System Reliability Old

SYSTEM RELIABILITY

ENN530 Slide <2> System Reliability

SYSTEM RELIABILITY • A system is formed by a number of

components connected together. The reliability of a system is the probability that a device or system will operate for a given period of time and under given operating conditions.

• There are categories of failure modes, which are studied through two simplest systems: series and parallel.


Series Reliability • A set of components, connected together so

as to form a system, is said to be in series reliability if the failure of any one component causes the failure of the total system (The system can be regarded as being “fault-intolerant”).

• Note: The components themselves need not be physically or topologically in series, what relevant is only the fact that all of them must succeed for the system to succeed.


Reliability Block Diagram • Reliability Block Diagram (RBD) is used to

model the effect of component failures on system performance.

Figure 1: Series RBD

Unit 1 Cause Unit 2 Unit n Effect


Series Reliability

What is reliability of such system?


Series Reliability Function

• Dependent

Where Rs represents system reliability, n is the total number of components in a system. ti is time to failure of i th item.

• Independent

12121312121 ......)(},...{ −== nnns ttttPtttPttPtPtttPR

∏∏==

≡==n

ii

n

iins RtPtttPR

1121 )(},...{


Characteristics of Reliability function of a series system

1. The value of the reliability function of the system, Rs(t), for a series configuration is less than or equal to the minimum value of the individual reliability function of the constituting items, which is:

)}({)(0,...,2,1

tRMintR iis =≤


Characteristics of Reliability function of a series system

2. If λi (t) represent the failure rate of item i, then the system reliability of a series system can be written as:

dxxtRt n

iis )]}([exp{)(

0 1∫ ∑

=

−= λ


Example: A system consists of four items, each of them are necessary to maintain the required function of the system. The time to failure distribution and their corresponding parameter values are given in the following table. Find the reliability of the system for 500 and 700 hours of operation.

Item Time to failure distribution

Parameter values

Item 1 Exponential λ= 0.001 Item 2 Weibull η= 1200 h β= 3.2 Item 3 Normal μ= 800 h σ= 350

Item 4 Weibull η =2000 h β=1.75


Solution: From the information given in the table, the reliability function of various items can be written as:

])2000

(exp[)(

)350

800()(

])1200

(exp[)(

)001.0exp()(

75.14

3

2.32

1

ttR

ttR

ttR

ttR

−=

−Φ=

−=

×−=

])2000

(exp[)(

)350

800()(

])1200

(exp[)(

)001.0exp()(

75.14

3

2.32

1

ttR

ttR

ttR

ttR

−=

−Φ=

−=

×−=


Solution (Cont.): Since the items are connected in series, the reliability function of the system is given by:

])2000

(exp[)350

800(])1200

(exp[)001.0exp()( 75.12.3 tttttRs −×−

Φ×−××−=

Substituting t = 500 and 750 in the above equation, we get:

1759.08355.05568.08003.04723.0)750(4202.09154.08043.09410.06065.0)500(

=×××==×××=

RR


Failure Rate of series configuration with n items

)()(1

ttn

iis ∑

=

= λλ)exp()(

tlExponentia

ii λλ −

))(exp()(

)(

1 ii

iii

i ttWeibull

ββ

ηηηβ

−−1

1))(()( −

=∑= i

i

n

i i

is

tt β

ηηβλ

})(21exp{

21

)(

2

i

i

i

tNormal

σµ

πσ−

− )(/)()(1 i

in

iis

ttftσ

µλ −Φ= ∑

=

Probability density function of i-th item, fi(t)

Failure rate of the system, λs(t)


Mean time to failure (MTTF) of a series system

The mean time to failure, MTTF, of a series configuration, denoted by MTTFs, can be written as:

dttRdtRMTTF i

n

iss )(0

10

∫∫∞

=

∞

Π==


Characteristics of MTTFs of series system

The mean time to failure of a system with series RBD will be less than the mean time to failure of any its constituting items, where MTTFi denote the mean time to failure of the item i.

1.

2. For complex repairable systems, MTTFs represents the mean time to first failure

}{0,...2,1 iis MTTFMinMTTF

=≤


Example: A system consists of three items connected in series. The time-to-failure distribution and their corresponding parameter values are given at the following table. Find the mean time to failure of the system. Compare the value of MTTFs with mean time to failure of individual items.

