formulas for reliability, Bathtub hazard rate curve, MTTF ,reliability networks, series network, parallel network, Hybrid network, reliability evaluation technique, network reduction approach, decomposition approach, parts count method, Markov Method

1. 1
REPAIRABLE AND NON-REPAIRABLE ITEMS
For a Non-repairable item such as a light bulb, a transistor, a rocket
motor or an unmanned spacecraft, reliability is the survival
probability over the item’s expected life, or for a period during its
life, when only one failure can occur.
During the item’s life the instantaneous probability of the first and
only failure is called the hazard rate.
Life values such as the mean life or mean time to failure (MTTF)
are other reliability characteristics that can be used.
When a part fails in a non-repairable system, the system fails
(usually) and system reliability is, therefore, a function of the time
to the first part failure.

2. For items which are repaired when they fail, reliability is the
probability that failure will not occur in the period of interest,
when more than one failure can occur. It can also be expressed as
the failure rate.
However, the failure rate expresses the instantaneous probability
of failure per unit time, when several failures can occur in a time
continuum.
Repairable system reliability can also be characterized by the mean
time between failures (MTBF), but only under the particular
condition of a constant failure rate.
We are also concerned with the availability of repairable items,
since repair takes time. Availability is affected by the rate of
occurrence of failures (failure rate) and by maintenance time.
2

3. Chapter-2 Reliability of Systems
GENERAL RELIABILITY ANALYSIS RELATED FORMULAS
There are a number of formulas often used in conducting reliability
analysis. This section presents four of these formulas based on the
reliability function.
Failure density function: This is defined by dRt/dt=-f (t )...(2)
where: R(t) is the item reliability at time t, f(t) is the failure (or
3
probability) density function.
Hazard rate function: This is expressed by λ(t)=f(t)/R(t)…(3)
where: λ(t) is the item hazard rate or time dependent failure rate.
Substituting Equation (2) into Equation (3) yields
λ(t)= - 1/R(t)x d R(t)/dt …(4)
General reliability function: This can be obtained by using Equation (4).
Thus, we have 1/R(t) x dR(t)=- λ (t)dt …..(5)
Integrating both sides of Equation (5) over the time interval [o, t], we get
1 t
  
dR(t) (t)dt...(6)
since at t = 0, R (t) = 1.
R(t)
0
R( t )
1

4. GENERAL RELIABILITY ANALYSIS RELATED FORMULAS
Evaluating the left-hand side of Equation (6) yields
t
  
From Equation (7), we get
t
 ( t )dt
The above equation is the general expression for the
reliability function. Thus, it can be used to obtain
reliability of an item when its times to failure follow any
known statistical distribution, for example, exponential,
Rayleigh,Weibull, and gamma distributions.
4
ln R(t) (t)dt...(7)
0 
R(t) 
e 0
...(8)
 

5. GENERAL RELIABILITY ANALYSIS RELATED FORMULAS
Mean time to failure: This can be obtained by using any of the
following three formulas:
MTTF  E t 
tf t dt
( ) ( ) ...(9)
or
MTTF R t dt
( ) .............(10)

or
where:
MTTF is the item mean time to failure,
E(t) is the expected value,
s is the Laplace transform variable,
R(s) is the Laplace transform for the reliability function, R (t).
is the failure rate
5
...(11)
1
( )
0
0
0

 





MTTF LimitR s
s


6. Mean time between failure MTBF
where MTBF stands for mean operating time between failures.
MTBF should be confined to the case of repairable items with
constant failure rate
6
GENERAL RELIABILITY ANALYSIS RELATED FORMULAS
1

MTBF
is the failure rate 

7. Review Questions:
• Define the following terms: Reliability, Failure, Downtime, Maintainability,
Redundancy, Active redundancy, Availability, Mean time to failure (exponential
distribution, Useful life, Mission time, Human error, Human reliability.
• Discuss the need for reliability.
• Draw the bathtub hazard rate curve and discuss its three important regions.
7

8. Bathtub Hazard Rate Curve
• Bathtub hazard rate curve is a well known concept to
represent failure behavior of various engineering
items/products because the failure rate of these items
changes with time. Its name stem from its shape resembling a
bathtub as shown in Figure 1. Three distinct regions of the
curve are identified in the figure: burn-in region(early
failures), useful life region, and wear-out region. These
regions denote three phases that a newly manufactured
product passes through during its life span.
• During the burn-in region/period, the product hazard rate
(i.e., time dependent failure rate) decreases and some of the
reasons for the occurrence of failures during this period are
poor workmanship, substandard parts and materials, poor
quality control, poor manufacturing methods, …….
8

