Method of paths and cuts


good for complex systems with independent subsystems
is based on system graph
o graph has at least n edges, where n is number of subsystems
Definition:
order-preserving sequence is sequence of edges, that correspond to sequence of
nodes that actual node is input node of edge and following node is output node of edge.
Definition:
path is order-preserving sequence that has all nodes in sequence different, it
means that all edges are different too.
Lemma:
path is minimal sequence (it consists of minimal number of edges)
Definition:
edge cut of graph is set of edges that for every path from input node to output
node contains at least one edge
How to calculate the reliability of system:
Suppose that in system exists i paths from input to output. The system is working if exists at
least one path with all nodes working without failure
Suppose that P1 … Pi are state without failure of paths 1 … i. Probability that the system is
working without failure is
R  P( P1  P2  ... Pi )
Vice versa, the failure is if at least one cut has all nodes with failure
Suppose that the system has j number of minimal cuts with states C1 , C2 , ... C j without failure
and states C1 , C 2 , ...C j with failure
Probability of failure is Q  P(C1  C 2  ...  C j )
Example of methods of paths and cuts:
Suppose system:
6 nodes and 6 edges
x1
x2
x5
x3
Order preserving sequence in T1=x1x2
the system:
T2=x3x4
T3=x1x6x4
x6
x4
T4=x3x5x2
T5=x1x5x6x2
T6=x3x5x6x4
Sequence T1, T2, T3 and T4 are paths.
Sequence T5 and T6 are not paths because they go twice through one node
The reliability of the system R =>
R  P(T1  T2  T3  T4 )  P( x1 x2  x3 x4  x1 x6 x4  x3 x5 x2 )
Failure of cut is if every subsystem corresponding to edge from cut has failure
State without failure for cut is if at least one subsystem corresponding to edge from cut is
working without failure
cuts of the systems are:
C1=x1x3
C4=x1x5x4
C2=x2x4
C5=x3x6x1
C3=x1x5x3
C6=x3x6x2
C1, C2, C4, C6 – are minimal cuts
C3, C5 – are not minimal because they contains the cut C1
from previous equation =>
reliability:
R  1  Q  1  P(C1  C 2  C 4  C 6 ) 
 1  P( x1 x3  x2 x4  x1 x5 x4  x3 x6 x2 )
Problems of this method:
(1) to find all path and cuts is complicated in complex systems =>
algorithms
(2) bad identification of minimal cuts leads to wrong result (or prolong the
calculation)
(3) to determine the probability of union of event is not simple, because these events
are not exclusive

if the events are exclusive => the probability is simple sum
Reliability estimation
because the probability of union of 3 events A, B, C is:
P( A  B  C )  P( A)  P( B)  P(C )  P( AB)  P( AC )  P( BC )  P( ABC )
if ABC are mutually exclusive => exclude the combination of events
in general it holds (for events A, B, C): P( A  B  C )  P( A)  P( B)  P(C )
two limit estimation
of reliability R
upper limit for case of dependent events
reliability R 
 P(T )  paths
i
i
prob. of failure Q   P(Ci )  cuts
upper limit
i
combination:
It has sence only if
R  1   P(C i )
i
lower limit
 P(Ti )  1 and  P(Ci )  1 , otherwise it brings no information
i
i
System from component with more states

