An inverse First-passage Problem for One-dimensional Diffusions reflected between
two boundaries
Mario Abundo Dipartimento di Matematica
Universit`a Tor Vergata, Roma, Italy (E-mail: abundo@mat.uniroma2.it)
Abstract. We study an inverse first-passage problem for a one-dimensional, time- homogeneous diffusionX(t) reflected between two boundaries aand b,which starts from a random position η. Let a ≤ S ≤ b be a given threshold, such that P(η ∈ [a, S]) = 1,andF an assigned distribution function. The problem consists of finding the distribution ofη such that the first-passage time ofX troughS has distribution F.
Keywords: First-passage-time, Inverse first-passage problem, Reflected diffusion.
1 Formulation of the problem and main results
Diffusion processes with reflecting boundaries appear in many applications in Economics and Finance (see e.g. Ball and Roma [7], Bertolla and Caballero [8], De Jong [11], Krugman [14], Svensson [26]), in Queueing (see e.g. Abate and Whitt [1], [2], Harrison [12], Srikant and Whitt [25], Ward and Glynn [28], [29]), and Mathematical Biology (see e.g. Ricciardi and Sacerdote [23]). In all these, the knowledge of the distribution of the first-passage-time (FPT) of the reflected diffusion through an assigned barrier is very important. Although FPT problems have been studied mostly for ordinary diffusions, i.e. without reflecting (see e.g. Abundo [6], Darling and Siegert [10], Ricciardi and Sato [24], and references therein), more recently some results appeared about the FPT of a one-dimensional reflected diffusion, through a threshold S (see e.g.
Chuancun and Huiqing [9], Lijunet al. [17], Qin Huet al. [22]). In this paper, we focus on FPT problems for a one-dimensional, temporally homogeneous reflected diffusion processX(t) with boundariesaandb,which is the solution of the stochastic differential equation with reflecting boundaries (SDER):
(dX(t) =µ(X(t))dt+σ(X(t))dBt+dLt−dUt
X(0) =η∈[a, b] (1)
where Bt is standard Brownian motion, the initial position η is a random variable, independent ofBt, L={Lt} andU ={Ut}, t≥0,are theregulators
3rdSMTDA Conference Proceedings, 11-14 June 2014, Lisbon Portugal C. H. Skiadas (Ed)
c 2014 ISAST
of points a and b, respectively, namely the local times ofX at a and b. The processesLandU are uniquely determined by the following properties (see e.g.
Harrison [12]):
(i) bothLt andUtare continuous nondecreasing processes withL0=U0= 0;
(ii)X(t)∈[a, b] for every timet≥0;
(iii) L and U increase only when X = a and X = b, respectively, that is, Rt
01{X(s)=a}dLs=LtandRt
01{X(s)=b}dUs=Ut,for anyt≥0.
We suppose that the coefficientsµ(·) andσ(·) are sufficiently regular (see e.g.
Lions and Sznitman [21]), so that, for fixed initial value the SDER (1) has a unique strong solutionX(t),which remains in the interval [a, b] for every time t≥0.For this reason,X(t) is also called aregulated diffusion betweenaand b. If S ∈ [a, b] is a threshold such that P(a ≤ η ≤ S) = 1, we consider the FPT of X through S, namely τS = inf{t > 0 : X(t) = S}, and we denote by τS(x) = inf{t > 0 : X(t) = S|η = x} the FPT of X through S with the condition that η=x.We assume thatτS(x) is finite with probability one
∀x∈[a, S],and that it possesses a density f(t|x).
Theinverse FPT problem for diffusions generally focuses on determining the barrier S,whenf(t|x) is given (see e.g. Abundo [6], Zucca and Sacerdote [27]). Since we assume that the initial position η is random, we consider a slight modification of the problem, that is the following inverse first-passage- time (IFPT) problem.
For a given distributionF,our aim isto find the density g of η(if it exists) for which it results P(τS ≤t) =F(t).
This IFPT problem has interesting applications in Mathematical Finance, in particular in credit risk modeling, where the FPT represents a default event of an obligor (see e.g. Jackson et al. [13]), in Biology, specially in the framework of diffusion models for neural activity (see e.g. Lansky and Smith [16]), and in Queueing theory (see e.g. Abate and Whitt [1], [2], Harrison [12]). For ordinary diffusions, it was studied in Jacksonet al. [13] in the case of Brownian motion, while some extensions to more general processes were obtained in Abundo [4], [5].
LetLh(x) =µ(x)h0(x)+12σ2(x)h00(x), x∈(a, b),the infinitesimal generator of X(t), acting onC2−functions hon (a, b).We recall the following result by Chuancun and Huiqing [9]:
Theorem 1 Let X be the solution of the SDER (1) with deterministic and fixed initial condition X(0) =x, and let S ∈[a, b]. For x∈[a, S] and θ ≥0, suppose thatu(x)satisfies the following equation:
(Lu(x) =θu(x), x∈(a, S)
u0(a) = 0 . (2)
Then, ifu(S)6= 0 forS∈[x, b], the Laplace transform ofτS(x)is explicitly given by:
E
e−θτS(x)
= u(x)
u(S) . (3)
By taking then−thderivative ofE e−θτS(x)
with respect toθ,and cal- culating it atθ= 0, we obtain (see Abundo [3]):
Proposition 2 Forn= 1,2, . . . ,the n−th order moments ofτS(x), if they exist finite, are the solutions to the problems:
(LTn(x) =−nTn−1(x), x∈(a, S)
Tn(S) = 0, Tn0(a) = 0 , (4)
whereT0(x)≡1.
