# Asymptotic Analysis of a Loss Model with Trunk Reservation I: Trunks Reserved for Fast Traffic

Authors: John A. Morrison; Charles  Knessl
Journal: Journal of Applied Mathematics and Stochastic Analysis
 (2008)

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## Abstract

We consider a model for a single link in a circuit-switched
network. The link hasC circuits, and the input consists
of offered calls of two types, that we call primary and secondary traffic. Of the C links, R are reserved for primary traffic. We
assume that both traffic types arrive as Poisson arrival streams.
Assuming that C is large and R=O(1), the arrival rate of primary traffic is O(C), while that of secondary traffic is smaller, of the order O(C). The holding times of the primary calls are
assumed to be exponentially distributed with unit mean. Those
of the secondary calls are exponentially distributed with a large
mean, that is, O(C). Thus, the primary calls have fast arrivals
and fast service, compared to the secondary calls. The loads for
both traffic types are comparable
(O(C)), and we assume that the system is “critically loaded”; that is, the system's capacity is approximately
equal to the total load. We analyze asymptotically
the steady state probability that n1 (resp.,
n2) circuits are occupied
by primary (resp., secondary) calls. In particular, we obtain
two-term asymptotic approximations to the blocking probabilities
for both traffic types.


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## Body

## 1. Introduction

A classic model in teletraffic is the Erlang loss
model. Here, we haveC servers (or
circuits), and customers (telephone calls) arrive as a Poisson process with rate parameter λ. The arriving customer takes one of the circuits if
one is available, and if they are all occupied then the call is blocked and
lost. When occupying a circuit, the customer has an exponentially distributed
holding time whose mean we take as the unit of time. It is well known that the steady
state probability that n circuits are
occupied is the truncated Poisson distribution, that is, Kλne−λ/n!, 0≤n≤C, with 1/K=∑n=0Cλne−λ/n!. This model dates back to circa 1918 [1]. When n=C, we obtain the steady state blocking probability. The
transient probability distribution is much more complicated, but it can be
computed in terms of special functions (see [2]).Over the years, many generalizations of the basic
model have been analyzed, including networks of such loss models (see [3, 4]).
One important extension is that of trunk reservation, which is fundamental in
the analysis of circuit-switched communication networks. Here, we consider a
model with C circuits that
are used by the two types of customers (or offered calls). We refer to these as
primary (or high-priority) calls and secondary (or low-priority) calls. They
arrive as Poisson arrival streams with respective rates λ and ν. Of the C circuits, R are reserved
for primary calls. Thus, if a high-priority call arrives, it is blocked if all C circuits are
busy, while a low-priority call is blocked if at least C−R circuits are
busy. All calls are assumed to have independent and exponentially distributed
holding times, with respective means 1 and 1/κ. The total load on the system is λ+ν/κ. If this exceeds C, then typically all the circuits are busy (an
overloaded link), while if λ+ν/κ<C, typically some circuits are free (an underloaded
link). An interesting situation is when C is large and λ+ν/κ≈C; we refer to this as “critical loading.”Previous work on this and related models includes
Mitra and Gibbens [5] who considered the asymptotic regime λ,C→∞ with C=λ−O(λ), R=O(λ), and ν=O(λ) (thus,
secondary calls are less frequent than primary calls). We thus have C/λ=1+O(1/λ); so this is an
example of critical loading. They analyzed a single link and used their results
to obtain approximations for more complicated loss networks with a distributed,
state-dependent, dynamic routing strategy. Related work appears in [6, 7], and
optimization and control policies for such problems were analyzed by Hunt and
Laws [8].Of fundamental importance in this model is the
probabilityB1 (resp., B2) that a
primary (resp., secondary) call is blocked and lost in the steady state.
Roberts [9, 10] obtained approximations to these blocking probabilities, which
are based on a certain recursion which is exact for special cases of the model
parameters, but not for all cases. Morrison [11] investigated this model for R=O(λ) and R=O(1), and obtained the blocking probabilities as
asymptotic series in powers of 1/λ. This led to a better understanding of the asymptotic
validity of Roberts' approximation(s). However, the coefficients in the
asymptotic series in [11] were not explicit, as their calculation still
involves recursively solving an infinite system of differential equations. But,
if it is further assumed that γ=ν/λ is small, the
blocking probabilities were obtained more explicitly in terms of parabolic
cylinder functions. Also, if R=O(1) rather than R=O(λ), explicit results are obtained without the small γ assumption.In [12], we analyzed the case R=O(1), with κ=O(1), and with the arrival
rates λ and ν  both O(C). Expressions for the blocking probabilities were
obtained for the overloaded and underloaded cases. In the first case, both
blocking probabilities remain O(1) as λ→∞, while in the second case they are exponentially
small. In this paper, we investigate the case of critical loading, where again C,λ→∞ but now with λ+ν/κ~C. We will also assume that ν=O(λ) (secondary calls
are less frequent than primary ones) but now with κ=O(1/λ); that is, secondary calls have large holding times.
