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Renewal analysis of a replacement process
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Operations Research Letters 31 (2003) 1 – 6
Operations Research Letters www.elsevier.com/locate/dsw
Renewal analysis of a replacement process Wolfgang Stadje ∗ Fachbereich Mathematik=Informatik, Universitat Osnabruck, 49069 Osnabruck, Germany Received 30 April 2001; received in revised form 10 May 2002; accepted 7 June 2002
Abstract We consider a unit with a random lifetime which is replaced at renewal times by a new identical one regardless of whether it has failed before or not. For this random periodic replacement policy, we derive exact formulas for the cycle length, de.ned as the time between the replacements of two successive failed units, the stationary probability of the current unit to have failed, and the stationary and the transient distributions of the residual lifetime of the current unit. c 2002 Elsevier Science B.V. All rights reserved. Keywords: Renewal process; Random replacement times; Cycle length; Residual lifetime
1. Introduction A technical unit with a random lifetime is replaced by a new identical one at time instants which form a renewal process. The i.i.d. lifetimes of the used number units and the renewal process of replacement times are assumed to be independent. The exchanges take place irrespective of the condition of the current unit, so that upon inspection either a failed unit has been in operation before being scrapped or a still functioning one is replaced preventively. In practice, the times of installing new units are often scheduled to be deterministic and periodic, i.e. to be multiples of some prespeci.ed T ¿ 0. However, in many situations some randomness must be taken into account, for example if the unit is not easily accessible (e.g. for satellites in orbit in outer space or in
∗
Fax: +49-541-9692770. E-mail address: [email protected] (W. Stadje).
medical applications where ‘inspection’ might involve surgery) or if the inspection team cannot exactly meet given dates. The above model, though of a very basic mathematical type, has apparently not been treated in the voluminous literature on stochastic replacement and maintenance. As representative papers on periodic replacement policies containing numerous references we mention [1– 6]. It is clear that the replacement process is regenerative, where as a cycle, we can take the time between two successive replacements of failed units. Such a cycle consists of a time interval during which only functioning units are used and a second interval during which a failed unit has not yet been replaced. We will derive the distribution of the cycle length and the stationary probability that the current unit is functioning. Our main results concern the residual lifetime of the unit in the system at time t, given that it will not be replaced. We will determine the stationary distribution of the remaining lifetime of the unit in the system, i.e. the time until the next failure if no more replacements
c 2002 Elsevier Science B.V. All rights reserved. 0167-6377/03/$ - see front matter PII: S 0 1 6 7 - 6 3 7 7 ( 0 2 ) 0 0 1 7 6 - 1
2
W. Stadje / Operations Research Letters 31 (2003) 1 – 6
occur; the transient distribution will be given in terms of its Laplace transform (LT) L.
2. Cycle length and stationary distribution Let A1 ; A2 ; : : : be the i.i.d. inter-replacement times, so that replacements take place at S0 = 0 and at any time Sn = A1 + · · · + An ; n ¿ 1. The i.i.d. lifetimes of the deployed units are denoted by B0 ; B1 ; B2 ; : : : : Let F and G be the distribution functions (d.f.’s) of An and Bn , respectively. To avoid trivialities, we assume that ∞ P(A1 ¿ B0 ) = G(u) dF(u) ∈ (0; 1)