Now, we report the explicit solutions of problems (2) and (4) for the Laplace transform and the moments ofτS(x),in the case of reflected Brownian motion with drift µ. By solving (2) by quadratures and using (3), we get that the Laplace transform ofτS(µ)(x) is:
E
e−θτS(µ)(x)
=e−(S−x)(
√
µ2+2θ−µ)· θe−2(x−a)
√
µ2+2θ+µ2+θ+µp µ2+ 2θ θe−2(S−a)
√
µ2+2θ+µ2+θ+µp µ2+ 2θ
. (5) Fora→ −∞the right-hand member of (5) tends toe−(S−x)(
√
µ2+2θ−µ),which is the well-known expression of the Laplace transform of the first-hitting time of ordinary Brownian motion with driftµto S,when starting fromx < S.
Taking the limit asµgoes to zero in (5), we obtain:
E
e−θτS(0)(x)
= e−x
√
2θ+e−(2a−x)
√ 2θ
e−S
√
2θ+e−(2a−S)
√
2θ . (6)
In the special casea= 0,the expression above writes:
e−x
√ 2θ+ex
√ 2θ
e−S
√ 2θ+eS
√
2θ = cosh(x√ 2θ) cosh(S√
2θ), x∈[0, S]. (7) Then, Laplace transform inversion yields that, for a = 0 and x ∈ [0, S] the density ofτS(0)(x) is, fort≥0 (cf. e.g. Darling and Siegert [10], Qin Huet al.
[22]):
f(0)(t|x) = π S2
∞
X
k=0
(−1)k(k+1 2) cos
(k+1
2)πx S
exp
−(k+1 2)2π2t
2S2
(8)
By solving (4) by quadratures, withn= 1 andn= 2,we obtain:
T1(µ)(x) = 1 2µ2
h
e2µ(a−S)−e2µ(a−x)i
+S−x
µ , x∈[a, S]. (9)
T2(µ)(x) = x2 µ2 − x
µ3
e2µ(a−S)+ 1 + 2Sµ+e2µ(a−x)
+c1+c2e−2µx, (10) where the constants c1 andc2 can be easily calculated. Letting µgo to zero, we obtain:
T1(0)(x) =−x2+ 2ax+S(S−2a), x∈[a, S]. (11) and
T2(0)(x) = x4 3 −4
3ax3−2S(S−2a)x2+Ax+B, x∈[a, S],
for certain constants A and B. In particular, for a = 0, we get T1(0)(x) =
−x2+S2, T2(0)(x) =x34 −2S2x2+53S4
Explicit formulae for the Laplace transform of the first-hitting time to a barrier S are known also for reflected OU process, reflected Bessel process and some other processes (see Chuancun and Huiqing [9], Lijun et al. [17]), but they involve special functions. An explicit spectral representation of the hitting time density was found in Qin Huet al. [22] for reflected BM, and in Linetsky [19], [20] for Cox-Ingersoll-Ross (CIR) and OU processes.
1.1 The IFPT problem for reflected Brownian motion with drift For a given barrierS∈[a, b],andX(0) =η∈[a, S],let us suppose thatτS(x) is a.s. finite for every x∈ [a, S], and it possesses a density f(t|x). Moreover, we suppose that the initial position η has a densityg(x) with support (a, S);
for θ ≥0 we denote by fb(θ|x) = R+∞
0 e−θxf(t|x)dt the Laplace transform of f(t|x) and bybg(θ) =RS
a e−θxg(x)dxthe (possibly bilateral) Laplace transform ofg.Then, the density ofτS is obtained asf(t) =RS
a f(t|x)g(x)dxand taking the Laplace transform on both sides we getfb(θ) =RS
a fb(θ|x)g(x)dx .
Now, we go to solve the IFPT problem, in the case when X = X(µ) is reflected BM with drift µ,between the boundaries aand b.For a given FPT distribution functionF (or equivalently for a given FPT density f =F0) our aim is to find the density g of the random initial position η, if it exists, such that P(τS≤t) =F(t).The following result holds (see Abundo [3]):
Theorem 3 ForS∈[a, b],letX(µ)be BM with drift µ,reflected between the boundaries a and b and starting from the random position η ∈[a, S]; suppose that the FPT of X(µ) through S has an assigned probability density f and denote by fb(θ) =R∞
0 e−θtf(t)dt, θ≥0, the Laplace transform of f. Then, if there exists a solution g to the IFPT problem for X(µ), its Laplace transform bg(θ),forθ≥0, must satisfy the equation:
fb(θ) = [θe2a
√
µ2+2θ
g(bp
µ2+ 2θ+µ) + (µ2+θ+µp
µ2+ 2θ )bg(µ−p
µ2+ 2θ )]
× θe−S(
√
µ2+2θ−µ)e2a
√
µ2+2θ+ (µ2+θ+µp
µ2+ 2θ )eS(
√
µ2+2θ+µ) −1
(12)
In particular, ifµ= 0,the above formula writes:
fb(θ) = bg(√
2θ) +g(−b √ 2θ)e−2a
√ 2θ
e−S
√
2θ+e(S−2a)
√
2θ , θ≥0. (13)
Furthermore, if µ = 0 and we require that the density g is symmetric with respect to (a+S)/2,then:
bg(θ) = e−Sθ+e−(2a−S)θ 1 +e(S−a)θ fb
θ2 2
, θ≥0. (14)
Iffb(θ) is analytic in a neighbor ofθ= 0,then thek−th order moments ofτS
exist finite and they are obtained in terms offb(θ) byE(τSk) = (−1)k ∂∂θkkf(θ)|b θ=0. The same thing holds for the moments ofη,ifbg(θ) is analytic. Thus, we obtain:
E(τS(µ)) = 1
µE(S−η)− 1
2µ2 e2µaE e−2µη−e−2µS
. (15)
Remark 4 LetX(t) be regulated BM, and suppose that τS has Gamma dis- tribution. Then, it can be shown that a solutiong to the IFPT problem, with g symmetric with respect to (a+S)/2 does not exist (see Abundo [3]).
