Expectation and Variance / Examples and

Computation of Mean Time to Failure

ECE 313
Probability with Engineering Applications
Lecture 28 - December 6, 1999
Professor Ravi K. Iyer
University of Illinois

Iyer - Lecture 28 ECE 313 - Fall 1999

Mean Time to Failure

¥ Let X denote the lifetime of a component so that its reliability

R(t) = P(X>t) and RÕ(t) = -f(t).
¥ The expected life or the mean time to failure (MTTF) of the
component is given by:
∞ ∞
E[ X ] = ∫ tf (t )dt = − ∫ tR′(t )dt
0 0

¥ Integrating by parts:
∞
E[ X ] = −tR(t ) 0 ∞ + ∫ R9t )dt
0

¥ Since R(t) approaches zero faster than t approaches ∞:

∞
E[ X ] = ∫ R(t )dt
0

Iyer - Lecture 28 ECE 313 - Fall 1999

 Mean Time to Failure (cont.)
¥ This latter expression for MTTF is in more common use in
reliability theory. Generally:
∞
E[ X k ] = ∫ t k f (t )dt
0
∞
= − ∫ t k R′(t )dt
0
∞ ∞
= −t k R(t ) + ∫ kt k −1 R(t )dt
0 0
¥ Thus:
∞
E[ X k ] = ∫ kt k −1 R(t )dt
0
Iyer - Lecture 28 ECE 313 - Fall 1999
Mean Time to Failure (cont.)
¥ In particular: 2
∞
∞ 
Var[ X ] = ∫ 2tR(t )dt −  ∫ R(t )dt 
0 0 
¥ If the component lifetime is exponentially distributed, then
R(t)=e-λt and:
∞ 1
E[ X ] = ∫ e − λt dt = ,
0 λ
∞ 1
Var[ X ] = ∫ 2te − λt dt − 2
0 λ
2 1 1
= − =
λ2 λ2 λ2
as derived earlier.
Iyer - Lecture 28 ECE 313 - Fall 1999
 Series System
¥ Assume that the lifetime of the ith component is exponentially
distributed with parameter λi. System reliability is given by:
R(t ) = ∏ Ri (t ) = ∏ e − λit = exp −( ∑ λi )t 
n n n
i =1 i =1  i =1 
¥ Thus, the lifetime of the
n
system is also exponentially distributed
with parameter λ = ∑ λi .
i =1
¥ The system series MTTF is:
1
n
∑ λi
i =1
¥ The MTTF of a series system is much smaller the the MTTF of
its components.
Iyer - Lecture 28 ECE 313 - Fall 1999
Series System (cont.)
¥ If Xi denotes the lifetime of component i, and X denotes the
series system lifetime, then we can show that:
0 ≤ E[ X ] ≤ min{RXi (t )}
i
¥ To prove inequality:
n ∞
RX (t ) = ∏ RXi (t ) ≤ min{∫ RXi (t )} since 0 ≤ RXi (t ) ≤ 1
i =1 i 0
¥ Then: ∞ ∞
E[ X ] = ∫ RX (t )dt ≤ min{∫ RXi (t )dt}
0 i 0
= min{E[ Xi ]}
i
Iyer - Lecture 28 ECE 313 - Fall 1999
 Parallel System
¥ Consider a parallel system of n independent components, with
Xi denoting the lifetime of component i and X denoting the
lifetime of the system:
¥ Then X = max{X1 , X2 ,..., Xn}, and
n
RX (t ) = 1 − ∏ [1 − RX1 (t )] ≥ 1 − [1 − RXi (t ) for all i
i =1
¥ Which implies that the reliability of a parallel redundant system
is larger than that of any of its components.
Iyer - Lecture 28 ECE 313 - Fall 1999
Parallel System (cont.)
∞ ∞
¥ Therefore: E[ X ] = ∫ RX (t )dt ≥ max{∫ RX (t )dt}
i i
0 0
= max{E[ Xi ]}
i
¥ Assume that Xi is exponentially distributed with parameter λ (all
components have the same parameter). Then:
RX (t ) = 1 − (1 − e − λt ) n
and
∞
E[ X ] = ∫ [1 − (1 − e − λt ) n ]dt
0
Iyer - Lecture 28 ECE 313 - Fall 1999
 Parallel System (cont.)
¥ Let u = 1 − e − λt , then dt = 1 λ (1 − u)du.
1 1 1 − un
¥ Thus: E[ X ] = ∫ du
λ 0 1− u
¥ Since the integrand above is the sum of a finite geometric
series:
11 n
E[ X ] = ∫ ( ∑ u i −1 )du
λ 0 i =1
1 n−1 1 i −1
= ∑ ∫ u du
λ i=0 0
Iyer - Lecture 28 ECE 313 - Fall 1999
Parallel System (cont.)
1
i −1 ui 1 1
¥ Note that: ∫ u du = =
0 i 0 i
¥ Thus, with the usual exponential assumptions, the MTTF of a
parallel redundant system is given by:
1 n1 H 1n(n)
E[ X ] = ∑ = n ≈
λ i =1 i λ λ
¥ The next figure shows the expected life of a parallel system as a
function of n. It should be noted that beyond n = 2 or 3, the gain
in expected life is not very significant. Notes that the rate of
increase in the MTTF is 1/(nλ).
Iyer - Lecture 28 ECE 313 - Fall 1999
 Parallel System (cont.)
¥ The variation in the expected life with the degree of (parallel)
redundancy (simplex failure rate λ=10-6)
Iyer - Lecture 28 ECE 313 - Fall 1999
Parallel System (cont.)
¥ Alternatively the formula for E[X] can be derived by noting that X
is exponentially distributed with parameters nλ , (n − 1)λ , L, λ .
n
¥ In other words, X = ∑ Yi where Yi is exponentially distributed
with parameter iλ. i =1
¥ Then, using the linearity property of expectation:
n n Hn
E[ X ] = ∑ E[Yi ] = ∑ =
i =1λ i =1
¥ Also, since the YiÕs are mutually independent:
n n 1 1
Var[ X ] = ∑ Var[Yi ] = ∑ = Hn ( 2 )
i =1 i 2 λ2 λ2
i =1
¥ Note that CX < 1. Hence, not only does the parallel configuration
increase the MTTF, it also reduces the variability of the system
lifetime.
Iyer - Lecture 28 ECE 313 - Fall 1999
 Standby Redundancy
¥ Assume that the system has one component operating and (n-1)
cold (unpowered) spares.
¥ The failure rate of an operating component is λ, and a cold
spare does not fail.
¥ Furthermore, the switching equipment is failure free.
¥ Let Xi be the lifetime of the ith component, from the point at
which it is put into operation until its failure.
¥ Then the system lifetime, X, is given by: X = n X
∑ i
i =1
¥ Thus X has an n-stage Erlang distribution, and therefore:
n n
E[ X ] = and Var[ X ] =
λ λ2
Iyer - Lecture 28 ECE 313 - Fall 1999