Reliability Engineering T.veerarajan Sample Chapter

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In this chapter we shall consider a very important and growing area of applica-tion of some of the probability concepts discussed in the previous chapters,

namely, Reliability Engineering. Reliability considerations are playing an in-creasing role in almost all engineering disciplines. Generally the term ‘reliabil-ity’ of a component (or a system, that is, a set of components assembled to per-form a certain function) is understood as its capability to function without break-down. The better the component performs its intended function, the more reliableit is. In the broader sense, reliability is associated with dependability, with suc-cessful performance and with the absence of breakdown or failures. However,from the engineering analysis point of view, it is necessary to define reliabilityquantitatively as a probability. Technically, reliability may be defined as theprobability that a component (or a system) will perform properly for a specifiedperiod of time ‘t’ under a given set of operating conditions. In other words, it isthe probability that the component does not fail during the interval (0, t). We givebelow the formal mathematical definition of reliability and discuss the relatedconcepts.

��

The definitions given below with respect to a component hold good for a systemor a device also.

If a component is put into operation at some specified time, say t = 0, and if Tis the time until it fails or ceases to function properly, T is called the life length ortime to failure of the component. Obviously T (≥ 0) is a continuous random variablewith some probability density function f(t). Then the reliability or reliabilityfunction of the component at time ‘t’, denoted by R(t), is defined as

R(t) = P(T > t) or 1 – P(T ≤ t)= 1 – F(t) (1)

where F(t) is the cumulative distribution function of T, given by

F(t) =0

( )t

f t dt�

2 Probability, Statistics and Random Processes

Thus R(t) = 1 – 0

( )t

f t dt� = ( )t

f t dt�

� (2)

Since F(0) = 0 and F(∞) = 1 by the property of cdf, R(0) = 1 and R(∞) = 0

i.e., 0 ≤ R(t) ≤ 1. Also since d

dt F(t) = f (t),

we get f (t) = – ( )dR t

dt(3)

Now the conditional probability of failure in the interval (t, t + ∆t), given thatthe component has survived up to time t, viz.,

P{t ≤ T ≤ t + ∆t/T ≥ t} = ( )

1 – ( )

f t t

F t

�, by the definitions of pdf and cdf

= f (t) ∆t/R(t)

∴ The conditional probability of failure per unit time is given by ( )

( )

f t

R t and is

called the instantaneous failure rate or hazard function of the component, de-noted by l(t).

Thus l(t) = ( )

( )

f t

R t(4)

Now, using (3) in (4), we have

– ( )

( )

R t

R t

�= l(t) (5)

Integrating both sides of (5) with respect to t between 0 and t, we have

0

( )

( )

tR t

dtR t

�� = – 0

( )t

t dt��

i.e., {log R(t)}t0 = –

0

( )t

t dt��

i.e., log R(t) – log R(0) = – 0

( )t

t dt��

i.e., loge [R(t)] = – 0

( )t

t dt�� [� R(0) = 1]

∴ R(t) = 0

( )t

t dt

e��

(6)Using (6) in (4), we have

f (t) = l(t) 0

( )t

t dt

e��

(7)

The expected value of the time the failure T, denoted by E(T) and variance of T,

denoted by 2T� are two important parameters frequently used to characterise

reliability. E(T) is called mean time to failure and denoted by MTTF.

Reliability Engineering 3

Now MTTF = E(T) = 0

( )t f t dt�

�

= – 0

( ) ,tR t dt�

�� by (3)

= – [tR(t)]∞0 +

0

( )R t dt�

� (8)

on integration by parts

Now [tR(t)]t = ∞ 0

( )t

t dt

t

te��

��

� ��

� �

=

0

( )t

t dt

t

t

e�

��

� � � � �

= 0 (9)

and [tR(t)]t = 0 = 0 × R(0) = 0 × 1 = 0 (10)Using (9) and (10) in (8), we get

MTTF = 0

( )R t dt�

� (11)

Var(T ) s 2T = E{T – E(T )}2 or E(T2) – {E(t)}2

= 2

0

( )t f t dt�

� – (MTTF)2 (12)

Conditional reliability is another concept useful to describe the reliability of acomponent or system following a wear-in period (burn-in period) or after awarranty period.It is defined as

R(t/T0) = P{T > T0 + t/T > T0}

= 0

0

{ }

{ }

P T T t

P T T

�

= 0

0

( )

( )

R T t

R T

�(13)

=

0

0

0

0

( )

( )

T t

T

t dt

t dt

e

e

�

�

��

�

�

� =

00

0 0

( ) ( )

TT t

t dt t dt

e� �

��

=

0

0

( )

T t

T

t dt

e�

�

� �(14)

�� Some of the probability distributions describe the failure process and reliabilityof a component or a system more satisfactorily than others. They are the expo-nential, Weibull, normal and lognormal distributions. We shall now derive the


reliability characteristics relating to these failure distributions using the formu-las derived above.

�� If the time to failure T follows an exponential distribution with parameter λ, thenits pdf is given by

f(t) = le–lt, t ≥ 0 (15)Then, from (2),

R(t) = t

t

e dt��

�� = t t

te e� �� (16)

From (4), l(t) = ( )

( )

t

t

f t e

R t e

�

�

� �

�� = l (17)

This means that when the failure distribution is an exponential distribution withparameter l, the failure rate at any time is a constant, equal to l. Converselywhen l(t) = a constant l, we get from (7),

f (t) = l .

t

t

dt

e��

= le– lt, t ≥ 0Due to this property, the exponential distribution is often referred to as constantfailure rate distribution in reliability contexts. We have already derived in theearlier chapters, that

MTTF = E(T ) = 1

�(18)

and Var(T ) = s 2T = 2

1

�(19)

Also R(t/T0) = 0

0

( )0

0

( )

( )

T t

T

R T t e

R T e

�

�

� �

��

� by (16)

= e–lt (20)This means that the time to failure of a component is not dependent on how longthe component has been functioning. In other words the reliability of the compo-nent for the next 1000 hours, say, is the same regardless of whether the compo-nent is brand new or has been operating for several hours. This property is knownas the memoryless property of the constant failure rate distribution.

�� The pdf of the Weibull distribution was defined as

f(t) = abtb–1 e–atb, t ≥ 0 (21)

An alternative form of Weibull’s pdf is

f(t) = 1

exp ,t t

� ��

� � �

� � �� q > 0, b > 0, t ≥ 0 (22)

(22) is obtained from (21) by putting a = 1��

. b is called the shape parameter


and q is called the characteristic life or scale parameter of the Weibull’s distri-bution (22). If T follows Weibull’s distribution (22),

then R(t) = 1

. expt

t tdt

� ��

� � �

� � � ��

= ,x

t

e dx�

�� on putting t

�

�� = x

= e–x = expt

�

�

� �� (23)

l(t) = ( )

( )

f t

R t =

1

.t

��

� �

�� (24)

As derived in the chapter ‘some special probability distributions’, we have

MTTF = E(T ) = 11+��

�

� ��

(25)

and Var(T ) = s 2T = q2

22 1

1 1� ��

� �� (26)

Now R(t/T0) = 0

0

( )

( )

R t T

R T

�

=

0

0

exp

exp

t T

T

�

�

�

�

� ��

= 0 0expt T T

� �

� �

� �� (27)

�� If the time to failure T follows a normal distribution N(m, s ) its pdf is given by

f(t) = 2

2

1 ( )exp ,

2 2

t �

� � �

� ��

|–∞ < t < ∞ |

In this case, MTTF = E(T ) = m andVar (T ) = s2

T = s 2.

R(t) = ( )t

f t�

� dt is found out by expressing the integral in terms of standard

normal integral and using the normal tables.l(t) is then found out by using (4).


�� If X = log T follows a normal distribution N(m, s), then T follows a lognormaldistribution whose pdf is given by

f (t) = 2

2

1 1exp log

2 2 m

t

tst s�

� �� , t ≥ 0 (28)

where s = s is a shape parameter and tm, the median time to failure is the locationparameter, given by log tm = m.It can be proved that

MTTF = E(T ) = tm 2

exp2

s� ��

(29)

and Var(T ) = s 2T = t 2

m exp(s2) [exp (s2) – 1] (30)In this case,

F(t) = P(T ≤ t) = P{log T ≤ log t}

= Plog log

,T t� �

� �

� ��

since log T follows a N(m, s)

= P log log 1logm

m

T t t

s s t

� ��

= P 1log ,

m

tZ

s t

� �� where Z follows the standard

normal distribution N(0, 1).Then R(t) and l(t) are computed using (1) and (4).

�� The density function of the time to failure in years of the gizmos (for use

on widgets) manufactured by a certain company is given by f(t) = 3

200

( 10)t �, t ≥ 0.

(a) Derive the reliability function and determine the reliability for the firstyear of operation.

(b) Compute the MTTF.(c) What is the design life for a reliability 0.95?(d) Will a one-year burn-in period improve the reliability in part (a)? If so,

what is the new reliability?

��

(a) f (t) = 3

200

( 10)t �, t ≥ 0

R(t) = ( )t

f t dt�

� = 2 2

100 100

( 10) ( 10)t

t t

��

� � ��

∴ R(1) = 2

100

(1 10)� = 0.8264.


(b) MTTF = 0

( )R t dt�

� = 2

0

100

( 10)dt

t

�

��

= 0

100

10t

��

= 10 years.

(c) Design life is the time to failure (tD) that corresponds to a specifiedreliability. Now it is requeired to find tD corresponding to R = 0.95

∴2

100

( 10)t �= 0.95

i.e., (tD + 10)2 = 100.2632∴ tD = 0.2598 year or 95 days

(d) R(t/1) = ( 1)

(1)

R t

R

� =

2

100

( 11)t ��

2

100

11 =

2

121

( 11)t �

Now R(t/1) > R(t), if 2 2

121 100

( 11) ( 10)t t

� �

i.e., if2

2

( 10)

( 11)

t

t

��

> 100

121

i.e., if10

11

t

t

��

> 10

11i.e., 11t > 10t, which is true, as t ≥ 0

∴ One year burn-in period will improve the relaibility.

Now R(1/1) = 2

121

(1 11)� = 0.8403 > 0.8264.

�� The time to failure in operating hours of a critical solid-state

power unit has the hazard rate function ll(t) = 0.003 0.5

500

t� �� , for t ≥ 0.

(a) What is the reliability if the power unit must operate continuously for 50hours?

(b) Determine the design ofe if a reliability of 0.90 is desired.(c) Compute the MTTF.(d) Given that the unit has operated for 50 hours, what is the probability that

it will survive a second 50 hours of operation?

��

(a) R(t) = 0

exp ( )t

t dt��

∴ R(50) = 50 0.5

0

exp 0.003500

tdt

� ��

=50

3 2

0

0.003 2exp .

3500t

� ��


=0.003 2

exp 50 503500

� ��

= exp[–0.03162]= 0.9689

(b) R(tD) = 0.90

i.e.,0.5

0

exp 0.003500

Dt tdt

� �� = 0.90

i.e., 1 2

0

0.003

500

Dt

t dt� � = – 0.10536

i.e.,0.003

500× 3 22

3 Dt = 0.10536

∴ tD = 2 3

3 500 0.10536

2 0.003

� ��

= 111.54 hours.

(c) MTTF = 0

( )R t dt�

�

=3 20.003 2

3500

0

t

e dt

� ��

=3 2

0

,ate dt�

�� where a = 0.003 2

3 500

��

= 1 32 3

0

2.3

xe xa

�� dx, on putting x = at3/2

=2 3

2(2 3)

3a =

2 3

2 3(5 3)

23a

=2 3

0.9033

a, from the table of values of Gamma function.

= 45.65 hours.

(d) P(T ≥ 100/T ≥ 50) = ( 100)

( 50)

P T

P T

��

= (100)

(50)

R

R

=

100

50

exp ( )t dt��

= 3 2 3 20.002 0.002exp 100 50

500 500

� ��

= exp[{– 0.08944} – {– 0.03162}]= 0.9438


�� The reliability of a turbine blade is given by R(t) = 2

0

t1

t

� ��

, 0 ≤ t

≤ t0 , where t0 is the maximum life of the blade.(a) Show that the blades are experiencing wear out.(b) Compute MTTF as a function of the maximum life.(c) If the maximum life is 2000 operating hours, what is the design life for a

reliability of 0.90?

��

(a) R(t) = 2

0

1t

t

� ��

, 0 ≤ t ≤ t0

Now l(t)= – ( )

( )

R t

R t

�

= – 2

0 0 0

21 1

t t

t t t

� � ��

= 0

2

t t�

l′(t) = 2

0

2

( )t t� > 0 and so l(t) is an increasing function of t.

When the failure rate increases with time, it indicates that the blades areexperiencing wear out.

(b) MTTF = 0

( )R t dt�

� = 0 2

00

1t

tdt

t

� ��

=

030

00

13

t

t t

t

� ��

= 0

3

t

(c) When t0 = 2000, R(tD) = 0.90

i.e., 2

12000

Dt� �� = 0.90

∴ 1 – 2000

Dt = 0.9487

∴ tD = 102.63 hours.

�� Given that R(t) = 0.001te� , t ≥ 0

(a) Compute the reliability for a 50 hours mission.(b) Show that the hazard rate is decreasing.(c) Given a 10-hour wear-in period, compute the reliability for a 50-hour

mission.(d) What is the design life for a reliability of 0.95, given a 10-hour wear-in

period?


(a) R(t) = 0.001te� t ≥ 0

∴ R(50) = 0.001 50e� � = 0.9512

(b) l(t) = ( )

( )

R t

R t

� �= – 0.001 0.001 1

0.001t te et

� ��

=0.001

,t

which is a decreasing function of t.

(c) R(t/T0) = 0

0

( )

( )

R t T

R T

�

∴ R(50/10) = (60)

(10)

R

R =

0.001 60

0.001 10

e

e

� �

� � = 0.8651

(d) R(tD /10) = 0.95

i.e.,( 10)

(10)DR t

R

�= 0.95

i.e.,0.001 ( 10)Dte� � �

= 0.95 × 0.001 10e� �

∴ 0.001 ( 10)Dt� � = 0.15129

∴ tD = 12.89 hours.

