Reliability of Simple Systems

Definition of reliability

• Probability that a system will perform its intendedfunction for a specific period of time under a givenset of conditions

• T failure time (random variable) → R(t)=P{T>t}=1-FT(t)

In this lecture

• Objective:

• Computation of the system reliability R(t)

• Hypotheses:

• N = number of system components

• The components’ reliabilities Ri(t), i = 1, 2, …, N are known

• The system configuration is known

Series system

• All components must function for the system to function

1 2 3 N

Series system

• All components must function for the system to function

• For N exponential components:

)()(1

tRtRN

i

i=

=

etR t−=)(

1

1

N

i

i

E T

=

=

=

System failure rate

MTTF

1 2 3 N

0.960.9650.970.9750.980.9850.990.99510

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

component reliability

syste

m r

elia

bili

ty

N=1

N=10

N=50

N=100

N=200

N=300

N=400

Series system (identical components)

Parallel system

• All components must fail for the system to fail

Parallel system

• All components must fail for the system to fail

• For N exponential components:

( )1

1 1N

ti

i

R t e −

=

= − −

𝐹𝑇 𝑡 =ෑ

𝑖=1

𝑁

𝐹𝑇𝑖(𝑡) 𝑅 𝑡 = 1 − 𝐹𝑇 𝑡 = 1 −ෑ

𝑖=1

𝑁

1 − 𝑅𝑖(𝑡)

Parallel system

• All components must fail for the system to fail

• For N exponential components:

( )1

1 1N

ti

i

R t e −

=

= − − ( )

1

1 1 1

2 11

1 1 1

1

1 1

1 11

N N N

i i j ii i j

N N NN

Ni j i k j i j k

i

i

MTTF

−

= = = +

− −−

= = + = +

=

= − + +

+ − + −

+ +

𝐹𝑇 𝑡 =ෑ

𝑖=1

𝑁

𝐹𝑇𝑖(𝑡) 𝑅 𝑡 = 1 − 𝐹𝑇 𝑡 = 1 −ෑ

𝑖=1

𝑁

1 − 𝑅𝑖(𝑡)

0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 10.5

0.55

0.6

0.65

0.7

0.75

0.8

0.85

0.9

0.95

1

component reliability

syste

m r

elia

bili

ty

N=1

N=2

N=3

N=4

N=8

Parallel system (identical components)

Ex. 1: A system with 2 identical exponential components

• Failure rates:𝜆

• Questions:

1) find the MTTF of the parallel and of the series system

2) Compare the results with that of a single component

with the same failure rate

• For N identical exponential identical components:

• 𝑅𝑠𝑒𝑟𝑖𝑒𝑠 𝑡 = 𝑒−𝑁𝜆𝑡 𝑀𝑇𝑇𝐹𝑠𝑒𝑟𝑖𝑒𝑠 =1

𝑁𝜆

• 𝑅𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑡 = 1 −ෑ

𝑖=1

𝑁

1 − 𝑒−𝜆𝑡 𝑀𝑇𝑇𝐹𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 =

𝑛=1

𝑁1

𝑛𝜆

Generalization to the case of N identical components

r-out-of-N system

• N identical components function in parallel but only r are needed (r is the

minimum number of components which should work for the system to work)

• Popular configuration for

❑ Relief valves, pumps and motors that must provide a specific capacity to

meet design criteria (e.g. a pressure vessel is equipped with 6 pressure

valves, but a pressure transient can be controlled successfully by any 3 of

these valves)

❑ Instrumentation and control systems (e.g. an alarm is triggered if any two of

three temperature sensors record a temperature above a given threshold)

• Parallel system is a particular case of the r-out-of-N system with r = 1

• For N identical components:

r-out-of-N system

• N identical components function in parallel but only r are needed (r is the

minimum number of components which should work for the system to work)

• Popular configuration for

❑ Relief valves, pumps and motors that must provide a specific capacity to

meet design criteria (e.g. a pressure vessel is equipped with 6 pressure

valves, but a pressure transient can be controlled successfully by any 3 of

these valves)

❑ Instrumentation and control systems (e.g. an alarm is triggered if any two of

three temperature sensors record a temperature above a given threshold)

• Parallel system is a particular case of the r-out-of-N system with r = 1

• For N identical components with reliability 𝑅1(𝑡):

𝑅 𝑡 =

𝑘=𝑟

𝑁

𝑁𝑘

𝑅1 𝑡 𝑘 1 − 𝑅1 𝑡𝑁−𝑘

r-out-of-N system

• N identical components function in parallel but only r are needed

• Popular configuration for

❑ Relief valves, pumps and motors that must provide a specific capacity to

meet design criteria (e.g. a pressure vessel is equipped with 6 pressure

valves, but a pressure transient can be controlled successfully by any 3 of

these valves)