Thus, primary calls have faster arrivals and faster service. Note that the loads
due to the primary and secondary calls remain asymptotically comparable with
this scaling. In this asymptotic regime, we are able to obtain explicit
analytic expressions for the first two terms in the expansions in powers of 1/λ for the
blocking probabilities, which involve readily evaluated definite integrals.We are currently investigating situations where the
secondary calls are the ones with fast arrivals and service [13]. Here, the
asymptotic structure of the problem turns out to be quite different.We comment that the basic problem to be solved is a
two-dimensional difference equation (cf. (2.1)), with discontinuities of the
coefficient functions at various boundaries and an interface. Such a problem
appears to be very difficult or impossible to solve exactly. From a numerical
point of view, the problem corresponds to solving roughly N=C2/2 linear
equations. A good general method such as Gaussian elimination has computational
complexity O(N3) or O(C6). Some methods that use the sparseness of the system
and some iteration procedures may improve this to O(N2) or O(C4). The purpose of our asymptotic analysis is to obtain
reasonable approximations whose numerical evaluation has computational
complexity that is independent of C, and also to obtain explicit formulas that show the
dependence of the stationary distribution and blocking probabilities on the
model parameters (i.e., C, R, and the arrival and service rates).The paper is organized as follows. In Section2, the
problem is stated more precisely and the basic equations are obtained. Here, we
summarize our main results, which are derived in detail in Sections 3 and 5. In
Section 3, we obtain the leading terms for the blocking probabilities. In
Section 4, we relate the present results to the ones in [11, 12] using
asymptotic matching. The first-order correction terms to the blocking
probabilities are derived in Section 5, while in Section 6 we present some
numerical studies to assess the accuracy of the asymptotics.


## 2. Statement of the Problem and Summary of Results

We denote byN1(t) the number of
servers serving high-priority customers, and by N2(t) the number of
servers serving low-priority ones. The total number of servers (circuits) is C of which R are reserved
for the high-priority customers. Thus, if N1+N2=C, a newly arriving high-priority customer (call) is
lost; if N1+N2≥C−R, then a newly arriving low-priority call is lost. The
high- and low-priority customers arrive as independent Poisson processes, with
respective rates λ and ν. The service times are exponentially distributed with
respective means 1 and 1/κ. Thus, the unit of time is taken as the service rate
of the high-priority customers.We denote the steady state joint distribution of the
numbers of servers used by the two priority classes byp(n1,n2)=limt→∞Pr[N1(t)=n1,N2(t)=n2]. We let I{A} be the
indicator function on the event A. Then, from the description of the model, we obtain
the following balance equation:(2.1)[λI{n1+n2+1≤C}+νI{n1+n2+1≤C−R}+n1+κn2]p(n1,n2)=λI{n1≥1}p(n1−1,n2)+νI{n1+n2≤C−R}I{n2≥1}p(n1,n2−1)+I{n1+n2+1≤C}(n1+1)p(n1+1,n2)+κI{n1+n2+1≤C}I{n2+1≤C−R}(n2+1)p(n1,n2+1). This applies
over the domain
(2.2){n1≥0,0≤n2≤C−R,⁢n1+n2≤C}. Thus, we may
view the problem as solving a second-order difference equation in two
variables, over the triangle {(n1,n2):0≤n1+n2≤C−R} and the oblique
strip {(n1,n2):C−R≤n1+n2≤C,0≤n2≤C−R}, with the two subdomains separated by the “interface” {(n1,n2):n1+n2=C−R}. There are also boundary conditions inherent in
(2.1), along n2=0, n1=0, n1+n2=C(n2≤C−R), and n2=C−R(n1≤R). The normalization condition is(2.3)∑n2=0C−R∑n1=0C−n2p(n1,n2)=1. Of particular
interest are the blocking probabilities for the high-priority customers,
defined by
(2.4)B1=∑n1=RCp(n1,C−n1), and for the
low-priority ones, defined by
(2.5)B2=∑ℓ=0R∑n1=ℓC−R+ℓp(n1,C−R+ℓ−n1). Note that we
clearly have 0<B1<B2<1.We analyze the problem in the asymptotic limit where(2.6)C⟶∞,R=O(1). We furthermore
assume that the arrival rate λ of
high-priority customers is large, of the same magnitude as C, and then scale the other rate parameters as(2.7)ν=γλ,κ=μλ,C−R=(1+γμ)λ−ωλ. Thus, the
arrival rate of low-priority customers is large, but only of the order O(λ). The service times of these customers however are
also large, and the total load due to low-priority customers is ν/κ=γλ/μ. Also, we have λ+ν/κ~C so that the
total load due to all customers is roughly equal to the capacity of the system.
Hence, this asymptotic limit may certainly be considered as “heavy traffic”
or “critical loading.”Once we input the model parametersλ, ν, κ, C, and R, we can compute γ, μ, and ω from (2.7). For