��! A one-year guarantee is given based on the assumption that nomore than 10% of the items will be returned. Assuming an exponential distribution,what is the maximum failure rate that can be tolerated?

If T is the time to failure of the item,

��

then P(T ≥ 1) ≥ 0.9 (� no more than 10% will be returned)

i.e., R(1) ≥ 0.9

i.e.,1

te dt��

�� ≥ 0.9 (� T follows an exponential distribution)

i.e., (–e–lt )∞1 ≥ 0.9

i.e., e –l ≥ 0.9

∴ – l ≥ – 0.1054

∴ l ≤ 0.1054/year

��" A manufacturer determines that, on the average, a television set isused 1.8 hours per day. A one-year warranty is offered on the picture tube havinga MTTF of 2000 hours. If the distribution is exponential, what percentage of thetubes will fail during the warranty period?


��Since the distribution of the time to failure of the picture tube is exponential,

R(t) = e–l t where l is the failure rateGiven that MTTF = 2000 hours

i.e.,0

te dt��

�� = 2000

i.e.,1

�= 2000 or l = 0.0005/hour

P(T � 1 year) = P(T ≤ 365 × 1.8 hours) [� the T.V. is operated for 1.8 hours/day]= 1 – P{T > 657}= 1 – R(657)= 1 – e–0.0005 × 657

= 0.28i.e., 28% of the tubes will fail during the warranty period.

��# Night watchmen carry an industrial flashlight 8 hours per night, 7nights per week. It is estimated that on the average the flashlight is turned onabout 20 minutes per 8-hour shift. The flashlight is assumed to have a constantfailure rate of 0.08/hour while it is turned on and of 0.005/hour when it is turnedoff but being carried.

(a) Estimate the MTTF of the light in working hours.(b) What is the probability of the light’s failing during one 8-hour shift?(c) What is the probability of its failing during one month (30 days) of

8-hour shifts?

�$��% In the earlier problems, we have not taken into account the possibilitythat an item may fail while it is not operating. Often such failure rates are smallenough to be neglected. However, for items that are operated only a small fractionof time, failure during non-operation may be quite significant and so should betaken into account. In this situation, the composite failure rate λc is computed aslc = fl0 + (1 – f )lN, where f is the fraction of time the unit is operating and l0 andlN are the failure rates during operating and non-operating times]

Since the flashlight is used only a small fraction of time (20 minutes out of 8hours), we compute the composite failure rate lc and use it for other calculations.

lc = fl0 + (1 – f )lN

= 1

24 × 0.08 + 23

24 × 0.005 = 0.008125/hour

(a) MTTF =1

c� =

1

0.008125 = 123 hours

(b) P(T ≤ 8) =

8

0

ctce �� dt = 1–e–8lc = 0.0629

(c) p = probability of failure in one day (night) = 0.0629


q = 1 – p = 0.9371n = 30

∴ Probability of failure in 30 days= 1 – Probability of no failure in 30 days= 1 – nC0 p

0 qn, by binomial law= 1 – (0.9371)30

= 0.8576

��& A relay circuit has an MTBF of 0.8 year. Assuming random failures,(a) Calculate the probability that the circuit will survive 1 year without failure.(b) What is the probability that there will be more than 2 failures in the first year?(c) What is the expected number of failures per year?

Since the failures are random events, the number of failures in an interval of

length t follows a Poisson process, given by P{N(t) = n} = ( )

,!

t ne t

n

� �� n = 0, 1, 2,

… ∞, where l, = failure rate.

Then the time between failures follows an exponential distribution with mean 1

�.

Now MTBF = Mean time between failures = 1

�

= 0.8 year

∴ l = 1

0.8 per year = 1.25 per year

(a) P{N (1) = 0} = 0

0!

e �� = e–1.25 = 0.2865

(b) P{N(1) > 2} = 1 – e–l 0 1 2

0! 1! 2!

� � ��

= 1 – 0.2865 2(1.25)

1 1.252

� ��

= 0.1315(c) E{N(t)} = lt

∴ E{Number of failures per year} = l = 1.25

��' For a system having a Weibull failure distribution with a shapeparameter of 1.4 and a scale parameter of 550 days, find

(a) R(100 days); (b) the B1 life; (c) MTTF; (d) the standard deviation, (e) thedesign life for a reliability of 0.90.

��

The pdf of the Weibull distribution is given by f(t) = 1

. exp ,t t

� ��

� � �

� � �� t ≥ 0. Now b = 1.4 and q = 550 days


(a) R(t) = expt

�

�

� ��

∴ R(100) = 1.4

100exp

550

� �� = 0.9122

(b)

�$��% The life time corresponding to a reliability of 0.99 is called the B1 life.Similarly that corresponding to R = 0.999 is called the B.1 life.

Let tR be the B1 life of the systemThen R(tR) = 0.99

i.e.,1.4

exp550

Rt� �� = 0.99

∴1.4

550Rt� �

� �� = 0.01005

∴ tR = 550 × 1

1.4(0.01005) = 20.6 days

(c) MTTF = 1. 1�

�

� ��

= 550 × 11

1.4� �� = 550 × (1.71)

= 550 × 0.91057, from the Gamma tables

= 500.8 days

(d) (S.D.)2 = q2

2

2 11 1

� �

� ��

= 5502

22 1

1 11.4 1.4

� ��

= 5502 � �2

(2.43) (1.71)� ��

= 5502 � �2

1.43 (1.43) (1.71)� ��

= 5502 [1.43 × 0.88604 × (0.91057)2]

∴ S.D = 550 × 0.66174 = 363.96 days


(e) Let tD be the required design life for R = 0.90

∴ R(tD) = 1.4

exp550

Dt� �� = 0.90

∴ 1.4

550Dt� �

� �� = 0.10536

∴ tD = 550 × 1

1.4(0.10536) = 110.2 days

�� ( A device has a decreasing failure rate characterised by a two-parameter Wibull distribution with q = 180 years and b = 0.5. The device isrequired to have a design life reliability of 0.90.

(a) What is the design life, if there is no wear-in period?(b) What is the design life, if there is a wear-in period of 1 month in the

beginning?

�� (a) Let tD be the required design life for R = 0.90

∴ R(tD) = 0.5

exp180

Dt� �� = 0.90

∴0.5

180Dt� �

� �� = 0.10536

∴ tD = 180 × (0.10536)2 = 1.998 � 2 years.(b) If T0 is the wear-in period, then

R(t/T0) = 0 0expt T T

� �

� �

� �� Let tW be the required design life with wear-in.

Then

0.5 0.51 1

12 12exp180 180

Wt� ��

= 0.90

i.e., – 0.5

12 1

12 180Wt� ��

� �� +

0.51

12 180

� ��

= – 0.10536

i.e.,0.5

12 1

12 180Wt� ��

� �� = 0.12688

∴ 12tW + 1 = 12 × 180 × 0.01610 = 34776i.e., tW = 2.18 yearsThus a wear-in period of 1 month increases the design life by nearly 0.81 year or10 months.


�� Two components have the same MTTF; the first has a constantfailure rate l0 and the second follows a Weibull distribution with b = 2.

(a) Find q in terms of l0.(b) If for each component the design life reliability must be 0.9, how much

longer (in percentage) is the design life of the second (Weibull)component?

(a) MTTF is the same for both the components.For the constant failure rate distribution (viz., exponential distribution),

MTTF = 0

1

�(1)

For the Weibull distribution with b = 2

(viz., Rayleight distribution), R(t) = 2 2te ��

MTTF =0

( )R t dt�

� = 2 2

0

te dt��

��

= q 2

0

,xe dx�

�� on putting t

� = x

= q 2

� (2)

From (1) and (2), we get

2

� � =

0

1

�

∴ q = 0

2

� �(3)

(b) Let t1 and t2 be the required design lives for the two components.

Then R(t1) = 0 1te �� = 0.90, for the first.

∴ l0t = 0.10536

∴ t1 =0

0.10536

�(4)

R(t2) =2 22te �� = 0.90, for the second.

∴ t22 = 0.10536 q2

or t2 = 0.32459 q

= 0.32459 × 0

2

� �, using (3)

=0

0.36626

�(5)

Increase in design life = t2 – t1 = 0.26090/l0


∴ % increase in design life = 2 1

1

( )t t

t

� × 100

=0.26090

0.10536 × 100

= 247.6.

�� The wearout (time to failure) of a machine part is normallydistributed with 90% of the failures occurring symmetrically between 200 and270 hours of use.

(a) Find the MTTF and standard deviation of failure times.(b) What is the reliability if the part is to be used for 210 hours and then

replaced?(c) Determine the design life if no more than a 1% probability of failure

prior to the replacement is to be tolerated.(d) Compute the reliability for a 10-hour use if the part has been operating

for 200 hours.Let the time to failure T follow N(m, s), where s is the MTTF and s is the S.D.

��(a) 90% of the failures lie symmetrically between 200 and 270

∴235

200

( )f t dt� = 270

235

( )f t dt� = 0.45, where m = 235 hours

Putting z = 235t

�

�, we get

35

0

( )z�

�� dz = 0.45

From the normal tables,35

�= 1.645

∴ s = 35

1.645 = 21.28 hours

(b) R(t) = ( )t

f t dt�

�

∴ R(210) = 210

( )f t dt�

� = 1.17

( )z dz��

��

= 0.5 + 1.17

0

( )z dz�� dz = 0.5 + 0.379 = 0.879

(c)If tD is the required design life, then

( )Dt

f t dt�� = 0.01 or

( 235) 21.28

( )Dt

z dz�

�

�� = 0.01

From the normal tables, 235

21.28Dt � = – 2.32

∴ tD = 185.6 hours


(d) 0

0

( )

( )

R t T

R T

�=

(210)

(200)

R

R = 210

200

( )

( )

f t dt

f t dt

�

�

�

�

= 1.17

1.64

( )

( )

z dz

z dz

�

�

�

��

�

�

� =

0.5 0.3790

0.5 04495

��

0.93

�� A cutting tool wears out with a time to failure that is normallydistributed. It is known that about 34.5% of the tools fail before 9 working daysand about 78.8 % fail before 12 working days.

(a) Compute the MTTF.(b) Determine its design life for a reliability of 0.99.(c) Determine the probability that the cutting tool will last one more day

given that it has been in use for 5 days.(a) Let T follow a N(m, s)

��

Given9

( )f t dt�� = 0.345

i.e., ( )

n

z dz

��

�

�

�� = 0.345, on putting z =

t �

�

�

i.e.,

9

0

( )z dz

�

�

�

�

� = 0.155

∴ 9�

�

�= 0.4, (1)

using the normal tables.

Also12

( )f t dt�� = 0.788

i.e.,

12

( )z dz

��

�

�

�� = 0.788

i.e.,12

0

( )z dz�

�

�

� = 0.288

∴ 12 �

�

�= 0.8 (2)


using the normal tables.Solving equations (1) and (2), we get

m = 10 and s = 2.5i.e., MTTF = 10 days.(b) Let tR be the required design life for R = 0.99

∴ ( )Rt

f t dt�

� = 0.99 or 10

2.5

( )Rt

z dz��

�� = 0.99

∴

10

2.5

0

( )

Rt

z dz�

�

� = 0.49

∴ 10

2.5Rt� = 2.32, using the normal tables.

∴ tR = 4.2 days

(c) P(T ≥ 6/T > 5) = ( 6)

( 5)

P T

P T

�

= 6

5

( )

( )

f t dt

f t dt

�

�

�

�

= 1.6

2

( )

( )

z dz

z dz

�

�

�

��

�

�

�=

1.6

02

0

0.5 ( )

0.5 ( )

z dz

z dz

�

�

�

�

�

�

= 0.94520

0.97725= 0.9672.

�� A complex machine has a high number of failures. The time tofailure was found to be log normal with s = 1.25. Specifications call for a reliabilityof 0.95 at 1000 cycles.

(a) Determine the median time to failure that must be achieved by engineeringmodifications to meet the specifications. Assume that design changes donot affect the shape parameter s.

(b) What are the corresponding MTTF and standard deviation?(c) If the desired median time to failure is obtained, what is the reliability

during the next 1000 cycles given that it has operated for 1000 cycles?

(a) R(t) = ( )t

f t�

� dt, where f (t) is the pdf of the lognormal distribution =

log log Mt t

s

�

��

� f(z) dz, where f(z) is the pdf of the standard normal distribution


Now R (1000) = 0.95

∴1 1000

log

( )

Ms t

z��

� ��

� dz = 0.95

From the normal tables, 1 1000log

1.25 Mt

� ��

= 1.645

∴ log 1000

Mt� ��

= 1.645 × 1.25

∴ tM = Median = 1000 × exp (1.645 × 1.25= 7817 cycles

(b) MTTF = tM × 2 2se = 7817 × e0.78125

= 17074 cycles

s2 = t2M

2se (es2

– 1)

= (7817)2 × e1.5625 × (e1.5625 – 1)

∴ s = 33155 cycles

(c) R(t/T0) = 0

0

( )

( )

R t T

R T

�

= (2000)

(1000)

R

R= 2000

1000

( )

( )

f t dt

f t dt

�

�

�

�

= 1 2000

log1.25 7817

( )z dz�

�

� ��

� ÷ 1 1000

log1.25 7817

( )z dz�

�

� ��

�

= 0.47

( )z dz��

�� ÷

0.71

( )z dz��

��

= 0.6808

0.7611 = 0.89

��! Fatigue wearout of a component has a log normal distributionwith tM = 5000 hours and s = 0.20.

(a) Compute the MTTF and SD.(b) Find the reliability of the component ofr 3000 hours.(c) Find the design life of the component for a reliability of 0.95.


