reliability 2013 1

EM 105 (B) Process Reliability Engineering

Module I

Definition of reliability key elements; failure analysis failure density failure rate probability of failure - bathtub curve - Basic reliability equations Reliability in terms of failure rate failure density - relation between reliability, failure density and hazard rate - Mean time to failure (MTTF) Integral equation of MTTF in terms of reliability

Module II

Hazard models constant hazard model linearly increasing hazard model expressions for reliability, failure density, and probability of failure of these models problems

Module III

System reliability components connected in series components connected in parallel mixed configuration reliability block diagrams (RBD) distinction between physical configuration and logical configuration problems

Module IV

Reliability improvement methods Redundancy unit redundancies element redundancies simplification of design parts derating operating environment; Cost of reliability factors to be considered for optimizing the reliability cost

References

1. L.S.Srinath, Reliability Engineering, Affiliated East-West Press Ltd., 1985 2. E. Balaguruswamy, Reliability Engineering, Tata McGraw Hill Publishing Co.,

1984

3. Charles E. Ebling, Reliability & Maintainability Engg., Tata McGraw Hill Publishing Co., 1997

4. Alessandro Birolini Reliability Engineering Theory and Practice, Springer, 2007.

5. Lewis, E., Introduction to Reliability Engineering, John Wiley & Sons, 1995

Lecture notes on Process Reliability Engineering Subject handled by Dr.Shouri P. V., Associate Professor in Mechanical Engineering, MEC, Cochin.

(for 1st year M.Tech. Mechanical Engineering batch)

2

EM 105 (B) Process Reliability Engineering

Definition of reliability

Reliability is the probability that an item will perform a required function for a specified

period of time under specified operating conditions.

This definition has four key elements:

1) The quantification of reliability in terms of probability.

2) A statement defining the required function as the function is defined in detail, it becomes more clear which product failures impair the success of the mission and

which do not.

3) A statement specifying the period of time deterioration of materials and parts with time is natural and consequently the performance level of the unit will also

go down with time. If the time period is not specified, probability is a meaningless

number for time oriented products.

4) A statement defining the operating condition and this could be with regard to temperature, humidity, shock, vibration, and so on.

Failure data analysis

Reliability is defined as the probability of a device giving satisfactory performance for a

specified period under specified operating conditions. When a unit or system does not

perform satisfactorily, it is said to have failed. The pattern of failure can be obtained from

life test results. That is, by testing a fairly large number of models until failure occurs,

and observing the failure rate characteristics as a function of time. The first step,

therefore, is to link reliability with experimental or field failure data. These data will also provide a basis for formulating or constructing mathematically a failure model for

general analysis.

The following data is related with a series of tests conducted under certain stipulated

conditions on 1000 electronic components. The total duration of the tests is 19 hours. The

number of components that fail during each hourly interval is noted and the results

obtained are tabulated below.



3

Time (t)

No. of

failures

( f)

Cumulative

failures (F)

No. of

survivors

(S)

Failure

density

(fd)

Failure

rate (Z)

Reliability

(R)

0 0 1000 1

130 0.130 0.139

1 130 870 0.870

83 0.083 0.100

2 213 787 0.787

75 0.075 0.100

3 288 712 0.712

68 0.068 0.100

4 356 644 0.644

62 0.062 0.101

5 418 582 0.582

56 0.056 0.101

6 474 526 0.526

51 0.051 0.101

7 525 475 0.475

46 0.046 0.101

8 571 429 0.429

41 0.041 0.100

9 612 388 0.388

37 0.037 0.100

10 649 351 0.351

34 0.034 0.101

11 683 317 0.317

31 0.031 0.103

12 714 286 0.286

28 0.028 0.103

13 742 258 0.258

64 0.064 0.283

14 806 194 0.194

76 0.076 0.486

15 882 118 0.118

62 0.062 0.714

16 944 56 0.056

40 0.04 1.110

17 984 16 0.016

12 0.012 1.200

18 996 4 0.004

4 0.004 2.000

19 1000 0 0

Total =1 Mean = 0.376



4

The failure density, failure rate, reliability and probability of failure can be defined as

follows.

Failure density (fd)

This is the ratio of number of failures during a given unit interval of time to the total

number of items at the very beginning of the test (also called as initial population). For

the given data the failure density associated with the first unit interval is

13.01000

1301d

f

Similarly, 083.01000

832d

f and so on.

