This document discusses redundancy allocation, which is a concept to increase system reliability by using components in parallel rather than series. It defines key reliability terms like MTBF and explains how parallel systems are more reliable than series systems. Specific redundancy concepts covered include standby redundancy, where extra components act as backups, and k-out-of-n systems, where the system succeeds if at least k out of n components work. The document also provides an example of applying redundancy to increase the reliability of an aircraft's emergency systems.

1. 1. Redundancy Allocation
ABSTRACT :
For a complex system the reliability of the whole system goes down for using a large numbers of
components. If we use the components in series mode then the reliability of whole system will be
least. To increase the reliability by an amount we have to use the components in parallel mode. By
Redundancy allocation we can design such a system where we use maximum possible components
parallely to increase the reliability of the whole system.
1. INTRODUCTION:
Now a days the consumer and capital goods industries, space research agencies like NASA1
, ISRO2
etc. , are facing a big problem of unreliability. These companies and research institutes can not
progress or be succeed in their mission without the knowledge of reliability engineering. Reliability
engineering is essential for companies of electronic goods to approximate warranty date, make idea
about wear out period. It helps companies to optimize the product cost, empower the processor
speed or making the product faster in less cost. To tackle the growing competition any company
need a clear idea about reliability.
Reliability is mostly important for any space research projects, like Mars Rover, Curiosity, Apollo
etc, or in any DRDO3
projects, atomic plants etc. Here high reliability is needed at every stage, little
mistake can cause a huge loss or destruction and the entire labor behind it will be meaningless.
Reliability is restricted for a certain condition and for a particular task. So Reliability is a
probability that the device will perform the task for the period of time under certain operating
condition, so it will not break down in that period. So it is the probability that the product will face
any failure4
. Reliability is thus also called as probability of survival. So probability that component
survive until some time t is, R(t) = P(X>t) = 1-F(t), where F(t) is called unreliability.
Redundancy allocation is a concept by which we can increase the reliability by using the
components in parallel mode. It helps us to develop the machine for our required reliability. It also
helps us to optimize the cost for required reliable system. Even for the series system also we can
increase the reliability of the whole machine by using some components in parallel; those concepts
are given by redundancy allocation. Sometime the reliability depends on the structure of the
machine; using redundancy on the components we can optimize the design for a proper function.
Any electronics goods should be reliable both in performance and cost; a company can decide the
warranty of their product by seeing its reliability. Redundancy allocation pays a big role behind that.
2. RELIABILITY, UNRELIABILITY, MTBF, FAILURE RATE:
Failure rate is a parameter. It is a frequency of malfunction. It is the measure of number of failure
per unit time.
And the reciprocal value of failure rate is called MTBF or mean time between failure.
Usually we denote the failure rate by λ. And the MTBF by m=1/λ.
The relation of the λ and is given below:
1 NASA: National Aeronautics and Space Administration
2 ISRO: Indian Space Research Organisation
3 DRDO: Defence Research & Development Organisation
4 If p be the probability of success the 1-p will be probability of failure

2. 2. Redundancy Allocation
The bathtub curve is the graph of the component failure rate as a function of time. This curve is the
mixture of 3 failure rate graph. One is early life failure, second is wearout failure, and third is
random failure.
Now from the Poisson distribution for parameter μ and for the random variable X having the
enumerable set {0, 1, 2,…} the probability mass function is given by, ( ) , 0,1,...
!
0,
x
e
f x for x
x
elsewhere
µ
µ−
= =
=
Now for the given time interval (0,t), and the failure rate λ we have μ=λt, and the probability mass
function is,
( )
( ) , 0,1,...
!
0,
t x
e t
f x for x
x
elsewhere
λ
λ−
= =
=
Now, if there is no failure up to time t, then the probability P(X=0) gives the reliability at time t as,
0
( )
( ) ( 0)
0!
t
te t
R t f x e
λ
λλ−
−
= = = =
And the probability that it fails during the time t, that is the unreliability is, Q(t) = 1 t
e λ−
− .
We can find it from exponential distribution like below.
The exponential density function is, ( ) t
f t e λ
λ −
= , λ be the constant failure rate.
This is drawn from the Poisson distribution5
invented by the French mathematician Poisson. We use
this distribution for finding the reliability as only the parameter λ or its reciprocal m describes
completely that distribution. And it is independent of the age of component as long as the constant
failure rate condition persists.
Hence, the reliability from the exponential distribution is, ( ) t t
t
R t e dt eλ λ
λ
∞
− −
= =∫
And unreliability is that the possibility that it may fail before time t is,
0
( ) 1
t
t t
Q t e dt eλ λ
λ − −
= = −∫
5 Mood, A. M. And Graybill, F. A. (1974). Introduction to the Theory of Statistics (page. 93). USA: McgGaw-Hill
fig:1