��

(a) MTTF = tM × 2 2se = 5000 × e0.02

= 5101 hours

s 2 = t2M ×

2 2

( 1)s se e �

= 50002 × e0.04 × (e0.04 – 1)∴ s = 1030 hours

(b) R(3000) = 3000

( )f t�

� dt, where f (t) is the pdf of the lognormal distribution

=1 3000

log0.2 5000

( )z dz�

�

� ��

� = 2.55

( )z dz��

�� = 0.9946

(c) If tD is the required design life,R(tD) = 0.95

i.e., 1

log0.2 5000

( )Dt

z dz�

�

� ��

� = 0.95

From the normal tables, 1

log0.2 5000

Dt� ��

= – 1.645

∴ tD = 3598.2 hours

�� (Short-answer questions)

1. What do you understand by reliability of a device?2. Write down the mathematical definition of reliability.3. Prove that 0 ≤ R(t) ≤ 1, where R(t) is the reliability function.4. What is hazard function? How is it related to reliability function?5. Derive the relation that expresses R(t) in terms of l(t).6. Derive the relation that expresses the pdf of the time to failure in terms of

hazard function.7. What is MTTF? How is it related to reliability function?8. Express the conditional reliability R(t/T0) in terms of the hazard function.9. Prove that the hazard function is a constant for the exponential failure distribution.

10. When the hazard function is a constant, prove that the failure timedistribution is an exponential distribution.

11. Write down the pdf of the constant failure rate distribution.


12. When the hazard function is a constant l, what are the MTTF and varianceof the time to failure?

13. Prove that, for an exponential distribution, R(t/T0) = R(t).14. What is memoryless property of the constant failure rate distribution?15. Write down the pdf of Weibull distribution with shape parameter m and

scale parameter n.16. Write down the formulas for R(t) and l(t) for a Weibull distribution with

shape parameter b and characteristic life q.17. When the time to failure T follows a Weibull distribution with parameters

b and q, what are MTTF and Var(T )?18. Derive the formula for R(t/T0) when T follows a Weibull distribution (b, q).19. Write down the density function of a lognormal distribution in terms of

its shape parameter and median time to failure.20. Write down the values of MTTF and s2

T for a lognormal distribution (s, tm).

��

21. The density function of the time to failure of an appliance is

f (t) = 32

( 4)t �; t(> 0) is in years.

(a) Find the reliability function R(t).(b) Find the failure rate l(t).(c) Find the MTTF.

22. A household appliance is advertised as having more than a 10-year life.If its pdf is given by

f (t) = 0.1(1 + 0.050t)–3; t ≥ 0determine its reliability for the next 10 years, if it has survived a 1-yearwarranty period.What is its MTTF before the warranty period?What is its MTTF after the warranty period assuming that it has stillsurvived?

23. A logic circuit is known to have a decreasing failure rate of the form

l(t) = 1

year, where is in years20

tt

(a) If the design life is one year, what is the reliability?(b) If the component undergoes wearin for one month before being put

into operation, what will the reliability be for a one-year designlife?

24. A component has the following hazard rate, where t is in years:l(t) = 0.4 4 t, t ≥ 0

(a) Find R(t).(b) Determine the probability of the component failing within the first

month of its operation.(c) What is the design life if a reliability of 0.95 is desired?

25. The pdf of the time to failure of a system is given by f(t) = 0.01, 0 ≤ t ≤100 days. Find.


(a) R(t); (b) the hazard rate function; (c) the MTTF; (d) the standarddeviation.

26. The failure-distribution is defined by

f(t) = 2

9

3

10

t, 0 ≤ t ≤ 1000 hours

(a) What is the probability of failure within a 100-hour warranty period?(b) Compute the MTTF.(c) Find the design life for a reliability of 0.99.(d) Compute the average failure rate over the first 500 hours.[Hint: Average failure rate of a component over the interval (t1, t2) isdefined as

AFR(t1, t2) = 2

12 1

1( )

t

t

t dtt t

�� = 1

2 1 2

( )1log

( )

R t

t t R t

� ��

27. The pdf of the time to failure of a new fuel injection system whichexperiences high failure rate is given byf(t) = 1.5/(t + 1)2.5, t ≥ 0, where t is measured in years. The reliabilityover its intended life of 2 years is unacceptable. Will a wear-in period of6 months significantly improve upon this reliability? If so, by how much?

28. A home computer manufacturer determines that his machine has a constantfailure rate of l = 0.4 per year in normal use. For how long should thewarranty be set, if no more than 5% of the computers are to be returned tothe manufacturer for repair?[Hint: Find to such that R(T0) ≥ 0.95]

29. A more general exponential reliability model is defined by R(t) = a–bt;a > 1; b > 0, where a and b are parameters. Find the hazard rate functionand show how this model is equivalent to R(t) = e–lt.

30. Show that, for an exponential distribution (l), the residual mean life is1

�, regardless of the length of the time the system has been operating.

[Hint: MTTF (t/T0) =

0

0( )T

R t T�

� dt]

31. The one-month reliability on an indicator lamp is 0.95 with the failurerate specified as constant. What is the probability that more than twospare bulbs will be needed during the first year of operation (Ignorereplacement time).

32. A turbine blade has demonstrated a Weibull failure pattern with adecreasing failure rate characterised by a shape parameter of 0.6 and ascale parameter of 800 hours.(a) Compute the reliability for a 100-hour mission.(b) If there is a 200-hour burn-in of the blades, what is the reliability

for a 100-hour mission?33. Two components have the same MTTF; the first has a constant failure

rate l = 0.141 and the second follows a Weibull distribution with shapeparameter equal to 2.


(a) Find the characteristic life of the second component.(b) If for each component the design life reliability must be 0.8, how much

longer (in percentage) is the design life of the second component thanthe first?

34. A pressure gauge has a Weibull failure distribution with a shape parameterof 2.1 and a characteristic life of 12,000 hours. Find (a) R(5000); (b) theBl and B.1 lives; (c) the MTTF and (d) the probability of failure in thefirst year of continuous operation.

35. A component has a Weibull failure distribution with b = 0.86 and q =2450 days. By how many days will the design life for a 0.90 reliabilityspecification be extended as a result of a 30-day burn-in period?

36. For a three-parameter Weibull distribution with b = 1.54, q = 8500 andt0 = 50 hours, find (a) R(150 hours); (b) MTTF; (c) SD and (d) the designlife for a reliability of 0.98.[Hint: The pdf of a 3-parameter Weibull distribution is f(t) =

10 0. exp

t t t t� �

�

� � �

� � �� ; t ≥ t0. All formulas for this distribution

can be had from the corresponding formulas for the 2-parameter Weibulldistribution by replacing t by (t – t0).

MTTF = t0 + 11�

�

� �� 37. A failure PDF for an appliance is assumed to be a normal distribution

with a mean of 5 years and an S.D. of 0.8 year. Find the design life of theappliance for (a) a reliability of 90%, (b) a reliability of 99%.

38. A lathe cutting tool has a life time that is normally distributed with anS.D. of 12.0 (cutting) hours. If a reliability of 0.99 is desired over 100hours of use, find the corresponding MTTF. If the reliability has to lie in(0.8, 0.9), find the range within which the tool has to be used.

39. A rotor used in A.C. motor has a time to failure that is lognormal with anMTTF of 3600 operating hours and a shape parameter s equal to 2.(a) Determine the probability that the rotor will survive the first 100 hours.(b) If it has not failed till the first 100 hours, what is the probability that it

will survive another 100 hours?40. The life time of a mechanical valve is known to be lognormal with median

= 2236 hours and s = 0.41.(a) Find the design life for a reliability of 0.98.(b) Compute the MTTF and S.D. of the time to failure.(c) The component is used in pumping device that will require 5 weeks of

continuous use. What is the probability of a failure occurring becauseof the valve?

��$)��*��*

As was pointed out, a system is generally understood as a set of components


assembled to perform a certain function. In the previous section we have consid-ered a number of important failure models (distributions) for the individual components. To evaluate the reliability of a complex system, we may apply a particular-failure law to the entire system. But it will be proper if we determine anappropriate reliability model for each component and then compute the reliabilityof the system by applying the relevant rules of probability according to theconfiguration of the components within the system. In this section, we shall discussthe system reliability in respect of a few simple but relatively important cases.

�� Series or nonredundant configuration is one in which the components of thesystem are connected in series (or serially) as shown in the following reliabilityblock diagram (Fig. 1.1). Each block represents a component.

��+��+�

In series configuration, all components must function for the system to function.In other words the failure of any component causes system failure.

Let R1(t), R2(t) and Rs(t) be the reliabilities of the components C1 and C2 andthe system (assuming that there are only 2 components in series),Then R1 = P(C1) = probability that C l functionsand R2 = P(C2) = probability that C2 functionsNow Rs = probability that both C1 and C2 function

= P(C1 ∩ C2) = P(C1), P(C2), assuming that C1 and C2 function independently.= R1 × R2

This result may be extended. If C1, C2, ..., Cn be a set of n independent componentsin series with reliabilities R1(t), R2(t), ..., Rn(t), then

Rs(t) = R1(t) × R2(t) × ... × Rn(t)≤ min {R1(t), R2(t), ..., Rn(t)} [� 0 < Ri(t) < 1]

i.e., the system reliability will not be greater than the smallest of the componentreliabilities.

��

1. If each component has a constant failure rate li, then

Rs(t) = 1 2. ntt te e e ��

= 1 2( )n te � � �� = ste ��

This means that the system also has a constant failure rate ls = 1

.n

ii

��

2. If the components follow the Weibull failure law the parameters bi andqi, then

Rs(t) = 1 2

1 2

exp exp expn

n

t t t��

� � �

� ��

�

= 1

expn

iii

t�

��

� ��


This means that the system does not follow Weibull failure law, even thoughevery component follows a Weibull failure distribution.

� �� Parallel or redundant configuration is one in which the components of the system areconnected in parallel as shwn in the following reliability block diagram (Fig. 1.2).

In parallel configuration, all components must fail for the system to fail. Thismeans that if one or more components function, the system continues to function.Taking n = 2 and denoting the system reliability by Rp(p‘p’ for parallelconfiguration), we have

Rp = P(C1 or C2 or both function)= P(C1 ∪ C2)= P(C1) + P(C2) – P(C1 ∩ C2)=P(C1) + P(C2) – P(C1) P(C2), since C1 and C2 are independent= R1 + R2 – R1 R2 = 1 – (1 – R1) (1 – R2)

Extending to n components, we haveRp = 1 – (1 – R1) (1 – R2) ... (1 – Rn)

≥ Max {R1, R2, ..., Rn}

�$��% For a two component system in parallel having constant failure rate,

Ri(t) = 1 – ite �� ; i = 1, 2.

Hence Rp(t) = 1 – 1 2(1 )(1 )t te e� ��

= 1 2t te e� �� – 1 2( )te � ��

and MTTF = 0

( )pR t dt�

�

= 1

0

te dt��

�� + 2

0

te dt��

�� – 1 2( )

0

te dt� ��

� ��

= 1 2 1 2

1 1 1

� � � ��

�.


� �� A system, in which m subsystems are connected in series where each subsystemhas n components connected in parallel as shown in Fig. 1.3, is said to be inparallel series configuration or low-level redundancy.

��+��+�

If R is the reliability of the individual component, the reliability of each of thesubsystems = 1 – (1 – R)n (In the diagram n = 2)Since m substystems are connected in series [In the Fig. 1.3, m = 3], the system

reliability for the low-level redundancy is given byRLow = {1 – (1 – R)n}m

�� A system, in which m subsystems are connected in parallel where each subsystemhas n components connected in series as in Fig. 1.4, is said to be in series parallelconfiguration or high-level redundancy.

��+��+

If R is the reliability of each component, the reliability of each of the subsystems= Rn (In Fig. 1.4, n = 2)

Since m subsystems are connected in parallel (In the diagram m = 3), the systemreliability for the high-level redundancy is given by

RHigh = 1 – (1 – Rn)m

�$��% When m = n = 2, RLow ≥ RHigh, since

RLow – RHigh = {1 – (1 – R)2}2 – {1 – (1 – (1 – R2)2}= (2R – R2)2 – (2R2 – R4)= 2R2 – 4R3 + 2R4

= 2R2 (1 – R)2 ≥ 0.

�� Markov analysis can be applied to compute system reliability. Fopr simplicity,we consider a system containing two components. To use Markov analysis, the


system is considered to be in one of the four states at any time as detailed inTable 1.1:

��

State Component 1 Component 2

1 operating operating

2 failed operating

3 operating failed

4 failed failed

Since the probability that the system undergoes a transition from one state toanother depends only on the present state but not on any of its previous states, wecan use Markov analysis to compute Pi(t), the probability that the system is instate i at time t and hence the system reliability.

If we assume that the components have constant failure rates l1 and l2, the statetransition diagram of the system will be as shown in Fig. 1.5: The nodes in the diagramrepresent the states of the system and the branches represent the transition rates fromone node to another which will be the same as instantaneous failure rates.

��+��+!

Now P1(t + ∆t) = P{the system is in state 1 at time t and neither Component 1 nor Component 2 fails during (t, t + ∆t)}= P1(t) × P{Component 1 does not fail in ∆t interval} ×P{Component 2 does not fail in ∆t interval}=P1(t) {1 – l1 Dt} {1 – l2 ∆t}, [since P{Component i fails in ∆t interval} = li – ∆t and P{Component i does not fail in ∆t interval} = 1 li ∆t]

= P1(t) {1 – (l1 + l2) ∆t}, omitting (∆t)2 term.

∴ 1 1( ) ( )P t t P t

t

� � ��

= – (l1 + l2) P1(t)

Taking limits on both sides as ∆t → 0, we get

1( )dP t

t� = – (l1 + l2) P1(t) (1)

�$��% Equation (1) and similar equations corresponding to the other nodesmay be obtained mechanically as follows:


L.H.S. of the equation is P¢i(t). To obtain the R.H.S. of the equation, we notedown the arrows entering and/or leaving the node i. If it is an entering arrow, wemultiply the transition rate corresponding to that arrow with the probabilitycorresponding to the node from which it emanates. If it is a leaving arrow, wemultiply the negative of the transition rate corresponding to that arrow with theprobability corresponding to the node from which it emanates (viz., the currentnode). Then the R.H.S. is obtained by adding these products.