Failure rate (Z)

This is the ratio of number of failures during a particular unit time interval to the average

population during that interval. The average population during an interval is the average

of populations at the beginning and at the end of the interval. For the given data, failure

rate during the first unit interval is

139.0

2

8701000

130)1(Z

Similarly, 100.0

2

787870

83)2(Z and so on.

Reliability (R)

This is the ratio of survivors at any given time to the total initial population. Probability of survival is another name for reliability. For the given data, reliability corresponding to first hour is

870.01000

870)1(R

787.01000

787)2(R and so on.

Probability of failure

Probability of failure can also be termed as unreliability factor. Since survival and failure

are complementary events, Probability of failure = 1- Probability of success (or

reliability).

For the given data, probability of failure corresponding to first hour = 1=R(1)

= 1-0.787

= 0.213

Similarly, probability of failure corresponding to fifth hour = 1=R(5)

= 1-0.582

= 0.418



5

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0 2 4 6 8 10 12 14 16 18

Time interval

Failu

re d

ensity

0

0.5

1

1.5

2

2.5

0 2 4 6 8 10 12 14 16 18

Time interval

Fai

lure

rat

e

0

0.2

0.4

0.6

0.8

1

1.2

0 2 4 6 8 10 12 14 16 18

Time

Relia

bili

ty

Variation of failure density, failure rate, and reliability with time



6

Conclusions

(i) Let 1d

f = failure density associated with the first unit time interval

2d

f = failure density associated with the first unit time interval

.

.

ld

f = failure density associated with the last unit time interval

Then, 121 lddd

fff

(ii) Let 1n number of failed components during the first unit interval

2n number of failed components during the second unit interval

.

.

tn number of failed components associated with the tth unit interval

N = Total initial population

Then, Reliability for the tth hour is the number of survivors till the t

th hour divided by the

initial population.

That is, N

n

N

n

N

n

N

nnnNtR tt

2121 1)(

)(

That is, tddd

ffftR 21

1)(

(iii) Probability of failure for tth hour =

tdddfff

21

(iv) Failure rate or hazard rate associated with the tth hour, Z(t) =

)()1(

)()1(2

2

)()1(

)()1()(

tRtR

tRtR

tNRtNR

tNRtNRtZ (for unit time interval)

)()(

)()(2)(

tRttR

tRttR

ttZ (for a time interval of t )

Problem

Following table gives the results of tests conducted under severe conditions on 1000

safety valves. Obtain the failure density and hazard rates for various time intervals.

Time

interval 0-4 4-8 8-12 12-16 16-20 20-24

No. of

failures 267 59 36 24 23 11



7

Solution

Time No. of failures Failure density Hazard rate

0

267 0.0668 0.00770

4

59 0.0150 0.00210

8

36 0.0090 0.0137

12

24 0.0060 0.0096

16

23 0.0058 0.0095

20

11 0.0028 0.0047

24

Sample calculation for time interval 0 4 is given below.

0668.010004

267df

Probability density function

Consider the life testing of N components. The total test last for T hours, at the end of

which all specimens will have failed.

Let 1n Number of components that failed during the 1st unit interval

2n Number of components that failed during the 2nd

unit interval

.

.

ln Number of components that failed during the last unit interval

Then, 121

N

nnn l

121 l

ddd fff

where .,21

etcff dd are the failure densities associated with the respective intervals.

When the total population is large and the time interval is very small, the variation of

failure density with time will be a smooth curve and the summation can be represented by

integrals.

0770.0

2

)2671000(10004

267Z



8

Therefore, with reference to the above figure, T

d df0

1)(

(where is a dummy variable)

That is, the probability that a specimen will fail in T hours is 1 (that is, a certainty).

In probability theory, the function )(df is known as the probability density function. In

order to make this function more general, the upper limit T is replaced by This means

that no specimen in any test will last for an infinite number of hours.

0

1)( dfd

Reliability and Probability of failure in terms of failure density

Reliability for the tth hour is the number of survivors till the t

th hour divided by the initial

population.

