29

Poisson DistributionCan be used to evaluate the probability of an isolated event occurring a specific number of times in a given time interval,

e.g. # of faults, # of lightning strokes time interval

Requirements:-Events must be random-Hazard rate must be constant

Only applies to the useful life period of a system component

Expression for Poisson Distribution: Px(t) = x!

e)( λtx −tλ

Px(t): probability of event occurring x times in time tλ : constant hazard rate (known as failure rate)

Expected value of Poisson distribution, E(x) = μ = λt

Probability of zero failures in time t, P0(t) = R(t) = =0!

et)( λt0 −λ λte−

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Poisson Distribution ExampleIf the average number of cable faults per year per 10 km of cable is 0.05,evaluate the probabilities of 0, 1, 2, .. faults occurring in(a) 20 year period(b) 40 year period

Px(t) = x!e 1x −)1(

Failure Rate, λ = 0.05 f/yr

(a) For a 20 year period, t = 20 yr

Expected # of failures, E(x) = λt = 0.05 x 20 = 1.0# of failures 0 1 2 3 4Probability 0.36788 0.36788 0.18394 0.06131 0.01533

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4# of failures

prob

abili

ty

Failure Distribution FunctionFailure Density Function

0

0.1

0.2

0.3

0.4

0 1 2 3 4# of failures

prob

abili

ty

Px(t) = x!

e)( λtx −tλ

32

Standby Systems - Poisson Distribution One or more components are in standby mode ready to operate when the main operating component fails.

A sensing and changeover device is required to bring the standby unit into operation when the main component fails.

Failure of standby component B is dependent on the failure or success of the main component A.

If the components are in their useful life period, their times to failure are exponentially distributed. Poisson Distribution can, therefore, be applied.

A

B

S A: main componentB: standby componentS: sensing and changeover device

33

Perfect SwitchingAssumptions:

- sensing & changeover device is 100% reliable- standby components do not fail in standby mode- components are identical with failure rate λ

A

B

S

The 2-component standby system is equivalent to a single unit which is allowed to fail once. If there is a 2nd failure, the equivalent unit also fails.

Applying Poisson’s Distribution:

Probability of x components failing in time t, Px(t) = x!e)( λtx −tλ

P (0 fail) = P0(t) =λte− P (1 fail) = P1(t) = λt. λte−

System Reliability, Rs (t) = P0 (t) + P1(t) = (1 + λt)λte−

∫∞

0

)( dttRs λ2

MTTF = =

34

Perfect Switching : 2-Standby Components

∫∞

0

)( dttRs λ3

MTTF = =

This system is equivalent to a single unit which is allowed to fail twice.

A

B

S

C Rs (t) = P0 (t) + P1(t) + P2(t) = (1 + λt + )λte−2)( 2tλ

For a system with ‘n’ standby components,

Rs (t) = ∑=

n

x 0 x!e)( λtx −tλ i.e. the sum of first n terms of the Poisson Distribution.

∫∞

0

)( dttRs λ1+n

MTTF = = The system MTTF is the sum of individual component MTTF values.(since each component gets its turn after complete operation of the previous component)

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ExampleCompare the reliability and MTTF of a 2-component system each having a failure rate of 0.02 f/hr after a time of 10 hr if they are

(a) parallel redundant (b) standby redundant with 100% reliable sensing and changeover device.

(a) Parallel system

Using product rule of unreliability,(b) Standby system

Qs(t) = Q1(t). Q2(t)

Qs(10h) = 0.032859

= (1- )(1- ) tλ−e tλ−e

= (1- )21002.0e x−

Rs(10h) = 1 - 0.032859 = 0.967141

MTTF = + + = 75 h1

1λ 2

1λ 21

1λλ +

Rs (t) = (1 + λt)λte−

Rs(10h) = 0.982477

λ2

MTTF = = 100 h

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Effect of Spare Components

Spare component system becomes similar to standby system (perfect switching) if

- replacement time of a failed component by a spare is insignificant- failed components are not repaired

With these assumptions, Poisson Distribution method can also be applied to evaluate the effect of spare components.

Applying Poisson’s Distribution:

Probability of x components failing in time t, Px(t) = x!e)( λtx −tλ

37

Standby Systems: Imperfect SwitchingSensing & changeover device (S) has a probability of successful operation, Ps

where Ps =A

B

S

For a 2-component standby system:

P (0 comp. fails) = P0(t) =λte−

P (1 fails AND S operates successfully) = P1(t). Ps = Ps .λt. λte−

System Reliability, Rs (t) = P0 (t) + Ps.P1(t) = (1 + λt. Ps)λte−

operations requested of #operations successful of #

∫∞

0

)( dttRsMTTF = = λ

sP+1

38

ExampleCompare the reliability and MTTF of a 2-component system each having a

failure rate of 0.02 f/hr after a time of 10 hr if they are

a) parallel redundant b) standby redundant with the sensing and changeover device reliability

between 1.0 and 0.9

(a) Parallel system

Using product rule of unreliability,

(b) Standby system

Qs(t) = Q1(t). Q2(t)

Rs(10h) = 0.967141

MTTF = 75 h

Rs (t) = (1 + λt. Ps)λte− Rs (10) = (1 + 0.2 Ps)

2.0e−

MTTF = =λ

sP+102.0

1 sP+

950.9661020.90

960.9693770.92

970.9726520.94

980.9759270.96

990.9792020.98

1000.9824771.00

MTTF (hr)Rs (10 hr)Ps