some of our numerical studies, it is desirable to fix C, R, γ, μ, and ω and then vary
the original rate parameters, which are computed by inverting (2.7) using(2.8)λ=(ω+ω2+4(1+γ/μ)(C−R))24(1+γ/μ)2, and ν and κ are obtained
from (2.7) once λ is known.We next scalen1 and n2 as(2.9)n1=λ+xλ,n1+n2=C−R+ℓ with
(2.10)p(n1,n2)=1λpℓ(x) and x and ℓ are taken as O(1). We consider the scaled state space with (X,Y)=C−1(n1,n2). Then, since C is large
and R is O(1), the domain in the (X,Y) plane is the
triangle 0≤X+Y≤1; X,Y≥0. The scaling (2.9) corresponds to a small
neighborhood of the point(2.11)X=μμ+γ,Y=γμ+γ. However, this
is where most of the probability mass accumulates in this asymptotic limit, and
the analysis of this range is sufficient to obtain the blocking probabilities Bj. We will obtain these as asymptotic series in powers
of 1/λ.Using (2.9) and (2.10) in (2.1), we obtain(2.12)[I{ℓ≤R−1}+1+γλI{ℓ≤−1}+x+γλ−μλ(x+ω)+μℓλ3/2]pℓ(x)=pℓ−1(x−1λ)+γλI{ℓ≤0}pℓ−1(x)+I{ℓ≤R−1}(1+xλ+1λ)pℓ+1(x+1λ)+I{ℓ≤R−1}(γλ−μλ(x+ω)+μ(ℓ+1)λ3/2)pℓ+1(x),ℓ≤R. Note that in
this asymptotic scaling, the indicator functions I{n1≥1}, I{n2≥1}, and I{n2+1≤C−R} may be replaced
by one, since these correspond to boundaries that are far from the point in
(2.11). However, the interface n1+n2=C−R and the
boundary n1+n2=C are evident in
(2.12) and will play a large part in the analysis. In Section 3, we will
analyze (2.12), and then also consider a second scale where ℓ is large and
negative. This second scale will lead to a diffusion equation in two variables.The main results are as follows. Forℓ=n1+n2−(C−R)=O(1) and n1=λ+xλ, we obtain the approximation p(n1,n2)=λ−1pℓ(x), with(2.13)pℓ(x)=F0(x)+1λpℓ(1)(x)+O(λ−1),ℓ≤R, where
(2.14)F0(x)=e−x2/2A0exp⁡[−μ2γ(ω+x)2](∫−x∞e−u2/2du)R=e−x2/2g0(x),(2.15)A0−1=∫−∞∞exp⁡[−μ2γ(ω+x)2](∫−x∞e−u2/2du)R+1dx. The correction
term in (2.13) takes the form pℓ(1)(x)=F1(x)−ℓ[xF0(x)+F0′(x)] for ℓ≤0 and the form pℓ(1)(x)=F1(x)−ℓ[(x+γ)F0(x)+F0′(x)] for 0≤ℓ≤R, where F1(x)=ϕ(1)(x,0). For the latter
function, we have ϕ(1)(x,y)=ψ(1)(x,x+y), where ψ(1) is given by
(5.5) in terms of g0 and g1. Then, g1 is given by
(5.66)–(5.68) (with Λ(ξ) and E(x) defined in
(5.6) and (5.56)), and the constant A1 in (5.66) can
be obtained by using (5.78) in (5.83).On the(x,y) scale, where x=(n1−λ)/λ and y=(C−R−n1−n2)/λ, we find that(2.16)p(n1,n2)=1λ[ϕ(0)(x,y)+1λϕ(1)(x,y)+O(λ−1)], where ϕ(0)(x,y)=e−x2/2g0(x+y) and ϕ(1) is as above.The analysis of these two scales leads to two-term
approximations to the blocking probabilities in (2.4) and (2.5). To leading
order, these are(2.17)(B1B2)∼1λ[∫−∞∞F0(x)dx](1R+1), where the
integral is evaluated using (2.14) and (2.15) (see also (3.49)). The O(λ−1) correction
terms follow from (5.80) and (5.81), using also (5.74).We note that the numerical evaluation of the leading
order asymptotic results involves only the integrals in (2.15) and (2.17). The
correction terms involve numerically evaluating some double or triple integrals
(cf. (5.78)), but the computational complexity of evaluating the asymptotic
results is independent of C (or λ).


## 3. Asymptotic Analysis: Leading Terms

We consider
(2.12) and assume that for λ→∞ the
probabilities have the expansion(3.1)pℓ(x)=pℓ(0)(x)+1λpℓ(1)(x)+1λpℓ(2)(x)+O(λ−3/2). In this
section, we focus on the leading term, but its calculation will necessitate
that we also analyze the problem for pℓ(1)(x). This correction term is calculated completely in
Section 5. We will also need to couple the analysis of the scale ℓ=O(1) to that where ℓ is large and
negative, with −ℓ=O(λ).Using (3.1) in (2.12) and equating coefficients of
powers of 1/λ, we obtain to leading order(3.2)[I{ℓ≤R−1}+1]pℓ(0)(x)=pℓ−1(0)(x)+I{ℓ≤R−1}pℓ+1(0)(x), and this
applies for all ℓ≤R. This is a simple difference equation with a boundary
condition at ℓ=R. Its most general solution is(3.3)pℓ(0)(x)=F0(x),ℓ≤R, where F0(x) is to be
determined.For1≤ℓ≤R, we obtain from (3.1), (3.3), and (2.12) the problem(3.4)[I{ℓ≤R−1}+1]pℓ(1)(x)+(x+γ)F0(x)=pℓ−1(1)(x)−F0'(x)+I{ℓ≤R−1}[pℓ+1(1)(x)+F0'(x)+(x+γ)F0(x)] whose solution
is
(3.5)pℓ(1)(x)=F1(x)−ℓ[(x+γ)F0(x)+F0'(x)],0≤ℓ≤R. Here, F1(x) is not yet
determined. For ℓ≤−1, the O(1/λ) terms in (2.12)
yield(3.6)2pℓ(1)(x)+(x+2γ)F0(x)=pℓ−1(1)(x)−F0'(x)+γF0(x)+pℓ+1(1)(x)+F0'(x)+(x+γ)F0(x),ℓ≤−1, which
simplifies to 2pℓ(1)(x)=pℓ−1(1)(x)+pℓ+1(1)(x) and hence(3.7)pℓ(1)(x)=F1(x)+ℓJ1(x),ℓ≤0. Here, we
imposed continuity between (3.5) and (3.7) along ℓ=0, and J1(x) is another
function not yet determined. By setting ℓ=0 in (2.12) and
comparing terms of order O(1/λ), we obtain(3.8)[I{R≥1}+1]p0(1)(x)+(x+γ)F0(x)=p−1(1)(x)−F0′(x)+γF0(x)+I{R≥1}[p1(1)(x)+F0′(x)+(x+γ)F0(x)]. Using (3.7) to
compute p−1(1) and (3.5) for p1(1) and p0(1), we obtain(3.9)J1(x)=−xF0(x)−F0′(x) so that (3.7)
becomes
(3.10)pℓ(1)(x)=F1(x)−ℓ[xF0(x)+F0′(x)]],ℓ≤0.