Now, we further investigate the question of the existence of solutions to the IFPT problem. Referring to regulated drifted BM, we will prove the existence of the density g of the initial position η ∈ [a, S] for a class of FPT densities f. For the sake of simplicity, we limit ourselves to the case when µ= 0, a = 0, S = 1 < b and g is required to be symmetric with respect to 1/2; in fact, forµ6= 0 the calculations involved are far more complicated.
For any integer k ≥ 0, set Ik(θ) = R1
−1e−θxxkdx; as easily seen, I0(θ) = 2 sinh(θ)/θand the recursive relationIk(θ) = (−1)keθθ−e−θ+kθIk−1(θ) allows to calculateIk(θ), for everyk.The following proposition holds (see Abundo [3]).
Proposition 5 Let X be regulated BM between the boundaries0 andb, and letS= 1< b;suppose that the Laplace transform off(t)has the form:
fb(θ) =fb2k(θ) := cosh(p θ/2) cosh(√
2θ)
1 + 1 2k
"r 2 θsinh
rθ 2
!
−I2k rθ
2
!#
,
(16) for some integer k > 0. Then, there exists the solution g =g2k of the IFPT problem for X,relative to the FPT density f,and it results:
g2k(x) =
1 + 1 2k
1−(2x−1)2k
, x∈(0,1). (17) As an application of the results for regulated BM, we consider now the piecewise-continuous process ξ(t), obtained by superimposing to BM a jump process, namely, for η ∈ [a, S] and t < T, we set ξ(t) = η+Bt, where T
is an exponentially distributed time with parameter λ > 0; we suppose that for t =T the process ξ(t) makes an upward jump and it crosses the barrier S, irrespective of its state before the occurrence of the jump. This kind of behavior is observed e.g. in the presence of a so called catastrophes. Next, let us consider the reflected diffusion X with boundaries a, b, associated to ξ. Then, for η ∈ [a, S] the FPT of X over S is τS = inf{t >0 : X(t)≥ S}.
Conditionally onη=x,we have:
P(τS(x)≤t) =P(τS(x)≤t|t < T)P(t < T) + 1·P(t≥T)
=P(τS(x)≤t)e−λt+ (1−e−λt).
Taking the derivative, we obtain the FPT density of X, conditional to the starting positionx:
f(t|x) =e−λtf(t|x) +λe−λt Z +∞
t
f(s|x)ds.
By straightforward calculations, we obtain its Laplace transform:
bf(θ|x) = Z ∞
0
e−θtf(t|x)dt= θ
λ+θf(λb +θ|x) + λ
λ+θ , θ≥0.
The following result holds (see Abundo [3]).
Proposition 6 For a = 0 < S < b, if there exists a function g, symmetric with respect toS/2,which is the solution to the IFPT problem ofX(t),relative toS and the FPT densityf , then its Laplace transform is given by:
bg(θ) = 2 cosh(Sθ) (θ2/2−λ)(1 +eSθ)
θ2 2
bf θ2
2 −λ
−λ
. (18)
Remark 7 Forλ= 0,namely when no jump occurs, (18) reduces to (14) with a= 0.
1.2 Reduction of reflected diffusions to reflected Brownian motion On the analogy of the definition holding for ordinary diffusions (see Abundo [5], [6]), we introduce the following:
DefinitionLetX(t)be a diffusion with reflecting boundariesaandb,which is driven by the SDER:
dX(t) =µ(X(t))dt+σ(X(t))dBt+dLt−dUt, X(0) =x∈[a, b].
We say that X(t) is conjugated to regulated BM if there exists an increasing differentiable functionV(x), withV(0) = 0,such that, for any t≥0 it results X(t) = V−1 Bt+V(x) +Lt−Ut
, where Lt =V0(a)Lt and Ut = V0(b)Ut
are regulators.
A class of reflected diffusions which are conjugated to regulated BM is given by processes which are solutions of SDERs such as:
dX(t) =1
2σ(X(t))σ0(X(t))dt+σ(X(t))dBt+Lt−Ut, X(0) =x (19) withσ(·)≥0.Indeed, if the integral V(x) :=Rx 1
σ(r)dris convergent, by Itˆo’s formula for reflected diffusions (see e.g. Harrison [12]), one gets V(X(t)) = Bt+V(x) +V0(a)Lt−V0(b)Ut.