That is, N

n

N

n

N

n

N

nnnNtR tt

2121 1)(

)(

That is, t

dddd dfffftR t0

11)(21

(for very small time intervals)

Probability of failuret

d dftR0

)(1

Reliability in terms of hazard rate

The hazard rate Z(t) can be expressed as

)()(

)()(2)(

tRttR

tRttR

ttZ



9

When t is very small and tends to zero, the value of R(t) approaches that of

)( ttR and we get

)(

)()(.)(0 ttR

tRttRLttZt

)(

)(

)(

1

td

tdR

tR Since

)(

)()(.

)(

)(

0 tR

tRttRLt

td

tdR

t

)(ln)( tRdt

dtZ Since

)(

)(

)(

1)(ln

td

tdR

tRtR

dt

d

For example, 22

1)2ln(

xx

dx

d

Integrating, t

CtRdZ0

)(ln)(

where C is a constant and is a dummy variable.

At ,0t Reliability 1 0C t

tRdZ0

)(ln)(

t

dZtR0

)(exp()(

Failure density in terms of failure rate and reliability

N

tRNttRN

ttf d

)()(1)(

When t is very small and tends to zero, dt

tdR

t

tRttRLttft

d

)()()(.)(0

Also we have, dt

tdRtRtZ

td

tdR

tRtZ

)()()(

)(

)(

)(

1)(

Combining above two equations we get,

)()()( tRtZtf d

Mean Time To Fail (MTTF) and Mean Time Between Failures (MTBF)

MTTF is the mean time to first failure and is used in case of components that are not

repaired when they fail, but are replaced by new components. On the other hand, MTBF



10

is the mean time between two successive component failures and is used with repairable

equipment or systems.

Note:

1) When the number of samples tested is small, it is possible to note the time to failure of each sample and the mean failure rate is given by the formula

)0(

)()0(1)(

N

TNN

TTZ

where, Z(T) is the mean failure rate for T hours

N(T) is the population remaining at time T

N(0) is the population at T = 0

2) As the number of specimens tested becomes large it is tedious to record the time to failure of each

specimen. Instead, the number which fail during specific intervals of time are recorded. Here

mean time to failure for N specimens will be

tnltntntnN

MTTF l.321

321

where, t is the time interval n1 is the number of specimens that failed during the 1

st interval n2 is the number of specimens that failed during the 2

nd interval

.

.

ln is the number of specimens that failed during the last interval

That is, in general, l

K

K tKnN

MTTF1

1

Problem

In the life testing of 10 specimens of a mini-mixer, the time to failure of each specimen is

recorded as given in the following table. Calculate the mean failure rate for 900 hours and

the mean time to failure for all ten specimens.

Specimen

Number

Time to Failure

(hours)

1 805

2 810

3 815

4 820

5 825

6 832

7 842

8 856

9 875

10 900



11

Solution

The mean failure rate can be calculated using the expression:

)0(

)()0(1)(

N

TNN

TTZ

Here, Z(900) is to be calculated and N(0)=10 ; N(900)=0

31011.1900

1

900

0900

900

1)900(Z (per hour)

Mean Time to Fail (MTTF)

83890087585684283282582081581080510

1)(MTTF hours

Problem

Ten transformers were tested for 500 hours each within the prescribed operating

conditions, and one transformer failed exactly at the end of the 500 hours exposure. What

is the failure rate for this type of transformer?

Solution

The mean failure rate can be calculated using the expression:

)0(

)()0(1)(

N

TNN

TTZ

0002.05000

1

10

910

500

1)500(Z failure / hour

Mean Time to Fail (MTTF) in integral form l

K

K tKnN

MTTF1

1

Also, by definition failure density fd can be expressed as tN

nf Kd , where Kn is the

number of failures associated with the thK time interval

tfN

nKd

K

Hence the expression for MTTF can be written as l

K

d ttKfMTTF K1

)(

Further tK is the elapsed time t and therefore the expression for the MTTF becomes l

K

d ttfMTTF1

For very small time intervals,

l

d dttfMTTF0



12

In the above equation, upper limit l is the number of hours after which there are no

survivors. It is customary to replace this by infinity since all components will have failed

at the end of an infinite test period.