We next consider the problem (2.12) for ℓ→−∞, with the scaling(3.11)n1=λ+xλ,n1+n2=C−R−yλ. Note that this
still corresponds to a local approximation near the point in (2.11). In terms
of (x,y), we let(3.12)pℓ(x)=ϕ(x,y),−∞<x<∞,⁢y>0, and (2.12),
upon multiplying by λ, becomes(3.13)[2λ+λ(x+2γ)−μ(x+y+ω)]ϕ(x,y)=λϕ(x−1λ,y+1λ)+γλϕ(x,y+1λ)+(λ+xλ+1)ϕ(x+1λ,y−1λ)+(γλ−μ(x+y+ω)+μλ)ϕ(x,y−1λ). We note that ℓ±1 corresponds to y∓1/λ, and that for y>0 the indicator
functions in (2.12) can all be replaced by one.We assume thatϕ has an
expansion in the form(3.14)ϕ(x,y)=ϕ(0)(x,y)+1λϕ(1)(x,y)+1λϕ(2)(x,y)+O(λ−3/2). Using (3.14) in
(3.13), we obtain to leading order the PDE
(3.15)ϕxx(0)−2ϕxy(0)+ϕyy(0)+x(ϕx(0)−ϕy(0))+ϕ(0)=0. This is a
parabolic PDE whose solution is facilitated by the change of variables
(3.16)x=ξ,x+y=η,ϕ(0)(x,y)=ψ(0)(ξ,η). Using (3.16),
(3.15) becomes
(3.17)ψξξ(0)+ξψξ(0)+ψ(0)=0,−∞<ξ<∞,⁢η>ξ. The most
general solution, that decays exponentially as ξ→±∞, is(3.18)ψ(0)(ξ,η)=e−ξ2/2g0(η), and hence
(3.19)ϕ(0)(x,y)=e−x2/2g0(x+y). We will
determine g0 shortly.We observe that on theℓ scale, with y=−ℓ/λ, expansion (3.14) becomes(3.20)ϕ(x,−ℓλ)=ϕ(0)(x,0)+1λ[ϕ(1)(x,0)−ℓϕy(0)(x,0)]+1λ[ϕ(2)(x,0)−ℓϕy(1)(x,0)+12ℓ2ϕyy(0)(x,0)]+O(λ−3/2). Comparing this
to (3.1), for ℓ<0 we conclude
from (3.3) that(3.21)ϕ(0)(x,0)=pℓ(0)(x)=F0(x)=e−x2/2g0(x) and, from
(3.10) that
(3.22)ϕ(1)(x,0)=F1(x),ϕy(0)(x,0)=xF0(x)+F0′(x). It follows that
(3.23)ϕy(0)(x,0)=ϕx(0)(x,0)+xϕ(0)(x,0) which is a
boundary condition for the PDE (3.15) along y=0. In terms of (ξ,η), this becomes ψξ(0)(ξ,ξ)+ξψ(0)(ξ,ξ)=0, but this holds automatically (for any g0) in view of
(3.18). To determine g0, we must analyze the correction term in (3.14).From (3.13) and (3.14), we find that the first
correction term ϕ(1) satisfies the
PDE(3.24)ϕxx(1)−2ϕxy(1)+ϕyy(1)+x(ϕx(1)−ϕy(1))+ϕ(1)=−[x2(ϕxx(0)−2ϕxy(0)+ϕyy(0))+γϕyy(0)+ϕx(0)−ϕy(0)+μ(ω+x+y)ϕy(0)+μϕ(0)]. Switching to
the (ξ,η) variables and
then using (3.18), we get(3.25)ψξξ(1)+ξψξ(1)+ψ(1)=ddξ[ψξ(1)+ξψ(1)]=-[ξ2ψξξ(0)+γψηη(0)+ψξ(0)+μ(ω+η)ψη(0)+μψ(0)]=-12ξψξξ(0)-ψξ(0)-e-ξ2/2[μg0(η)+μ(ω+η)g0′(η)+γg0′′(η)].
Integrating (3.25) with respect to ξ yields(3.26)ψξ(1)+ξψ(1)=−12[ξψξ(0)+ψ(0)]−(∫−ξ∞e−u2/2du)[γg0′′(η)+μ(ω+η)g0′(η)+μg0(η)]. Setting ξ=η=x in (3.26), we
obtain(3.27)ϕx(1)(x,0)+xϕ(1)(x,0)−ϕy(1)(x,0)=F1′(x)+xF1(x)−ϕy(1)(x,0)=−12(1−x2)e−x2/2g0(x)−(∫−x∞e−u2/2du)[γg0′′(x)+μ(ω+x)g0′(x)+μg0(x)]. We will show
that (3.27) leads to a differential equation for g0(x). But, we must first consider ϕy(1)(x,0). In view of (3.20), this term arises as a part of the O(1/λ) term in the
expansion on the ℓ scale. We
therefore return to (3.1) and (2.12).For1≤ℓ≤R, we obtain from (2.12) and (3.1), at order O(1/λ), the equation(3.28)[I{ℓ≤R−1}+1]pℓ(2)(x)+(x+γ)pℓ(1)(x)−μ(x+ω)pℓ(0)(x)=pℓ−1(2)(x)−ddxpℓ−1(1)(x)+12d2dx2pℓ−1(0)(x)+I{ℓ≤R−1}[pℓ+1(2)(x)+ddxpℓ+1(1)(x)+12d2dx2pℓ+1(0)(x)+xpℓ+1(1)(x)⁢+xddxpℓ+1(0)(x)+pℓ+1(0)(x)+γpℓ+1(1)(x)−μ(x+ω)pℓ+1(0)(x)]. Using (3.3) and
(3.5), we find that (3.28) has a solution in the form
(3.29)pℓ(2)(x)=F2(x)+ℓG2(x)+ℓ2H2(x),0≤ℓ≤R, where
(3.30)2H2(x)=[(x+γ)2+1]F0(x)+(2x+3γ)F0′(x)+F0′′(x),(3.31)G2(x)+(2R−1)H2(x)=R[F0′′(x)+2(x+γ)F0′(x)+((x+γ)2+1)F0(x)]+[μ(x+ω)−1]F0(x)−(x+γ)F0′(x)−12F0′′(x)−(x+γ)F1(x)−F1′(x). It follows that
(3.32)G2(x)+H2(x)=12F0′′(x)+[x+(2−R)γ]F0′(x)+[(x+γ)2+μ(x+ω)]F0(x)−(x+γ)F1(x)−F1′(x). We note that p0(2)(x)=F2(x) and(3.33)p1(2)(x)=F2(x)−(x+γ)F1(x)−F1′(x)+12F0′′(x)+[x+(2−R)γ]F0′(x)+[(x+γ)2+μ(x+ω)]F0(x).
