Handout on Poisson processes and MTTF

Poisson Process Derivation
The following discussion assumes that the failure rate is constant λ t = λ .
Furthermore, consider possible failures on the interval 0, t + dt and define P ! (t)
as the probability that ‘x’ failures occur in the interval 0, t . The special case would
be P ! (t), the probability that 0 failures occur in the interval 0, t , which is equal to
the reliability e !!" . The probability that one failure occurs in the interval 0, t +
dt is equal to:
P ! t + δt = P ! t P 1 failure in the interval t to t + dt
+ P ! t P 0 failures in the interval t to t + dt
(1)
Assuming that the interval dtis so small it can only contain one failure, the
probability that 1 failure occurs in the interval t, t + dt can be approximated by
λdt and the probability that no failure occurs is equal to 1 − λdt. Substituting these
probabilities into equation (1) yields:
P ! t + dt = P ! t λdt + P ! t 1 − λdt (2)
Equation (2) can be rewritten to obtain a first-­‐order differential equation:
P ! t + dt = P ! t + λdtP ! t − λdtP ! t ⟹
P ! t + dt − P ! t
= λP
dt
! t − λP ! t ⟹
dP ! t
= λP
dt ! t − λP ! t
(3)
Substituting P ! t = e !!" into equation (3) yields the following equation:
dP ! t
dt
= λe !!" − λP ! t (4)
A solution to differential equation (4) is given by:
P ! t = λte !!"
(5)
Following the above procedure, the expression for the probability of ‘x’ failures on
the interval 0, t + dt could be written as follows:
P ! t + dt = P ! t P x failures in the interval t to t + dt
+ P ! t P x − 1 failures in the interval t to t + dt
+ ⋯ + P !!! t P 1 failure in the interval t to t + dt
+ P ! t P 0 failures in the interval t to t + dt
(6)
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Since dt is so small it can only contain one failure, equation (6) can be reduced to:
P ! t + dt = P !!! t P 1 failure in the interval t to t + dt
+ P ! t P 0 failures in the interval t to t + dt
(7)
Which can be written as:
dP ! t
dt
= λP !!! t − λP ! t
(8)
Note that for x = 1 equation (8) equals Equation (3). Starting with x = 2, this
differential equation can be solved iteratively:
dP ! t
dt
= λP ! t − λP ! t = λ ! te !!" − λP ! t ⟹ P ! t = λt !
2 e!!"
(9)
dP ! t
dt
= λP ! t − λP ! t ⟹ P ! t = λt !
3 ⋅ 2 e!!"
(10)
Or, in general:
P ! t = λt !
e !!"
x!
(11)
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MTTF
The reliability is defined as the probability of having a working component or
system after an exposure time ‘t’, or in other words:
R t = P 0 failures in the interval 0 to t
(12)
Let’s now define the Failure Density Function (just like a probability density
function) as the probability of failure per unit of time. In that case, the unreliability
is given by the probability that a failure occurs within (0, t) or the area underneath
of the failure density curve:
F t =
!
!
f τ dτ
⟹ f t = dF(t)
dt
(13)
Similarly, the reliability is given by the probability of survival beyond time ‘t’.
R t =
!
!
f τ dτ = 1 − F t ⟹ f t =
dF t
dt
dR t
= −
dt
(14)
Substituting R(t) = e !!" in equation (14) yields a failure density equal to:
dR t
f t = −
dt
= λe !!" (15)
The Mean Time To Failure (MTTF) is given by the statistical mean (expected
value) of time ’t’ given the failure density function f(t), or
!
!
MTTF = E{t} = tf t dt = λte !!" dt = 1 λ
!
!
(16)
The solution to the integral in equation (16) uses integration by parts. A more
general expression for MTTF can be obtained by substituting equation (14) for f(t):
MTTF = E{t} =
!
!
tf t dt = −t
!
!
dR t
dt
dt
(17)
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= −tdR t
!
!
= tR(t) ! ! + R t dt
!
Since tR(t) ! ! = 0, equation (17) reduces to:
MTTF =
!
!
R t dt
! (18)
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