0

dttfMTTF d

Also we have,t

dfdtR0

)(1)(

Differentiating with respect to time, )()(

tfdt

tdRd

That is, dt

tdRtfd

)()( and substituting this in the equation for MTTF

00

)()(

ttdRdtdt

tdRtMTTF

Integrating by parts, dttRttRMTTF0

0 )()( vduuvudv

0

)( dttRMTTF {Since 1)0(R and 0)(R }



13

Hazard Models

The data obtained from failure tests can be analyzed to obtain reliability, failure density,

hazard rate and other necessary information. Obviously, the behavioural characteristics

exhibited by one class of components differ from those exhibited by another class of

components. In order to compare different behavioural characteristics and also to draw

general conclusions from behavioural patterns of similar components, a mathematical

model representing the failure characteristics of the components becomes necessary. The

procedure involves assuming a function for hazard rate and thereby obtaining reliability

and failure density by using this failure rate function. The assumed function for the

hazard rate will be the hazard model. Some of the common hazard models are discussed

below.

Constant hazard model

Here the failure rate is assumed to remain constant with time.

That is, )(tZ , a constant.

tddZtRt

tt

expexpexp)(exp)( 000

That is, for a constant hazard model, Reliability, tetR )(

Probability of failure, tetRtF 1)(1)(

Failure density, t

d etRtZtf )()()(

The variation of failure rate, reliability, probability of failure, and failure density with

respect to time for a constant hazard model is shown in the following figure.

Variation of failure rate, reliability, probability of failure, and failure density for a constant hazard model



14

It can be seen that, for a constant hazard model the mean time to failure is the reciprocal

of failure rate.

That is, 1

1011

)( 0

000

eee

dtedttRMTTFt

t

The constant hazard model is also known as exponential reliability case.

Linearly increasing hazard model

Here the hazard rate is assumed to increase linearly with time.

That is, KttZ )( , where K is a constant

2exp

2expexp)(exp)(

2

0

2

00

KtKdKdZtR

ttt

That is, for a linearly increasing hazard model, Reliability, 22

)(

Kt

etR

Probability of failure, 22

1)(1)(

Kt

etRtF

Failure density, KttRtZtf d )()()(2

2Kt

e

The variation of failure rate, reliability, probability of failure, and failure density with

respect to time for a linearly increasing hazard model is shown in the following figure.

Variation of failure rate, reliability, probability of failure, and failure density for a linearly increasing

hazard model

It can be seen from the failure density curve that the curve has a slope equal to K at

time 0t . Also the value of )(tf d reaches a maximum of e

Kat time

Kt

1, and tends

to zero as t becomes larger.



15

The Weibul Model

This model is expressed as 1,)( mKttZ m

Here K and m are parameters and if these are chosen appropriately, a variety of failure-

rate situations can be covered, including both the constant hazard and linearly increasing

hazard conditions.

If 0m ; KtZ )( - Constant hazard model

If 1m ; KttZ )( - Linearly increasing model

1exp

1expexp)(exp)(

1

0

1

00m

Kt

m

KdKdZtR

mt

mt

m

t

That is, in case of Weibul model, Reliability, 11

)( mKt m

etR

Probability of failure, 11

1)(1)( mKt m

etRtF

Failure density, m

d KttRtZtf )()()(1

1

m

Kt m

e

Following figure shows the variation of reliability in case of Weibul model for various

values of K and m

Variation of Reliability in case of Weibul model



16

Bath Tub Curve

Component failure rate as a function of age follows a curve that is concave upward, as

shown in the above figure. Because of its shape, this curve is also referred to as bath tub curve. This curve exhibits three distinct zones. The first is the short initial period called variously the early failure, infant mortality, or the burn in period. The decreasing but

greater failure rate early in life of the system is due to one or more of several potential

causes. The causes include inadequate testing or screening of components during

selection or acceptance, damage to components during production, assembly, or testing,

and choice of components which have too great a failure variability. It shall be a specific

goal of the supplier to ensure that the early failure period is rigorously controlled and

covered by a suitable warranty.

Bath tub curve

The failures in the second zone are termed service failures. During this period, the failure

or hazard rate is constant and it represents the effective life of the product.

The failures in the third zone are the wear-out failures. The incidence of failure in this

zone is high since most of the components will have exceeded their service life, and

consequently would have deteriorated. Hence, they are appropriately called wear-out

failures.

Note: Failure (death) rates for human beings are different by sex, race, nationality, and other factors but

all failure rate for humans appear to exhibit this distinctive bath tub curve. The failure rate for infants is extremely high for the first few months, drops sharply, and remains fairly constant for many years and then

slowly climbs as the person ages.