=
0
x
(F(x − s) − L(x − s)) dL∗n−1 (s);
where we have conditioned on (Sn−1 ; An ) = (s; t) for the second equality and used Fubini’s theorem for the third. It follows that ∞ P(Ai 6 Bi−1 P(T 6 x) = P(An ¿ B0 ; A1 6 x) + n=2
=
0
Note that L(∞)=limx→∞ L(x)=P(A1 ¡ 0 ) ∈ (0; 1). B∞ By ∗, we denote convolution; let M (x)= n=1 L∗n (x), where L∗n is the n fold convolution of H with itself. Theorem 1. The d.f. K(x) = P(T 6 x) is given by K = F − M + F ∗ M:
(2.1)
The expected cycle length is E(T ) = a=(1 − L(∞)):
(2.2)
Proof. Clearly; for n ¿ 2; P(Ai 6 Bi−1 for i = 1; : : : ; n − 1; An ¿ Bn−1 ; Sn 6 x) P(Ai 6 Bi−1 for i = 1; : : : ; n − 1; = s+t6x
Sn−1 ∈ ds)P(Bn−1 ¡ t)P(An ∈ dt) x x−s G(t) dF(t) dL∗n−1 (s) = 0
0
x
G(t) dF(t)
0
+
0
so that neither only failed units nor only functioning units are replaced. Moreover, let E(An ) = a ∈ (0; ∞), E(Bn ) = b ∈ (0; ∞). For simplicity, we suppose that F and G are continuous on [0; ∞). We start by determining the d.f. of the cycle length T , i.e. of the time between the replacement of two successive failed units. Let x L(x) = (1 − G(u)) dF(u); x ¿ 0:
for i = 1; : : : ; n−1; An ¿Bn−1 ; Sn 6x)
∞ n=2
x
(F(x−s)−L(x−s)) dL∗n−1 (s)
0
= F(x) +
0
x
(F(x − s) − 1) dM (s)
= F(x) − M (x) + (F ∗ M )(x): For the expected cycle length; we have the renewal equation E(T ) = E(A1 ) + P(A1 ¡ B0 )E(T ); since T = A1 if A1 ¿ B0 ; or T = A1 + T if A1 ¡ B0 ; D where T = T and T is independent of A1 and B0 . Eq. (2.2) follows immediately. We remark that M (x) is equal to the expected number of indices n ∈ N satisfying Sn 6 x and Ai 6 Bi−1 for i = 1; : : : ; n − 1, i.e., the number of replacements of functioning units up to time x before the .rst failure. Similarly, the expected ∞ number of units that failed in [0; x] is given by n=1 K ∗n (x). Let p(t) be the probability that the unit in the system at time t has failed, and let p0 be the corresponding steady-state probability, i.e., p0 = limt→∞ p(t). Theorem 2. p0 = a−1
∞ 0
0
t
(t − s) dG(s) dF(t):
(2.3)
Proof. Split the cycle [0; T ] in two parts: [0; T ] = [0; T0 ] ∪ [T0 ; T ]; where the units in [0; T0 ] are all functioning and the unit in place during [T0 ; T ] has failed. Then p0 = E(T − T0 )=E(T ) by the renewal theorem
W. Stadje / Operations Research Letters 31 (2003) 1 – 6
for alternating renewal processes. Since
−e t
t
− e t
and (1 − L(∞))E(T ) = a by (2.2); the result follows.
−F(t)
Now let X (t) be the residual lifetime of the unit present in the system at time t, i.e., the remaining time after t in which the system is functioning if no further replacement takes place. (If the unit present at time t has failed, we set X (t) = 0.) Let X (∞) be a random variable having the stationary distribution of X (t).
−
e− x d(U ∗ G)(x)
Theorem 3. P(X (∞) = 0) = p0 and
0
t
0
P(X (∞) ¿ t) ∞ −1 =a (1 − G(t + s))(1 − F(s)) ds:
e− x dG(x)
0
E(T − T0 ) = E(A1 − B0 | A1 ¿ B0 ) ∞t (t − s) dG(s) dF(t) = 0 0 1 − L(∞)
t
∞
t
∞
t−x
e− u dG(u)
t
0
(2.4)
P(X (∞) ¿ t) ∞ 1 − F(s) = (1 − G(t + s)) ds: a 0
t ( )
= E(e− X (t) 1{X (t)¿0} );
+(1 − F(t))(1 − G(t + x)): By (3.2); the LT satis.es t ( ) =
t
0
×
∞
Theorem 4. t − x ( ) = 1 + e dU (x) e t ’B ( ) t 0
+ (1 − F(t))
e− x dx G(t + x):
0
(3.2)
= E(e− X (t) 1{X (t)¿0} )
t ( )
t−u ( ) dF(u)
(3.3)
∞ Let ( ; ) = 0 e− t t ( ) dt be the ordinary LT of the function t → t ( ). By (3.3); ( ; ) = ( ; )’A ( ) +
¿0
in ∞terms∗n of G, the renewal function U (x) = n=1 F (x) corresponding to F, and the LTs ’A ( ) = E(e− A1 ); ’B ( ) = E(e− B0 ) of the interreplacement times and the lifetimes.