For ℓ≤−1, the O(1/λ) terms in
(2.12), with (3.1), yield(3.34)2pℓ(2)(x)+(x+2γ)pℓ(1)(x)−μ(x+ω)pℓ(0)(x)=pℓ−1(2)(x)−ddxpℓ−1(1)(x)+12d2dx2pℓ−1(0)(x)+γpℓ−1(1)(x)+pℓ+1(2)(x)+ddxpℓ+1(1)(x)+12d2dx2pℓ+1(0)(x)+xpℓ+1(1)(x)+xddxpℓ+1(0)(x)+pℓ+1(0)(x)+γpℓ+1(1)(x)−μ(x+ω)pℓ+1(0)(x). Using (3.3) and
(3.10), we find after some calculation that (3.34) simplifies to
(3.35)2pℓ(2)(x)−pℓ−1(2)(x)−pℓ+1(2)(x)=−[F0′′(x)+2xF0′(x)+(x2+1)F0(x)], and hence
(3.36)pℓ(2)(x)=F2(x)+ℓJ2(x)+12ℓ2[F0′′(x)+2xF0′(x)+(x2+1)F0(x)],ℓ≤0. Setting ℓ=0 in (2.12), we
obtain at order O(1/λ) the interface
relation(3.37)[1+I{R≥1}]p0(2)(x)+(x+γ)p0(1)(x)−μ(x+ω)p0(0)(x)=p−1(2)(x)−ddxp−1(1)(x)+12d2dx2p−1(0)(x)+γp−1(1)(x)+I{R≥1}[p1(2)(x)+ddxp1(1)(x)+12d2dx2p1(0)(x)+xp1(1)(x)+xddxp1(0)(x)+p1(0)(x)+γp1(1)(x)−μ(x+ω)p1(0)(x)]. Using (3.36),
with ℓ=0 and ℓ=−1, and (3.33), we obtain from (3.37) after some
calculation(3.38)J2(x)=−[F1′(x)+xF1(x)]+12(x2+1)F0(x)−RγF0′(x)+γ[F0′(x)+xF0(x)]+μ(x+ω)F0(x).
Now, comparing the −ℓ/λ terms in (3.20)
and (3.1) with (3.36), we conclude that(3.39)−ϕy(1)(x,0)=J2(x). But then F1′(x)+xF1(x)−ϕy(1)(x,0) (cf. (3.27))
involves F0′(x) only, and since F0 and g0 are related via
(3.21), we obtain from (3.27) the following ODE for g0(x):(3.40)e−x2/2[γg0′(x)+μ(ω+x)g0(x)]−Rγddx[e−x2/2g0(x)]+(∫−x∞e−u2/2du)[γg0′′(x)+μ(ω+x)g0′(x)+μg0(x)]=0. This equation
may be written as a perfect derivative as
(3.41)ddx[∫−x∞e−u2/2du[γg0′(x)+μ(ω+x)g0(x)]−Rγe−x2/2g0(x)]=0. Integrating
once and requiring g0 to vanish as x→∞ yield(3.42)g0'(x)g0(x)=−μγ(ω+x)+Re−x2/2∫−x∞e−u2/2du, and hence
(3.43)g0(x)=A0exp⁡[−μ2γ(ω+x)2](∫−x∞e−u2/2du)R, where A0 is a constant,
which will be fixed by normalization.We use (2.10) and (3.12) in the normalization sum
(2.3), and then use the Euler-MacLaurin formula to approximate sums by
integrals. To leading order, this yields(3.44)∑n2=0C−R∑n1=0C−n2p(n1,n2)∼∫0∞∫−∞∞ϕ(0)(x,y)dxdy=1. We use (3.19)
and (3.43) and evaluate one of the two integrals in (3.44) using integration by
parts, with the result
(3.45)A0=[∫−∞∞exp⁡[−μ2γ(ω+x)2](∫−x∞e−u2/2du)R+1dx]−1.
To summarize the calculation, we have shown that on
the (x,y) scale in (3.11)
we have(3.46)p(n1,n2)∼A0λe−x2/2exp⁡[−μ2γ(ω+x+y)2](∫−x−y∞e−u2/2du)R with A0 given by
(3.45). Note that p(n1,n2) is O(1/λ), but the probabilities are spread out over an O(λ)×O(λ) range near the
point (CX,CY) given by
(2.11). On the (x,ℓ) scale in (2.9),
we have obtained(3.47)p(n1,n2)∼A0λe−x2/2exp⁡[−μ2γ(ω+x)2](∫−x∞e−u2/2du)R, which applies
for −∞<ℓ≤R and is
independent of ℓ. To calculate correction terms to (3.46) and (3.47),
we would need to find F1(x) in (3.5) and
(3.10), and solve (3.24) for ϕ(1)(x,y). This ultimately involves calculating the O(1/λ) and O(1/λ3/2) terms in (3.1)
and the O(1/λ) term in (3.14);
this is done in Section 5.We next calculate the blocking probabilities in (2.4)
and (2.5). Evaluating these sums requires the expansion on the (x,ℓ) scale. Again,
approximating sums by integrals and using the scaling (2.9), we obtain(3.48)B1∼1λ∫−∞∞pR(0)(x)dx=1λ∫−∞∞F0(x)dx,B2∼1λ∫−∞∞∑ℓ=0Rpℓ(0)(x)dx∼R+1λ∫−∞∞F0(x)dx as pℓ(0)(x)=F0(x). From (3.21), (3.43), and (3.45), we then obtain(3.49)∫−∞∞F0(x)dx=∫−∞∞e−x2/2exp⁡[−(μ/2γ)(ω+x)2](∫−x∞e−u2/2du)Rdx∫−∞∞exp⁡[−(μ/2γ)(ω+x)2](∫−x∞e−u2/2du)R+1dx. The numerical
accuracy of (3.48) is investigated in
Section 6. This completes the analysis of the leading terms.