9. incorrect installation and start-up human error, inadequate
debugging, incorrect packaging, inadequate processes, and
poor handling methods. Other names used for the “burn-in
region” are “debugging region,” “infant mortality region,” and
“break-in region.”
• During the useful life region, the product hazard rate remains
constant and the failures occur randomly or unpredictably.
Some of the reasons for their occurrence are undetectable
defects, abuse, low safety factors, higher random stress than
expected, unavoidable conditions, and human errors.
• During the wear-out region, the product hazard rate increases
and some of the reasons for the occurrence of “wear-out
region” failures are as follows: Poor maintenance, Wear due to
friction, Wear due to aging, Corrosion and creep, Wrong
overhaul practices, and Short designed-in life of the product.
9

10. Figure 1: Bathtub hazard rate curve.
10

11. 11

12. Example 1 :
• Assume that a railway engine’s constant failure rate λ is 0.0002
failures per hour. Calculate the engine’s mean time to failure.
1
1
MTTF  
Thus, the railway engine’s expected time to failure is 5000 h.
• Assume that the failure rate of an automobile is 0.0004 failures/h.
Calculate the automobile reliability for a 15-h mission and mean
time to failure.
Using the given data in Equation
R t e
( ) ...(8)
12
5000h
0.0002
λ
(0.0004)(15)

e
0.994
( )
0







e
t
t dt
t


Example 2 :

13. 13
Similarly, inserting the specified data for the automobile failure
rate into Equation MTTF, we get
MTTF R t dt
( ) .............(10)

0

MTTF 
e dt


MTTF e dt
h
t
t
1
0.0004
2,500
..
..
0
(0.0004)
0









Thus, the reliability and mean time to failure of the automobile
are 0.994 and 2,500 h, respectively.

14. Reliability Networks
An engineering system can form various different configurations in
conducting reliability analysis. If the reliability factor or the
probability of failure of the system is to be determined, we will find
that it is very difficult to analyze the system as a whole.
The failure of the system as a whole can be attributed to the failure
of one or more components of the system not functioning in the
stipulated manner.
Depending on the type in which the sub-system and elements are
connected to constitute the given system, the combinatorial rules of
probability are applied to obtain the system reliability.
14

15. Series Network
• Each block in the diagram represents a unit/component.
• Diagram represents a system with m number of units acting in
series.
• If any one of the units fails, the system fails.
• In other words, all units must operate normally for the systems
success.
• The reliability of series systems network is expressed by:
where,
Rs=series system reliability or probability of success,
xi=event denoting the success of unit i, for i=1,2,3,…,m and
P(x1,x2,x3,..xm)=probability of occurrence of events x1,x2,x3,…,xm
15
( ... )......... .(1) s 1 2 3 m R  P x x x x

16. Series Network Diagram
16

17. For independently failing units, eq. (1) becomes
where P(x) is the occurrence probability of event xi, for i=1,2,3,…,m
If we let Ri=P(xi) in eq. (2) it becomes:
m
i s 
where Ri= is the unit i reliability, for i=1,2,3,…,m.
For Ri>0.95 in eq. (3), system reliability Rs can be approximated by
using eq.
m
  
s i 
For identical units (i.e., Ri=R) eq. (4) becomes
where, R is the unit reliability.
17
R P(x )P(x )P(x )......... .P(x )......( 2) s 1 2 3 m 
R R ......( 3)
i 1


R 1 (1 R )......( 4)
i 1

R 1 m(1 R)......( 5) s   

18. Example 1:
• Assume an automobile has four independent and identical tires.
The tire reliability is 0.97. If any one of the tries is punctured, the
automobile cannot be driven. Calculate the automobile reliability
with respect to tires by using eq. (3) and eq. (5). Comment on the
end result.
Similarly, using the given data values in eq. (5) yields:
• Both the above reliability results are very close. More specifically,
the system reliability value obtained through using eq. (5) is lower
than when the exact eq. (3) was used.
18
R (0.97)(0.97)(0.97)(0.97) 0.8853 s  
R 1 4(1 0.97) 0.88 s    

19. Parallel Network
• This is a widely used network and it represents a system with m units
operating simultaneously. At least one unit must operate normally for
the system success.
• Each block in the diagram denotes a unit. The failure probability of
the parallel system/network is given by:
where: Fp=failure probability of the parallel system,
= event denoting the failure of unit i; for i=1,2,3,…,m
=probability of occurrence of events
i x
P(x1 x2 x3...... xm )
x1 x2 x3...... xm
For independently failing units, eq. (6) becomes
where: is the probability of occurrence of failure event xi , for
i=1,2,3,…,m
19
Fp P(x1x2x3......xm) ...(6) 
Fp Px1x2 x3 ......xm  ...(7) 
P(x1)