(3.1)
P(X (t − u) ¿ x) dF(u)
3. The LT of the transient residual lifetime Now, we will explicitly derive the LT
dU (x):
Proof. Conditioning on whether A1 = u ∈ [0; t] or A1 ¿ t we obtain P(X (t) ¿ x) =
Proof. Consider a stationary process for the replacements. Let A be the waiting time for the next replacement after time 0; which is independent of the lifetime B0 and has the density function a−1 (1 − F). D Then clearly X (∞)=max[B0 − A; 0]; so that
e (t−x) dG(x)
e (t−x) F(t − x)
0
×
3
×
∞
0
e−
t− x
∞
0
dx G(t + x) dt
= ( ; )’A ( ) ∞ u + e− u−( 0
−
0
0
∞
F(t)
t
(1 − F(t))
∞
− )t
dt dG(u)
e− u−(
− )t
dG(u) dt
4
W. Stadje / Operations Research Letters 31 (2003) 1 – 6
’B ( ) − ’B ( ) − ∞ − )t F(t) e− u dG(u) dt;
= ( ; )’A ( ) + −
∞
0
e
−(
where we have substituted u = t + x and used Fubini’s theorem for the second equality. It follows from (3.4) that ’B ( ) − ’B ( ) ( − )(1 − ’A ( )) ∞ 1 − e−( − )t F(t) 1 − ’A ( ) 0 ∞ − u × e dG(u) dt:
(3.5)
t
It turns out that the Laplace inverses of the two summands on the right-hand side of (3.5); considered as functions of for .xed ; can be given in closed form. The .rst term on the right-hand side of (3.5) is the LT of t (1) − x t → t ( ) = 1 + e dU (x) e t ’B ( ) −e t −e t
0
t
0
t
0
e− x dG(x)
=’B ( )
∞
0
+ ’B ( ) −
∞
0
−
0
∞
e( −
∞
0
e( − e( −
)t
where we have used integration by parts and the ∞ relation 0 e− x dU (x) = ’A ( )=(1 − ’A ( )). Furthermore; III =
B ( ) ; −
IV =
A ( )’B ( ) : (1 − ’A ( ))( − )
Thus; a short calculation yields I + II − III − IV =
(3.6)
t
t
0
)t
0
e− x dU (x)
0
t
=− dt
∞
0
×
dt
=−
e−(
∞
e− x d(U ∗ G)(x)
F(t − x)
∞
t−x
e
− u
dG(u)
dU (x);
(2) t ( ) dt
0
−
t − x e dG(x) dt
=I + II − III − IV:
t
e
(t−x)
(3.7)
0
’B ( ) − ’B ( ) ( − )(1 − ’A ( ))
as claimed. Similarly, the second term on the right-hand side of (3.5) is the LT of ∞ (2) e (t−x) dG(x) t → t ( ) = −F(t) −
e− x d(U ∗ G)(x):
dt
0
)t
’B ( )’A ( ) ; (1 − ’A ( ))( − )
because ∞ e− t
e( −
)t
II =
’B ( ) ; −
To see this; write ∞ e− t t(1) ( ) dt 0
I=
t
(3.4)
( ; ) =
Clearly;
∞
t−x
0
e
∞
− )t
− t
F(t)
t
0
− )t
t
e− x dG(x)
dt
(t−x) e F(t − x)
e− u dG(u)
e−(
∞
dU (x) dt
F(t)
t
∞
e− x dG(x)
dt
W. Stadje / Operations Research Letters 31 (2003) 1 – 6
∞ ’A ( ) e−( − )t F(t) 1 − ’A ( ) 0 ∞ × e− u dG(u) dt
4. Example: Poisson inter-replacement times
−
Let F(x) = 1 − e−x=a , x ¿ 0, so that ’A ( ) = ( a + 1)−1 . Then L(x) has the LT x ∞ − x −1 −u=a e dx (1 − G(u))a e du
t
∞ 1 e−( − )t F(t) 1 − ’A ( ) 0 ∞ − x × e dG(x) dt:
=−
0
=a−1
t
=
It follows that t ( )
=
(1) t ( )
(2) t ( ):
+
(3.8)