## 4. Consistency with Previous Results

In [11], Morrison studied the current model with the
scaling C=λ−O(λ) and R being either O(λ) or O(1). For the latter case, we define β from C−R=λ−βλ (see
[11, equations (7.16)–(7.18)]) and then(4.1)(B1B2)∼1λ(1R+1)1W0(β+γ/κ), where
(4.2)W0(Z)=∫0∞e−u2/2e−Zudu.
We show that (4.1) matches asymptotically to (3.48), in the limit where μ→∞. When μ→∞, we can simplify the integrals in both the numerator
and denominator in (3.49), as the factor exp[−μ(ω+x)2/2γ] has the effect
of freezing the remaining integrands at x=−ω. Thus, by the Laplace method, we
obtain(4.3)∫−∞∞F0(x)dx∼e−ω2/2∫ω∞e−u2/2du. But, in view of
(2.7), C−R=λ+λ(γ/κ−ω) so that β=ω−γ/κ. Since W0(ω)=eω2/2∫ω∞e−u2/2du, (4.1) agrees with (3.48)-(3.49) and (4.3).In [12], Knessl and Morrison analyzed the model in the
limit λ→∞ with ν=(ρ−κ)λ=O(λ) and C−R=σλ/κ=O(λ). The total load is thus λ+ν/κ=ρλ/κ and the cases ρ<σ (resp., ρ>σ) correspond to
an underloaded (resp., overloaded) link. The case of critical loading was not
considered in [12].We consider first the asymptotic matching of
(3.48)-(3.49) to the underloaded case in [12]. We note that the parameters ρ and σ are related to
the current ones by(4.4)σ=γ+μλ−μωλ,ρ=γ+μλ,κ=μλ. For the
matching, we must thus let ω→−∞ with |ω|≪λ, and ρ/σ↑1. The results in [12] for B1 and B2 were as follows:(4.5)(B1B2)∼κ2πσλe(σ−ρ)λ/κ(ρσ)σλ/κ(aR1−aR+11−a), where
(4.6)a=ρσ(1+ρ−κ)<1. In view of
(4.4), we have, in the matching region, ρ~σ, a~1, κ/σ~μ/(μ+γ), and(4.7)1−ρσ∼−μω(γ+μ)λ. Hence,
(4.8)λκ[σ−ρ+σlog⁡(ρσ)]∼−μω22(γ+μ) and (4.5) becomes
(4.9)(B1B2)∼μ2πλ(γ+μ)exp⁡[−μω22(γ+μ)](1R+1).
We show that (4.9) agrees with (3.48)-(3.49) when the
latter is expanded for ω→−∞. In (3.49), the main contribution to the integral in
the denominator comes from x=−ω=|ω|, and in the numerator from x=μ|ω|/(γ+μ), where (d/dx)[x2/2+(μ/2γ)(ω+x)2]=0. Thus, we can approximate ∫−x∞e−u2/2du by 2π everywhere, and
obtain(4.10)∫−∞∞F0(x)dx∼12π∫−∞∞e−x2/2exp⁡[−(μ/(2γ))(ω+x)2]dx∫−∞∞exp⁡[−(μ/(2γ))(ω+x)2]dx=μ2π(γ+μ)exp⁡[−μω22(γ+μ)]. With (4.10),
(3.48) agrees with (4.9).Next, we consider the overloaded case in [12], for
which we obtained(4.11)(B1B2)∼[1b−1+1−aR+11−a]−1(aR1−aR+11−a), where
(4.12)a=1σ+ζ−κζ,b=1+ρ−κσ+ζ−κζ>a, and ζ is the solution
of(4.13)ζ[1b−1+1−aR+11−a]=1b−1+1−aR1−a. We note that by
using (4.12) in (4.13), ζ is a particular
root of a polynomial. For the matching, we will take ρ/σ↓1 in (4.11) and ω→+∞ in
(3.48)-(3.49). As ρ/σ→1, we will have ζ→1 and this will
allow us to simplify (4.11). Let us set(4.14)ζ=1−θλ with which a−1=σ+ζ−κζ=1+(γ−θ)/λ+μ(θ−ω)/λ, and thus(4.15)a=1+θ−γλ+1λ[(θ−γ)2+μ(ω−θ)]+O(λ−3/2). Furthermore,
(4.16)b=(1+ρ−κ)a=(1+γλ)a=1+θλ+1λ[θ(θ−γ)+μ(ω−θ)]+O(λ−3/2). Using
(4.14)–(4.16) in (4.13), we find after some calculation that
(4.17)θ∼μω(R+1)γ+μ and then (4.11),
for ρ/σ↓1, simplify to(4.18)(B1B2)∼θλ(1R+1).