till now:
system has two states
failure

general situation:
correct working
correct working
failure
different types
Example 1.: diode failure states
short-circuit
break
Example. 2.: transistor – combination of short-circuit and break of CB, BE transits
Diode
working without failure: x
mutually
excluded, only
one state can be
in one moment
short circuit: x S
break: x 0
P( x  xS  x0 )  P( x)  P( xS )  P( x0 )  1
a) probability of working without failure of diode is:
P( x)  1  P( x S  x0 )  1  P( x S )  P( x0 )
b) system of two diodes in serial connection:
1
summary:
2
System without failure: 1) both diode O.K.
2) one diode short circuit
Failure state: 1) both diode short circuit
2) at least one diode with
break
path of the system (positive combinations):
x1 x2
x1 x 2 S
x1 S x 2
cuts of the system (negative combinations)
x1 x20 , x1S x2S , x1S x20 , x10 x2 , x10 x 2S , x10 x20
probability of working without failure (using paths):
R  P( x1x2  x1 x2S  x1S x2 )
or (with cuts)
R  (1  P( x1 x20  x1S x 2S  x1S x20  x10 x2  x10 x 2S  x10 x 20 )
enumeration:
probability of short circuit
R  p 2  2 pqS
probability of working without failure
together:
p  qS  q0  1
probability of break failure
c) 2 diodes in parallel:
1
2
paths: x1x2 , x1 x 20 , x 10 x2
R  P( x1x2  x1 x 20  x 10 x2 )
for independent failures, both diode (substitution):
probability of failure break
R  p 2  2 pq0
Remark: ► this method can be used for more state of components
► complicated construction of graphs for more states
► more simple is to directly enumerate paths and cuts
Systems with time dependent failure probabilities
steps:
1. to evaluate the probability of working without failure of system
2. to substitute the probability distribution to probability of components
variation:
1. failures are independent – simple (approximation for big R)
2. failures are dependent:

►it is not possible to isolate the influence of the system structure and
influence of the reliability of components
solution
► cumulated probability distributions
► stochastic Markov models
Serial system (time dependent probability distributions)
 the whole system is working if all components are working
 system of n independent components
n
R (t )   P ( xi )
i 1
suppose the components has the exponential distribution
 i t
P ( xi )  e
 substitution
n
n
common used equation R(t )   e  i t  e
  i t
i 1
i 1
Remarks
1. probability distribution of the system has again the exponential distribution
n
   i
i
2. commonly used Weibull distribution for serial system (early stage (m<1) + normal
work (m>1))
n mi
tm
t

n 
t0 i
i 1 t0 i
R (t )   e
e
´
i 1
t
in general case if Zi (t )   i ( x ) dx then reliability of serial system is:
0
n
n
R (t )   e  Z i ( t )  e
 Zi ( t )
i 1
i 1
Parallel system
 working without failure if at least one component is working (there is not failure of all
components)
n
R (t )  1   P ( xi )
i 1
 using exponential distribution
n
R(t )  1   (1  e  i t )
i 1
P( xi )  1  e it
P( xi )  1  e
for Weibull distribution:

tm
t0
n
R(t )  1   (1  e

tm
t0
)
i 1
Remark
 serial system  lower limit of working without failure
 parallel system  upper limit of working without failure
=> for system with defined components
 complicated computation of working without failure with general
distributions => simplification to compute numerically for short time
interval
 substitution of complicated distributions with Taylor formula
 usually it satisfies to compute only the mean time to failure

TS   R(t ) dt
0
for serial system MTTF:
exponential distribution:
TS 
1
n
 i
i 1
1
n 1 
1
TS  (  1)   ( ) m
m
i 1 t0i
Weibull distribution with the
same parameter m for all
component distribution
for parallel system (it is more complicated):
► exponential distribution:
(2 components in parallel): TS 
1
1

1
2

1
1  2
(n – components in parallel):
1 1 1 1 
TS     ...  
 1 2 3 n 
 1

1
1

 

 ... 
n 1  n 
 1  2 2  3


1
1

 
 ... 
n  2  n 1  n 
 1  2  3
  1
1
n 1
n

i
i
► for identical components in parallel and exponential distribution:
1  2  ...  n   
 n  n  n
 n
  
      
1  1   2   3 
n 1  n  
TS 


 ...   1
 1
2
3
n 




Remark: in practice the computation of reliability for serial and parallel system is done only
for exponential distribution – other distribution are very complicated to express!
System with backup
 increase reliability  backup of systems (or parts of the system)
 types of backups:
— permanent
— with switch
— majority
1.
Permanent backup: (with running backup or parallel backup)
 common failure – "break" => it exist backup (parallel) path
 common failure – "short-circuit" => backup is connecting in serial
 combination of previous
x11
x11
x11
x21
x21
x21
xm1
xmn
backup of system in global
x11
x12
x1n
x21
x22
x2n
xm1
xmn
backup of components
suppose: the system is created from identical independent components, probability of working
without failure is p
backup of the whole system

