  A21  A24  A26  A27  A28  A29  A30  A31  A32 )

because states A1…A32 are mutually excluded => the probability of the union is sum of the
probabilities
two
possibilities
for
computation
prob.
of
working
without
failure:
R  1  Q  1  [ P( A16 )  P( A17 )  ...  P( A32 )]
Σ of correct independent
states from index
according to probability of
failure.
Special case: the system with identical subsystems with probability of working without failure p
R  1  Q  1  [ p 3 (1  p) 2  6 p 2 (1  p)3  5 p(1  p) 4  (1  p)5 ]
Remark.: (1) Probability of working without failure can be calculated as R=P(union of events of
correct working)
(2) in practice is selected computation of P or Q depending on number of failure states
Disadvantage of this approach is the combinatory explosion for bigger systems