For the other states, the corresponding equations are

2 ( )dP t

dt= l1.P1(t) – l2. P2(t) (2)

3 ( )dP t

dt= l2 P1(t) – l1 p3(t) (3)

4 ( )dP t

dt= l2 P2(t) + l1 P3(t) (4)

Solving equation (1), we get

P1(t) = c1 1 2( )te � �� (5)

Initially, P1(0) = P(the system is in state 1 at t = 0)= P (both the components function at t = 0)= 1

Using this initial condition in (5), we get c1 = 0

∴ P1(t) = 1 2( )te � �� (6)Using (6) in (2), we have

P ′2(t) + l2 P2(t)= l1 1 2( )te � �� (7)

I.F. equation (7) = 2te�

∴ Solution of equation (7) is

22. ( )te P t� = c2 – 1te �� (8)

Using the initial condition P2(0) = 0 in (8), we get c2 = 1∴ Solution of equation (2) becomes

P2(t) = 2te �� – 1 2( )te � �� (9)Similarly, the solution of equation (3) is

P3(t) = 1te �� – 1 2( )te � �� (10)To get P4(t), we need not solve equation (4), but we may use the relation

P1(t) + P2(t) + P3(t) + P4(t) = 1 (11)as the system has to be in any one of the four mutually exclusive and exhaustivestates.

Now the system reliability depends on the configuration.For series configuration,

Rs(t) = P(both components are functioning)= P(the system is in state 1)


= P1(t) = 1 2( )te � �� , which agrees with the result derived already. For parallel configuration,

Rp(t) = P(at least one of the components function)= P(the system is in state 1, 2 or 3)= P1(t) + P2(t) + P3(t)

= 1te �� + 2te �� – 1 2( )te � �� which agrees with the result derived already.

�� In parallel configuration with two components, we have assumed so far that boththe components are operating from the start. Such a system is called an activeparallel (redundant) system. Now we shall consider the parallel configurationwith two components in which only one component will be operating from thestart and the other, called the standby or backup component, will be kept in reserveand brought into operation only when the first fails. Such a system is called apassive parallel system or standby redundant system.Let us now compute the reliability of such a standby system with the simplifyingassumption that the standby unit does not fail while it is kept in standby mode (inreserve)Let l1 and l2 be the constant failure rates of the active unit and the standby unit(when operative) respectively. This system will be in any of three states describedbelow:State 1 Main unit is functioning and the other in standby modeState 2 Main unit has failed and the standby unit is functioningState 3 Both units have failed.The state transition diagram for the standby system is given in Fig. 1.6:

��+��+"

From Fig. 1.6, we get the following Markov equation easily as explained earlier.P¢1(t) = – l1P1(t) (1)P ′2 (t) = l1P1(t) – l2P2(t) (2)P ′3 (t) = l2P2(t) or P1(t) + P2(t) + P3(t) = 1 (3)

Solving equation (1), we get P1(t) = 11

tc e �� (4)

Using the initial condition P1(0) = 1, we get C1 = 1

∴ Solution of equation (1) in P1(t) = 1te �� (5)Using (5) in (2), it becomes


P ′2(t) + l2P2(t) = l1 1te �� (6)

I.F. of equation (6) = 2te ��

∴ Solution (2) is

2te� P2(t) = 2 1( )12

2 1

te c� ��

� �

� ��

(7)

Using the initial condition P2(0) = 0 in (7),

we get c2 = – 1

2 1

�

� ��

Using this value in (7), The solution of (2) is got as

P2(t) = 1 21

2 1

( )t te e� ��

� �

� ��

(8)

Using (5) and (8) in (3), we can get the value of P3(t).Now the reliability of the standby system is given by

R(t) = P(main unit or standby unit is functioning)= P(system is in state 1 or 2)= P1(t) + P2(t)

= 1 1 21

2 1

( )t t te e e� � ��

� �

� � ��

= 1 212 1

2 1

( )t te e� ��

� �

� ��

�$��% 1. When the failure rates of the main and standby units are equal, viz.,when l1 = l2 = l,

R(t) = � �( )

0

1lim ( ) t h t

hh e e

h� ��

!

� ��

= 0

1lim 1

htt

h

ee

h� �

��

!

� ��

= 2 2

0lim 1 .

2!t

h

h te ht

h� ��

!

� ��

�

= (1 + lt) e–lt

2. When l1 = l2 = l,(i) MTTF (for active redundant system)

= 2

0

(2 )t te e dt� ��

� �� = 3

2�


(ii) MTTF (for standby redundant system)

= 0

(1 ) tt e dt��

�� = 2

�.

�� An electronic circuit consists of 5 silicon transistors, 3 silicon diodes,10 composition resistors and 2 ceramic capacitors connected in seriesconfiguration. The hourly failure rate of each component is given below:

Silicon transistor : lt = 4 ¥ 10–5

Silicon diode : ld = 3 ¥ 10–5

Composition resistor : lr = 2 ¥ 10–4

Ceramic capacitor : lc = 2 ¥ 10–4

Calculate the reliability of the circuit for 10 hours, when the components fol-low exponential distribution.

��Since the components are connected in series, the system (circuit) reliability

is given byRs(t) = R1(t) . R2(t) . R3(t) . R4(t)

= 31 2 4. . .tt t te e e e��

= (5 3 10 2 )t d r c te � � � ��

∴ Rs(10) = 5 5 4 4(20 10 9 10 20 10 4 10 ) 10e� � � ��

= 4(20 9 200 40) 10e��

= e–0.0269 = 0.9735.

�� There are 16 components in a non-redundant system. The averagereliability of each component is 0.99. In order to achieve at least this systemreliability by using a redundant system with 4 identical new components, whatshould be the least reliability of each new component?For the non-redundant system,

��Rs = R16 = (0.99)16 = 0.85

Let the new components have a reliability of R′ each.Then for the redundant system with 4 components, Rp ≥ 0.85i.e., 1 – (1 – R′)4 ≥ 0.85i.e., (1 – R′)4 ≤ 0.15

∴ 1 – R ′ ≤ 1

4(0.15) or 0.62∴ R ′ ≥ 0.38i.e., the reliability of each of the new components should be at least 0.38.

�� Thermocouples of a particular design have a failure rate of 0.008per hour. How many thermocouples must be placed in parallel if the system is torun for 100 hours with a system failure probability of no more than 0.05? Assumethat all failures are independent.


��If T is the time to failure of the system, it is required that

P(T ≤ 100) ≤ 0.05i.e., 1 – Rp(100) ≤ 0.05 (1)Let the number of thermocouples to be connected in parallel be n.Then Rp(t) = 1 – (1 – R)n (2)where R is the reliability of each couple.

The failure rate of each couple = 0.008 (constant)∴ R = e–0.008t (3)Using (3) in (2), we have

1 – Rp(t) = (1 – e–0.008t)n

∴ 1 – Rp(100) = (1 – e–0.8)n (4)Using (4) in (1), we have

(1 – e–0.8)n ≤ 0.05i.e., (0.55067)n ≤ 0.05 (5)By trials, we find that (5) is not satisfied when n = 0, 1, 2, 3, 4 and 5.When n = 6, (0.55067)n = 0.02788 < 0.05Hence 6 thermocouples must be used in the parallel configuration.

�� Two parallel, identical and independent components have constantfailure rate. If it is desired that Rs(1000) = 0.95, find the component and systemMTTF. In addition to the independent components if a common mode componentwith a constant failure rate of 0.00001 is connected to the system, find the newsystem MTTF.

��Let R and l be the reliability and failure rate of each of the two independent

components. Then R(t) = e–lt

Since the two components are connected in parallel,Rs(t) = 1 – {1 – R(t)}2

= 1 – {1 – e–lt}2

i.e., 2e –lt – e–2lt = Rs(t)∴ 2e–1000l – e–2000l = Rs(1000) = 0.95i.e., x2 – 2x + 0.95 = 0, on putting x = e–1000l

∴ x = 2 4 3.8

2

" �= 1.22361 or 0.77639

i.e., e–1000l = 1.22361 or 0.77639∴ – 1000l = 0.20181 or – 0.25310Rejecting the value 0.20191, we get

l = 0.0002531

∴ Component MTTF = 1

� =

1

0.0002531 = 3951

and system MTTF = 0

( )sR t dt�

�


= 2

0

(2 )t te e dt� ��

� ��

= 3

2� =

3

0.0005062 = 5926.5.

�$��% Common Mode Failure Component

We have assumed that the components connected in series and parallelconfigurations are independent. This assumption of independence of failuresamong the components within a system may be easily violated. For example, thecomponents in the system may share the same power source, environmental dust,humidity, vibration, etc., resulting in what is known as ‘common mode failure’.This common mode failure is represented by including an extra (common mode)component in series with those components that share the failure mode.

Let l′ be the failure rate of common mode component connected in series withthe already existing redundant system. Then the system reliability becomes

R′s(t) = Rs(t) ⋅ R′(t), where R′(t) is the reliability of thecommon mode component.

= 2e–lt – e–2lt) e–l′t

∴ R′s(1000) = (2e–0.2531 – e–0.5062) × e–0.01

(� l = 0.0002531 and l′ = 0.00001) = 0.94Now the new system MTTF is given by

(MTTF)′ = ( ) (2 )

0

[2 ]te e dt� � � ��

� � � ��

= 2

� �� –

1

2� ��

= 2

0.0002631 –

1

0.0005162 = 5664.4.

��! It is known that the reliability function for a critical solid state

power unit for use in a communication satellite is R(t) = 10

10 t�, t ≥ 0 and t

measured in years.(a) How many units must be placed in parallel in order to achieve a reliability

of 0.98 for 5 year operation?(b) If there is an additional common mode with constant failure rate of 0.002

as a result of environmental factors, how many units should be placed inparallel?

��(a) Let n units be placed in parallel.

Then Rs(t) = 1 – 10

110

n

t� ��


As Rs(5) ≥ 0.98, we have

1 – 2

13

n� �� ≥ 0.98

i.e.,1

3n ≤ 0.02 or 3n ≥ 50 ∴ n = 4

(b) When the additional common mode unit is also used.

Rs(t) = 10

110

n

t

� ��

× e–0.002t

As Rs(5) ≥ 0.98, we have

11

3

n� �� e–001 ≥ 0.98

i.e., 1 – 1

3n≥ 0.98

0.99

i.e.,1

3n≤

0.981

0.99� ��

i.e., 3n ≥ 99 (nearly) ∴ n = 5.

,�-��" If two identical components having a guaranteed life of 2 months anda constant failure rate of 0.15 per year are connected in parallel, what is thesystem reliability for 10,000 hours of continuous operation?

If Rc(t) is the reliability of each component, then Rc(t) = e–0.15t (t in years)

Each component has a guaranteed life (wear-in period) of 2 months or 1

6 year.

∴ = Rc(t/t0) = 0

0

( )

( )c

c

R t t

R t

� =

10.15( )

6

10.15

6

te

e

� �

� � = e–0.15t

If Rs(t) is the system reliability, thenRs(t) = 2Rc(t/t0) – {Rc(t/t0)}

2

= 2 e–0.l5t – e–0.30t

∴ Rs(10000 hours) = Rs(1.l4 year)= 2e–0.15×1.14 – e–0.30×l.14

= 0.975.

��# For a redundant system with n independent identical componentswith constant failure rate l, show that the MTTF is equal to

1

1

1 ( 1)n i

ii

nCi�

�

�

�


If A = 0.01/hour, what is the minimum value of n so that the MTTF is at least 200hours?

��If Rs(t) is the system reliability,

Rs(t) = 1– (1– e–lt)n, since the component reliability = e–lt

= 1 – 0

( 1) .n

i i ti

i

nC e ��

��

= 1

1

( 1) .n

i i ti

i

nC e ��

��

∴ MTTF = 0

( )sR t dt�

� = 1

1 0

( 1)n

i i ti

i

nC e dt��

� �

��

= 1

1

1 ( 1)n i

ii

nCi�

�

�

�

When n = 2, MTTF = 100 2 1

1

( 1)2

i

ii

Ci

�

�

�

= 100 1

22

� �� = 150

When n = 3, MTTF = 100 3 1

1

( 1)3

i

ii

Ci

�

�

�

= 100 2 1

32 3

� �� = 100 × 11

6 = 183

When n = 4, MTTF = 100 4 1

1

( 1)4

i

ii

Ci

�

�

�

= 100 × 25

12 = 208

Hence the required minimum value of n = 4.

��& A jet engine consists of 5 modules (connected in series) each ofwhich was found to have a Weibull failure distribution with a shape parameter of1.5. Their characteristic lives are (in operating cycles) 3600, 7200, 5850, 4780and 9300. Find the reliability function of the engine and the MTTF.

If R(t) is the reliability function of the engine (system) thenR(t) = R1(t) . R2(t) ... R5(t)

= 1.51.5

51 21.5( )( ) ( ) tt te e e �� # �

= 1.5 1.5 1.5 1.5

1 2 5( )t

e� � ��


= 1.50.000012642te�

=

1.5

1842.7

t

e� ��

This shows that the engine failure time also follows a Weibull’s distribution withb = 1.5 and q = 1842.7.

MTTF of the engine = 11�

�

� ��

= 1842.7 × 1

11.5

� ��

= 1842.7 × 0.9033= 1664.5 cycles.

��' A signal processor has a reliability of 0.90. Because of the lowerreliability a redundant signal processor is. to be added. However, a signal splittermust be added before the processors and a comparator must be added after thesignal processors. Each of the new components has a reliability of 0.95. Doesadding a redundant signal processor increase the system reliability?

��+��+#

R1 = R(first subsystem) = R(S1) ⋅ R(P1) ⋅ R(C1)= 0.95 × 0.90 × 0.95= 0.81225

R2 = R(second subsystem) = R(S2) ⋅ R(P2) ⋅ R(C2)= (0.95)3

= 0.857375R(system) = 1 – (1 – R1) (1 – R2)

= 1 – 0.0268 = 0.9732The addition of a redundant signal processor increases the reliability.

��( Six identical components with constant failure rates are connectedin (a) high level redundancy with 3 components in each subsystem (b) low levelredundancy with 2 components in each subsystem. Determine the componentMTTF in each case, necessary to provide a system reliability of 0.90 after 100hours of operation.


Let l be the constant failure rate of each component. Then R = e–l t, for eachcomponent. For high level redundancy,

Rs(t) = 1 – [1 – {R(t)}3]2

= 1 – (1 – e–3lt )2

∴ Rs(100) = 1 – (1 – e–300l)2 = 0.90i.e., (1 – e–300l)2 = 0.1i.e., 1– e –300l = 0.31623∴ e–300l = 0.68377∴ 300l = 0.38013

∴ MTTF of each component = 1 300

0.38013�� = 789.2 hours.