System reliability

A system or a complex product is an assembly of a number of parts or components. The

components may be connected in series or in parallel, or it may be a mixed system, where

the components are connected in series as well as in parallel.

Series configuration

If the components of an assembly are connected in series the failure of any component

causes the failure of the assembly or system.



17

Let us consider a system consisting of n units which are connected in series as shown in

the following figure.

Reliability block diagram of a system having n components connected in series

Let the successful operation of these individual units be represented

by nXXX ,,, 21 and their respective probabilities

by )(,),(),( 21 nXPXPXP . For the successful operation of the system, it is

necessary that all n units function satisfactorily. Hence, the probability of the

simultaneous successful operation of all units is ).( 21 nXandandandXXP .

Therefore according to multiplication rule,

).()( 21 nXandandandXXPtR

)()()()()( 121213121 nn andXandXXXPandXXXPXXPXPtR

In this expression, )( 12 XXP represents the probability of the successful operation of unit

2 under the condition that unit 1 operates successfully. Similarly,

)/( 121 nn XandandXXXP represents the probability of the successful operation

of unit n under the condition that all the remaining units 1,2,3,,n-1 are working successfully.If the successful operation of each unit is independent of the successful

operation of the remaining units, then events nXXX ,,, 21 are independent and

the above equation becomes

)()()()( 21 nXPXPXPtR

That is, nRRRRtR 321)( ,

where nRRRR 321 are component reliabilities.

Parallel configuration

Several systems exist in which successful operation depends on the satisfactory

functioning of any one of their n sub-systems or elements. They are said to be connected

in parallel. Let us consider a system consisting of n units which are connected in parallel

as shown in the following figure.



18

Reliability block diagram of a system having n components connected in parallel

Let nXXX ,,, 21 represent the successful operation of units 1,2,.,n

respectively. Similarly, let nXXX ,,, 21 represent the unsuccessful operation.

If )( 1XP is the probability of successful operation of unit 1, then )( 1XP is the probability

of its failure. Further, )(1)( 11 XPXP

For the complete failure of the system, all n units have to fail simultaneously. If F(t) is

the probability of failure of the system, then

).()( 21 nXandandXandXPtF

)()()()()( 121213121 nn XandXandXXPXandXXPXXPXPtF

In this expression, )/( 213 XandXXP represents the probability of failure of unit 3 under

the condition that units 1 and 2 have failed. The other terms can also be interpreted in the

same manner. If the unit failures are independent of each other, then

)(1)()()()( 21 tRXPXPXPtF n

nXPXPXPtR 11(11)( 21

That is, )1()1(11)( 21 nRRRtR

where nRRRR 321 are component reliabilities



19

Mixed configuration

If a system is having a mixed configuration, then it will have components connected in

parallel as well as in series and the following figure indicates a system having

components in series and parallel.

Reliability block diagram of a system having n components connected in parallel

If 54321 ,,,, RRRRR and 6R are the respective component reliabilities of the components in

the above configuration, then system reliability can be expressed as

654321 )1)(1)(1(1)( RRRRRRtR

Problem

A certain type of electronic component has a uniform failure rate of 0.00001 per hour.

What is the reliability for a specified period of service of 10,000 hours?

Solution

000001.0 per hour

10000t hours

90483.0)1000( 1.010000)00001.0( eeR (or) 90.483%

Problem

Given a MTTF of 5000 hours and a uniform failure rate. What is the probability that the

system failure occurs within 200 hours?

Solution

5000

1 (per hour)

t = 200 hours

96079.0)200(200

5000

1

eR (or) 96.079%

03921.096079.01)200(F (or) 3.921%



20

Problem

The following reliability requirements have been set on the subsystems of a

communication system.

Subsystem Reliability (for a 4 hour period)

Receiver 0.970

Control System 0.989

Power Supply 0.995

Antenna 0.996

What is the expected reliability of the overall system?

Solution

Assuming that all the four components are necessary for the successful operation of the

system, 4321)( RRRRtR

950.0)996.0)(995.0)(989.0)(970.0()(tR (or) 95%

Problem

An element has a probability of successful operation of 60% over a given period of time.

If 4 such components are connected in parallel estimate the improvement factor

Solution

Reliability when 4 components (each having a reliability of 0.6) are connected in parallel

can be expressed as:

9744.0)6.01)(6.01)(6.01)(6.01(1)(tR

If x is the improvement factor, then 624.16.0

9744.09744.06.0 xx

Problem

For the Reliability Block Diagram (RBD) shown in the following figure calculate the

system reliability. The respective component reliability values are also indicated in the

figure.