❑ Instrumentation and control systems (e.g. an alarm is triggered if any two of

three temperature sensors record a temperature above a given threshold)

• Parallel system is a particular case of the r-out-of-N system with r = 1

• For N identical exponential components:

( )eek

NtR t

kNkt

N

rk

−−

−

=

−

= 1)(

1N

k r

MTTFk=

=

Standby system

• One component is functioning and when it fails it is replaced

immediately by another component (sequential operation of one

component at a time)

• Two types of standby:

• Cold: the standby unit cannot fail until it is switched on

• Hot: the standby unit can fail also while in standby

Cold standby - only one standby unit

𝑓1(𝑡)

𝑓2(𝑡)

• Compute the probability density function 𝑓𝑇(𝑡) of

the system failure time T

Cold standby: Exercise 2

• Consider a “cold” standby system of two units with constant

failure rates

• The on-line unit has a MTTF of 2 years

• When it fails, the standby unit comes on line and its MTTF

is 3 years

• Assume that each component has an exponential failure

times distribution

(1) What is the probability density function of the system

failure time? What is the MTTF of the system?

(2) Repeat assuming that the two components are in

parallel

(3) Compare the results between cold standby and parallel

Laplace Transform

−==0

)()(~

)(111

dttfesftfL T

ts

TT

)(~

)(~

)(*)()(~

2121 sfsftftfLsf TTTTT ==

)(~

)( 1 sfLtf TT

−=

s

dtesfeLtst

+===

+−−

1

1

0

)(

111)(

~

sssfT

+

+=

2

2

1

1)(~

)()( 12

21

21 tt

T eetf

−−−

−=

From tables …

Example

( ) ( )( )( )

( )( )( )ss

BABAs

ss

sBsA

eBLeALs

B

s

A

sssf

tt

++

+++=

++

+++=

+=+

++

=

+

+=

−−−−

21

2121

21

1221

2

1

1

1

2

2

1

1

2

2

1

1

)(

)(~

21

Laplace Transform: Computation (N=2)

( ) ( )( )( )

( )( )( )ss

BABAs

ss

sBsA

eBLeALs

B

s

A

sssf

tt

++

+++=

++

+++=

+=+

++

=

+

+=

−−−−

21

2121

21

1221

2

1

1

1

2

2

1

1

2

2

1

1

)(

)(~

21

0 1

=+

=+

1

021

BA

BA

21

2

21

1

−−=

−=

A

B

Laplace Transform: Computation (N=2)

)(

)(~

)(

21

21

12

21

21

1

tt

tt

TT

ee

eBeA

sfLtf

−−

−−

−

−−

=

+=

=

Laplace Transform Table

Hot standby (1)

• The standby unit can fail also while in standby

• The convolution theorem can no longer be used to calculate the

reliability of the system, because there is no independence of failures

any more: the failure of the first unit changes the failure behavior of the

second unit

Hot standby (1)

• The convolution theorem can no longer be used to calculate the

reliability of the system, because there is no independence of failures

any more: the failure of the first unit changes the failure behavior of the

second unit

• Simple case of two components (𝑅1 𝑡 , 𝑅𝑠 𝑡 , 𝑅2(𝑡)): the system will

perform its task in the interval (0, t) in either of the two mutually

exclusive ways:

• the online component 1 does not fail in (0, t)

[probability = R1(t) ]

• the online component fails in (, + d ) [probability = fT1()d];

the standby component 2 does not fail in (0, ) [probability Rs( ) and it

operates successfully from to t [probability R2(t-)]

0 𝜏 𝜏 + 𝑑𝜏 t

Ex 2: Time-dependent systems

• When both A and B are fully energized they share the total

load and the failure densities are fA(t) and fB(t)

• If either one fails, the survivor must carry the full load and its

failure density becomes gA(t) or gB(t)

A

B

Find the reliability R(t) of the system if

( ) ( ) t

A Bf t f t e −= = ( ) ( ) 1k t

A Bg t g t k e k −= =

Objectives of This Lecture (Reliability of Simple System)

◼ Compute the reliability of the following types of systems:

◼ Parallel

◼ Series

◼ Cold Standby

◼ Hot Standby

Lecture 5: Where to study?

◼ Slides

◼ Red book (‘An introduction to the basics or reliability and risk analysis’, E. Zio):

◼ All Chapter 5

◼ Exercises on Green Book (‘basics of reliability and risk analysis – Workout Problems and Solutions, E. Zio, P. Baraldi, F. Cadini)

◼ All problems in Chapter 5