For low level redundancyRs(t) = [1 – {1 – R(t)}2]3

= [1 – {1 – e –l t}2]3

∴ Rs(100) = [1 – {1 – e–100l}2]3 = 0.90i.e., 1 – (1 – e –100l)2 = 0.96549∴ (1 – e–100l)2 = 0.03451∴ 1 – e –100l = 0.18577∴ e –100l = 0.81423∴ 100l = 0.20551

∴ MTTF of each component = 1 300

0.20551�� = 486.6 hours.

�� A system consists of two subsystems in parallel. The reliability of

each sub system is given by (Weibull failure) R(t) =

2t

e �

� �� . Assuming that commonmode failure may be neglected, determine the system MTTF.

��The system reliability is given by

Rs(t) = –

2 2

1

t

e �

� ��

= 1 – 2 2 2 221 2 t te e� ��

= 2 2 2 222 t te e� ��

∴ System MTTF = 0

( )sR t dt�

� = 22 2 2 22

0 0

t te dt e dt� ��

� ��

= 2 2

0 0

2 ,2

x ye dx e dy�

��

� �# � #� �


(on putting t = q x in the first integral and t = 1

2 q y in the second integral.)

= 2q × 2 2 2

� � ��

= 1.15 q.

�� A system is designed to operate for 100 days. The system consistsof three components in series. Their failure distributions are (1) Weibull withshape parameter 1.2 and scale parameter 840 days; (2) longnormal with shapeparameter(s) 0.7 and median 435 days; (3) constant failure rate of 0.0001.

(a) Compute the system reliability.(b) If two units each of components 1 and 2 are available, determine the

high level redundancy reliability. Assume that the components 1 and 2can be configured as a sub-assembly.

(c) If two units each of components 1 and 2 are available, determine the lowlevel redundancy reliability.

��(a) For the first (Weibull) component,

R1(t) = expt

�

�

� �� =

1.2

exp840

t� ��

∴ R1(100) = 1.2

100exp

840

� �� = 0.9252

For the second (longnormal) component,

R2(t) = ( ) ,t

f t dt�

� where f(t) is the lognormal pdf

∴ R2(100) = 100

( )f t dt�

�

= 1 100

log0.7 435

( ) ,z dz�

�

� ��

� wher ( ) is the standard

normal density

f z

= 2.10

( )z dz��

�� = 0.9821

For the third (constant failure rate) component,R3(t) = e–lt = e–0.0001t

∴ R3(100) = e–0.01 = 0.9900Since the three components are connected in series,

Rs(100) = R1(100) × R2(100) × R3(100)= 0.9252 × 0.9821 × 0.9900= 0.8996


(a)

��+��+&

(b)

��+��+'

RHL = 1 – (1 – R1R2)2

= 1 – {1 – 0.9252 × 0.9821}2

= 0.9917RLL = {1 – (1 – R1)

2} × {1 – (1 – (1 – R2)2}

= 0.9944 × 0.9997= 0.9941

�� Find the variance of the time to failure for two identical units,each with a failure rate l, placed in standby parallel configuration. Compare theresult with the variance of the same two units placed in active parallelconfiguration. Ignore switching failures and failures in the standby mode.

��For standby parallel configuration:

R(t) = (1 + lt) e–lt

f(t) = – R′(t) = – [le–l t – l(1 + lt) e–lt]= l2t e–l t

MTTF = E(T ), where T is the time to failure of the system

= 0 0

( ) or ( )tf t dt R t dt� ��

� �

� �

= l2 22 3

0

2 2t t te e e

t t� � �

� � �

��

= 2

�

E(T 2) = 2

0

( )t f t dt�

� = 2 3

0

tt e dt��

��

= l2 3 22 3 4

0

3 6 6t t t te e e e

t t t� � � �

� � � �

��


= 2

6

�

Variance of T = E(T 2) – E 2(T )

= 2

6

� =

2 2

4 2

� �� (1)

For the active parallel configuration:R(t) = 2e–lt – e–2lt

∴ f (t) = – R ′(t) = 2le–lt – 2le–2lt

MTTF = E(T) = 0 0

( ) or ( )R t dt t f t dt� ��

� �

� �

= 2

0

2

2

t te e� �

� �

��

= 2 1 3

2 2� � ��

E(T 2) = 2

0

( )t f t dt�

�

= 2l 2 2 2

0 0

t tt e dt t e dt� ��

� ��

� � ��

= 2l 22 3

0

2 2t t te e e

t t� � �

� � �

��

– 2 2 2

22 3

0

2 22 4 8

t t te e et t

� � �

� � �

��

= 2l 3 3

2 1

4� �

� �� =

2

7

2�

Var (T) = E(T 2) – E 2(T)

= 2

7

2� –

2

9

4� =

2

5

4�(2)

Comparing (1) and (2), we see that the variance is greater for standby parallelsystem

�� A computerised airline reservation system has a main computer onlineand a secondary standby computer. The online computer fails at a constant rate of0.001 failure per hour and the standby unit fails when online at the constant rate of0.005 failure per hour. There are no failures while the unit is in the standby mode.


(a) Determine the system reliability over a 72 hours period.(b) The airline desires to have a system MTTF of 2000 hours. Determine the

minimum MTTF of the main computer to achieve this goal, assumingthat the standby computer MTTF does not change.

(a) l1 = 0.001/hour; l2 = 0.005/hour

Rs(t) = 1 22 1

2 1

1{ }t te e� ��

� �

� ��

= 1

0.004 {0.005 e–0.001t – 0.001 e– 0.005t}

∴ Rs (72) = 1

4 {5e–0.072 – e–0.360}

= 0.9887

(b) MTTF = 0

( )R t dt�

�

= � �1 22 1

2 1 0

1 t te e dt� ��

��

� �

= 2 1

2 1 1 2

1 � �

� � � �

� ��

= 2 1

1 2 1 2

1 1� �

� � � �

��

The requirement is 1 2

1 1

� �� = 2000

∴1

1

� = 2000 –

1

0.005 (� MTTF and hence l2 do not change

for standby unit)= 1800

i.e., MTTF of the main computer should be increased to 1800 hours.

��! A fuel pump with an MTTF of 3000 hours is to operate continuouslyon a 500 hour mission.

(a) What is the mission reliability?(b) Two such pumps are put in standby parallel configuration. If there are no

failures of the back up pump while in standby mode, what are the systemMTTF and the mission reliability?

(c) If the standby failure rate is 75% that of the main pump (whenoperational), what are the system MTTF and the mission reliability?

��

(a) MTTF = 1

1

� = 3000 ∴ l =

1

3000R(t) = e–lt ∴ R(500) = e–(500/3000) = 0.8465


(b) l1 = l2 = l = 1

3000Rs(t) = (1 + lt)e–lt

∴ Rs(500) = 500

13000

� �� e–(500/300) = 0.9876

MTTF = 2

� = 6000 hours

(c) l1 = 1

3000; l2 =

3

4 ×

1 1

3000 4000�

Rs(t) = � �1 22 1

2 1

1 t te e� ��

� ��

= 12,000 × 1 1

4000 30001 1

3000 4000

t te e� ��

��

∴ Rs(500) = (4e –1/8 – 3e–1/6)= 0.9905

MTTF = 1 2

1 1

� �� = 7000 hours.

�� (Short answer questions)

1. What do you mean by non-redundant and redundant configurations?2. Derive the reliability of a system consisting of two components in series

in terms of reliabilities of the components.3. Derive the reliability of a system consisting of two components in parallel

in terms of the reliabilities of the components.4. Prove that the reliability of a non-redundant system cannot exceed the

least component reliability.5. Prove that the reliability of a redundant system is not less than the

maximum component reliability.6. Find the MTTF of (a) a non-redundant system (b) a redundant system

consisting of two constant failure rate components.7. Explain low level redundancy with a diagram.8. Obtain the reliability of a low level redundancy with m subsystems each

of which consists of n components.9. Explain high level redundancy with a diagram.

10. Obtain the reliability of a high level redundancy with m subsystems eachof which consists of n components.


11. What is the difference between active and standby redundant systems?12. Write down the formulas for the reliability of a standby redundant system,

when the main and standby components have(i) unequal constant failure rates and

(ii) equal constant failure rates.13. What are the MTTF’s for active and standby redundant systems when the

main and the standby components have unequal constant failure rates.14. What are the MTTF’s for active and standby redundant systems when the

main and standby components have equal constant failure rates.15. Find the reliability of an aircraft electronic system consisting of sensor,

guidance, computer and fire control subsystems with respectivereliabilities 0.80, 0.90, 0.95 and 0.65 are connected in series.

16. In a lead there are 16 glow bulbs connected in series and the averagereliability of each bulb is 0.99. Compute the reliability of the lead.

17. An equipment consists of 3 components A, B and C in parallel and therespective reliabilities are RA = 0.92, RB = 0.95 and RC = 0.96. Calculatethe equipment reliability.

18. If the reliability of each of 3 components connected in parallel is 0.9,calculate the reliability of the system.

19. If RA = 0.8, RB = 0.9 and RC = 0.85 and if A, B, C are connected in seriesto form a subsystem, what is the reliability of the high level redundantsystem with two such subsystems.

20. If two A’s, two B’s and two C’s in Question (19) are connected in paralleland the three subsets are connected to form a low level redundant system,find the reliability of the system.

��

21. The time to failure (in years) of a certain brand of computer has the pdff (t) = 1/(t + 1)2; t > 0

(a) If three of these computers are placed in parallel aboard the proposedspace station, what is the system reliability for the first 6 months ofoperation?

(b) What is the system design life in days if a reliability of 0.999 required?22. Ten Weibull components, each having a shape parameter of 0.80 must

operate in series. Determine a common characteristic life in order thatthey have a design life of 1 year with a reliability of 0.99.

23. If three components, each with constant failure rate are placed in parallel,determine the system reliability for 0.1 year and the MTTF. Their failurerates are 5 per year, 10 per year and 15 per year.

24. Specifications for a power unit consisting of three independent and seriallyconnected components require a design life of 5 years with a 0.95reliability.

(a) If the constant failure rates l1, l2, l3 are such l1 = 2l2 and l3 =3l2, what should be the MTTF of each component of the system?


(b) If two identical power units are placed in parallel, what is thesystem:i reliability at 5 years and what is the system MTTF?

25. A power supply consists of three rectifiers in series. Each rectifier has aWeibull failure distribution with b = 2.1. However they have differentcharacteristic lifetimes given by 12,000 hours, 18,500 hours and 21,500hours. Find the MTTF and the design life of the power supplycorresponding to a reliability of 0.90.

26. What is the maximum number of identical and independent Weibull com-ponents having a scale parameter of 10000 operating hours and a shapeparameter of 1.3 that can be put in series if a reliability of 0.95 at 100operating hours is desired? What is the resulting MTTF?

27. Which of the following systems has the higher reliability at the end of100 operating hours?(i) Two constant failure rate components in parallel each having an MTTF

of 1000 hours.(ii) A Weibull component with a shape parameter of 2 and a characteristic

life of 10,000 hours in series with a constant failure rate componentwith a failure rate of 0.00005.

28. Find the minimum number of redundant components, each having areliability of 0.4, necessary to achieve a system reliability of 0.95. Thereis a common mode failure probability of 0.03.

29. Three communication channels in parallel have independent failure modesof 0.1 failure per hour. These components must share a commontransceiver. Determine the MTTF of the transceiver in order that the systemhas a reliability of 0.85 to support a 5 hour mission. Assume constantfailure rates.

30. The reliability of a communication channel is 0.40. How many channelsshould be placed in parallel redundancy so as to achieve the reliability ofreceiving the information is 0.80. If these channels are used to configurehigh level and low level redundant systems, what are the correspondingsystem reliabilities?

31. 2n identical constant failure rate components are used to configureredundant systems either with two subsystems in parallel or with nsubsystems in series. Which system will give higher reliability?

32. Compute the reliability of the system for the connection given in Fig.1.10, if the reliability of A, B, C, D are 0.95, 0.99, 0.90 and 0.96respectively:

��+��+�(


33. Compute the reliability of the system for the connection given in Fig.1.11, if the reliabilities of A, B, C, D are 0.90, 0.95, 0.96 and 0.98respectively:

��+��+��

34. A non-redundant system with 100 components has a design life (L)reliability of 0.90. The system is redesigned so that it has only 70components. Estimate the design life of the redesigned system, assumingthat all the components have constant failure rates of the same value.

3.5. Find the variance of the time to failure of the following systems, assuminga constant failure rate l for each component:

(a) a system with two components in series(b) a system with two components in parallel.

36. How many identical components with constant failure rate must beconnected in parallel to at least double the MTTF of a single component?[Hint: Use Example (7)]

37. Calculate the reliability of the Fig. 1.12 (a) and (b) systems:

(a) (b)��+��+��

38. Nine components each with a reliability 0.9 are used to configure (a) ahigh level redundancy with 3 subsystems and (b) a low level redundancywith 3 subsystems. In order to equalise the reliabilities of two systems, towhich system a common mode failure component is to be added. What isits reliability?

39. Two stamping machines with constant failure rate operate in passiveparallel positions on an assembly line. If one fails the other takes up theload by doubling its operating speed. When this happens, however, thefailure rate also doubles. Compare the design lives of the main machineand the standby system for a reliability of 0.99 and 0.95.

40. Two nickel-cadmium batteries provide electrical power to operate asatellite transceiver.If the batteries are operating in standby parallel positions, they have anindividual failure rate of 0.1 per year. If one fails the other can operatewith a failure rate of 0.3 per year. Find the system reliability for 2 years.What is the system MTTF?


-��.��/��

No equipment (system) can be perfectly reliable in spite of the utmost care andbest effort on the part of the designer and manufacturer. In fact, very few systemsare designed to operate without maintenance of any kind. For a large number ofsystems, maintenance is a must, as it is one of the effective ways of increasingthe reliability of the system.

Usually two kinds of maintenance are adopted. They are preventive mainte-nance and corrective or repair maintenance. Preventive maintenance is mainte-nance done periodically before the failure of the system, so as to increase thereliability of the system by removing the ageing effects of wear, corrosion, fatigueand related phenomena. On the other hand, repair maintenance is performed oncethe failure has occurred so as to return the system to operation as soon as possible.