20. Parallel Network Diagram
20

21. Parallel Network
• If we let Fi=P(xi) in eq. (7) it yields:
where: Fi is the failure probability of unit i for i=1,2,3,….,m
Subtracting eq. (8) from unity yields the following expression for
parallel network reliability:
i p 
m
i p p 
where Rp is the parallel system reliability.
For identical units, eq. (9) becomes
m
where: F is the unit failure probability.
Since R+F=1, eq. (10) is rewritten to the following form:
where: R is the unit reliability.
21
F F ...(8)
i 1


R 1 F 1 F ...(9)
i 1

   
R 1 F ...(10) m
p  
R 1 (1 R) ...(11) m
p   

22. The plots of eq. (11) shown in Figure 1 (parallel system reliability
plots) clearly demonstrates that as the unit reliability and the
number of redundant units increase, the parallel system
reliability increases accordingly.
22

23. Example 2:
• A computer has two independent and identical Central
Processing units (CPUs) operating simultaneously. At least
one CPU must operate normally for the computer to
function successfully. If the CPU reliability is 0.96, calculate
the computer reliability with respect to CPUs.
• By substituting the specified data values into eq. (11), we get
• Thus, the computer reliability with respect to CPUs is 0.9984.
23
R 1 (1 0.96) 0.9984 2
p    

24. Series-Parallel Network
This network represents a system having m number of subsystems in
series. In turn, each subsystem contains k number of active (i.e.,
operating) units in parallel. If any one of the subsystems fails, the
system fails. Each block diagram in the diagram represents a unit.
Figure 2 (below) shows series-parallel network/system.
24

25. For independent units, using eq. (9) we write the following equation
for ith subsystem’s reliability,Figure 2 .
k
ij pi 
where Rpi is the reliability of the parallel subsystem i and Fij is the ith
subsystem’s jth unit’s failure probability.
Substituting eq. (13) into eq. (3) yields the following expression for
series-parallel network/system reliability:
m

 
sp   ij
 

where Rsp is the series-parallel network/system reliability.
For identical units, eq. (14) becomes (where R is the unit reliability)
Where F is the unit failure probability. Since R+F=1, eq. (15) is
rewritten to the following form:
25
R 1 F ...(13)
j 1
 
R 1 F ...(14)
i 1
k
j 1






R 1 F k  m
...(15)
sp  
m
R  1  1 
R k  ...(16)
sp

26. For R=0.8, the plots of eq. (16) are shown in Figure 3 (below).
These plots indicate that as the number of subsystems m
increase, the system reliability decreases, accordingly. On the
other hand, as the number of units k increases, the system
reliability also increases.
26

27. Example 3:
• Assume that a system has four active, independent, and
identical units forming a series-parallel configuration (i.e.,
k=2, m=2). Each unit’s reliability is 0.94. Calculate the system
reliability.
• By substituting the given data values into eq. (16) yields:
• Thus, the system reliability is 0.9928.
27
2
R  1  1  0.94 2  
0.9928
sp

28. Parallel-Series Network
• This network represents a system having m number of subsystems
in parallel. In turn, each subsystem contains k number of active (i.e.,
operating) units in series. At least one subsystem must function
normally for the system success. The network/system block diagram
is shown in Figure 4. Each block in the diagram denotes a unit.
• For independent and identical units, using eq. (3), we get the
following equation for the i ’th subsystems reliability, in Figure 4 :
k
ij si 
where Rsi is the reliability of the series subsystem i and Rij is the ith
subsystems jth units reliability. By subtracting eq. (17) from unity, we
get
k
ij si si 
where Fsi is the failure probability of the series subsystem i.
28
R R ...(17)
j 1


F 1 R 1 R ...(18)
j 1
   

29. 29
Figure 4 Parallel-series network system.