Inserting (3.6) and (3.7) in (3.8) yields (3.1). The LT ( ) = E(e− X (∞) 1{X (∞)¿0} ) of the stationary distribution can be computed from (2.4); it is given by ∞ 1 − ’B ( ) ( ) = e x F(x) − a−1 ’A ( ) 0 ∞ × e− u dG(u) d x: (3.9) x
(1) t ( ) =
as t → ∞:
(1) t ( )
∞
t
t
−1
)x
(1 − G(x)) d x
n=1
=
1 − ’B ( + a−1 ) : a + ’B ( + a−1 )
Now it follows from (2.1) that T has the LT ∞ E(e− T ) = ’A ( ) + (’A ( ) − 1) e− x dM (x)
∞
t
∞
=
−1
’B ( + a ) : a + ’B ( + a−1 )
In particular, E(T ) = −
e− (x−t) dG(x)
−’B ( ) +
0
e−( +a
Thus, the LT of M (x) is given by n ∞ ∞ 1 − ’B ( + a−1 ) e− x dM (x) = a + 1 0
(3.10)
can be rewritten as
∞
0
→ ( )
Indeed,
0
1 − ’B ( + a−1 ) : a + 1
Starting from (3.1) it can be proved directly that t ( )
5
d a : E(e− T )| =0 = d ’B (a−1 )
To compute p0 from Theorem 2, we also need ∞ L(∞) = P(A1 ¡ B0 ) = (1 − G(x))a−1 e−x=a d x
e− (x−t) dU (x)
e− (x−t) d(U ∗ G)(x)
0
(3.11)
and it is not diIcult to see that the three terms on the right-hand side of (3.11) converge to 0, −’B ( )( a)−1 and ( a)−1 , respectively, as t → ∞, (2)
by the renewal theorem. Regarding t ( ), the .rst integral on the right-hand side of (3.7) tends to 0, while the second ∞ is equal to −(r ∗ U )(t), where r(x) = e x F(x) x e− u dG(u), and thus converges to ∞ −a−1 0 r(x) d x. Combining these limiting relations yields (3.10).
= 1 − ’B (a−1 ) and ∞ 0
=
0
0
t
(t − s) dG(s)
∞ ∞ s
=a’B (a−1 ):
(t − s)a
dF(t) −1 −t=a
e
dt
dG(s)
6
W. Stadje / Operations Research Letters 31 (2003) 1 – 6
A short calculation now yields p0 = ’B (a
−1
It follows that the stationary distribution of X (t) has the density q(x) on (0; ∞).
):
Finally let us determine ( ) and Q. By Theorem 4, ∞ 1 ( ) = (1 − ’B ( )) − a−1 e x (1 − e−x=a ) a 0 ∞ × e− u dG(u) d x x
=
1 (1 − ’B ( )) a ∞ u e x d x e− u dG(u) −a−1 0
+ a−1 =
’B (a
−1
0
∞
0
0
u
e( −a
) − ’B ( ) : a − 1
−1
)x
d x e− u dG(u) (4.1)
The right-hand side of (4.1) is the ordinary LT of the function ∞ −1 q(x) = a e−(u−x)=a dG(u); x ¿ 0: x
References [1] M. Berg, B. Epstein, A modi.ed block replacement policy, Naval Res. Logist. 23 (1976) 15–24. [2] J. Franz, On estimation problems in random censored repair models, Econom. Quality Control 9 (1994) 125–142. [3] X. Liu, V. Makis, A.K.S. Jardine, A replacement model with overhauls and repairs, Naval Res. Logist. 42 (1995) 1063–1079. [4] T. Nakagawa, A summary of imperfect maintenance policies with minimal repair, RAIRO Rech. OpNer. 14 (1980) 249–255. [5] T. Nakagawa, A summary of periodic replacement with minimal repair at failure, J. Oper. Res. Soc. Japan 24 (1981a) 213–228. [6] T. Nakagawa, Modi.ed periodic replacement with minimal repair at failure, IEEE Trans. Reliability R-30 (1981b) 165–168.