Let us consider a diffusion X with reflecting boundaries a and b, which is conjugated to regulated BM via the function V. Then, the process Y(t) :=
V(X(t)) is regulated BM between the boundariesV(a) andV(b),starting from V(x),that isY(t) =V(x)+Bt+Lt−Ut,whereLt=V0(a)LtandUt=V0(b)Ut are the regulators ofY(t),which increase only whenY =V(a) andY =V(b), respectively. Thus, forx∈[a, S] :
τS(x) = inf{t≥0 :X(t) =S|X(0) =x}=τSY0(V(x)),
where S0 = V(S) and the superscript refers to the process Y.Therefore, the solutiong to the IFPT problem for X,relative to the FPT densityf and the barrier S, can be written in terms of the solution eg to the IFPT problem for regulated BMY(t) relative to the FPT densityf and the barrierV(S).If one seeks e.g. thategis symmetric with respect to (V(a) +V(S))/2,then (see (14)) the Laplace transform of egturns out to be:
be
g(θ) = e−V(S)θ+e−(2V(a)−V(S))θ 1 +e(V(S)−V(a))θ fb
θ2 2
, θ≥0. (20)
2 A few examples
Example 1
Let X(t) be regulated BM with boundaries a, b (a < S < b), starting from η∈[a, S] and consider the FPT density:
f(t) = 1 (S−a)2
∞
X
k=0
exp
"
− k+122 π2t 2(S−a)2
#
, (21)
or the corresponding FPT Laplace transform:
fb(θ) =tanh((S−a)√ 2θ) (S−a)√
2θ . (22)
Then, the solution g to the IFPT problem for X(t) is the uniform density in (a, S).In particular, fora= 0, S= 1,(21) and (22) become:
f(t) =
∞
X
k=0
exp
"
−1 2
k+1
2 2
π2t
#
andfb(θ) = tanh(√
√ 2θ)
2θ (23)
and the solutiong to the IFPT problem is the uniform density in (0,1).
Example 2
For a = 0 < S < b, letX(t) be regulated BM starting from η ∈ [a, S], and consider the FPT density whose Laplace transform is:
fb(θ) = π2 2
1 + cosh(S√ 2θ) cosh(S√
2θ)(2θS2+π2)
Then, the solution to the IFPT problem forXisg(x) = 2Sπ sin πxS
, x∈(0, S).
Example 3
Take a = 0, S = 1, and let X(t) be regulated BM starting from η ∈ [0,1];
consider the FPT density whose Laplace transform is:
fb(θ) = 1 +e
√2θ e
√2θ−2e
√
θ/2+ 1 θcosh(√
2θ)e
√2θ
Then, the solution to the IFPT problem forXis the triangular density in [0,1] : g(x) =
(4x, x∈[0,12] 4x(1−x), x∈(12,1] .
Example 4
Take a = 0, S = 1, and let X(t) be regulated BM starting from η ∈ [0,1];
consider the FPT density whose Laplace transform is:
fb(θ) = 3
1 +e
√
2θ e−
√ 2θ(√
2θ+ 2) +√ 2θ−2 θ√
2θ e
√ 2θ+e−
√ 2θ
Then, the solutiong to the IFPT problem forX is a Beta density in [0,1],i.e.
g(x) = 6x(1−x).Notice thatfbandgare obtained as special cases offb2k and g2k of Proposition 5, fork= 1.
Example 5
For 0 =a < S < bletXbe the jump-process considered at the end of subsection 1.1, and let:
bf(θ) = 1 λ+θ
"
θ·tanh(Sp
2(λ+θ)) Sp
2(λ+θ) +λ
# .
By Laplace inversion, one obtains:
f(t) =e−λt
"∞ X
k=0
1
S2+ 2λS2 (k+12)2π2
exp
−(k+ 1/2)2π2t 2S2
# ,
which can be written as
f(t) =e−λtφ(t) +λe−λt Z ∞
t
φ(s)ds, (24)
whereφ(t) is the FPT density considered in Example 1 witha= 0,i.e.:
φ(t) = 1 S2
∞
X
k=0
exp
−(k+ 1/2)2π2t 2S2
. (25)
Then, the solution g to the IFPT problem forX(t),relative toS ∈(0, b) and bf ,which is symmetric with respect toS/2, is the uniform density in (0, S).
Example 6(Reflected geometric Brownian motion) Let 0< a < S < b,andX(t) the solution of the SDER:
dX(t) =rX(t)dt+σX(t)dBt+dLt−dUt, X(0) =η∈[a, S],
where r andσ are positive constant. The equation without reflecting is well- known in the framework of Mathematical Finance, since it describes the time evolution of a stock price. As easily seen, lnX(t) = lnη+µt+σBt+ ¯Lt−U¯t, whereµ=r−σ2/2 and ¯Lt,U¯tare regulators; thus, lnX(t)/σis regulated BM with driftµ/σ,between the boundaries lnσa, lnσb. Then, the IFPT problem for X(t) relative toSand the FPT densityf,is reduced to the IFPT problem for regulated drifted BM, starting from lnση, relative to lnσS and the same FPT densityf.
Example 7
LetX be a reflected diffusion in [a, b],which is conjugated to regulated BM via the functionV; then, examples of solutions to the IFPT problem forX can be obtained from Examples 1-4 regarding regulated BM.
For instance, let us consider the FPT density
f(t) = 1
(V(S)−V(a))2
∞
X
k=0
exp
"
− k+122
π2t 2(V(S)−V(a))2
#
, (26)
Then, the solution to the IFPT problem for X relative to the barrierS (a <
S < b) is
g(x) = V0(x)
V(S)−V(a)·1(a,S)(x). (27)
As explicit examples of reflected diffusionsX which are conjugated to reg- ulated BM, we mention the following.
(i)The process driven by
(dX(t) = 13X(t)1/3dt+X(t)2/3dBt+dLt−dUt
X(0) =η∈[a, b] , (28)
which is conjugated to regulated BM via the functionV(x) = 3x1/3i.e. X(t) = η1/3+13Bt+Lt−Ut3
.Here, as well as in the next examples,Lt=V0(a)Lt andUt=V0(b)Ut.
(ii)Forc >0,the process driven by
(dX(t) =3c82(X(t))1/2dt+c(X(t))3/4 dBt+dLt−dUt
X(0) =η∈[a, b] (a≥0) , (29)
which is conjugated to regulated BM via the functionV(x) = 4cx1/4i.e. X(t) = η1/4+c4Bt+Lt−Ut4
.