The amount and the type of maintenance that is used depends on the respectivecosts and safety consideration of system failure. It is generally assumed that apreventive maintenance action is less costly than a repair maintenance action.

��!� � �� Let R(t) and RM(t) be the reliability of a system without maintenance and withmaintenance.

Let the preventive maintenance be performed on the system at intervals of T.Since RM(t) = P{the maintained system does not fail before t}, we have

RM(t) = R(t), for 0 ≤ t < T= R(T), for t = T

After performing the first maintenance operation at T, the system becomes asgood as new.Hence, if T ≤ t < 2T,

RM(t) = P{the system does not fail up to T and it survives for a time (t –T) without failure}= R(T) · R(t – T), for T ≤ t < 2T

Similarly after two maintenance operations,RM(t) = {R(T)}2 · R(t – 2T), for 2T ≤ t < 3T

Proceeding like this, we get in general,RM(t) = {(R(T)}n · R(t – nT), for nT ≤ t < (n + 1 )T

(n = 0, 1, 2, ...)MTTF of a system with preventive maintenance is given by

MTTF = 0

( )MR t dt�

�

=( 1)

0

( ) ,n T

Mn nT

R t dt��

� � by dividing the range into intervals of length T

= ( 1)

0

{ ( )} ( ) ,n T

n

n nT

R T R t nT dt��

��


= 0 0

{ ( )} ( ) ,T

n

n

R T R t dt�

�� on putting t – nT = t ′

= 0

( )

1 ( )

T

R t dt

R T�

�

�$��% 1. If the system has a constant failure rate l, then R(t) = e–lt and RM(t)=(e–lT )n · e–l(t – nT), for nT £ t < (n + 1)T,

where n = 0, 1, 2, ...= e–nlT · e–lt + nlT

= e–lt = R(T), for 0 £ t < •This means that, in the case of constant failure rate, preventive maintenance doesnot improve the reliability of the system.2. If the failure distribution of the system is Weibull with parameters b and q, then

R(t) = ,t

exp�

�

� �� for non-maintained system.

For the maintained system,

RM(t) = ,

n

T t nTexp exp

� �

� �

� � � �� for nT £ t £ (n + 1)T and n = 0, 1, 2, ...

= ,T t nT

exp n exp n� �

� �

� � � �� To examine the effects of preventive maintenance, we find RM(t)/R(t) at the timeof preventive maintenance t = nT, where n = 1, 2, 3, ...

( )

( )MR nT

R nT=

Texp n

nTexp

�

�

�

�

� ��

= T nT

exp n� �

� �

� �� > 1

if – nT n T� � �

� ��

#� > 0

i.e., if nb –1 – 1 > 0i.e., if b > 1This means that RM(t) > R(t), viz., preventive maintenance will improve thereliability of the Weibull system, only when the shape parameter b > 1.


�� A measure of how fast a component (system) may be repaired following failure isknown as maintainability. Repairs require different lengths of time and even thetime to perform a given repair is uncertain (random), because circumstances, skilllevel, experience of maintenance personnel and such other factors vary. Hence thetime T required to repair a failed component (system) is a continuous R.V.

Maintainability is mathematically defined as the cumulative distribution func-tion (cdf) of the R.V.T. representing the time to repair and denoted as M(t).

i.e., M(t) = P{T ≤ t} = 0

( )t

m t dt� (1)

where m(t) is the pdf of T.The expected value of repair time T is called the mean time to repair (MTTR)and is given by

MTTR = E(T) = 0

( )t

t m t dt�� (2)

If the conditional probability that the (component) system will be repaired (madeoperational) between t and t + ∆t, given that it has failed at t and the repair startsimmediately, is m(t) ∆t, then m(t) is called the instantaneous repair rate or simplythe repair rate and denotes the number of repairs in unit time.

i.e., m(t) ∆t = { }

( )

P t T t t

P T t

� � � �

= ( )

1 ( )

m t t

M t

��

∴ m(t) = ( )

1 ( )

m t

M t�(3)

From (1), on differentiation, we get

m(t) = ( )d

M tdt

(4)

Using (4) in (3), we have

m(t) = ( )

1 ( )

M t

M t

��

(5)

Integrating both sides of (5) with respect to t between 0 and t, we get

0

( )t

t dt�� = 0

( )

1 ( )

tM t

M t

�� dt

i.e., [–log {1 – M(t)}]t0 =

0

( )t

t dt��

i.e., 1 – M(t) = 0

( )t

t dt

e��

[� M(0) = 0]


or M(t) = 0

( )

1

t

t dt

e��

� (6)Using (5) and (6) in (4), we get

m(t) = m(t) · 0

( )t

t dt

e��

(7)

�$��% If m(t) = m(a constant), then from (7),

m(t) = me–mt, t > 0i.e., the time to repair T follows an exponential distribution with parameter m.Conversely if m(t) = me–m t, t > 0, then

M(t) = 1 – e–m t, using (6)

∴ m(t) = t

t

e

e

�

�

� �

� using (3)

= mFor the constant repair rate distribution

MTTR = 0

( )tm t dt�

� = 0

tt e dt��

�#�

= m 2

0

t te et

� �

� �

��

= 1

�.

�� "�� Let us consider a two component redundant system in which both the failure rateand repair rate are constant.

If we assume that repair can be completed for a failed unit before the otherunit has failed, system failure is ruled out and the system reliability is improved.Otherwise both units may fail resulting in less reliability.

The corresponding Markov state-transition diagram is given in Fig. 1.13, in whichl1 and l2 represent the transition rates from states 1 and 2 to states 2 and 3 respectively.

�� They do not necessarily represent the failure rates of the two components.

m2 represents the transition rate from state 2 to state 1.State 1 represents the situation in which both the components are operating, state

2 represents the situation in which one component is operating and the other is beingrepaired and state 3 represents the situation in which both components have failed.

��+��+��


Proceeding as in the discussion of system reliability without repair using Markovanalysis, we get the following differential equations for the state probabilities:

P1′(t) = – l1 P1(t) + m2 P2(t) (1)P2′(t) = l1 P1(t) – (l2 + m2) P2(t) (2)P3′(t) = l2 P2(t) (3)

Equation (1) is (D + l1) P1(t) – m2 P2(t) = 0 (1)′Equation (2) is l1 P1(t) – (D + l2 + m2) P2(t) = 0 (2)′

where D = d

dt. Equations (1)′ and (2)′ are simultaneous differential equations in

P1(t) and P2(t).Eliminating P2(t) from (1)′ and (2)′, we get

{(D + l1) (D + l2 + m2) – l1 m2} P1(t) = 0i.e., {D2 + (l1 + l2 + m2) D + l1 l2} P1(t) = 0 (3)A.E. corresponding to equation (3) is

m2 + (l1 + l2 + m2) m + l1 l2 = 0 (4)Solving equation (4), we get

m = 2

1 2 2 1 2 2 1 2( ) ( ) 4

2

� � � � � � � �� " � � �

= m1, m2 (say)∴ Solution of equation (3) is

P1(t) = Aem1t + Bem2t (5)Initiall both the components are operating, viz., the system is in state 1.∴ P1(0) = 1∴ we get A + B = 1 (6)

[from (5)]Using (5) in (1), we have

P2(t) = 2

1

� [m1Aem1t + m2 Bem2t + l1 Aem1t + l1Bem2 t] (7)

Initial condition is P2(0) = 0∴ m1A + m2B + l1A + l1B = 0 [from (7)]i.e., (m1 + l1)A + (m2 + l1) B = 0 (8)Solving equations (6) and (8), we get

A = – 2 1

1 2

( )m

m m

��

and B = 1 1

1 2

m

m m

��

(9)


P1(t) = – 2 1

1 2

( )m

m m

��

em1t + 1 1

1 2

m

m m

��

em2t (10)


P2(t) = 2

1

� 2 11 1 2 1

1 2

( )( )( )m t m tm me e

m m

� ��


=2 1 2

1

( )m m� �[{m1m2 + l1(m1 + m2) + l2

1}(em2t – em1t)]

=2 1 2

1

( )m m� � [{l1l2 – l1(l1 + l2 + m2) + l2

1} (em2t – em

1t)]

[� m1, m2 are roots of (4)]

= 1

1 2m m

�

�(em

1t – em

2t) (11)

If R(t) is the system reliability, thenR(t) = P{one or both components are operating}

= P{system is in state 1 or 2}= P1(t) + P2(t)

= 21

1 2

m tme

m m� – 12

1 2

m tme

m m� on using (10) and (11) (12)

�� 1. Let the system be a two-component active redundant system under repair.

In this case, both the components may operate simultaneously.Let us make a simplifying assumption that the rate of failure for eachcomponent is l. and the rate of repair for each component is m.

∴ l1 = Rate of transition of the system from state 1 to state 2.= Rate of failure of either component 1 or component 2 ( � state 2 corresponds to either component operating)= l + l ( � simultaneous failures of components are ruled out).= 2l

Also l2 = l and m2 = Rate of repair of one of the components= m

Hence R(t), the reliability of the system is given by (12), where m1, m2 are theroots of the equation

m2 + (3l + m)m + 2l2 = 0 (13)

i.e., m1, m2 = � �2 21(3 ) 6

2� � � �� " � � (14)

In this case,

MTTF = 2 11 2

1 2 1 20 0 0

( ) m t m tm mR t dt e dt e dt

m m m m

� � ��

� ��

= 1 2

1 2 2 1

1 m m

m m m m

� ��

(� m 1, m 2 < 0)

= – 1 2

1 2

( )m m

m m

� = 2

3

2

� �

�

�(15)


Note: If we put m = 0, we get MTTF = 3

2�, which we have derived already for

a 2 compnent active redundant system without repair.2. Let the system be a two-component standby redundant system under repair.

System changes from state 1 to state 2 due to the failure of the maincomponent. As before the standby component is assumed not to fail inthe standby mode. Hence the rate of transition l1 from state 1 to state 2 isthe same as the rate of failure of the main component viz., lSimilarly l2 can be considered as the rate of failure of the standbycomponent viz., lThe rate of transition from state 2 to state 1 is the same as the rate ofrepair of the failed main component, viz., m2 = m, say.In this case R(t) is given by (12), where m1, m2 are given by the equation(4), in which m2 is replaced by m.

Also MTTF = 1 22 2

1 2

( ) 2 2m m

m m

� � �

��

� �� (16)

�$��% If we put m, = 0, we get MT TF = 2

�, which we have derived already for

a 2 component standby redundant system without repair.

�/��

Closely associated with the reliability of repairable (maintained) systems is con-cept of availability. Like reliability and maintainability, availability is also aprobability.

Availability is defined as the probability that a component (or system) is per-forming its intended function at a given time ‘t’ on the assumption that it isoperated and maintained as per the prescribed conditions. This is referred to aspoint availability and denoted by A(t).

It is to be observed that reliability is concerned with failure-free operation upto time t, whereas availability is concerned with the capability to operate at thepoint of time t.

If A(t) is the point availability of a component (or system), then

A(t2 – t1) = 2

12 1

1( )

t

t

A tt t� �

is called the interval availability or mission availability.In particular, the interval availability over the interval (0, T) is given by

A(T) = 0

1( )

T

A t dtT �

Now limT!� A(T) is called the steady-state or asymptotic or long-run availability

and denoted by A or A(∞).


��#� ��$Let us assume that the component will be in one of the two possible states: oper-ating (state 1) or under repair (state 2). Let the component have constant failurerate l and constant repair rate m. The corresponding Markov state transition dia-gram is given in Fig. 1.14:

��+��+�

The differential equation for the state probability of the component is

1( )dP t

dt= – lP1(t) + µP2(t) (1)

P1(t) + P2(t) = 1 (2)Using (2) in (1), we have

P1′(t) (l + m) P1(t) = m (3)I.F. of equation (3) = e(l + m)t

Solution of equation (3) is

e(l + m)t ⋅ P1(t) = m ( )te dt c� ��

= �

� �� e(l + m)t + c (4)

Since the component is in state 1 initially, P1(0) = 1.

Using this in (4), we get c = 1 – �

� �� =

�

� ��.

Using this value of c in (4), we get

P1(t) = � �

� � � ��

� � e–(l + m)t (5)

Since state 1 is the available state,A(t) = P1(t)

i.e., A(t) = � �

� � � ��

� � e–(l + m)t (6)

The interval availability over (0, T ) is given by

A(T ) = 0

1( )

T

A t dtT �

= 2( ) T

� �

� � � ��

� � {1 – e–(l + m)T} (7)

The steady-state availability is then given by

A(∞) = 2( )

� �

� � � ��

� �

( )1lim

T

T

e

T

� ��

!�

� ��


= �

� ��(8)

= 1

1 1

�

� �� =

MTTF

MTTF + MTTR(9)

�$��: Since failures of the component are random events, then the nmber N(t)of failures in interval ‘t’ follows a Poisson process given by

P{N(t) = r} = e–lt◊ ( )

!

rt

r

� r = 0, 1, 2, ...,

where l represents the constant failure rate of the component. Then the timebetween failures is a continuous R.V. that follows an exponential distribution

with parameter l. The expected value of the time between failures is given by 1

�and is called mean time between failures and denoted as MTBF.

Hence equation (9) is also given as

A(∞) = MTBF

MTBF + MTTR(10)

Formulas (9) or (10) may be used even if failure and repair distributions are notexponential.

�$��% In most situations, repair tares are much larger than failure rates. Hence

�

� is very small.

From (8), we have A(•) = 1

11

1

�

� �

�

��

� ��

= 1 – �

� +

2�

�

� ��

– ... •

i.e., A(•) � 1 �

�, omitting higher power of

�

�.

Now the component unavailability is given by

( )A t = 1 – A(t) or P2(t)

= 1 – � �

� � � ��

� � e–(l + m)t

=�

� �� {1 – e–(l + m)t}

∴ ( )A � =�

� ��


��If we consider a non-redundant system consisting of n independent repetitioncomponents connected in series, then the system reliability As(t) is given by

As(t) = A1(t) ⋅ A2(t) ... An(t)[� all the components must be available for the system to be available]For a standby system consisting of n independent components connected in

parallel, the system availability As(t) is given byAs(t) = 1 – {1 – A1(t)} {1 – A2(t)}...{1 – An(t)}

[� all the components must be unavailable for the system to be unavailable],where Ai(t) is the availability of the ith component.