30. Using eq. (18) in eq. (9) yields:

 
 

R   
R
ps ij where Rps is the parallel-series network/system reliability. For
identical units eq. (19) simplifies to
where R is the unit reliability.
30
1 1 ...(19)
1 1
 

 

m
i
k
j
m
1 1 k  ...(20)
ps R    R

31. For R=0.8, the plots the eq. (20) are shown in Figure (below). The
plots show that as the number of units k increases, the
system/network reliability decreases accordingly. On the other
hand, as the number of subsystems m increases, the system
reliability also increases.
31

32. Example 4:
A system is composed of four active, independent, and
identical units forming a parallel-series configuration
(i.e., k=m=2). Calculate the system reliability, if each
units reliability is 0.94.
By substituting the given data into eq. (20), we get
Thus, parallel-series system reliability is 0.9865.
32
2
R  1  1  0.94 2  
0.9865
ps

33. Review Questions:
• Compare series and parallel networks.
• Compare series-parallel and parallel-series networks.
• Prove the reliability of a series and parallel network/system.
• Prove the reliability of a parallel-series network.
• A system has three independent, identical, and active units. At
least two units must operate normally for the system success. The
reliability of each unit is 0.91. Calculate the system reliability.
• An aircraft has four active, independent, and identical engines. At
35000 ft above ground at least one engine must operate
normally for the aircraft to fly successfully. Calculate the reliability
of the aircraft flying at 35000 ft, if the engine probability of failure
is 0.05.
• Assume that an automobile has four independent and identical
tires. The tire reliability is 0.93. If any one of the tires is
punctured, the automobile cannot be driven. Calculate the
automobile reliability with respect to tires.
33

34. 34
Reliability Allocation
The process by which the failure allowance for a system
is allocated in some logical manner among its sub-systems
and elements is termed as reliability allocation.
Reliability allocation may simply be described as the
process of assigning reliability requirements to
individual parts or components to achieve the specified
system reliability.

35. The reliability allocation problem is bit complex and not straight
forward.
Some of the associated reasons are as follows:
• Role the component plays for the operation of the system.
• Component complexity.
• The chargeable or assignable component reliability with the
type of function to be conducted.
• Approaches available for accomplishing the given allocation
task.
• Lack of detailed information on many of the above factors in
the early design stage.
35

36. Nonetheless, there are many benefits of the reliability
allocation because
it forces individuals involved in design and
development to clearly understand and develop the
relationships between reliabilities of components,
subsystems, and systems,
it forces the design engineer to seriously consider
reliability equally with other design parameters such
as performance, weight, and cost, and it ensures
satisfactory design, manufacturing approaches, and
test methods.
36

37. 37
Two reliability allocation methods are described as follows:
(1) HYBRID METHOD: This method is the result of combining
two approaches: similar familiar systems and factors of
influence. The resulting method incorporates benefits of these
two methods.
The basis for the similar familiar systems reliability allocation
approach is the familiarity of the designer with similar systems
or sub-systems. In addition, failure data collected on similar
systems from various sources can also be used during the
allocation process.
The main drawback of this approach is to assume that
reliability and life cycle cost of previous similar designs were
adequate.

38. The factors of influence method is based upon the following factors that are
considered to effect the system reliability:
• Complexity/Time: The complexity relates to the number of subsystem
parts and the time to the relative operating time of the item during the
entire system functional period.
• Failure criticality: This factor considers the criticality of the item failure on
the system. For example, some auxiliary instrument failure in an aircraft
may not be as critical as the engine failure.
• Environment: This factor takes into consideration the susceptibility or
exposure of items to environmental conditions such as temperature,
humidity, and vibration.
• State-of-the-Art: This factor takes into consideration the advancement in
the state-of-the-art for a specific item.
In using the above influence factors, each item is rated with respect to each
of these influence factors by assigning a number from 1 to 10. The
assignment of 1 means the item under consideration is least affected by
the factor in question and 10 means the item is most affected by the same
influence factor. Ultimately, the reliability is allocated by using the weight
of these assigned numbers for all influence factors considered.
38

39. (2) FAILURE RATE AllOCATION METHOD: This method is concerned
with allocating failure rates to system components when the
system required failure rate is known. The following assumptions
are associated with this method:
• System components form a series configuration.
• System components fail independently.
• Time to component failure is exponentially distributed.
Thus, the system failure rate is


s i  
where: n is the number of components in the system.
λs is the system failure rate.
λi is the failure rate of system component I; for i=1, 2, 3, …,n.
If the system required failure rate is λsr, then allocation component
failure rate such that:
39
...(1.3)
1

n
i

40. n

   i 
1
*
where: λ*
...(1.4)
sr
i is the failure rate allocated to component i; for i=1, 2, 3,…,n.
The following steps are associated with this method:
1. Estimate the component failure rates λi for i=1, 2, 3, …,n, using the
past data.
2. Calculate the relative weight, θi , of component i using the preceding
step failure rate data and the following equation:

for i n n
It is to be noted that θi denotes the relative failure
vulnerability of the component i and
n
 