(iii) (Feller process or CIR model) Forb > a≥0,the process driven by
(dX(t) =14dt+p
X(t)dBt+dLt−dUt
X(0) =η∈[a, b] , (30)
which is conjugated to regulated BM via the functionV(x) = 2√
xi.e. X(t) =
1
4(Bt+ 2√
η+Lt−Ut)2.Notice that the process is always≥0.
(iv)(Wright & Fisher-like process) For 0≤a < b≤1,the process driven by:
(dX(t) = 14−12X(t) dt+p
X(t)(1−X(t))dBt+dLt−dUt
X(0) =η∈[a, b] ,
which is conjugated to regulated BM via the functionV(x) = 2 arcsin√ x.This equation is used for instance in the Wright-Fisher model for population genetics and in certain diffusion models for neural activity (see e.g. Lanskaet al. [15]);
it resultsX(t) = sin2(Bt/2 + arcsin√
η+Lt−Ut) and soX(t)∈[0,1] for all t≥0. Notice that, if we takea= 0 andb= 1,both boundaries are attainable and there is no need for reflection in a and b, because the process without reflecting cannot never exit the interval [0,1] (see e.g. Abundo [5]).
If the FPT density is given by (26), from (27) we obtain that the solutions to the IFPT problems for the processes (i)–(iv) above, relative to the barrier S,are explicitly given by:
g(x) =
3x2/3 S1/3−a1/3−1
·1(a,S)(x)(i), g(x) = 14
x3/4 S1/4−a1/4−1
·1(a,S)(x)(ii), g(x) = 12h√
x√ S−√
ai−1
·1(a,S)(x)(iii), and g(x) = 12h
arcsin√
S−arcsin√ ap
x(1−x)i−1
·1(a,S)(x)(iv).
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Stability analysis of multiserver discrete-time queueing systems with interruptions and
regenerative input flow
Larisa G. Afanasyeva1 and Andrey Tkachenko2
1 Lomonosov Moscow State University, Department of Mathematics and Mechanics, 119991, Leninskie Gory, Moscow, Russia
(E-mail: l.g.afanaseva@yandex.ru)
2 National Research University Higher School of Economics, Department of Applied Economics, 101000, Myasnitskaya Ulitsa, Moscow, Russia
(E-mail: tkachenko av@hse.ru)
Abstract. This paper is focused on a discrete-time multichannel queueing system with heterogeneous servers, regenerative input flow, and interruptions. The break- downs of the servers may occur at any time even if they are not occupied by customers.
Consecutive moments of breakdowns are defined by a renewal process. We consider the preemptive repeat different service discipline. Exploiting coupling method the necessary and sufficient stability condition for this system is established..
Keywords: Multichannel system, Regenerative flow, Stability, Interruption, Vaca- tion, Unreliable servers.
1 Introduction
Queueing systems in which servers may be not available for operation arise naturally as models of many computer, communication and manufacturing systems. Service interruptions may result from resource sharing, server break- downs, priority assignment, vacations, some external events, and other. For instance, if we concern the system with customers priority, the working period with secondary customers is equivalent to the service interruption of the pri- mary customers during this period. Therefore, there is significant interest in the investigation of queueing systems with server interruptions.
This study is focused on a queueing system with a regenerative input flow and heterogeneous servers that suffer independent interruptions. Service times are generally distributed. The breakdowns of the servers may occur at any time even if they are not occupied by customers. Consecutive moments of breakdowns are defined by a renewal process. We consider the preemptive repeat different service discipline where service is repeated from the beginning with different independent service time after restoration of the server [5].
Systems with unreliable servers have been intensively investigated for a long time. The main point was focused on the single-server case. There are some
3rdSMTDA Conference Proceedings, 11-14 June 2014, Lisbon Portugal
review papers, that cover most of the literature in these sphere. Concerning sys- tems with servers vacations it should be mentioned [3] and [7]. The framework of problems dealing with servers breakdowns and their solutions are presented in [8]. The are some other articles with extensive literature survey as well [11], [9].
One of the powerful approach to obtain stability conditions for systems with interruptions is synchronization method combined with regenerative theory.
Basing on this method in [9] multichannel queueing systems with identically distributed service times by different servers, renewal input flow, alternating renewal-type servers’ interruption in discrete time was considered. Authors established some sufficient conditions of stability for the preemptive repeat dif- ferent and preemptive resume service disciplines. In the paper [1] the same method was implemented for asymptotic analysis of single-server system with regenerative input. The similar approach was applied to study stability con- dition of the multichannel system with heterogeneous servers and regenerative input flow in a random environment [13], where random environment inter- rupted all the servers simultaneously.
In this paper we consider the multichannel model with interruption in dis- crete time cases. The necessary and sufficient condition of stability is estab- lished. The key element of our analysis is synchronization of processes under consideration. This method is based on the regeneration property of the input flow and renewal structure of processes describing the servers’ breakdowns [1].
The article is organised as follows. In the next section the model is described in detail. In the third section auxiliary service flows are introduced and traffic rate is established. In the fourth section we construct synchronization of input and service flows. Two final sections are devoted to the (in)stability problem.
2 Model description
We consider a system, which hasmheterogeneous servers and a common queue.