Reliability block diagram



21

Solution

8183.0)90.0()94.01)(95.01(1)96.0)(95.0()(tR (or) 81.83%

Problem

The MTBF of equipment is 500 hours. What is the failure rate expressed in

a) Failures / hour b) Failures / 106 hours c) % failures / 1000 hours

Is MTBF a guaranteed failure free period?

Solution

Failure rate, 002.0500

11

MTBFfailures / hour {Answer (a)}

200010002.0 6 failures / 106 hours {Answer (b)}

21000002.0 failures / 1000 hours

= 200 % failures / 1000 hours {Answer (c)}

MTBF cannot be regarded as a guaranteed failure free period as it is only a mean value of

operating times between failures.

Reliability Increasing Techniques

One way of achieving high reliabilities is by introducing redundant parts. For example we

may have two parts in parallel such that the system operates if at least one part operates.

Here the probability that the system fails is equal to the probability both parts fail. If the

failures are assumed to be independent, then the system reliability will be R(t) = 1- (1-

R1)(1-R2), where R1 and R2 are the reliability of the two parts respectively. If the

reliability of each part is 0.95 at time t, then the reliability of the system is

R(t) = 1-(1-0.95)(1-0.95) = 0.9975

By adding a redundant part we have increased the reliability of system at time t from 0.95

to 0.9975

We have been assuming that both parts are operating whenever the system is on and the

failure of one part does not affect the operation of other part. This is some times called

hot standby and is not always practical. We may need to provide a cold stand by where

the second part is switched into service when the first one fails. Then we must also take

into account the reliability of the switch. If we assume we have, as before, two

components with reliability 0.95 at time t and a switching device with reliability 0.98 at

time t we have the system reliability at time t as

R(t) = 0.95 + (0.05)(0.98)(0.95) = 0.9966



22

The above equation is just the probability the first part is operating plus the probability

the first part fails times the probability the switch operates times the probability the

second part operates.

It is to be noted that there is a point of diminishing returns in using redundancy

configurations; an increase in the level of parallel redundancy employed increases size,

weight, cost and volume of the equipment and often requires complicated failure-sensing

devices whose reliability need to be considered.

As this figure illustrates, as the amount of dollars invested in reliability increases,

Operation and Support (OS) costs decrease. When the reliability is combined with OS,

the result is total cost. The objective is to reach the lowest point in the total cost curve at

which the benefits of reliability (expressed as total operating cost) are optimized with the

cost of obtaining that level of reliability.

It is usually necessary to perform trade- off calculations to determine the advisability of

parallel redundancy versus improvement of the reliability of the basic subsystem by other

means. Methods for improvement could include the following considerations:

1) Review the users needs to see if the function of the unreliable part is really necessary to the user. If not eliminate these parts from the design. That is,

decrease the number of component parts for a system and believe in the vital few.

For a system that contains items connected in series, assuming independence of

their individual failures, the reliability of the system is the product of the

reliabilities of the individual items. If a product has 5 parts, each with reliability

0.9, then system reliability is 59.09.05

whereas, if there would have been only

three parts, the reliability of the system would be 73.09.03

2) Review the selection of any parts that are relatively new and unproven. Use standard parts whose reliability has been proved. (However, be sure that the

conditions of previous use are applicable to the new product.)



23

3) Use derating to assure that stresses applied to the parts are lower than the stresses the parts can normally withstand.

4) Use robust design methods that enable a product to handle unexpected environments.

5) Control the operating environment to provide conditions that yield lower failure rates Common examples are (a) potting electronic components to protect them

against climate and shock, and (b) use of cooling equipments to keep down

ambient temperatures.

6) Specify replacement schedules to remove and replace low-reliability parts before they reach the wear-out stage.

7) Prescribe screening tests to detect infant-mortality failures and to eliminate substandard components. The tests include burn in, accelerated life tests etc.

8) Conduct research and development to attain an improvement in the basic reliability of those components which contribute most of the unreliability.

reliability 2013 1

Documents

quantification of reliability

terms of reliability

system reliability components

reliability cost references

pattern of failure

failure rate characteristics

field failure data

time period