3. Allocate failure rate to component i using the following relationship:
λ*
i = θi λsr , for i=1, 2, …,n …(1.7)
It must be remembered that eq. (1.7) is subject to the condition that the equality
holds in eq. (1.4). 40
i
, 1,2,..., ...(1.5)
1
i
i
i
i  




1 ...(1.6)
1
i
i 

41. Example: Assume that a military system is composed of five
independent subsystems in series and its specified failure rate is
0.0006 failures/h. The estimated failure rates from past experience for
subsystems 1, 2, 3, 4, and 5 are λ1=0.0001 failures/h, λ2=0.0002
failures/h., λ3=0.0003 failures/h., λ4=0.0004 failures/h., and λ5=0.0005
failures/h. respectively. Allocate the specified system failure rate to
five subsystems.
Answer: Using eq. (1.3) and the given data, we get the following
estimated military system failure rate:
5
s i       
λ λ (0.0001) (0.0002 ) (0.0003 ) (0.0004 ) (0.0005 ) 0.0015 failures / h
i 1

Thus, utilizing eq. (1.5) and calculated and given values, we get the
following relative weights for subsystems 1, 2, 3, 4, and 5,
respectively:
θ1=(0.0001÷0.0015)=0.0667, θ2=(0.0002÷0.0015)=0.1333,
θ3=(0.0003÷0.0015)=0.2, θ4=(0.0004÷0.0015)=0.2667,
θ5=(0.0005÷0.0015)=0.3333 41

42. Using eq. (1.7) and calculated and given values, the subsystems 1,
2, 3, 4, and 5 allocated failure rates, respectively, are as follows:
λ*
1=θ1 λsr =(0.0667)(0.0006) =0.00004 failures/h
λ*
2=θ2 λsr =(0.1333)(0.0006) =0.00007 failures/h
λ*
3=θ3 λsr =(0.2)(0.0006) =0.00012 failures/h
λ*
4=θ4 λsr =(0.2667)(0.0006) =0.00016 failures/h
λ*
5=θ5 λsr =(0.333)(0.0006) =0.00019 failures/h
42

43. SAME LIKE ABOVE PROBLEM
Problem:
An aerospace system is made up of seven independent
subsystems in series and it specified failure rate 0.009 failures/h.
Subsystems 1, 2, 3, 4, 5, 6, and 7 estimated failure rates from
previous experience are 0.001 failures/h, 0.002 failures/h, 0.003
failures/h, 0.004 failures/h, 0.005 failures/h, 0.006 failures/h, and
0.007 failures/h, respectively. Allocate the specified system failure
rate to seven subsystems.
43

44. Reliability Evaluation Methods
• Introduction: Reliability evaluation is an important activity for
ensuring the reliability of engineering products. It normally
begins right from the conceptual design stage of products with
specified reliability. Over the years, many reliability evaluation
methods and techniques have been developed.
• Some examples of these methods and techniques are fault tree
analysis (FTA), failure modes and effect analysis (FMEA),
Markov method, network reduction method, and
decomposition method.
• The use of these methods for a particular application depends
on various factors including the specified requirement, the type
of project under consideration, the specific need, and the
inclination of the parties involved. For example, FMEA is often
required in aerospace/defense related projects and FTA in
nuclear power generation projects.
44

45. The ease of use and the requirement of specific experience of
users (analysts) may vary from one method to another. For
example, in the real world application the network reduction
method is probably the easiest to use and it does not really
require any specific experience from its users.
In contrast, FMEA and FTA are relatively more demanding to
perform and require considerable experience of analysts in the
area of design.
Different types of reliability evaluation methods are discussed
below.

46. Network Reduction Method
• This is probably the simplest method for evaluating the reliability of systems
composed of independent series and parallel subsystems. It sequentially
reduces the parallel and series subsystems to equivalent hypothetical single
units until the complete system itself becomes a single hypothetical unit. The
bridge configurations or subsystems (if any) in the system can be converted to
series and parallel equivalents by using delta-star conversions or the
decomposition method.
• The main advantage of this approach is that it is easy to understand and apply.
The method is demonstrated through the following example:
Example 1: A network representing an engineering system with independent
units is shown in Figure (1). Each block in the figure denotes a unit. The
reliability RJ of unit j, for j = 1, 2, 3, ... , 7 is given. Determine the network
reliability by using the network reduction method.
• First we have identified subsystems A, B, and C of the network as shown in
Figure 1(i). The subsystem A has three units in series; thus we reduce them to a
single hypothetical unit as follows:
RA(reliability of subsystem A)=R1R2R3=(0.5)(0.6)(0.7)=0.21
Thus, subsystem A has been reduced to a single hypothetical unit having
reliability 0.21. 46