Operations Research Letters 31 (2003) 1 – 6
Operations Research Letters www.elsevier.com/locate/dsw
Renewal analysis of a replacement process Wolfgang Stadje ∗ Fachbereich Mathematik=Informatik, Universitat Osnabruck, 49069 Osnabruck, Germany Received 30 April 2001; received in revised form 10 May 2002; accepted 7 June 2002
Abstract We consider a unit with a random lifetime which is replaced at renewal times by a new identical one regardless of whether it has failed before or not. For this random periodic replacement policy, we derive exact formulas for the cycle length, de.ned as the time between the replacements of two successive failed units, the stationary probability of the current unit to have failed, and the stationary and the transient distributions of the residual lifetime of the current unit. c 2002 Elsevier Science B.V. All rights reserved. Keywords: Renewal process; Random replacement times; Cycle length; Residual lifetime
1. Introduction A technical unit with a random lifetime is replaced by a new identical one at time instants which form a renewal process. The i.i.d. lifetimes of the used number units and the renewal process of replacement times are assumed to be independent. The exchanges take place irrespective of the condition of the current unit, so that upon inspection either a failed unit has been in operation before being scrapped or a still functioning one is replaced preventively. In practice, the times of installing new units are often scheduled to be deterministic and periodic, i.e. to be multiples of some prespeci.ed T ¿ 0. However, in many situations some randomness must be taken into account, for example if the unit is not easily accessible (e.g. for satellites in orbit in outer space or in
∗
Fax: +49-541-9692770. E-mail address: [email protected] (W. Stadje).
medical applications where ‘inspection’ might involve surgery) or if the inspection team cannot exactly meet given dates. The above model, though of a very basic mathematical type, has apparently not been treated in the voluminous literature on stochastic replacement and maintenance. As representative papers on periodic replacement policies containing numerous references we mention [1– 6]. It is clear that the replacement process is regenerative, where as a cycle, we can take the time between two successive replacements of failed units. Such a cycle consists of a time interval during which only functioning units are used and a second interval during which a failed unit has not yet been replaced. We will derive the distribution of the cycle length and the stationary probability that the current unit is functioning. Our main results concern the residual lifetime of the unit in the system at time t, given that it will not be replaced. We will determine the stationary distribution of the remaining lifetime of the unit in the system, i.e. the time until the next failure if no more replacements
c 2002 Elsevier Science B.V. All rights reserved. 0167-6377/03/$ - see front matter PII: S 0 1 6 7 - 6 3 7 7 ( 0 2 ) 0 0 1 7 6 - 1
2
W. Stadje / Operations Research Letters 31 (2003) 1 – 6
occur; the transient distribution will be given in terms of its Laplace transform (LT) L.
2. Cycle length and stationary distribution Let A1 ; A2 ; : : : be the i.i.d. inter-replacement times, so that replacements take place at S0 = 0 and at any time Sn = A1 + · · · + An ; n ¿ 1. The i.i.d. lifetimes of the deployed units are denoted by B0 ; B1 ; B2 ; : : : : Let F and G be the distribution functions (d.f.’s) of An and Bn , respectively. To avoid trivialities, we assume that ∞ P(A1 ¿ B0 ) = G(u) dF(u) ∈ (0; 1)