Item Distribution Parameter Values Item 1 Weibull η1= 10, β1=2.5

Item 2 Exponential λ= 0.2

Item 3 Weibull 2= 20, β2= 3


Solution: Mean time to failure of the system is given by:

dtttt ))(exp()exp()(exp( 21

20 1

ββ

ηλ

η−×−×−= ∫

∞

dttRMTTF ii

s )(0

3

1∫Π∞

=

=

hrdtttt 48.3))20

(exp()2.0exp()10

(exp( 35.2

0

=−×−×−= ∫∞

Compare with MTTF for Item1: 8.87 hr, Item2: 5 hr, Item3:17.86 hr. Note that the formula can only be evaluated using numerical integration.


• Consider a series system of n components, each has a constant failure rate

• since:

Constant failure rate in series reliability

∏=

=n

iis RR

1

])...(exp[)(),exp(

21 ttRthentR

and

ns

ii

λλλλ

+++−=−=


Constant failure rate in series reliability

• The system’s failure rate is: • MTTF:

∑=

λ=λn

iis

1

∑=

λ=

λ=θ n

ii

ss

1

11


Example • The pump system in Slide 5: Assume failure

rate of the motor is 4.0X10-6/hr, the gear box is 3.2X10-6/hr, and the pump is 9.8X10-6/hr.

• What is system failure rate?

• At what probability the system can operate for at least 1000 hours?

663

1101710)8.92.34( −−

=

×=×++== ∑i

is λλ

983.0)( == − ts

setR λ


• A set of components are involved in a system. The system fails only when all components have failed.

Figure 2: Parallel RBD • Such a system is regarded as being “fault-tolerant”.

Parallel Reliability

Unit 1

Unit 2

Unit n

Effect Cause


Pump system


Parallel Reliability Function

}...{ 21 ns tttPR +++=

∏=

==n

iins tFtFtFtFF

121 )()()...()(

∏∏==

−−=−=−=n

ii

n

iiss tRtFFR

11

)}(1{1)(11


• A limit switch has a probability of failure of 17.8% over a one year period. If two switches are installed in parallel, what is the probability that both fail in a one year period.

• Fs= F1 xF2 = .178 x .178 = .032 = 3.2% • Thus the system is much more reliable with two

switches.

• With three switches, Fs=0.1783 = 0.56%.

Parallel Systems - Example


Characteristics of a parallel system

1. The system reliability, Rs(t), is more than reliability of the any of the consisting items. That is,

)}({)(,...,1

tRMaxtR inis =≥

)}({)(,...,1

tRMaxtR inis =≥

• If λi(t) represents the failure rate of item i, then the reliability function of a parallel configuration can be written as:

)])(exp(1[1)(0

1dtttR

t

i

n

is ∫−−Π−==

λ


Example:

A fly-by-wire aircraft has four flight control system electronics (FCSE) connected in parallel. The time-to-failure of FCSE can be represented by Weibull distribution with scale parameter η = 2800 and β = 2.8. Find the reliability of flight control system for 1000 hours of operation.


Solution: Reliability function for a parallel system with four identical items is given by:

44

1)](1[1)](1[1)( tRtRtR iis −−=−Π−=

=

9455.0))28001000(exp())(exp()( 8.2 =−=−= β

ηttR

999991.0]9455.01[1)1000( 4 =−−=sR

Where R(t) is the reliability function of each item. For t=1000, R(t) is given by:

Thus, the reliability of flight control system for 1000 hours of operation is given by:


Failure Rate of a parallel system

n

i

i

n

ji

n

jiij

s

tR

tFtft

1

1 ,1

)](1[1

)}()({)(

=

=≠=

−Π−

Π×=

∏

∑λ ∏

∑

=

=≠=

−−

Π×=

n

ii

n

ji

n

jiij

s

tR

tFtft

1

1 ,1

)](1[1

)}()({)(λ

Where, fi(t) is the probability density function of item i


Example:

A fly-by-wire aircraft has four flight control system electronics (FCSE) connected in parallel. The time-to-failure of FCSE can be represented by Weibull distribution with scale parameter η = 2800 and β = 2.8. Find the failure rate of the system at time t = 100.