47. Network Reduction Method…
• The reduced network is shown in Figure 1(ii). Now, this
network is made up of two parallel subsystems B and C acting in series.
Thus, we reduce subsystem B to a single hypothetical unit as follows:
RB(reliability of subsystem B)=1-(1-R5)(1-R6)(1-R7)=1-(1-0.7)(1-0.8)(1-0.9)=0.994
Thus, subsystem B has been reduced to a single hypothetical unit having
reliability 0.994. The reduced network is shown in Figure 1(iii). This net-work
contains subsystem C and a hypothetical unit, representing subsystem
B, in series. In similar manner to subsystem B, we reduce subsystem C to a
single hypothetical unit:
RC(reliability of subsystem C)=1-(1-RA)(1-R4)=1-(1-0.21)(1-0.8)=0.842
Similarly, the reduced network is shown in Figure 1(iv). This network is
composed of two hypothetical units, representing subsystems B and C, in
series. The reliability of this network is given by
Rn=RCRC=(0.994)(0.842)=0.8369
where Rn is the reliability of the whole network shown in Figure 1 (i). All in
all, by using the network reduction method, the Figure 1(i) network was
reduced to a single hypothetical unit having reliability 0.8369 (Figure 1(v));
which is the whole network's reliability. 47

48. Figure 1: Diagrammatic steps of the network reduction: (i) original network;
(ii) reduced network; (iii) reduced network; (iv) reduced network; (v) single
hypothetical unit.
48

49. Decomposition Method
This method is used to evaluate reliability of complex systems. It
decomposes complex systems into simpler subsystems by
applying the conditional probability measures of subsystems.
The method begins by first selecting the key element or unit to be
used to decompose a given network/system. The poor choice of
this key element leads to poor efficiency of computing system
reliability. Nonetheless, the past experience usually plays an
instrumental role in selecting the right key element.
First, the method assumes that the key element/unit, say x, is
replaced by another element that never fails (i.e., 100% reliable)
and then it assumes that the key element is 100% unreliable (i.e.,
it is completely removed from the system or network). Under this
scenario, the overall system/network reliability is given by
Rs=P(x)P(system good/x good)+P( )P(system good/ x fails) …(10)
49
X

50. Decomposition Method…
where: Rs=system reliability
P(system good/x good)=reliability of the system when x is
100% reliable.
P(system good/ x fails)=reliability of the system when x is
100% unreliable
P(x)=reliability of the key element x
P( )=unreliability of the key element x
Similarly, the overall system/network unreliability is expressed by:
URs=P(x)P(system fails/x good)+P( )P(system fails/x fails)
where: URs=system unreliability
P(system fails/ x good)=unreliability of the system when x is 100%
reliable
P(system fails/x fails)=unreliability of the system when x is 100%
unreliable
50
X
X

51. Example5: A five independent unit bridge network is shown in Figure 6.
Each block in the diagram denotes a unit and each unit’s reliability is
denoted by Ri, for i=1,2,3,…,5. Develop an expression for the network by
utilizing the decomposition method.
• First of all, in this example
we identify the Figure 6 unit
with reliability R3 as our key
element, say x. thus, by
replacing the key element in
Figure 6 with 100% reliable
unit and then with 100%
unreliable unit results in
Figure 7(a) and Figure 7(b)
diagrams, respectively.
51

52. Figure 7: Reduced networks of Figure 6 diagram: (a) For a 100% reliable key
element, (b) For 100% unreliable key element.
Using the network reduction method, we obtain the following reliability
expression for Figure 7(a):
Rsp=[1-(1-R1)(1-R4)][1-(1-R2)(1-R5)] …..(12)
where: Rsp is the series=parallel network reliability (i.e., the system reliability
when the key element is 100% reliable)
For identical units (i.e., R1=R2=R4=R5=R) eq. (12) becomes
Rsp=[1-(1-R)2]2=(2R-R2)2 …..(13)
where: R is the unit reliability. Similarly, by utilizing the network reduction
approach, we get the following reliability expression for Figure 7(b):
Rps=1-(1-R1R2)(1-R4R5) …..(14) 52

