BASIC RELIABILITY “bottom-up” MODELS
1.Series model
The series model is used to go from individual components to the entire
system, assuming the system fails when the first component fails and all
components fail or survive independently of one another.
R (t )   R (t )
n
sys
i
i 1
h (t )   h (t )
n
sys
i 1
i
Example 1
What is the reliability of the series system shown below?
A
= .001 fph,
B
= .002 fph , mission time t = 50 hours
For the system to work, both devices must work. If one
device fails, the system fails.
RA = reliability of device A = probability that device A will
work for at least 50 hours
RA =
= e -(.001)(50) = .9512
RB = reliability of device B = probability that device B will
work for at least 50 hours
RB =
= e -(.002)(50) = .9048
RS = reliability of system = probability that the system will
work for at least 50 hours
RS = RA X RB
RS = .9512 X .9048 = .8606
2.Parallel or redundant mode
The parallel model assumes all n components that make up a system
operate independently and the system works as long as at least one
component still works. If there are n components, any (n-1) of them may
be considered redundant to the remaining one (even if the components
are all different). When the system is turned on, all the components
operate until they fail. The system reaches failure at the time of the last
component failure.
For a parallel model, the CDF Fs(t) for the system is just the product of
the CDF's Fi(t) for the components or
F (t )   F (t )
n
sys
i 1
i
So, Rs(t) and hs(t) can be evaluated using basic definitions, once
we have Fs(t).
The schematic below represents a parallel system with 5
components and the (reliability) equivalent 1 component system
with a CDF Fs equal to the product of the 5 component CDF's.
Example 2
What is the reliability of the parallel system shown below?
A
= .001 fph,
B
= .002 fph ,mission time t = 50 hrs
For the system to work, one or both devices must work. The
system will fail when both devices fail.
From example 1, RA = .9512 and RB = .9048
FA = 1 - RA = probability that device A fails and FB = 1 - RB =
probability that device B fails.
Fsys = FAFB = (1-RA)(1-RB) ,so:
Rsys = reliability of system = RA + RB - RARB = .9512 +.9048 –
(.9512)(.9048) = .9954
For n identical components:
1
1 1
1
MTTF  (1        )

2 3
n
3.
r out of n model
An r out of n model is a system that survives when
at least r of its components are working (any r)
An "r out of n" system contains both the series system model and the
parallel system model as special cases. The system has n components that
operate or fail independently of one another and as long as at least r of
these components (any r) survive, the system survives. System failure
occurs when the (n-r+1)th component failure occurs.
When r = n, the r out of n model reduces to the series model. When r = 1,
the r out of n model becomes the parallel model.
We treat here the simple case where all the components are identical.
Formulas and assumptions for r out of n model (identical components):
1. All components have the identical reliability function R(t).
2. All components operate independently of one another (as far as failure is
concerned).
3. The system can survive any (n-r) of the components failing. The system fails
at the instant of the (n-r+1)th component failure.
System reliability is given by adding the probability of exactly r
components surviving to time t to the probability of exactly (r+1)
components surviving, and so on up to the probability of all components
surviving to time t. These are binomial probabilities (with p = R(t)), so
the system reliability is given by:
n 
i
ni
Rsys(t )      R(t )  1  R(t )
i 
n
i r
Note: If we relax the assumption that all the components are identical,
then Rs(t) would be the sum of probabilities evaluated for all possible
terms that could be formed by picking at least r survivors and the
corresponding failures. The probability for each term is evaluated as a
product of R(t)'s and F(t)'s. For example, for n = 4 and r = 2, the system
reliability would be (abbreviating the notation for R(t) and F(t) by using
only R and F)
Rsys = R1R2F3F4 + R1R3F2F4 + R1R4F2F3 + R2R3F1F4
+ R2R4F1F3 + R3R4F1F2 + R1R2R3F4 + R1R3R4F2
+ R1R2R4F3 + R2R3R4F1 + R1R2R3R4
Example 3
2
4
What is the reliability of the
out of
system if: all λi =
0.001 fph and the mission time t = 50 hours ?
Solution
All
Ri  0.9512
,so:
 4
i
4i
Rsys(t )      0.9512  0.0488
 2
4
i 2
R  0.9996
sys
4. Standby model
The Standby Model evaluates improved reliability when backup
replacements are switched on when failures occur.
A Standby Model refers to the case in which a key component (or
assembly) has an identical backup component in an "off" state until
needed. When the original component fails, a switch turns on the
"standby" backup component and the system continues to operate.
In the simple case, assume the non-standby part of the system has CDF
F(t) and there are (n-1) identical backup units that will operate in
sequence until the last one fails. At that point, the system finally fails.
The total system lifetime is the sum of n identically distributed random
lifetimes, each having CDF F(t).
In other words, Tn = t1 + t2+ ... + tn, where each ti has CDF F(t) and Tn
has a CDF we denote by Fn(t). This can be evaluated using convolution
formulas:
In general, convolutions are solved numerically. However, for the special
case when F(t) is the exponential model, the above integrations can be
solved in closed form.
Special Case: The Exponential (or Gamma) Standby Model
If F(t) has the exponential CDF (i.e., F(t) = 1 - e-λt ), then
R e
sys
(t ) 2 (t ) 3
(t )
 [1  t 