=
0
x
(F(x − s) − L(x − s)) dL∗n−1 (s);
where we have conditioned on (Sn−1 ; An ) = (s; t) for the second equality and used Fubini’s theorem for the third. It follows that ∞ P(Ai 6 Bi−1 P(T 6 x) = P(An ¿ B0 ; A1 6 x) + n=2
=
0
Note that L(∞)=limx→∞ L(x)=P(A1 ¡ 0 ) ∈ (0; 1). B∞ By ∗, we denote convolution; let M (x)= n=1 L∗n (x), where L∗n is the n fold convolution of H with itself. Theorem 1. The d.f. K(x) = P(T 6 x) is given by K = F − M + F ∗ M:
(2.1)
The expected cycle length is E(T ) = a=(1 − L(∞)):
(2.2)
Proof. Clearly; for n ¿ 2; P(Ai 6 Bi−1 for i = 1; : : : ; n − 1; An ¿ Bn−1 ; Sn 6 x) P(Ai 6 Bi−1 for i = 1; : : : ; n − 1; = s+t6x
Sn−1 ∈ ds)P(Bn−1 ¡ t)P(An ∈ dt) x x−s G(t) dF(t) dL∗n−1 (s) = 0
0
x
G(t) dF(t)
0
+
0
so that neither only failed units nor only functioning units are replaced. Moreover, let E(An ) = a ∈ (0; ∞), E(Bn ) = b ∈ (0; ∞). For simplicity, we suppose that F and G are continuous on [0; ∞). We start by determining the d.f. of the cycle length T , i.e. of the time between the replacement of two successive failed units. Let x L(x) = (1 − G(u)) dF(u); x ¿ 0:
for i = 1; : : : ; n−1; An ¿Bn−1 ; Sn 6x)
∞ n=2
x
(F(x−s)−L(x−s)) dL∗n−1 (s)
0
= F(x) +
0
x
(F(x − s) − 1) dM (s)
= F(x) − M (x) + (F ∗ M )(x): For the expected cycle length; we have the renewal equation E(T ) = E(A1 ) + P(A1 ¡ B0 )E(T ); since T = A1 if A1 ¿ B0 ; or T = A1 + T if A1 ¡ B0 ; D where T = T and T is independent of A1 and B0 . Eq. (2.2) follows immediately. We remark that M (x) is equal to the expected number of indices n ∈ N satisfying Sn 6 x and Ai 6 Bi−1 for i = 1; : : : ; n − 1, i.e., the number of replacements of functioning units up to time x before the .rst failure. Similarly, the expected ∞ number of units that failed in [0; x] is given by n=1 K ∗n (x). Let p(t) be the probability that the unit in the system at time t has failed, and let p0 be the corresponding steady-state probability, i.e., p0 = limt→∞ p(t). Theorem 2. p0 = a−1
∞ 0
0
t
(t − s) dG(s) dF(t):
(2.3)
Proof. Split the cycle [0; T ] in two parts: [0; T ] = [0; T0 ] ∪ [T0 ; T ]; where the units in [0; T0 ] are all functioning and the unit in place during [T0 ; T ] has failed. Then p0 = E(T − T0 )=E(T ) by the renewal theorem
W. Stadje / Operations Research Letters 31 (2003) 1 – 6
for alternating renewal processes. Since
−e t
t
− e t
and (1 − L(∞))E(T ) = a by (2.2); the result follows.
−F(t)
Now let X (t) be the residual lifetime of the unit present in the system at time t, i.e., the remaining time after t in which the system is functioning if no further replacement takes place. (If the unit present at time t has failed, we set X (t) = 0.) Let X (∞) be a random variable having the stationary distribution of X (t).
−
e− x d(U ∗ G)(x)
Theorem 3. P(X (∞) = 0) = p0 and
0
t
0
P(X (∞) ¿ t) ∞ −1 =a (1 − G(t + s))(1 − F(s)) ds:
e− x dG(x)
0
E(T − T0 ) = E(A1 − B0 | A1 ¿ B0 ) ∞t (t − s) dG(s) dF(t) = 0 0 1 − L(∞)
t
∞
t
∞
t−x
e− u dG(u)
t
0
(2.4)
P(X (∞) ¿ t) ∞ 1 − F(s) = (1 − G(t + s)) ds: a 0
t ( )
= E(e− X (t) 1{X (t)¿0} );
+(1 − F(t))(1 − G(t + x)): By (3.2); the LT satis.es t ( ) =
t
0
×
∞
Theorem 4. t − x ( ) = 1 + e dU (x) e t ’B ( ) t 0
+ (1 − F(t))
e− x dx G(t + x):
0
(3.2)
= E(e− X (t) 1{X (t)¿0} )
t ( )
t−u ( ) dF(u)
(3.3)
∞ Let ( ; ) = 0 e− t t ( ) dt be the ordinary LT of the function t → t ( ). By (3.3); ( ; ) = ( ; )’A ( ) +
¿0
in ∞terms∗n of G, the renewal function U (x) = n=1 F (x) corresponding to F, and the LTs ’A ( ) = E(e− A1 ); ’B ( ) = E(e− B0 ) of the interreplacement times and the lifetimes.