Service times of customers by the ith server constitute a sequence {ηi,n}∞n=1 of independent identically distributed (i.i.d.) random variables that do not depend on input flow and service times by the other servers. Let Bi(t) be a distribution function of ηi,n andbi =Eηi,n <∞ (i= 1, m). We assume that servers may be unavailable for service from time to time. The breakdowns of the servers may occur at any time even if they are not occupied by customers. Let {s(2)i,n}∞n=0be moments of breakdowns and{s(1)i,n}∞n=1be moments of restoration for the ith server. Here 0 = s(2)i,0 < s(1)i,1 < s(2)i,1 < s(1)i,2. . .. Then u(1)i,n = s(1)i,n−s(2)i,n−1andu(2)i,n=s(2)i,n−s(1)i,n denote the length of thenth blocked andnth available period of the ith server respectively (i = 1, m). The{u(1)i,n, u(2)i,n}∞n=1 constitute sequence of i.i.d. random vectors (for all i = 1, m) that do not depend on input flow and service times. However, for each nand i, u(1)i,n and
(2) (1) (2)
ai = a(1)i +a(2)i (i = 1, m). Server is free if it is neither serving a customer nor interrupted. If server becomes free and there are customers in the queue, a new customer enters the server. It is possible that more than one server becomes free simultaneously. Then customer in the queue choose an idle server according to some algorithm, possibly random. For definiteness we assume that a customer choose a free server with least number. It is possible that an unavailable period may start while a customer receiving service. Then service of the customer immediately interrupted. There are various disciplines for continuation of the service after server restoration may be investigated [5]. We consider preemptive repeat different service disciplines. This means that service is repeated from the start and the service time after restoration is independent of the original service time. Besides, customers remain with the same server until service completion. In order to ensure the service process for theith server we have to assume thatP(ηi,1≤u(2)i,1)>0 for alli= 1, m. If this condition is not fulfilled for some serveri, then theith server has to be excluded since it is always busy by service of the single customer.
We assume that input flowX(t) is a regenerative one.
Definition 1. A stochastic flow X(t) is called regenerative if there is an in- creasing sequence of random variables{θi, i≥0}, θ0= 0 such that the sequence {κi}∞i=1 ={X(θi−1+t)−X(θi−1), θi−θi−1, t∈ [0, θi−θi−1)}∞i=1 consists of independent identically distributed random elements.
The random variableθiis said to be theith regeneration point ofX(t) andτi= θi−θi−1is theith regeneration period (i= 1,2, . . .). Letξi=X(θi)−X(θi−1) be the number of arrived customers during theith regeneration period. Assume that τ = Eτ1 < ∞, a = Eξ1 < ∞. The limit λX = limt→∞X(t)t w.p.1 is called the intensity of X(t). It is easy to prove that λX = aτ (e.g., see [1]).
The class of regenerative flows contains most of fundamental flows that are exploited in queueing theory. Firstly, the doubly stochastic Poisson process [6] with stochastic regenerative intensity is regenerative one. There are many other examples of the regenerative flows, for instance, semi-markovian, Markov- modulated, Markov-arrival, and other processes [2]. Important properties of regeneration flows are given in [1].
We consider the operation of the described system in discrete-time case, i.e.
time is divided into fixed length intervals or slots and all arrivals, departures, interruptions (restoration) are synchronized with respect to slot boundaries.
Moreover, in case of synchronization of some events at one slot these events are ordered as follows: arrival, departure, and interruption (reconstruction).
System is observed at the end of a slot, when all events of the slot are realized.
3 Auxiliary processes
In this section we define auxiliary processes Yi(t) (i = 1, m) that will be ex-
the processesYi(t) we introduce the familyn
{η(j)i,n}∞n=1o∞ j=1
of independent se- quences {ηi,n(j)}∞n=1 consisting of i.i.d. random variables with d.f. Bi(x). Let Ki,j(t) be the counting process associated with the sequence {η(j)i,n}∞n=1, i.e.
Ki,j(t) = max{k :Pk
n=1ηi,n(j) ≤ t} (Ki,j(0) = 0) and µi(t) be the number of cycles for the ith server during [0, t], i.e. µi(t) = max{j : Pj
n=1ui,n ≤ t}
(µi(0) = 0). Then the processYi(t) is defined by the relation Yi(t) =st
µi(t)
X
j=1
Ki,j u(2)i,j
+Ki,µi(t)+1 maxh
0, t−s(1)i,µ
i(t)+1
i. (1)
By Hi(t) denote the renewal function forKi,j(t), i.e. Hi(t) =EKi,j(t).
Lemma 1. There exists the limit
t→∞lim Yi(t)
t =EHi(u(2)i,n) ai
=λYiw.p.1.
Proof. Puttinggi(n) =Pn
j=1Ki,j(u(2)i,j) we get from (1) the evident inequality gi(µi(t))≤Yi(t)≤gi(µi(t) + 1). (2) SinceKi,j(u(2)i,j) is a sequence of i.i.d. random variables with a finite mean by SLLN we have n−sgi(n) −−−−→w.p.1
n→∞ EHi(u(2)i,j). Besides, it follows from renewal theory that t−1µi(t)−−−→w.p.1
t→∞ a−1i . In view of independence {{η(j)i,n}∞n=1}∞j=1 and {u(1)i,n, u(2)i,n}∞n=1 one can easy obtain the convergence
gi(µi(t)) µi(t)
w.p.1
−−−→t→∞ a−1i EHi(u(2)i,1).
Thus the proof is follows from (2).
LetY(t) = Pm
i=1Yi(t) be the number of customers served by the system during [0, t] under assumption that the queue is not empty within this interval.
From Lemma 1 we have λY = lim
t→∞
Y(t)
t =
m
X
i=1
EHi(u(2)i,1) ai
w.p.1. (3)
We think ofλXandλY as the arrival and service rate respectively. Intuitively, it is clear that traffic rateρof the system has to be determined as
ρ= λX
λY
= λX
Pm
i=1a−1i EHi(u(2)i,1)
. (4)
4 Synchronization of renewal points for input and service flows.
First we prove a lemma for general regenerative aperiodic flows in discrete-time case. LetZ1(t) andZ2(t) be independent, regenerative, flows with regeneration points {θ1,j}∞j=1 and {θ2,j}∞j=1 respectively (θi,0 = 0;i = 1,2). As usually aperiodicity means that
GCD{k:P(θi,1=k)>0}= 1, i= 1,2. (5) Define common points of regeneration for Z1(t) andZ2(t) by
Tk= min{θ1,j > Tk−1:
∞
[
l=1
{θ2,l =θ1,j}}, T0= 0.