53. where: Rps is the parallel-series network reliability (i.e., the system reliability
when the key element is 100% unreliable).
For identical units, eq. (14) becomes:
Rps=1-(1-R)2=2R2-R4 ...(15)
The reliability and unreliability of the key element x, respectively, are given by:
P(x)=R3 …(16) and P( X
)=1-R3 …(17)
For R3=R, eq. (16) and eq. (17) become:
P(x)=R …(18) and P( )=(1-R) …(19)
Substituting eq. (12), eq. (14), eq. (16), and eq. (17) into eq. (10) yields:
Rs=R3[1-(1-R1)(1-R4)][1-(1-R2)(1-R5)]+(1-R3)[1-(1-R1R2)(1-R4R5)]…(20)
For identical units, inserting eq. (13), eq. (15), eq. (18) and eq. (19) into eq. (10),
we get:
Rs=R(2R-R2)2+(1-R)(2R2-R4)=2R2+2R3-5R4+2R5 …(21)
Thus, eq. (20) and eq. (21) are reliability expressions for Figure 6 network with
non-identical and identical units, respectively.
53
X

54. Delta-Star Method
• This is the simplest and very practical approach to evaluate reliability of
bridge networks. This technique transforms a bridge network to its
equivalent series and parallel form. However, the transformation process
introduces a small error in the end result, but for practical purposes it
should be neglected.
• Once a bridge network is transformed to its equivalent parallel and series
form, the network reduction approach can be applied to obtain network
reliability. The delta-star method can easily handle networks containing
more than one bridge configurations. Furthermore, it can be applied to
bridge networks composed of devices having two mutually exclusive
failure modes.
• Figure 8 shows delta to star equivalent reliability diagram. The numbers
1,2, and 3 denote nodes, the blocks the units, and R(.) the respective unit
reliability.
• In Figure 8, it is assumed that three units of a system with reliabilities R12,
R13, and R23 form the delta configuration and its star equivalent
configuration units' reliabilities are R1, R2, and R3.
• Using Equations (3) and (9) and Figure 8, we write down the following
equivalent reliability equations for network reliability between nodes 1, 2;
2, 3; and I, 3, respectively:
54

55. Figure 8. Delta to star equivalent reliability diagram.
55

56. R1R2=1-(1-R12)(1-R13R23) …(49)
R2R3=1-(1-R23)(1-R12R13) …(50)
R1R3=1-(1-R13)(1-R12R23) …(51)
Solving eqs. (49) through (51), we get
AC
where:
A=1-(1-R12)(1-R13R23) …(53)
B=1-(1-R23)(1-R12R13) …(54)
C=1-(1-R13)(1-R12R23) …(55)
56
...(52)
B
R1 
...(57)
AB
BC
A
R
...(56)
C
R
2
3



57. Example: A five independent unit bridge network with
specified unit reliability Ri; for i=a, b, c, d, and e is shown in
Figure 9. Calculate the network reliability by using the delta-star
method and also use the specified data in eq. (3) and (9) to
obtain the bridge network reliability. Compare both results.
Figure 9. A five unit bridge network with specified unit reliabilities.
57

58. In Figure 9 nodes labeled 1, 2, and 3 denote delta configurations.
Using eqs. (52), (56) and (57) and the given data, we get the
following star equivalent reliabilities:
AC
R1  
where: A=B=C=1-(1-0.8)[1-(0.8)(0.8)]=0.9280
R2=0.9633 and R3=0.9633
Using the above results, the equivalent network to Figure 9 bridge
network is shown in Figure 10.
The reliability of Figure 10 network, Rbr, is
Rbr=R3[1-(1-R1Rd )(1-R2Re)]=0.9126
By substituting the given data into eq. (21), we get
Rbr=2(0.8)5-5(0.8)4+2(0.8)3+2(0.8)2=0.9114
Both the reliability results are basically same, i.e., 0.9126 and
0.9114. All in all, for practical purposes the delta-star approach
is quite effective
58
0.9633
B

59. Figure 10. Equivalent network to bridge configuration of Figure 9.
59

60. Similar to above problem
• Calculate the reliability of the Figure A network using the delta-star
approach. Assume that each block in the figure denotes a
60
unit with reliability 0.8 and all units fail independently.
Figure A

61. Parts Count Method
This is a very practically inclined method used during bid proposal and
early design phases to estimate equipment failure rate. The
information required to use this method includes generic part types
and quantities, part quality levels, and equipment use environment.
Under single use environment, the equipment failure rate can be
estimated by using the following equation:
m
λ Q λ F  ...(58)
g q i
i E 
i 1


where: λE is the equipment failure rate, expressed in failure/106h.
m is the number of different generic part/component classifications in
the equipment under consideration.
λg is the generic failure rate of generic part I expressed in failure/106h.
Qi is the quantity of generic part i.
Fq is the quality factor of generic part i.
(Note: From table you will get λg and Fq values) 61