(3.1)
P(X (t − u) ¿ x) dF(u)
3. The LT of the transient residual lifetime Now, we will explicitly derive the LT
dU (x):
Proof. Conditioning on whether A1 = u ∈ [0; t] or A1 ¿ t we obtain P(X (t) ¿ x) =
Proof. Consider a stationary process for the replacements. Let A be the waiting time for the next replacement after time 0; which is independent of the lifetime B0 and has the density function a−1 (1 − F). D Then clearly X (∞)=max[B0 − A; 0]; so that
e (t−x) dG(x)
e (t−x) F(t − x)
0
×
3
×
∞
0
e−
t− x
∞
0
dx G(t + x) dt
= ( ; )’A ( ) ∞ u + e− u−( 0
−
0
0
∞
F(t)
t
(1 − F(t))
∞
− )t
dt dG(u)
e− u−(
− )t
dG(u) dt
4
W. Stadje / Operations Research Letters 31 (2003) 1 – 6
’B ( ) − ’B ( ) − ∞ − )t F(t) e− u dG(u) dt;
= ( ; )’A ( ) + −
∞
0
e
−(
where we have substituted u = t + x and used Fubini’s theorem for the second equality. It follows from (3.4) that ’B ( ) − ’B ( ) ( − )(1 − ’A ( )) ∞ 1 − e−( − )t F(t) 1 − ’A ( ) 0 ∞ − u × e dG(u) dt:
(3.5)
t
It turns out that the Laplace inverses of the two summands on the right-hand side of (3.5); considered as functions of for .xed ; can be given in closed form. The .rst term on the right-hand side of (3.5) is the LT of t (1) − x t → t ( ) = 1 + e dU (x) e t ’B ( ) −e t −e t
0
t
0
t
0
e− x dG(x)
=’B ( )
∞
0
+ ’B ( ) −
∞
0
−
0
∞
e( −
∞
0
e( − e( −
)t
where we have used integration by parts and the ∞ relation 0 e− x dU (x) = ’A ( )=(1 − ’A ( )). Furthermore; III =
B ( ) ; −
IV =
A ( )’B ( ) : (1 − ’A ( ))( − )
Thus; a short calculation yields I + II − III − IV =
(3.6)
t
t
0
)t
0
e− x dU (x)
0
t
=− dt
∞
0
×
dt
=−
e−(
∞
e− x d(U ∗ G)(x)
F(t − x)
∞
t−x
e
− u
dG(u)
dU (x);
(2) t ( ) dt
0
−
t − x e dG(x) dt
=I + II − III − IV:
t
e
(t−x)
(3.7)
0
’B ( ) − ’B ( ) ( − )(1 − ’A ( ))
as claimed. Similarly, the second term on the right-hand side of (3.5) is the LT of ∞ (2) e (t−x) dG(x) t → t ( ) = −F(t) −
e− x d(U ∗ G)(x):
dt
0
)t
’B ( )’A ( ) ; (1 − ’A ( ))( − )
because ∞ e− t
e( −
)t
II =
’B ( ) ; −
To see this; write ∞ e− t t(1) ( ) dt 0
I=
t
(3.4)
( ; ) =
Clearly;
∞
t−x
0
e
∞
− )t
− t
F(t)
t
0
− )t
t
e− x dG(x)
dt
(t−x) e F(t − x)
e− u dG(u)
e−(
∞
dU (x) dt
F(t)
t
∞
e− x dG(x)
dt
W. Stadje / Operations Research Letters 31 (2003) 1 – 6
∞ ’A ( ) e−( − )t F(t) 1 − ’A ( ) 0 ∞ × e− u dG(u) dt
4. Example: Poisson inter-replacement times
−
Let F(x) = 1 − e−x=a , x ¿ 0, so that ’A ( ) = ( a + 1)−1 . Then L(x) has the LT x ∞ − x −1 −u=a e dx (1 − G(u))a e du
t
∞ 1 e−( − )t F(t) 1 − ’A ( ) 0 ∞ − x × e dG(x) dt:
=−
0
=a−1
t
=
It follows that t ( )
=
(1) t ( )
(2) t ( ):
+
(3.8)
