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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
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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.
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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.
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Chapter-2 Reliability of SystemsGENERAL 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 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) yieldsλ(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
since at t = 0, R (t) = 1.
)6...(dt)t()t(dR)t(R
1 t
0
)t(R
1
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GENERAL RELIABILITY ANALYSIS RELATED FORMULAS
Evaluating the left-hand side of Equation (6) yields
From Equation (7), we get
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.
)7...(dt)t()t(Rlnt
0
)8...(e)t(R
t
0
dt)t(
5
GENERAL RELIABILITY ANALYSIS RELATED FORMULAS
Mean time to failure: This can be obtained by using any of the following three formulas:
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
)11...(1
)(
)10...(..........)(
)9...()()(
0
0
0
sRLimitMTTF
or
dttRMTTF
or
dtttftEMTTF
s
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Mean time between failure MTBFwhere MTBF stands for mean operating time between failures. MTBF should be confined to the case of repairable items with constant failure rate
GENERAL RELIABILITY ANALYSIS RELATED FORMULAS
1
MTBF
is the failure rate
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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.
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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, …….
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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.
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Figure 1: Bathtub hazard rate curve.
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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.
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
h50000002.0
1
λ
1MTTF
994.0
)8...()(
)15)(0004.0(
)(0
ee
etRt
dttt
Example 2 :
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Similarly, inserting the specified data for the automobile failure rate into Equation MTTF, we get
h
dteMTTF
dteMTTF
dttRMTTF
t
t
500,20004.0
1
..
..
)10...(..........)(
0
)0004.0(
0
0
Thus, the reliability and mean time to failure of the automobile are 0.994 and 2,500 h, respectively.
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Reliability NetworksAn 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.
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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
)1.()............( 321 ms xxxxPR
16
Series Network Diagram
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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:
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.
For identical units (i.e., Ri=R) eq. (4) becomes
where, R is the unit reliability.
)2)......(x(P.).........x(P)x(P)x(PR m321s
)3......(RRm
1iis
)4)......(R1(1Rm
1iis
)5)......(R1(m1R s
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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.
8853.0)97.0)(97.0)(97.0)(97.0(R s
88.0)97.01(41R s
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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
For independently failing units, eq. (6) becomes
where: is the probability of occurrence of failure event xi , for i=1,2,3,…,m
)6...()x......xxx(PF m321p
)7...(x......xxxPF m321p
ix
)......( 321 mxxxxPmxxxx ......321
)( 1xP
20
Parallel Network Diagram
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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:
where Rp is the parallel system reliability.
For identical units, eq. (9) becomes
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.
)8...(FFm
1iip
)9...(F1F1Rm
1iipp
)10...(F1R mp
)11...()R1(1R mp
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.
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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 0.9984.9984.0)96.01(1R 2
p
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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.
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For independent units, using eq. (9) we write the following equation for ith subsystem’s reliability,Figure 2 .
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:
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:
)13...(F1Rk
1jijpi
)14...(F1Rm
1i
k
1jijsp
)15...(F1Rmk
sp
)16...(R11Rmk
sp
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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.
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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.
9928.094.011R22
sp
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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 :
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
where Fsi is the failure probability of the series subsystem i.
)17...(RRk
1jijsi
)18...(R1R1Fk
1jijsisi
29
Figure 4 Parallel-series network system.
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Using eq. (18) in eq. (9) yields:
where Rps is the parallel-series network/system reliability. For identical units eq. (19) simplifies to
where R is the unit reliability.
)19...(111 1
m
i
k
jijps RR
)20...(11mk
ps RR
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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.
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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.
9865.094.011R22
ps
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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.
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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.
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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.
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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.
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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.
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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.
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(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
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:
)3.1...(1
n
iis
40
where: λ*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:
It is to be noted that θi denotes the relative failure vulnerability of the component i and
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).
)4.1...(1
*sr
n
ii
)5.1...(,...,2,1,
1
niforn
ii
ii
)6.1...(11
n
ii
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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:
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
h/failures0015.0)0005.0()0004.0()0003.0()0002.0()0001.0(λλ5
1iis
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
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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.
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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.
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.
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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.
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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.
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Figure 1: Diagrammatic steps of the network reduction: (i) original network; (ii) reduced network; (iii) reduced network; (iv) reduced network; (v) single hypothetical unit.
49
Decomposition MethodThis 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)X
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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% unreliableP(x)=reliability of the key element xP( )=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
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.
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)
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( )=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.
X
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:
55
Figure 8. Delta to star equivalent reliability diagram.
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
where: A=1-(1-R12)(1-R13R23) …(53)
B=1-(1-R23)(1-R12R13) …(54)
C=1-(1-R13)(1-R12R23) …(55)
)52...(B
ACR1
)57...(A
BCR
)56...(C
ABR
3
2
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.
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:
where: A=B=C=1-(1-0.8)[1-(0.8)(0.8)]=0.9280R2=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 getRbr=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
9633.0B
ACR1
59
Figure 10. Equivalent network to bridge configuration of Figure 9.
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 unit with reliability 0.8 and all units fail independently.
Figure A
61
Parts Count MethodThis 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:
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)
)58...(FλQλiqg
m
1iiE
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.
63
For many electronic parts, the base failure rate, λb, is calculated by using the following equation:
where: C is a constantE is the activation energy for the process.K is the Boltzmann’s constant.T is the absolute temperature.
)60...(kT
EexpCλb
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.
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.
66
In the limiting case, eq. (3) and eq. (4) become
and
At time t=0, P0(0)=1 and P1(0)=0
Solving eq. (5) and eq. (6), we getP0(t)=Rs(t)=e –λt …(7)
P1(t)=URs(t)=(1-e –λt ) …(8)
)5...(tPλdt
)t(dP
t
)t(P)tt(P0
000
0tLimit
)6...(tPdt
)t(dP
t
)t(P)tt(P1
111
0tLimit
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:
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.
)9...(λ
1dtedt)t(RMTTF
0
tλ
0
ss
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
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