We expand (3.49) for ω→+∞. Scaling x=ωt, we obtain(4.19)∫−∞∞F0(x)dx=∫−∞∞exp⁡{−(ω2/2)[t2+(μ/γ)(t−1)2]}(∫ωt∞e−u2/2du)Rdt∫−∞∞exp⁡[−(ω2μ/(2γ))(t−1)2](∫ωt∞e−u2/2du)R+1dt. For t>0, we use the asymptotic approximation(4.20)∫ωt∞e−u2/2du∼1ωte−ω2t2/2,ω⟶∞, and conclude that both integrals in (4.19) have their
major contribution from where
(4.21)ddt[(R+1)t22+μ2γ(t−1)2]=0⟹t=t*=μ(R+1)γ+μ. But then by the
Laplace method, we have
(4.22)∫−∞∞F0(x)dx∼ωt*∼θ, and (3.48) agrees with (4.18).This completes the matching verifications.


## 5. Correction Terms

We will compute theO(1/λ) terms in
expansions (3.1) and (3.14), and then obtain O(1/λ) corrections to the
blocking probabilities.We first consider (3.13). Using the relations ϕ(x,y±1/λ)=ψ(ξ,η±1/λ) and ϕ(x±1/λ,y∓1/λ)=ψ(ξ±1/λ,η), we rewrite (3.13) in terms of (ξ,η). Defining(5.1)ϕ(x,y)=ψ(ξ,η)=ψ(0)(ξ,η)+1λψ(1)(ξ,η)+1λψ(2)(ξ,η)+O(λ3/2), we obtain from
(3.13)
(5.2)ψξξ+(ξψ)ξ+1λ[12ξψξξ+ψξ+γψηη+μ(ω+η)ψη+μψ]+1λ[112ψξξξξ+16ξψξξξ+12ψξξ−12μ(ω+η)ψηη−μψη]=O(λ−3/2) so that ψ(2) satisfies(5.3)ψξξ(2)+ξψξ(2)+ψ(2)=−ℒ1ψ(1)−ℒ2ψ(0), where ℒi are the
operators(5.4)ℒ1=12ξ∂ξ2+∂ξ+γ∂η2+μ(ω+η)∂η+μ,ℒ2=112∂ξ4+16ξ∂ξ3+12∂ξ2−12μ(ω+η)∂η2−μ∂η.
Before analyzing (5.3), we obtain a more complete
description of ψ(1). We recall that ψ(0) is known
completely, in view of (3.18), (3.43), and (3.45). But, we computed ψ(1) only partially
as the combination ψξ(1)+ξψ(1) in (3.26). We
integrate (3.26) to get(5.5)ψ(1)(ξ,η)=e−ξ2/2[16ξ(ξ2−3)g0(η)+g1(η)]+e−ξ2/2Λ(ξ)[γg0′′(η)+μ(ω+η)g0′(η)+μg0(η)], where
(5.6)Λ(ξ)=∫0−ξev2/2∫v∞e−u2/2dudv=∫0∞e−t2/21−eξttdt. We observe that Λ(ξ) satisfies(5.7)Λ′′(ξ)−ξΛ′(ξ)=−1 and Λ′(ξ) decays as ξ→−∞. We also have(5.8)ddξ[e−ξ2/2Λ′(ξ)]=e−ξ2/2. The function g1 will be
determined shortly (actually, not very shortly, but only after a lengthy
calculation).We evaluate the right side of (5.3) more explicitly.
Some terms are expressible as derivatives with respect to ξ, while the ones involving derivatives in η may be
evaluated using (3.18) and (5.5). Then, (5.3) becomes(5.9)(ψξ(2)+ξψ(2))ξ=−[12(ξψξ(1)+ψ(1))+112ψξξξ(0)+ξ6ψξξ(0)+13ψξ(0)]ξ−e−ξ2/2[γg1′′(η)+μ(ω+η)g1′(η)+μg1(η)]−e−ξ2/216ξ(ξ2−3)[γg0′′(η)+μ(ω+η)g0′(η)+μg0(η)]−e−ξ2/2Λ(ξ)[γ∂η2+μ(ω+η)∂η+μ]2g0(η)+e−ξ2/2[12μ(ω+η)g0′′(η)+μg0′(η)]. From (3.18), we
have
(5.10)112ψξξξ(0)+16ξψξξ(0)+13ψξ(0)=112(ξ3−3ξ)e−ξ2/2g0(η) and by direct
calculation
(5.11)(ξ∂ξ+1)[(ξ3−3ξ)e−ξ2/2]=−e−ξ2/2[ξ5−7ξ3+6ξ],(ξ∂ξ+1)[e−ξ2/2Λ(ξ)]=−ξ∫−ξ∞e−u2/2du−(ξ2−1)e−ξ2/2Λ(ξ),(ξ∂ξ+1)[e−ξ2/2]=−e−ξ2/2(ξ2−1). With the above, we
integrate (5.9) to get
(5.12)ψξ(2)+ξψ(2)=e−ξ2/2{112(ξ5−8ξ3+9ξ)g0(η)⁢⁢+12(ξ2−1)g1(η)+12(ξ2−1)[Λ(ξ)+13]Dg0(η)}−[∫−∞ξe−u2/2Λ(u)du]D2g0(η)+[∫−ξ∞e−u2/2du]⋅{ξ2Dg0(η)−Dg1(η)+μ[12(ω+η)g0′′(η)+g0′(η)]}, where
(5.13)D=γ∂η2+μ(ω+η)∂η+μ. We will show
that the calculation of g1 will require