Solution: Since all the four items are identical, the failure rate of the system can be written as:

4

3

)]([1)]([)(4)(

tFtFtfts −

××=λ

))(exp()()( 1 ββ

ηηηβ tttf −= − ))(exp(1)( β

ηttF −−=

Substituting t = 100, we get λs(t) = 8.0 x 10-8

Where:


MTTFs of a parallel system

The mean time to failure system can be written as:

dttRdtRMTTF i

n

iss )]}(1[1{0

10

−Π−== ∫∫∞

=

∞


Constant failure rate in parallel reliability

• For each component, failure rate is a constant. • If the system has two units, we assume that

the failure of the two units are independent. • Assume time-to-failure distribution of

component i is exponential with mean λi. Then MTTFs is given by:

dttdtRMTTF i

n

ii

n

is )]}exp(1[1{0

10

1λ−−Π−=Π= ∫∫

∞

=

∞

=


For particular values of n, we can simplify the equation to derive the expression for the MTTFs as following:

Assume n = 2 (Two components):

2121

111λλλλ +

−+=sMTTF

Assume n = 3 (Three components) :

321323121321

1111111λλλλλλλλλλλλ ++

++

−+

−+

−++=sMTTF


REDUNDANCY • Systems in which some components may fail

without causing system failure are “fault tolerant”. To achieve this, some measure of hardware redundancy must be introduced. There are two main types of redundancy: active and passive (or standby) redundancy.

• In Active redundancy, two or more components are placed in parallel reliability. Load is normally shared between the components. If one component fails, then the remaining units carry the full load.

• In Passive (Standby) redundancy, one component carries full load. If this component fails, another standby component (not necessarily identical) is switched on to take the place.


Pump system


Active Redundancy

• System failure function:

a

b

[ ][ ][ ] [ ] [ ][ ])()()()(1

)(1)(1)()()(

tRtRtRtRtRtR

tFtFtF

baba

ba

bas

+−−=−−=

=


Active Redundancy • Reliability function:

• If a and b are constant failure rate components, what is Reliability function?

• Is the resulted system a constant failure rate system?

)()()()()(1)(

tRtRtRtRtFtR

baba

ss

−+=−=

No

ttts

baba eeetR )()( λλλλ +−−− −+=


“m-out-of-n” Good: • In some cases, m out of n identical units must

stay “up” for the system to be operational. • Rs(m,n,p) – System reliability when there are

“n” number of components in a system, the system is considered being functioning properly if m or more units (m, m+1, m+3,.…, n-1,n) are functioning. Let p be the probability that a unit is functioning. The probability of exactly m functioning units (out of n):

( ) mnm ppmn

pnmp −−

= )1(,,


Therefore, the system reliability:

( ) ∑=

=n

mrs pnrppnmR ),,(,,

∑=

−−

=

n

mr

rnr pprn

)1(


Example:

A system has 4 components, what is the reliability of the system being operational with at least 3 units being operational:

∑=

−−

=

4

3

4)1(34

r

rr pp

43

04343

34

)1(44

)1(34

pp

pppp

−=

−

+−

= −


Standby Redundancy

1

2

ττττ

dtRftRRt

s )()()( 20

11 −⋅+= ∫=


Standby Redundancy • R1(t) – the reliability of unit 1 at time t

• R2(t - τ) − the reliability of the unit 2 after it started to operate at time τ

• F1(t) – the pdf for the time to failure of unit 1(τ)


Constant Failure Rate Special case 1: Time to failure distribution of all units follows an exponential distribution

tetR 1)(1λ−= tetR 2)(2

λ−=

τλ

τλ

τ

λλλλ

τλτλ

τ

λ

deee

deeetR

tttt

tt

tsb

∫

∫

=

−−−−

−−−

=

−

+=

+=

0

)(1

)(

01

2121

211)(


)(

)1(

121

212

1

21

1

)(

21

1

ttt

tt

t

eee

eee

λλλ

λλλ

λ

λλλ

λλλ

−−−

−−−

−

−−

+=

−−

+=

We assume that switch is perfect, no standby failure.

Constant Failure Rate (Cont.)


Series – Parallel system

)](1(1[ ,11 ji

m

i

n

js xRR −Π−Π===

R(xi,j) – Reliability of component i in sub-system j n = no. of sub-system, j = 1,2,…..,n; i=1,…m.

m

.

.

.

.

.

.

.

.

…..

m

.

.

.

.

.

.

.

.

m

.

.

.

.

.

.

.

.

1 2 n


Special case: all units are identical. R(xi,j) = p Rs = [ 1 – (1-p)m]n

Example: n = 3, m = 2

Rs = [ 1 – (1-p)2]3

If p = 0.85, Rs = 0.934007


Increasing System Reliability • During design stage

– Simplifying the system – Use better quality components – Improve environments, e.g. use cooling fans,

reduce vibration, etc. – Use Fault-Tolerance in the system


Increasing System Reliability • During development and production stage

– Design improvement – Burn-in components or system – Improve quality control during manufacturing – Use of preventive maintenance – use of corrective maintenance

Any Questions?

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System Reliability Old