Lemma 2. Let condition (5) is fulfilled and E(θi,1) <∞ (i = 1,2) then the sequence {Tk}∞k=1 consist of regeneration points forZ1(t)andZ2(t)and
ET1<∞. (6)
Proof. Since the first statement follows from the construction of Tk we prove (6). Define
νk = min{j > νk−1:
∞
[
l=1
{θ1,j =θ2,l}}, ν0= 0,
so thatTk =θ1,νk. Then{νk−νk−1}∞k=1is a sequence of i.i.d. random variables and in accordance with Wald identity [4]ET1=Eθ1,1·Eν1. Therefore, we need to prove the finiteness of Eν1. Let h2(t) (h(t)) be the probability that there is a renewal at time t for the renewal processes {θ2,n}∞n=1 ({νk}∞k=1), so that h2(t) =P∞
l=1P(θ2,l=t) andh(t) =P∞
k=1P(νk =t). Taking into account (5) from the Blackwell theorem [12] we get
h2(t)−−−→
t→∞
1 Eθ2,1
, h(t)−−−→
t→∞
1 Eν1
. (7)
In view of independenceZ1(t) andZ2(t) we have h(j) =P{
∞
[
l=1
{θ1,j =θ2,l}}=E
∞
X
l=1
P{θ1,j =θ2,l|θ1,j}
!
=Eh2(θ1,j). (8)
Sinceθ1,j −−−→w.p.1
j→∞ ∞, in view of independenceZ1(t) andZ2(t) we geth2(θ1,j)−−−→w.p.1
j→∞
1
Eθ2,l. Thus from (7), (8) and Lebesgue’s dominated convergence theorem we
To exploit Lemma 2 for synchronization of these processes we introduce the following counting processes
N0(t) =max{k:θk≤t},
Ni(t) =max{k:s(2)i,k ≤t}, i= 1, m.
To construct synchronization of the mentioned counting processes we need additional assumptions.
Conjecture 1. The counting processesN0(t) andNi(t) (i= 1, m) are aperiodic.
Let us define subsequence {Tk}∞k=0 of the sequence {θj}∞j=1 by the recurrent relation
Tk=min{θj> Tk−1:
m
\
j=1
{Ni(θj)−Ni(θj−1) = 1}}, (T0= 0). (9)
In the other words {Tk}k≥0 are the common regeneration points of the input flow X(t) and Ni(t) (i = 1, m). Since we consider the preemptive repeat different service discipline then {Tk}k≥0 is a sequence of regeneration points forYi(t) andY(t) as well. Moreover, from Lemma 2 we obtain
ET1= 1 Eτ1
m
Y
i=1
a−1i <∞.
Thus we constructed the sequence of common regeneration points for the pro- cessesX(t) andY(t). Denote∆Xk =X(Tk)−X(Tk−1),∆Yk =Y(Tk)−Y(Tk−1).
Lemma 3. Let Conjecture 1 is fulfilled. Then the traffic rate of the system defined by (4) is equal to
ρ=E∆Xk E∆Yk .
Proof. Since {Tk}k≥0 is a sequence of regeneration points for X(t) andY(t) and ET1< inf ty it follows that sequences{∆Xk }k≥0 and {∆Yk}k≥0 consist of i.i.d. random variables with finite means. Letµ(t) = max{k:Tk ≤t}. From the renewal theory and SLLN we have
λX= lim
t→∞
X(t) t = lim
t→∞
µ(t)
t 1 µ(t)
µ(t)
X
k=1
(X(Tk)−X(Tk−1)) +X(t)−XTµ(t)
t
=
= E∆Xk
, w.p.1.
5 Stability theorem
Let Q(t) be the number of customers in the system (including customers in servers and queue) at instant t.
Definition 2. The process{Q(t), t≥0}is called stochastically bounded if for anyε >0 there existsy <∞such that for anyt >0
P{Q(t)< y}>1−ε.
This definition is close to the notion oftightness [10] .
Theorem 1. Let Conjecture 1 be fulfilled and traffic coefficientρis defined by (4). Then
• Ifρ >1, then Q(t)−−−→w.p.1
t→∞ ∞.
• Ifρ= 1, then Q(t)is stochastically unbounded.
• Ifρ <1, then Q(t)is stochastically bounded.
Proof. We start from the caseρ >1. It is evident that the following stochastic inequality holds
Q(t)≥Q(0)−Y(t) +X(t), t≥0. (10) From (4), we getλX > λY. Thus in view of definitions ofλX,λY and (10) we obtain the first statement of the theorem.
Letρ= 1. Consider the embedded process qn =Q(Tn) and denote Zk = Pk
j=1(∆Xj −∆Yj) (Z0= 0). From the stochastic inequality (10) we get qk≥stQ(0) +Zk, k >0. (11) Whenρ= 1, it follows from Lemma 3 thatE∆Xj =E∆Yj . Therefore,{Zk}k≥0 is a random walk with zero drift. Except when ∆Xj = ∆Yj =c w.p.1 (c is a constant) there existsα >0 such that for anyM ≥0
lim inf
k→∞ P(Zk> M)≥α
(see, e.g. [4]). It means thatqk andQ(t) are not stochastically bounded.