Inserting (3.6) and (3.7) in (3.8) yields (3.1). The LT ( ) = E(e− X (∞) 1{X (∞)¿0} ) of the stationary distribution can be computed from (2.4); it is given by ∞ 1 − ’B ( ) ( ) = e x F(x) − a−1 ’A ( ) 0 ∞ × e− u dG(u) d x: (3.9) x
(1) t ( ) =
as t → ∞:
(1) t ( )
∞
t
t
−1
)x
(1 − G(x)) d x
n=1
=
1 − ’B ( + a−1 ) : a + ’B ( + a−1 )
Now it follows from (2.1) that T has the LT ∞ E(e− T ) = ’A ( ) + (’A ( ) − 1) e− x dM (x)
∞
t
∞
=
−1
’B ( + a ) : a + ’B ( + a−1 )
In particular, E(T ) = −
e− (x−t) dG(x)
−’B ( ) +
0
e−( +a
Thus, the LT of M (x) is given by n ∞ ∞ 1 − ’B ( + a−1 ) e− x dM (x) = a + 1 0
(3.10)
can be rewritten as
∞
0
→ ( )
Indeed,
0
1 − ’B ( + a−1 ) : a + 1
Starting from (3.1) it can be proved directly that t ( )
5
d a : E(e− T )| =0 = d ’B (a−1 )
To compute p0 from Theorem 2, we also need ∞ L(∞) = P(A1 ¡ B0 ) = (1 − G(x))a−1 e−x=a d x
e− (x−t) dU (x)
e− (x−t) d(U ∗ G)(x)
0
(3.11)
and it is not diIcult to see that the three terms on the right-hand side of (3.11) converge to 0, −’B ( )( a)−1 and ( a)−1 , respectively, as t → ∞, (2)
by the renewal theorem. Regarding t ( ), the .rst integral on the right-hand side of (3.7) tends to 0, while the second ∞ is equal to −(r ∗ U )(t), where r(x) = e x F(x) x e− u dG(u), and thus converges to ∞ −a−1 0 r(x) d x. Combining these limiting relations yields (3.10).
= 1 − ’B (a−1 ) and ∞ 0
=
0
0
t
(t − s) dG(s)
∞ ∞ s
=a’B (a−1 ):
(t − s)a
dF(t) −1 −t=a
e
dt
dG(s)
6
W. Stadje / Operations Research Letters 31 (2003) 1 – 6
A short calculation now yields p0 = ’B (a
−1
It follows that the stationary distribution of X (t) has the density q(x) on (0; ∞).
):
Finally let us determine ( ) and Q. By Theorem 4, ∞ 1 ( ) = (1 − ’B ( )) − a−1 e x (1 − e−x=a ) a 0 ∞ × e− u dG(u) d x x
=
1 (1 − ’B ( )) a ∞ u e x d x e− u dG(u) −a−1 0
+ a−1 =
’B (a
−1
0
∞
0
0
u
e( −a
) − ’B ( ) : a − 1
−1
)x
d x e− u dG(u) (4.1)
The right-hand side of (4.1) is the ordinary LT of the function ∞ −1 q(x) = a e−(u−x)=a dG(u); x ¿ 0: x
References [1] M. Berg, B. Epstein, A modi.ed block replacement policy, Naval Res. Logist. 23 (1976) 15–24. [2] J. Franz, On estimation problems in random censored repair models, Econom. Quality Control 9 (1994) 125–142. [3] X. Liu, V. Makis, A.K.S. Jardine, A replacement model with overhauls and repairs, Naval Res. Logist. 42 (1995) 1063–1079. [4] T. Nakagawa, A summary of imperfect maintenance policies with minimal repair, RAIRO Rech. OpNer. 14 (1980) 249–255. [5] T. Nakagawa, A summary of periodic replacement with minimal repair at failure, J. Oper. Res. Soc. Japan 24 (1981a) 213–228. [6] T. Nakagawa, Modi.ed periodic replacement with minimal repair at failure, IEEE Trans. Reliability R-30 (1981b) 165–168.