3. 3. Redundancy Allocation
And the mean time between failure (MTBF) is,
t
0
, is the reliability of the systems sR dt R∫ .
3. SERIES SYSTEM:
In this kind of arrangement we use the components in a series like below,
In that system, if one component fails then the entire machine goes down. The least value of
reliability of the components used in the machine is the maximum possible reliability of the
machine.
1 2 n
s
s 1 2 n
That is let the reliablity of n components are R ,R ,...,R .
And the reliability of the system is R .
Then, R min{R ,R ,...,R }.≤
To get the reliability of the system we have to find the probability that all components are working
up to time t. And hence the reliability of the system is,
s 1 2 n 1 2R =R R ... R (1 ) (1 ) ... (1 )nQ Q Q× × × = − × − × × − . Here we assume that the failure possibility is
independent one from another.
In other sense if 1 2 nA ,A ,...,A be the events that the corresponding components will works then the
reliability can be describes as, .
4. PARALLEL SYSTEM:
In this kind of system we use the components in parallel mode like below.
fig:2
fig:3
1 2( ... )nP A A A∩ ∩ ∩

4. 4. Redundancy Allocation
In this kind of system the entire machine works until the single component works. Hence the
maximum value of the reliabilities of the components is the minimum possible value of the
reliability of the machine.
1 2 n
s
s 1 2 n
That is let the reliablity of n components are R ,R ,...,R .
And the reliability of the system is R .
Then, R max{R ,R ,...,R }.≥
Now the reliability of the system we have to find the probability that at least one system will work
at that particular time. Which will be, 1 2 nP(A A ... A )U U U .
In another sense the machine will fail if all the components fail simultaneously. That probability is
1 2 nQ Q ... Q× × × . Hence the reliability of the system will be, s 1 2 nR =1-Q Q ... Q× × × .
5. PARALLEL SYSTEM VERSUS SERIES SYSTEM :
Now we will see that the parallel system is more reliable than series system.
We can show this by 2 ways,
1> By showing, 1 2 1 2 n(1 ) (1 ) ... (1 ) 1-Q Q ... QnQ Q Q− × − × × − ≤ × × ×
2> Or,
As, the events are independent then (2) becomes, 1 2 n 1 2 n(A ) (A )...P(A ) P(A A ... A )P P ≤ U U U .
For (1) we can prove like below.
(2) can be proved as,
1 2 1 2
1 2 1 2 1 2 1 2 1 2 1 2
1 2
1 2
, ( ) : (1- )(1- )...(1- ) 1- ...
(1) .
, (1- )(1- ) 1- - 1 -
1-
, (2) .
( )
(1- )(1- )
n nLet P n Q Q Q Q Q Q
Then P is true
Now Q Q Q Q Q Q Q Q Q Q Q Q
Q Q
So P is true
Let P m is true
Then Q Q
≤
= + ≤ − +
=
1 1 2 1
1 1 2 1 2 1
1 2 1
...(1- )(1- ) (1- ... )(1- )
[ sin ( )]
1- - ... ...
1 ...
( )
m m m m
m m m m
m m
Q Q Q Q Q Q
u g P m
Q Q Q Q Q Q Q Q
Q Q Q Q
Hence P m is tr
+ +
+ +
+
≤
≤ +
≤ −
( ) . (1), (2) .
( )
.
ue if P m is true Again P P are true
Hence by principle of mathematical induction P n is true for
all n N∈
1 2 1 2( ... ) ( ... )n nP A A A P A A A∩ ∩ ∩ ≤ U U U