62. Failure rate estimation of an electronic part:
As the design matures, more information becomes available, the
failure rates of equipment components are estimated. Usually, in
the case of electronic parts, the MIL-HDBK-217 is used to estimate
the failure rate of electronic parts. The failure rates are added to
obtain total equipment failure rate. This number provides a better
picture of the actual failure rate of the equipment under
consideration than the one obtained through using eq. (58).
An equation of the following form is used to estimate failure rates of
many electronic parts:
λp=λbθeθq……(59)
where: λp is the part failure rate.
λb is the base failure rate and is normally defined by a model
relating the influence of temperature and electrical stresses on the
part under consideration.
θe is the factor that accounts for the influence of environment.
θq is the factor that accounts for part quality level.
62

63. For many electronic parts, the base failure rate, λb, is
calculated by using the following equation:
E

 
λ Cexp b 
where:
C is a constant
E is the activation energy for the process.
K is the Boltzmann’s constant.
T is the absolute temperature.
63
...(60)
kT





64. Markov Method
• This is a widely used method in industry to perform various types of
reliability analysis. The method is named after a Russian mathematician,
Andrei Andreyevich Markov (1856-1922). Markov method is quite useful to
model systems with dependent failure and repair modes and is based on the
following assumptions:
• The probability of transition from one system state to another in
the finite time interval Δt is given by λΔt, where λ is the transition
rate (e.g., constant failure or repair rate of an item) from one system
state to another.
• The probability of more than one transition in time interval Δt from
one state to the next state is negligible (e.g., (λΔt) (λΔt)→0).
• The occurrences are independent of each other.
Example: An engineering system can either be in a working state or a failed
state. The system state space diagram is shown in Figure 5. The numerals in
boxes denote the system state. The system fails at a constant failure rate λ.
Develop expressions for system reliability, unreliability, and mean time to
failure.
64

65. With the aid of Markov method, we write down the following
equations for the Figure 5 diagram for state 0 and state 1,
respectively,:
P0(t+Δt) = P0(t)(1-λΔt) ….(3) and
P1(t+Δt) = P1(t)+(λΔt)P0(t)….(4)
where: Pi(t+Δt)=probability that at time (t+Δt) the system is in state i,
i=0 (working normally), i=1 (failed),
Pi(t)=probability that at time t the system is in state i, i=0 (working
normally) i=1 (failed),
λ=system constant failure rate,
λΔt=probability of system failure in finite time interval Δt,
(1-λΔt)=probability of no failure in time interval Δt when the
system is in state 0.
65

66. 66
In the limiting case, eq. (3) and eq. (4) become
  
Limit   
and
dP (t)
P (t t) P (t)
dP (t)
P (t t) P (t)
Limit   
At time t=0, P0(0)=1 and P1(0)=0
Solving eq. (5) and eq. (6), we get
P0(t)=Rs(t)=e –λt …(7)
P1(t)=URs(t)=(1-e –λt ) …(8)
λP t ...(5)
dt
t
0
0 0 0
t 0

 
P t ...(6)
dt
t
1
1 1 1
t 0
  
 

67. Figure 5. System state space diagram.
where Rs(t) is the system reliability at time t and URs (t) is the
system unreliability at time t. The system mean time to failure is
given by:
1
s s     
where MTTFs is the system mean time to failure. Thus, expressions
for system reliability, unreliability, and mean time to failure are
given by eq. (7), eq. (8), and eq. (9), respectively.
67
...(9)
λ
MTTF R (t)dt e dt
0
λt
0




68. Review Questions:
• Calculate the reliability of the
Figure A network using the
delta-star approach. Assume
that each block in the figure
denotes a unit with reliability
0.8 and all units fail
independently.
• An aerospace system is
made up of seven
independent subsystems in
series and it specified
failure rate 0.009
failures/h. Subsystems 1, 2,
3, 4, 5, 6, and 7 estimated
failure rates from previous
experience are 0.001
failures/h, 0.002 failures/h,
0.003 failures/h, 0.004
failures/h, 0.005 failures/h,
0.006 failures/h, and 0.007
failures/h, respectively.
Allocate the specified
system failure rate to seven
subsystems.
Figure A 68

69. 69

70. • A bridge network is composed of five independent and identical
units. The constant failure rate of a unit is 0.0005 failures/h. Calculate
the network reliability for a 300-h mission and mean time to failure.
Similar to above problem
70