�� "�� Let us consider a two component standby system in which the failure rates l1 andl2 and common repair rate m are constant.

As before we assume that the standby component does not fail in standbymode, but the repair of the standby unit is also permitted. Also we assume thatonly one repair, viz., the repair of the main unit or the standby unit is possible. Inother words we assume that there is a single repair person, who can repair themain unit before the standby unit fails.

The corresponding Markov state transition diagram is given in Fig. 1.15, where,in state 1 main unit is operating while the standby unit is not, in state 2 main unithas failed while the standby unit has become operative and in state 3 standbyalso has failed.

��+��+�!

The differential equations for the state probabilities areP ′1(t) = – l1 P1(t) + mP2(t) (1)P ′2(t) = l1 P1(t) + mP3(t) – (l2 + m) P2(t) (2)

and P1(t) + P2(t) + P3(t) = 1 (3)The availability of the system As(t) is given by As(t) = P1(t) + P2(t).If we solve the equations (1), (2) and (3) simultaneously either directly or byusing Laplace transforms, we can get

As(t) = 1 2

1 2 1 2

1 2 1 2 1 2

1m t m te e

m m m m m m

� � � � � �� where m1, m2 are the roots of the equation

m2 + (l1 + l2 + 2m) m + (l1 l2 + l1 m + m2) = 0The initial conditions for solving the above equations are P1(0) = 1 and P2(0) =P3(0) = 0.


�� The solution of the above equations has been avoided, as only the steady-state availability of the standby system is frequently required.

Now in the steady-state Pi(t) does not depend on t and P¢i (t) = 0. Hence theabove equations become

– l1 P1 + mP2 = 0 (4)l1 P1 + mP3 – (l2 + m)P2 = 0 (5)

and P1 + P2 + P3 = 1 (6)Using (6) in (5), we have

(m – l1) P1 + (2m + l2) P2 = m (7)Using (4) in (7), we have

(m2 + l1 m + l1 l2) P1 = m2

∴ P1 = 2

21 1 2

�

� � � � �� or 1 1 2

21

� � �

� �

� ��

(8)

P2 = 11P

�

�(9)

and P3 = 1 – P1 – 11P

�

�

= 2

1

� [m2 – m2 P1 – l1 mP1]

= 2

1

� [(m2 + l1m + l1 l2)P1 – m2 P1 – l1 mP1]

= 1 212

P� �

�(10)

The steady-state availability of the standby system is then given byAs(∞) = P1 + P2

= 2

12

1 1 2

� � �

� � � � �

��

We can alternatively obtain this value of As(∞) from that of As(t) by lettingt → ∞ and using m1 m2 = m2 + l1 m + l1 l2.

�� If a device has a failure ratel(t) = (0.015 + 0.02t)/year, where t is in years,

(a) Calculate the reliability for a 5 year design life, assuming that nomaintenance is performed.

(b) Calculate the reliability for a 5 year design life, assuming that annualpreventive maintenance restores the device to an as-good-as newcondition.


(c) Repeat part (b) assuming that there is a 5% chance that the preventivemaintenance will cause immediate failure.

��

(a) R(t) = 0

( )t

t dt

e��

∴ R(5) =

5

0

(0.015 0.02 )t dt

e� ��

(1)= e– (0.015 × 5 + 0.01 × 25)

= e–0.325 = 0.7225(b) Since annual preventive maintenance is performed, there will be 4

preventive maintenances in the first 5 years. RM(t) = {R(T)}n × R(t – nT), after n maintenances

Here t = 5, T = 1 and n = 4∴ RM(5) = {R(1)}4 × R(5 – 4)

= {R(1)}5

= {e–0.025}5, using (1)= 0.8825

(c) P{preventive maintenance causes immediate failure} = 0.05∴ P{the device survives after each preventive maintenance} = 0.95As there are 4 maintenances,

RM(5) = RM(5) without breakdown × probability of no breakdown in

5 years= 0.8825 × (0.95)4= 0.7188.

�� The time to failure (in hours) of a piece of equipment is uniformlydistributed over (0, 1000) hours.

(a) Determine the MTTF.(b) Determine the MTTF if preventive maintenance will restore the system to

as good as new condition and is performed every 100 operating hours.(c) Compare the reliability with and without preventive maintenance at 225

operating hours. Assume the 100 hour maintenance interval and amaintenance-induced failure probability of 0.01 each time preventivemaintenance is performed.

(d) Is there a significant improvement in reliability if a 50 hour preventivemaintenance interval is assumed?The pdf of the time to failure is given by

f (t) = 1

1000, in 0 ≤ t ≤ 1000

��

(a) MTTF = 1000

0

1

1000t dt� =

10002

0

1

1000 2

t� ��

= 500 hours.


(b) (MTTF)M = 100

0

( )R t dt� ÷ {1 – R(100)}, where

R(t) = 1000

1

1000t

dt�= 1 – 0.001t

i.e., (MTTF)M =

100

0

(1 0.001 )

1 (1 0.001 100)

t dt�

� � �

�

= 1002

0

10.001

0.1 2

tt� ��

= 950 hours.

(c) R(225) = 1 – 0.001 × 225 = 0.775RM(225) = {R(T)}2 × R(t – 2T) × (1 – p)2 [see (c) in the previous example],where p is the probability of maintenance induced failure.

�� There will be n = 2 preventive maintenances in (0, 225)

= {R(100)}2 × R(25) × (1 – 0.01)2

= (0.9)2 × (0.975) × (0.99)2

= 0.774R(225) � RM(225)

(d) If T = 50, n = the number of preventive maintenance = 4∴ R′M(225) = {R(5)}4 × R(225 – 200)

= (0.95)4 × 0.975 = 0.794If there is no maintenance induced failure probability, the reliability isimprved when T = 50.

�� A reliability engineer has determined that the hazard rate functionfor a milling machine is l(t) = 0.0004521t0.8, t ≥ 0, where t is measured inyears. Determine which of the following options will provide the greatest reliabilityover the machine’s 20 years operating life.Option A: Do nothing-operate the machine until it fails.Option B: An annualpreventive maintenance program (with no maintenance-induced failures)

Option C: Operate a second machine in parallel with the first (active redundant).

��Option A

RA(t) = 0.8

0

exp 0.0004521t

t dt� ��


= 1.80.0004521exp

1.8t

� ��

∴ RA(20) = 1.80.0004521exp 20

1.8� ��

= 0.9463

Option BRB(t) = {R(T)}n ⋅ R(t – nT)

= {R(1)}19 × R(20 – 19) [� there will be 19maintenances in (0, 20)]

= 20

0.0004521exp

1.8

� � ��

= 0.9950

Option CRC (20) = 1 – {1 – R(20)}2

= 1 – {1 – 0.9463}2 = 0.9971Option C will give the greatest reliability.

�� The time to repair a power generator is best described by its pdf

m(t) = ,2t

333 1 ≤ t ≤ 10 hours

(a) Find the probability that a repair will be completed in 6 hours.(b) What is the MTTR?(c) Find the repair rate.

��(a) P(T < 6) = P(1 ≤ T < 6), where T is the time to repair

= 6

1

( )m t dt�

= 6 2

1333

tdt� =

63

1999

t� ��

= 0.2152

(b) MTTR = 0

( )tm t dt�

� = 610 3 4

1 1333 4 333

t tdt

� ��

= 7.5 hours

(c) Repair rate = m(t) = ( )

1 ( )

m t

M t�

= 2

10 2

333

333t

t

tdt�

= 2

3 3

3331

(10 )999

t

t�

= 2

3

3

1000

t

t� per hour.


��! The time to repair of an equipment follows a lognormal distributionwith a MTTR of 2 hours and a shape parameter of 0.2.

(a) Find the median time to repair.(b) Determine the repair time such that 95% of the repairs will be

accomplished within the specified time.(c) Determine the probability that a repair will be completed within 100

minutes.

��(a) MTTR of a lognormal distribution is given by

MTTR = tM exp (s2/2), where tM is the median and s is the shape parameter

∴ tM = MTTR × 2

exp2

s� ��

= 2 × e–0.02 = 1.96 hours.(b) Let T represent the time to repair.

P{T ≤ TR} = 0.95

i.e.,0

( )RT

f t dt� = 0.95, where f (t) is the pdf of the lognormaldistribution

i.e.,

1log

1 0log

( )

R

M

M

T

s t

s t

z dz�

� ��

� ��

� = 0.95, where f(z) is the standard normal density

function

i.e.,

5 log1.96

( )

RT

z dz�

� ��

�� = 0.95

∴ 5 log 1.96

RT� ��

= 1.645, from the normal tables

∴1.96

RT= e0.329

∴ TR = 2.72 hours.

(c)100

60P T� ��

= 5 3

0

( )f t dt�

=

1.675 log

1.96

( )z dz�

� ��

��


=0.8

( )z dz�

�

�� = 0.212.

��" A mechanical pumping device has a constant failure rate of 0.023failure per hour and an exponential repair time with a mean of 10 hours. If twopumps operate in an active redundant configuration, determine the system MT TFand the system reliability for 72 hours.

Refer to deduction (1) under the discussion of reliability of a two componentredundant system with repair (using Markov analysis).

For the active redundant configuration,

R(t) = 2 11 2

1 2 1 2

m t m tm me e

m m m m�

� �(1)

where m1, m2 = � �2 21(3 6

2� � � �� " � �

Here l = 0.023 and m = 1

10 = 0.1

∴ m1, m2 = � �2 21(0.069 0.1) (0.023) 6 0.023 0.1 (0.1)

2� � " � � � �

= – 0.0065, – 0.1625Using these values in (1), we have

R(t) = 0.0065

0.156

� e–0.1625t +

0.1625

0.156 e–0.0065t

∴ R(72) = – 0.0417 × e–11.7 + 1.0417 × e–0.468

= 0.6524

MTTF = 2

3

2

� �

�

� =

2

3 0.023 0.1

2 (0.023)

� ��

= 159.7 hours.

��# An engine health monitoring system consists of a primary unit anda standby unit. The MTTF of the primary unit is 1000 operating hours and theMTTF of the standby unit is 333 hours when in operation. There are no failureswhile the backup unit is in standby.

��If the primary unit may be repaired at a repair rate of 0.01 per hour, while thestandby unit is operating, estimate the design life for a reliability of 0.90.

Refer to the discussion of reliability of a two component redundant systemwith repair (using Markov analysis).For the standby redundant system,

R(t) = 2 11 2

1 2 1 2

m t m tm me e

m m m m�

� �(1)

where m1 and m2 are the roots of the equationm2 + (l1 + l2 + m)m + l1 l2 = 0 (2)


Here l1 = 1

1000 = 0.001/hour; l2 =

1

333 = 0.003/hour and m = 0.01/hour.

Using these values in (2), we have

m2 + 0.014 m + 0.000003 = 0

∴ m1, m2 = 20.014 (0.014) 4 0.000003

2

� " � �

= – 0.00022, – 0.01378Using these values in (1), we get

R(t) = – 0.01622 × e–0.01378t + 1.01622 × e–0.00022t

When the reliability is 0.90, the design life D is given by1.01622 × e–0.00022D – 0.01622 × e–0.01378D = 0.90

Solving this equation by trials, we getD = 550 hours.

��& Reliability testing has indicated that a voltage inverter has a 6month reliability of 0.87 without repair facility. If repair facility is made availablewith an MTTR of 2.2 months, compute the availability over the 6-month period.(Assume constant failure and repair rates)

For constant failure rate l, reliability is given by R(t) = e–lt.As R(6) = 0.87, e–6l = 0.87∴ l = 0.0232/month

MTTR = 1

� = 2.2 ∴ m = 0.4545/month

Interval availability over (0, T ) is given by

A(T) = 2( ) T

� �

� � � ��

� � {1 – e–(l + m)T}

∴ A(6) = 2

0.4545 0.0232

0.4777 (0.4777) 6�

� {1 – e–0.4777 × 6}

��' A critical communications relay has a constant failure rate of 0.1per day. Once it has failed, the mean time to repair is 2.5 days (the repair rate isconstant).

(a) What are the point availability at the end of 2 days, the interval availabilityover a 2-day mission, starting from zero and the steady-state availability?

(b) If two communication relays operate in series, compute the availabilityat the end of 2 days.

(c) If they operate in parallel, compute the steady-state availability of thesystem.

(d) If one communication relay operates in a standby mode with no failure instandby, what is the steady-state availability?

��

l = 0.1 per day; 1

� = 2.5 ∴ m = 0.4 per day


(a) AP(t) = � �

� � � ��

� � e–(l + m)t

∴ Ap(2) = 0.4 0.1

0.5 0.5� = e–0.5 × 2 = 0.8736

AI(T) = 2( ) XT

� �

� � � ��

� � {1 – e–(l + m)T}

∴ AI(2) = 2

0.4 0.1

0.5 (0.5) 2�

� {1 – e–0.5 × 2} = 0.9264

A(∞) = �

� �� =

0.4

0.5 = 0.8

(b) As(2) = {Ap(2)}2 = (0.8736)2 = 0.7632(c) As(∞) = 1 – {1 – A(∞)}2

= 1 – {1 – 0.8}2 = 0.96(d) For the standby redundant system,

As(∞) = 2

12

1 1 2

� � �

� � � � �

��

; Here l1 = l2 = 0.1 and m = 0.4

i.e., As(∞) = 0.04 0.16

0.16 0.04 0.01

��

= 0.20

0.21 = 0.9524.

�� ( A new computer has a constant failure rate of 0.02 per day(assuming continuous use) and a constant repair rate of 0.1 per day.

(a) Compute the interval availability for the first 30 days and the steady-state availability.

(b) Determine the steady-state availability if a standby unit is purchased.Assume no failures in standby.

(c) If both units are active, what is the steady-state availability?l = 0.02/day; m = 0.1/day.

