ESTIMATION OF SYSTEM RELIABILITY IN STRESS-STRENGTH
MODELS FOR DISTRIBUTIONS USEFUL IN LIFE TESTING
DAVID D. HANAGAL
Department of Statistics, University of Poona, Pune-411007, India.
Abstract
In this paper, we estimate the reliability of a parallel system with two components. We assume the strengths of two components are subjected to a common stress which is independent of the strengths of the components. If (X1 , X2 )
are strengths of two components subjected to a common random stress X3 , then
the reliability of a system or system reliability (R) is given by R = P [X3 <
M ax(X1 , X2 )]. We estimate R when (X1 , X2 , X3 ) follow three independent gamma
or weibull or pareto distributions. We also obtain the asymptotic normal (AN) distributions of these estimates.
1
Introduction
Estimation of reliability of a component i.e., P [Y < X] when X is random strength
and Y is independent random stress of a component has been discussed extensively in
the literature. Enis and Geisser(1971), Tong(1974), Kelley, Kelley and Schucany(1976)
and Beg(1980) have obtained the estimate of P [Y < X] when X and Y are independent
exponential variables. The situation where X and Y follow independent normal distributions are tackled by Church and Harris(1970), Enis and Geisser(1971), Downton(1973),
Reiser and Guttman(1986) and recently Weerahandi and Johnson(1992). Estimation of
P [Y < X] when (X, Y ) have independent gamma distributions has been discussed by
Basu(1981) and Reiser and Rocke(1993). Beg and Singh(1979) obtained the estimate
1
of P [Y < X] when X and Y are independent pareto distributions. Bhattacharyya and
Johnson(1974, 1975, 1977) and Johnson(1988) have discussed estimation of system reliability by nonparametric method as well as assuming stress and strength are exponential
distributions.
In this paper, we consider the estimation of system reliability, R = P [X3 < M ax(X1 ,
X2 )] when (X1 , X2 , X3 ) are mutually independently distributed. Here (X1 , X2 ) are random strtengths of two components subjected to a common stress (X3 ). There are some
applications described below in stress-strength context.
1) The welding machine which gives stress on both eyes of the operator of welding
machine. If the number of hours operating the welding machine is less than the maximum
of the strength of two eyes, then a person can able to work successfuly. Here the strength
of two eyes is the maximum number of hours working with welding machine and R is
the probability that a person can able to work successfuly.
2) Two filaments in an electric bulb fail due to high voltage. If the voltage is less than
the maximum of the strength of the two filaments, then the electric bulb will function.
Here strength of the two filaments is maximum voltage allowable and R is the probability
that an electric bulb is functioning.
Let (X1i , X2i , X3i ), i = 1,...,n be i.i.d. random sample of size n and n1 = number of
observations with X3i < M ax(X1i , X2i ) in the sample of size n. Then the distribution of
n1 is binomial (n,R). The natural estimate of R is R̃ = n1 /n which is AN[R,R(1-R)/n].
Here R is the function of the parameters of the distributions of (X1 , X2 , X3 ).
We consider maximum likelihood estimate of R when (X1 , X2 , X3 ) are independent
gamma distributions, independent weibull distributions and independent pareto distributions in Section 2.
2
2
Estimation of System Reliability
In this Section, we first consider the distribution of stress (X3 ) and strengths (X1 , X2 )
as
fi (xi ) = µαi i xαi i −1 exp(−µi xi )/Γ(αi ),
xi > 0,
µi > 0,
i = 1, 2, 3
where αi , i = 1,2,3 are known integers. When αi = 1, i=1,2,3; the distribution of Xi ,
i=1,2,3 will be independent exponential random variables.
The system reliability is given by R = P [X3 < M ax(X1 , X2 )]. We first obtain the
distribution of Z = M ax(X1 , X2 ) i.e.,
H(z) = P [Z < z]
= P [X1 < z].P [X2 < z]
= (1 −
αX
1 −1
µi1 z i exp(−µ1 z))(1
−
αX
2 −1
i=0
where P [Xi < z] = (1 −
Pαi −1
j=0
µi2 z i exp(−µ2 z))
i=0
µji z j exp(−µi z)), i = 1,2,3.
Hence,
H(z) = 1 −
αX
1 −1
µi1 z i exp(−µ1 z) −
i=0
+
µj2 z j exp(−µ2 z)
j=0
αX
1 −1 αX
2 −1
i=0
αX
2 −1
µi1 µj2 z i+j exp(−(µ1 + µ2 )z)
j=0
H(z) = P [Z > z]
=
αX
1 −1
µi1 z i exp(−µ1 z) +
i=0
−
µi1 µj2 z i+j exp(−(µ1 + µ2 )z)
j=0
Now the system reliability R is given by
R =
Z
0
µj2 z j exp(−µ2 z)
j=0
αX
1 −1 αX
2 −1
i=0
αX
2 −1
∞
H(x3 )dF3 (x3 )
3
=
αX
1 −1
i=0
−
αX
2 −1
µi1 µα3 3 Γ(α3 + i)
µj2 µα3 3 Γ(α3 + j)
+
α3 +j Γ(α )Γ(j + 1)
(µ1 + µ3 )α3 +i Γ(α3 )Γ(i + 1)
3
j=0 (µ2 + µ3 )
αX
1 −1 αX
2 −1
i=0
j=0
µi1 µj2 µα3 3 Γ(α3 + i + j)
(µ1 + µ2 + µ3 )α3 +i+j Γ(α3 )Γ(i + 1)Γ(j + 1)
The estimate of R based on MLEs of (µ1 , µ2 , µ3 ) is given by
R̂ =
αX
1 −1
i=0
−
αX
2 −1
µ̂i1 µ̂α3 3 Γ(α3 + i)
µ̂j2 µ̂α3 3 Γ(α3 + j)
+
α3 +j Γ(α )Γ(j + 1)
(µ̂1 + µ̂3 )α3 +i Γ(α3 )Γ(i + 1)
3
j=0 (µ̂2 + µ̂3 )
αX
1 −1 αX
2 −1
i=0
j=0
µ̂i1 µ̂j2 µ̂α3 3 Γ(α3 + i + j)
(µ̂1 + µ̂2 + µ̂3 )α3 +i+j Γ(α3 )Γ(i + 1)Γ(j + 1)
The MLEs of the parameters (µ1 , µ2 , µ3 ) are µ̂i = nαi /
Pn
j=1
xij , i = 1,2,3.
The asymptotic distribution of R̂ is AN[R, G01 Λ1 G1 ] where G01 = (∂R/∂µ1 , ∂R/∂µ2 ,
∂R/∂µ3 ) and Λ1 = n1 diag(µ21 /α1 , µ22 /α2 , µ23 /α3 ).
We next consider the distribution of stress and strengths are independent Weibull
distributions. The cdf of Xi , i = 1,2,3 are
Fi (xi ) = 1 − exp{−(xi /θi )c },
xi > 0, θi , c > 0, i = 1, 2, 3
Here we assume (X1 , X2 , X3 ) have common known shape parameter. When c = 1, Xi , i
= 1,2,3 will be independent exponential random variables. when c = 2, we have Xi , i=
1,2,3 independent Rayleigh distributions.
Now the system reliability R is
R =
1
1
1
+
−
1 + (θ3 /θ1 )c 1 + (θ3 /θ2 )c 1 + (θ3 /θ1 )c + (θ3 /θ2 )c
The estimate of R based on MLEs of (θ1 , θ2 , θ3 ) is
R̂ =
1
1 + (θ̂3 /θ̂1 )c
+
1
1 + (θ̂3 /θ̂2 )c
The MLEs of (θ1 , θ2 , θ3 ) are given by θ̂i = (
Pn
j=1
−
1
1 + (θ̂3 /θ̂1 )c + (θ̂3 /θ̂2 )c
xcij /n)1/c ,
i = 1, 2, 3.
The distribution of R̂ is AN[R, G02 Λ2 G2 ] where G02 = (∂R/∂θ1 , ∂R/∂θ2 , ∂R/∂θ3 ) and
Λ2 = n1 diag(θ12 /c21 , θ22 /c22 , θ32 /c23 ).
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Now we next consider the estimate of R when (X1 , X2 , X3 ) have independent Pareto
distribution with different lower bound i.e., k1 , k2 , k3 respectively. Now the cdf of Xi ,
i=1,2,3 are
Fi (xi ) = 1 − (ki /xi )ai ,
ki > 0, ai > 0; xi ≥ k,
i = 1, 2, 3.
Now the system reliability (R) is given by
R=