 
]
2!
3!
(n  1)!
n 1
t
MTTF 
n

and the PDF fn(t) is the well-known gamma distribution.
For n = 2:
 t
Rparallel
 2e
Rstandby
 (1  t )  e
For
t  1 ;
e
 2 t
 t
Fparallel
2
Fstandby
Example 4:
An unmanned space probe sent out to explore the solar system has
an onboard computer with reliability characterized by the exponential
distribution with a Mean Time To Failure (MTTF) of 1/ = 30 months
(a constant failure rate of 1/30 = .033 fails per month). The probability of
surviving a two year mission is only e-24/30 = .45. If, however, a second
computer is included in the probe in a standby mode, the reliability at 24
months (using the above formula for F2) becomes .8 × .449 + .449 = .81.
The failure rate at 24 months (f2/[1-F2]) reduces to [(24/900) ×.449]/.81 =
.015 fails per month. At 12 months the failure rate is only .0095 fails per
month, which is less than 1/3 of the failure rate calculated for the nonstandby case.
Standby units (as the example shows) are an effective way of increasing
reliability and reducing failure rates, especially during the early stages of
product life. Their improvement effect is similar to, but greater than, that
of parallel redundancy. The drawback, from a practical standpoint, is the
expense of extra components that are not needed for functionality.
5.Complex systems
Often the reliability of complex systems can be evaluated by
successive applications of Series and/or Parallel model
formulas.
Many complex systems can be diagrammed as combinations of Series
components, Parallel components, R out of N components and Standby
components. By using the formulas for these models, subsystems or
sections of the original system can be replaced by an "equivalent" single
component with a known CDF or Reliability function. Proceeding like
this, it may be possible to eventually reduce the entire system to one
component with a known CDF.
Below is an example of a complex system composed of both components
in parallel and components in series is reduced first to a series system and
finally to a one-component system.
Note: The reduction methods described above will work for
many, but not all, systems. Some systems with a complicated
operational logic structure will need a more formal structural
analysis methodology. For example let’s solve the system
bellow by the help of Baye’s theorem.
About gamma distribution
There are two ways of writing (parameterizing) the gamma distribution
that are common in the literature. In addition, different authors use
different symbols for the shape and scale parameters. Below we show
three ways of writing the gamma, with a = = , the "shape" parameter,
and b =1/ , the scale parameter.
EXCEL uses
.
Note: When α = 1, the gamma reduces to an exponential distribution with
β = 1/λ.
For the standby model α = n and β =1/λ.
EXCEL has built-in functions to evaluate the gamma pdf and cdf. The
syntax is:
=GAMMADIST(t,α,β,FALSE) for the PDF
=GAMMADIST(t,α,β,TRUE) for the CDF