Now let ρ <1. Consider the ith server. We assume that service times of customers processing during the kth available period [s(1)i,k, s(2)i,k] (k = 1,2, . . .) are consequently selected from the sequence of i.i.d.r.v.’s {η(k)i,n}n≥1. Let us recall that processYi(t) is designed by the same sequence on thekth cycle with the help of the formula (1). Denote by Yei(t) the number of served customers by theith server during the interval [0, t]. Introduce the event
An={Q(t)≥mfor allt∈[Tn−1, Tn]}. (12)
where ∆eYn = Ye(Tn)−Ye(Tn−1). Now we introduce the embedded process xn = (qn, e1(n), . . . , em(n)), n ≥ 1, whereei(n) = 1 if service of a customer was interrupted by breakage of theith server andei(n) = 0 otherwise. Hence ei(n) = 1 for all i = 1, m if qn ≥ m. In view of interruption discipline and properties of the moments of synchronization {Tn}n≥1 the process {xn}n≥1 is a Markov chain with countable set of states K = {{0}; (j, e1, . . . , em), j = 1, m−1, ei ∈ {0; 1};j, j≥m}. LetK0 be the set of unessential states and Kj
(j = 1, r) an irreducible classes of communicating states. From the condition E∆Xj <E∆Yj it follows that for anyj0∈Kone can findt0such thatP(Q(t0)<
m|Q(0) = j0) >0. Therefore there exist k0 andn0 such that for any j0 ∈K P(qn0< m+k0|q0=j0)>0. It provides the finiteness of number of classesr, so we have
K=K0 r
[
j=1
Kj.
Consider the first classK1. Assume that it is aperiodic, then there exists
n→∞lim P(xn=x|x0=y) =π(1)x , (14) wherex∈K1,y∈K1. If we prove that
X
x∈K1
π(1)x = 1 (15)
thenqn is stochastically bounded. Let us show that (15) is fulfilled employing Foster’s criterion. We define the function f(q, e1, . . . , em) =q. It is sufficient to show that for someε1>0 there existsMε1 > msuch that
E(f(xn)−f(xn−1)|xn−1=x)<−ε1 (16) for allx= (q, e1, . . . , em) withq > Mε1. Takin into account (13) we get w.p.1
qn=qn−1+∆Xn −∆eYn =qn−1+∆Xn −∆eYnI(An)−∆eYnI(An)≤
≤qn−1+∆Xn −∆eYnI(An) =qn−1+∆Xn −∆Yn +∆YnI(An).
(17) Fromρ <1 we have that there existsδ >0 such that
E∆Xk −E∆Yk =−δ. (18)
Since ∆Xn and∆Yn are integrable, it follows that we can chooseMδ such that ifqn−1> Mδ, thenE∆YnI(An)< E∆Yn−E∆2 Xn = δ2. Therefore, we obtain from (17) and (18)
E(f(xn)−f(xn−1)|xn−1=x)< E∆Xn −E∆Yn +δ 2 =−δ
2
chain {ex(l)n }n≥1 is irreducible and aperiodic. Arguing as above we prove that qe(l)n =qnh+l is stochastically bounded asn→ ∞.
We can similarly prove stochastically boundedness for qn for initial state x0= (q0, e1(0), . . . , em(0) from any other classKi(i= 2, r). Since the number of classes r is finite we conclude thatqn is stochastically bounded as n→ ∞ for any initial state of Markov chainx0∈K. Therefore, the processQ(t) is also stochastically bounded.
Remark 1. Let us note that the first statement of Theorem 1 is not based on regenerative property of X(t). It is sufficient to assume that there exists a positive finite limit limt→∞t−1X(t) =λX w.p.1.
Remark 2. So far we consider zero-delayed regenerative flow X(t) and Yi(t) assuming that
P(θ0= 0) =P(s(2)i,0 = 0) = 1, (i= 1, m).
Let this condition does not hold and we have delayed regenerative floes. Note that results of Lemmas 1 — 3 on which the proof of Theorem 1 is based are fulfilled for delayed regenerative flows. We need only to claim
P(θ0<∞) =P(s(2)i,0 <∞) = 1, (i= 1, m).
For the following theorem we need definition of the ergodicity (stability) Definition 3. Process {Q(t), t≥ 0} is called ergodic if for any initial state Q(0) there exists
t→∞lim P{Q(t)≤x}=F(x),
whereF(x) is a distribution function and it does not depend on Q(0).
Theorem 2. Let Conjecture 1 be fulfilled and Markov chain {xn}n≥1 is irre- ducible and aperiodic. Then
• Ifρ >1, then Q(t)−−−→w.p.1
t→∞ ∞.
• Ifρ= 1, then Q(t)−−−→P
t→∞ ∞.
• Ifρ <1, then Q(t)is ergodic.
Proof. The first statement of the theorem is proved in Theorem 1. The set of states K of Markov chain {xn}n≥1 may have some unessential states but all the essential states organize the unique class K1 of communicating states.
It follows from Theorem 1 that there exists the limit (14), where πx(1) > 0 for x ∈ K1 and (15) is fulfilled, i.e. the Markov chain {xn}n≥1 is ergodic.
Let us take a state j0 ∈ K1, j0 ≥ m and assume that x0 = j0. Denote νj0 = min{n >0 :xn =j0}, so thatνj0 is the time of the return to the state j0. Since Markov chain {xn}n≥1 is ergodic, it follows that Eνj0 < ∞. Now consider Q(t). Note that Q(t) is a regenerative process and Tn is a point of regeneration of Q(t) ifqn =Q(Tn) =j0. Letτj be the time of return to the