��

(a) AI(T) = 2( ) T

� �

� � � ��

� � � {1 – e–(l + m)T}

∴ AI(30) = 2

0.1 0.02

0.12 (0.12) 30�

� {1 – e–0.12 × 30}

= 0.8784

A(∞) = 0.1

0.12

�

� ��

� = 0.8333

(b) For the standby redundant system,

As(∞) = 2

2 2

0.002 0.01

0.01 0.002 0.0004

��

� ��

� ��

� ��


= 0.9677For the active redundant system,

As(∞) = 1 – {1 – A(∞)}2

= 1–{1 – 0.8333}2

= 0.9722.

�� An office machine has a time to failure distribution that is lognormalwith a shape parameter s = 0.86 and a scale parameter tM = 40 operating hours.The repair distribution is normal with a mean of 3.5 hours and a standard deviationof 1.8 hours. Find the steady-state availability of the machine.

��For the lognormal failure distribution, mean = MTTF = tM exp(s2/2)

= 40 exp {(0.86)2/2} = 40 × e0.3698 = 57.9 hoursMean of the normal repair distribution

= MTTR = 3.5 hours

A(∞) = MTTF

MTTF + MTTR =

57.9

57.9 + 3.5

= 0.943

�� The distribution of the time to failure of a component is Weibullwith b = 2.4 and q = 400 hours and the repair distribution is lognormal with tM= 4.8 hours and s = 1.2. Find the steady-state availability. For the Weibull failuredistribution,

Mean = MTTF = 11+�

�

� ��

= 1

400 1+2.4

� ��

= 400 × (1.42)

= 400 × 0.88636= 354.5 hours

for the lognormal repair distribution,Mean = MTTR = tM exp(s2/2)

= 4.8 × exp{(1.2)2/2}= 9.86 hours

A(∞) = MTTF

MTTF + MTTR

= 354.5

354.5 9.86� = 0.9729


�� A system may be found in one of the three states: operating,degraded or failed. When operating, it fails at the constant rate of 1 per day andbecomes degraded at the rate of 1 per day. If degraded, its failure rate increasesto 2 per day. Repair occurs only in the failed mode and restores the system to theoperating state with a repair rate of 4 per day. If the operating and degradedstates are considered the available states, determine the steady-state availability.

�� A system is said to be in degraded state, if it continues to perform itsfunction but at a less than specified operating level.

For example, a copying machine may not be able to automatically feed originalsand may require manual operation or a computer system may not be able toaccess all of its direct access storage, devices.

��The situation of this problem may be represented by the Markov state transi-

tion diagram as in Fig. 1.16, in which states 1, 2 and 3 represent respectively theoperating, degraded and failed states of the system:

��+��+�"

The differential equations for the state probabilities areP ′1(t) = – (l1 + l2) P1(t) + m ⋅ P3(t) (1)P ′2(t) = l2 P1(t) – l3 P2(t) (2)P1(t) + P2(t) + P3(t) = 1 (3)

When the system is in steady-state,P ′i (t) = 0 and Pi (t) = Pi (constant); i = 1, 2, 3

∴ The above equations become– (l1 + l2) P1 + m ⋅ P3 = 0 (4)

l2 P1 – l3 P2 = 0 (5)P1 + P2 + P3 = 1 (6)

Solving the equations (4), (5) and (6), we get

P1 = 3

2 3 3 1 2( ) ( )

� �

� � � � � �� and P2 = 2

2 3 3 1 2( ) ( )

� �

� � � � � �� Now A(∞) = P1 + P2

= 2 3

2 3 3 1 2

( )

( ) ( )

� � �

� � � � � �

��

= (1 2) 4

,(1 2) 4 2 (1 1)

� ��

using the given values

= 0.75.


�� For a two component standby system with repair permitted foreither component with a single repair crew and no failures in standby, the failureand repair rates are given by

l1 = rate of failure of main unit = 0.002l2 = rate of failure of standby unit = 0.001m = repair rate for either unit = 0.01

Compute the steady-state availability of the system.

�� The steady-state probabilities are given by

P1 = 1

1 1 22

1� � �

� �

��

and P2 = 11P

�

�

[Refer to the discussion of availability of a two component standby system withrepair by Markov analysis]

∴ P1 = 1

0.002 0.0000021

0.01 0.0001

�� = 0.8197

P2 = 0.002

0.01 × 0.8197 = 0.1639

Now As(∞) = P1 + P2 = 0.9836.

��! A generator system consists of a primary unit and a standby unit.The primary unit fails at a constant rate of 2 per month and the standby unit failsonly when online at a constant rate of 4 per month. Repair can begin only whenboth units have failed. Both units are repaired at the same time with an MT TR of

2

3 month.

Derive the steady-state equations for the state probabilities and solve for thesystem availability.

�� The Markov state transition diagram for this problem is givenby Fig. 1.17:

In state 1, main unit is operating and the other is in standby mode.In state 2, main unit has failed and the standby unit operates.In state 3, standby unit has also failed in addition to the main unit.

��+��+�#

The steady-state probabilities of the three states are–l1P1 + mP3 = 0 (1)


l1P1 – l2P2 = 0 (2)l2P2 – mP3 = 0 (3)

or P1 + P2 + P3 = 1 (4)Solving equations (1), (2) and (4), we get

P1 = 2

1 2 1 2( )

��

� � � � �� and P2 = 1

1 2 1 2( )

��

� � � � ��

The system steady-state availability is given byAs(∞) = P1 + P2

= 1 2

1 2 1 2

( )

( )

� � �

� � � � �

��

=

3(2 4)

23

(2 4) 82

� �

� � � =

917

= 0.5294.

�� (Short answer questions)

1. What do you mean by preventive maintenance and correctivemaintenance?

2. How are the reliabilities with and without maintenance related?3. Derive the formula for MTTF of a system with preventive maintenance.4. Show that preventive maintenance does not improve the reliability of a

system with constant failure rate.5. When will the reliability of a Weibull system improve due to preventive

maintenance?6. What do you mean by maintainability?7. How are the maintainability and repair rate related?8. Find the MTTR of a system with constant repair rate.9. What is the value of MTTF of a two component redundant system with

constant failure rate l and constant repair rate m, when the system is (i)active redundant (ii) standby redundant?

10. What is availability? How is it different from reliability?11. Define point, mission and steady-state availabilities.12. Express steady-state availability in terms of mean times to failure and

repair.13. Express the system availability in terms of component availabilities, when

the system is (i) non-redundant (ii) redundant.14. Write down the formula for the steady-state availability of a two

component standby redundant system with repair. State the conditionsunder which this formula may be used.


��

15. A flange bolt wears out because of fatigue in accordance with thelognormal distribution with MTTF = 10,000 hours and s = 2. If preventivemaintenance consists of periodically replacing the bolt, what is thereliability at 550 hours with and without preventive maintenance? Assumethat the bolts are replaced every 100 operating hours.

16. Preventive maintenance is to be performed every 5 days on a systemhaving a rectangular failure distribution in (0, 100) days. Derive thereliability function for the system under preventive maintenance. Comparethe MTTF and the reliability at 17 days with and without preventivemaintenance.

17. A compressor has a Weibull failure process with b = 2 and q = 100 days.If preventive maintenance is resorted to every 20 days, find the reliabilityfor 90 days. Find also the design lives at 0.90 reliability without and withpreventive maintenance.

18. For a component having a lognormal failure distribution with s = 1 andmedian 5000 hours, prove that the reliability at 5000 hours increases byabout 70%, if we perform preventive maintenance at intervals of 500hours.

19. The pdf of the time to failure in years of the drive train on a Rapit TransitAuthority bus is given by

f(t) = 0.2 – 0.02 t, 0 ≤ t ≤ 10 years(a) If the bus undergoes preventive maintenance every 6 months that

restores it to an as-good-as-new condition, determine its reliabilityat the end of a 15 month warranty.

(b) Compute the MTTF under the preventive maintenance plan givenabove.

20. Find the MTTF for an active two-component redundant system withconstant failure rate l under preventive maintenance done at regularintervals of T. Deduce the same when there is no preventive maintenance.[Hint: R(t) = 2e–lt – e–2lt]

21. The time to repair an engine module is lognormal with s = 1.21.Specifications require 90% of the repairs to be accomplished within 10hours. Determine the necessary median and mean time to repair.

22. If the failure time of an equipment is lognormal with the shape parameter0.7 and if it is designed to have a 90% chance of operating for 5 years,find its MTTF.When it fails, the time to repair follows a lognormal distribution with amean of 2 hours and a shape parameter of 1, what is the probability thatit is repaired within 4 hours?

23. A computer system consists of two active parallel processors each havinga constant failure rate of 0.5 failure per day. Repair of a failed processor


requires an average of 1

2 a day (exponential distribution). Find the system

reliability for a single day and the system MTTF. Find the same if there isno repair facility.

24. An on-board computer system has, through the use of built-in testequipment, the capability of being restored when a failure occurs. Astandby computer is available for use whenever the primary fails.Assuming that l1 = 0.0005 per hour, l2 = 0.002 per hour and m = 0.1 perhour, determine the system reliability at 1000 hours and at 2000 hours.

25. An airlines maintains an online reservation system with a standbycomputer available if the primary fails. The on-line system fails at theconstant rate of once per day while the standby fails (only when online)at the constant rate of twice per day. If the primary unit may be repairedat a constant rate with an MTTR of 0.5 of a day, what is the single dayreliability?

26. A computer has an MTTF = 34 hours and an MTTR = 2.5 hours. What isthe steady-state availability? If the MTTR is reduced to 1.5 hours, whatMTTF can be tolerated without decreasing the steady-state availabilityof the computer?

27. A component has MTBF = 200 hours and MTTR = 10 hours with bothfailure and repair distributions exponential. Find the availability of thecomponent in the interval (0, 10) hours, at the end of 10 hours and after along time.

28. Given two components, each having a constant failure rate of 0.10 failureper hour and a constant repair rate of 0.20 repair per hour, compute pointand interval availability for a 10 hour mission and steady-state availabilityfor both series and parallel configurations.

29. For the standby redundant system given in problem (25) above, find theavailability of the system at the end of one day.

30. In Example (9), determine how many days the relay must be operating inorder for the point availability to be within 0.001 of the steady-stateavailability.

31. If the system availability for a standby system with repair allowed foreither component with a single repair crew and no failures while in standbymode is 0.9, what is the maximum acceptable value of the failure to repair

rate ratio �

�? (Assume that l1 = l2 = l)

32. A system has a steady-state availability of 0.90. Two such systems, eachwith its own repair crew, are placed in parallel. What is the steady-stateavailability (a) for a standby parallel configuration with perfect switchingand no failure of the unit in standby; (b) for an active parallelconfiguration?


33. In Example (13), if l1 = 1, l2 = 2, l3 = 3 and m = 10, find the steady-stateavailability.

34. A two-component active redundant system can be repaired only afterboth units have failed. Only one unit can then be repaired at a time. Theunit have constant failure rates of l1 and l2 respectively and the mean

repair time to complete both repairs is 1

�. Find an expression for steady-

state availability of the system.35. A radar unit consists of 2 active redundant primary components: a power

supply and a transceiver. The power supply has a constant failure rate of0.0031 failure per operating hour and takes an average of 5 hours torepair (constant repair rate). The transceiver fails on the average every100 operating-hours (constant failure rate) and takes 3 hours to repair(constant repair rate). What is the point availability of the radar unit after10 hours of use? How much does it differ from the steady-stateavailability?

21. (a) 16/(t + 4)2; (b) 2/(t + 4); (c) 6422. (a) 0.46; (b) 0.045 year; (c) 19.955 years23. 0.905; 0.92824. e–t2/5; 0.0014; 3.32 months

25. 1

100 (100 – t); 1

100 t�; 50; 50

26. 0.001; 750 hours; 215.4 hours; 0.00027/hour.27. yes; by 0.09 28. 47 days 29. b log a31. 0.0246 32. 0.75; 0.89 33. 8; 138.8%34. 0.85; 1342 hours; 447 hours; 10629 hours; 0.394535. 15.9 days 36. 99.9; 7701; 507537. 4 years; 3.16 years38. 128 years; (112.6, 117.9) years39. 0.7852; 0.857940. 964.8 hours; 2432; 1604.5; 0.0084

15. 0.4446 16. 0.85 17. 0.999818. 0.999 19. 0.8495 20. 0.929021. 0.9630; 40.6 days 22. 5588 years23. 0.8068; 0.24 year


24. 585, 292, 195 years; 0.9975; 147 years25. 8837.5 hours; 3417 hours 26. n = 20; 922 hours27. R1 = 0.9909; R2 = 0.9949; system (ii) has higher reliability28. 8 29. 50 hours 30. 4; 0.2944; 0.409631. n subsystems in series 32. 0.9649

33. 0.9856 34. 0.7L 35. (a) 2

1

4�; (b)

2

5

4�36. 4 37. (a) 0.9769; (b) 0.999838. To be added to the low level redundancy system; 0.983039. (i) system design life = 10 × machine design life;

(ii) system design life = 5 × machine design life40. 0.9537; 13.3 years

15. 0.6736; 0.570716. RM(t) = (0.95)n × {(1 + 0.05 n) – 0.01 f}; RM(17) = 0.8402;

R(17) = 0.83; (MTTF)M = 97.5 days; MTTF = 50 days17. 0.8437; 32.5 days; 55.9 days19. 0.7743; 6.3 months

20. 3 3;

2 2

Te �

� �

��

21. Median = 0.6731 hour; MTTR = 1.3996 hours22. MTTF = 2.5984 years; 0.999523. R = 0.90; MTTF = 7 days, with repair

R = 0.845; MTTF = 3 days, without repair24. 0.9904; 0.9808 25. 0.712526. 0.9315; 20.4 hours 27. 0.981; 0.969; 0.95228. 0.468; 0.596 and 0.445 for series configuration 0.900; 0.948 and 0.889

for parallel configuration29. 0.7125 30. 11 days 31. 0.39332. 0.989; 0.990 33. 0.8475

34. A(∞) = 1 2

2 1

1� �

� �

� ��

P1 where P1 = 1

1 2 1 2

2 1

1� � � �

� � �

��

35. As(10) = 0.9966; As(∞) = 0.9996

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Reliability Engineering T.veerarajan Sample Chapter