R  1 1 pn
backup of components:


m
R  1  1  p 

m n
with two components (backup of system (a), backup of components (b))
a)
x1
x2
Input
b)
x'1
x'2
x1
x2
Input
x'1
the ration:

Output
Ra  2 p 2  p 4  p 2 2  p 2
Output
Rb  p 2 2  p 

2
x'2
Rb
Ra
Rb 2  p 2
21  p 2

 1
and 0  p  1
Ra
2  p2
2  p2
2  p2  0
it’s true, that
Remark
2.
Rb
 1 => parallel backup of components is better
Ra
 the example can be generalized for n and m components in serial and
parallel backup
 parallel backup of components is better (it holds for components with two
state and failure is break general case can lead to opposite conclusion)
Backup with switch (backup with immediate repair) – failure leads to switch bad
component). Switch can recognize the failure state of the system.
x1
input
s
x2
assumption: system has m identical independent
components, where components in
backup are not used if the system
works properly => Poisson distribution
f m; a  
a m a
e
m!
m  0,1,2, ...
for 2 components (no failure or one failure)
RZ  e a  a  e a  1  a e a
 if the probability of working without failure is p => according to Poisson
distribution ~ probability of no failure is (m=0) =>
P  e a

substitution:
RZ  1 ln p
a  ln  p 
R
1
p(1-ln p) (with switch)
0,8
2p-p2 - (with parallel backup)
0,6
p - (without backup)
0,4
0,2
1
0,75
0,5
0,25
0
p
Comparison of distribution without backup, with parallel backup and backup with switch
Remark Parallel backup is worse then the backup with switch (in parallel system the backup
system is working together with the original system => there is bigger chance for
failure of backup system)
- parallel backup ~ maximum (time to failure of component, time to
failure of backup system)
MTTF
- with switch ~ sum of times of both components and backup system
 t
 using exponential distribution Px1   Px2   e
:
 t
 Pswitch
switch: RZ  1  t e
 t
 e 2 t
parallel backup: RZ  2e
 influence of failure of the switch ! => probability of working without failure depends on
the reliability of switch. => (failure of switch => failure of system)  not exact
1. failure of the switch depends on failure of the system
2. switch has failure only when it is switching, if there is switch of the
correct working subsystem, it is not failure


a)
x11
x12
x1n
x21
x22
x2n
xm1
xm2
xmn
x11
x12
x1n
x21
x22
x2n
xm1
xm2
xmn
b)
a) whole seriál system
b) components
Backup with switch
Remark: using ideal switch it is better to backup components then the whole system. Real
switch decrease reliability and increase the price of the product
 computation of complex system (more then constant probability distribution)
are too complicated => there are used method of statistical modeling or
Markov models
3.
Majority backup
 improvement of reliability of discrete (digital) systems (compared with analog
systems it is more easy to compare the outputs of the system)!

backup of digital system must increase the reliability for all
possible outputs
Computation of majority from odd number of systems
suppose we have 3 systems:
x11
input
x21
M
xm1
Majority backup
output
Suppose the majority comparator component (M) is working without failure and x1, x2, x3 are
identical:
from binomial distribution:
R3  p3  3 p 2 1  p   p 2 3  2 p  , p is the probability of working component x
for general majority from (2n+1) is: n N
2 n 1 2n  1

 i
 p 1  p 2 n 1i
R2 n 1   
i 
i  n 1
 to compare the system of one component
R3  p 3  3 p 2 1  p 
ratio
R1  p and three components
R3
 p 3  2 p 
R1
R3
 1  p  0,5
R1
 it means that majority backup improve the reliability only if every component has
reliability bigger then 0.5. This holds for every n  1 that only if p > 0.5 then
R2 n1
1
R1
The figure shows the majority backup for components with exponential
 t
distribution p  e
, for =1. The time t=0.69 represents time when R(t)=0.5
n
R2n+1
1
n=0
n=1
n=2
n=4
0,5
n=0
n=1
n=2
n=4
e-t
0
0,5
0,69
1
1,5
t