5. 5. Redundancy Allocation
1 2 1 2
1 2 1 2 1 2
1 2 1 2 1 2
( ) : ( ) ( )... ( ) ( ... )
.
(1) .
2;
( ) ( ) ( ) - ( ) ( )
( ) ( ) ( ) ( ) - ( ) ( )
n nLet P n P A P A P A P A A A
be the proposition
Then P is true
Now for n
P A A P A P A P A P A
P A P A P A P A P A P A
≤ + + +
=
+ = +
≥ +
1 2
1 2 1
1 1 1 1
1 2 1
( ) ( )
( ) .
( ... )
( ... ) ( ) - ( ... )
( ) ( ).... ( )
( 1) ( ) , (1)
(2)
m m
m m m m
m
P A P A
Let P m also holds
Now P A A A A
P A A P A P A A A
P A P A P A
Hence P m is true if P m is true again P and
P is tru
+
+ +
+
=
+ + + +
≥ +
≥
+
,
( )
e so by principle of mathematical induction
P n is true for all n N∈
Again (2) can be prove form the image below,
The graph below tells us about the reliability of series, parallel and normal system.
fig:4
1 2 1 1 2( ... ) ( ) ( ... )n nP A A A P A P A A A∩ ∩ ∩ ≤ ≤ U U U

6. 6. Redundancy Allocation
6. STAND-BY SYSTEM:
Sometime it is difficult to run all the components in parallel system together. In this system the
other components in parallel mode waits until a particular component break down. For example
there is n+ 1 components, then the system will fail if only n+1 components fails together.
As we know that R(t)+Q(t)=1
Again,
2
( )
1 (1 ...)
2
t t t t
e e e tλ λ λ λ
λ− −
= = + + +
Then,
2
( )
( ) ( ) 1 (1 ...)
2
t t t t
R t Q t e e e tλ λ λ λ
λ− −
+ = = = + + +
If there is n equal components are supporting in stand-by mode then reliability is,
2
( ) ( )
( ) (1 ... )
2 !
n
t t t
R t e t
n
λ λ λ
λ−
= + + + +
And MTBF =
0
( )R t
∞
∫
7. K-OUT OF N SYSTEM:
A ‘k out of n’ system is the special case of parallel redundancy. It succeeds if at least k components
out of n parallel components work properly. He diagram is given below:
fig:5

7. 7. Redundancy Allocation
The reliability of this kind of system is derived from the binomial (n,p) distribution as they are iid.
The reliability is,
n
r=k
( , ) n i n i
s rR n k C p q −
= ∑
Here p=R, then this becomes,
n
r=k
( , ) (1 )n i n i
s rR n k C R R −
= −∑ , this holds when all components have
same reliability.
8. RELIABILITY IN AIRBORNE SYSTEM:
This is an example how redundancy is necessary in case of a complex system. We will discuss the
reliability of an airborne system. There are few factors in airbus maintenance system to run that
smoothly. Like the machinery product works properly or not, the maintenance of technical team is
proper or not, the pilots and ATC6
control the plane properly or not. Based on these propositions the
reliability of the system can be computed as,
Reliability = machinery reliability × control reliability × maintenance reliability.
In this kind of cases we need very high reliability. So we have to increase these individual
reliabilities as much as possible or very close to 1.
6 ATC: air traffic control
fig:6

8. 8. Redundancy Allocation
For that we will see our first case of reliability that is machinery reliability. That depends on few
factors like, airbus structure, emergency controls, emergency propulsion, landing gear, emergency
power, navigational instruments etc. hence the machinery reliability is the product of all these
reliability. Now to increase the individual reliability as well as system reliability we have to use
redundant components at possible places. For example let us consider the partial reliability
sR '=R(emergency propulsion) R(electric power)×
Let us consider the parallel model for this partial reliability,
Let the reliability of emergency propulsion is 0.9999984 and let the failure rate of the electric power
be 0.00049. Then for using the parallel system we have the reliability for 10hour flight is,
2 .00049 10 2
s e pR '=1-(1-R R ) 1 (1 .9999984 ) .999964e− ×
= − − × =
9. CONCLUSION AND FUTURE WORK:
Redundancy is very much useful in our daily life to increase reliability as well as availability. For
example in our IIT we use several proxy server, if by chance one fails then we can use another.
Several server systems store their data in more than one place to increase the availability,
operational speed and to save their self from the data loss situation. But redundancy does not come
in free of cost. Sometimes it costs huge. I want to analyze that how the maximum reliability can be
achieved using optimum cost. The designing part of a particular system for specific configuration
and reliability with optimal cost will also be my point of view.
fig:7

9. 9. Redundancy Allocation
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Application (pp. 283-290, 309-324). Englewood: Prectice-Hall, ISBN: 0-13-711564-4.
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0-486-43867-8.
 Zuo, M. J., Huang, J. and Kuo, W (2002). Multi-State k-out-of-n Systems. London:
Springer-Verlag.
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