a3
( k1 )a1
a1 +a3 k3
+
a3
( k2 )a2
a2 +a3 k3
1 − ( kk13 )a3 {1 −






k3 a3

1 − ( k2 ) {1 −
a
−
a
k1 1 k2 2
a3
,
a1 +a2 +a3 k3 a1 +a2
M ax(k1 , k2 ) < k3
a3
a1 +a3
−
a3
( k2 )a2
a2 +a3 k1
+
a3
( k2 )a2 }
a1 +a2 +a3 k1
k3 < k 1 < k 2
a3
a2 +a3
−
a3
( k1 )a1
a1 +a3 k2
+
a3
( k1 )a1 }
a1 +a2 +a3 k2
k3 < k 2 < k 1
The estimate of R based on MLEs of (a1 , a2 , a3 , k1 , k2 , k3 ) is
R̂ =








â3
( k̂1 )â1
â1 +â3 k̂3
+
â3
( k̂2 )â2
â2 +â3 k̂3
1 − ( k̂k̂3 )â3 {1 −

1





k̂3 â3

1 − ( k̂ )
2
{1 −
−
(k̂1 )â1 (k̂2 )â2
â3
,
â1 +â2 +â3 (k̂3 )â1 +â2
M ax(k̂1 , k̂2 ) < k̂3
â3
â1 +â3
−
â3
( k̂2 )â2
â2 +â3 k̂1
+
â3
( k̂2 )â2 }
â1 +â2 +â3 k̂1
k̂3 < k̂1 < k̂2
â3
â2 +â3
−
â3
( k̂1 )â1
â1 +â3 k̂2
+
â3
( k̂1 )â1 }
â1 +â2 +â3 k̂2
k̂3 < k̂2 < k̂1
The MLEs of (a1 , a2 , a3 , k1 , k2 , k3 ) are âi = 1/[log(Gi /k̂i )], i = 1,2,3
where k̂i = Xi(1) = M in(Xi1 , ...., Xin ), i=1,2,3
Gi = (
Qn
j=1
Xij )1/n , i = 1,2,3.
The distribution of Xi(1) is Pareto with parameters (nai , ki ), i=1,2,3 and MLEs (âi , k̂i )
are independent.[See Johnson and Kotz(1970)].
Now the asymptotic distribution of R is AN (R, B40 Λ4 G5 ) where B40 = (∂R/∂a1 ,
∂R/∂a2 , ∂R/∂a3 , ∂R/∂k1 , ∂R/∂k2 , ∂R/∂k3 ) and Λ4 = n1 diag(a21 , a22 , a23 , n2 a1 k12 /[(na1 −
2)(na1 − 1)2 ], n2 a2 k22 /[(na2 − 2)(na2 − 1)2 ], n2 a3 k32 /[(na3 − 2)(na3 − 1)2 ]).
Ackowledgement
I thank the referee for his valuable and constructive suggestions which resulted in an
improved shorter version of the earlier manuscript.
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