only that we evaluate (5.12) along η=ξ. However, we must first reconsider the scale ℓ=O(1) and analyze at
least partly the term pℓ(3)(x) in (3.1) (i.e.,
the coefficient of λ−3/2 in the series).Returning to (2.12) with the expansion (3.1), we find
that for 1≤ℓ≤R, pℓ(3) satisfies(5.14)[I{ℓ≤R−1}+1]pℓ(3)(x)+(x+γ)pℓ(2)(x)−μ(x+ω)pℓ(1)(x)+μℓpℓ(0)(x)=pℓ−1(3)(x)−ddxpℓ−1(2)(x)+12d2dx2pℓ−1(1)(x)−16d3dx3pℓ−1(0)(x)+I{ℓ≤R−1}[pℓ+1(3)(x)+ddxpℓ+1(2)(x)+12d2dx2pℓ+1(1)(x)+16d3dx3pℓ+1(0)(x)]+I{ℓ≤R−1}[xpℓ+1(2)(x)+xddxpℓ+1(1)(x)+12xd2dx2pℓ+1(0)(x)+pℓ+1(1)(x)⁢+ddxpℓ+1(0)(x)+γpℓ+1(2)(x)−μ(x+ω)pℓ+1(1)(x)+μ(ℓ+1)pℓ+1(0)(x)]. We recall that
for 0≤ℓ≤R, pℓ(0) is given by
(3.3), pℓ(1) by (3.5), and pℓ(2) by (3.29). Let
us write(5.15)pℓ(1)(x)=F1(x)+ℓG1(x),G1(x)=−(x+γ)F0(x)−F0'(x),0≤ℓ≤R. We then rewrite
(5.14) as
(5.16)pℓ(3)(x)−pℓ−1(3)(x)−I{ℓ≤R−1}[pℓ+1(3)(x)−pℓ(3)(x)]=I{ℓ≤R−1}[(x+γ)pℓ+1(2)(x)−μ(x+ω)pℓ+1(1)(x)+μpℓ+1(0)(x)]−[(x+γ)pℓ(2)(x)−μ(x+ω)pℓ(1)(x)+μℓpℓ(0)(x)]−16F0′′′(x)+12[F1′′(x)+(ℓ−1)G1′′(x)]−F2′(x)−(ℓ−1)G2′(x)−(ℓ−1)2H2′(x)+I{ℓ≤R−1}{16F0′′′(x)+12xF0′′(x)+F0′(x)+12[F1′′(x)+(ℓ+1)G1′′(x)]⁢+x[F1′(x)+(ℓ+1)G1′(x)]+F1(x)+(ℓ+1)G1(x)+F2′(x)⁢+(ℓ+1)G2′(x)+(ℓ+1)2H2′(x)}, which holds for 1≤ℓ≤R. We sum (5.16) for ℓ=1,2,…,R and use the
identities(5.17)∑ℓ=1RI{ℓ≤R−1}(ℓ+1)=∑ℓ=1R−1(ℓ+1)=12(R−1)(R+2),∑ℓ=1R[I{ℓ≤R−1}(ℓ+1)2−(ℓ−1)2]=R2−1. After some
rearrangement, we obtain
(5.18)p1(3)(x)−p0(3)(x)+(x+γ)p1(2)(x)−μ(x+ω)p1(1)(x)+μp1(0)(x)=−16F0′′′+(R−1)[12xF0′′+F0′]+(R−12)F1′′+(R−1)[xF1′+F1]−F2′+12(R2−1)G1′′+12(R−1)(R+2)[xG1′+G1]+(R−1)G2′+(R2−1)H2′, where all
derivatives are with respect to x.Settingℓ=0 in (2.12) and
using again expansion (3.1), we obtain at O(λ−3/2) the relation(5.19)[I{R≥1}+1]p0(3)(x)+(x+γ)p0(2)(x)−μ(x+ω)p0(1)(x)=p−1(3)(x)−ddxp−1(2)(x)+12d2dx2p−1(1)(x)−16d3dx3p−1(0)(x)+γp−1(2)(x)+I{R≥1}[p1(3)(x)+ddxp1(2)(x)+12d2dx2p1(1)(x)+16d3dx3p1(0)(x)]+xp1(2)(x)+xddxp1(1)(x)+12xd2dx2p1(0)(x)+p1(1)(x)+ddxp1(0)(x)+γp1(2)(x)−μ(x+ω)p1(1)(x)+μp1(0)(x).
Next, we consider (2.12) for ℓ<0, and recall that pℓ(0) is given by
(3.3) for all ℓ, pℓ(1) is given by
(3.7) or (3.10), and(5.20)pℓ(2)(x)=F2(x)+ℓJ2(x)+ℓ2K2(x),ℓ≤0, where J2 is in (3.27),
and (3.36) yields(5.21)K2(x)=12[F0′′(x)+2xF0′(x)+(x2+1)F0(x)]. We subtract I{R≥1}p0(3)+p−1(3)+γp−1(2) from both sides
of (5.19) and substitute (5.18) into the resulting equation. We then use (3.3),
(3.10), (5.20), (5.15), and (3.29), and after some calculation, obtain(5.22)p0(3)(x)−p−1(3)(x)+(x+γ)p0(2)(x)−μ(x+ω)p0(1)(x)−γp−1(2)(x)=−16F0′′′+12(F1′′−J1′′)−F2′+J2′−K2′+R{12xF0′′+F0′+F1′′+xF1′+F1+12(R+1)(xG1′+G1)+12RG1′′+G2′+RH2′}.
From (2.12), for ℓ≤−1, we obtain the following problem for pℓ(3):(5.23)2pℓ(3)+(x+2γ)pℓ(2)−μ(x+ω)pℓ(1)+μℓpℓ(0)=pℓ−1(3)−ddxpℓ−1(2)+12d2dx2pℓ−1(1)−16d3dx3pℓ−1(0)+γpℓ−1(2)+pℓ+1(3)+ddxpℓ+1(2)+12d2dx2pℓ+1(1)+16d3dx3pℓ+1(0)+xpℓ+1(2)+xddxpℓ+1(1)+12xd2dx2pℓ+1(0)+pℓ+1(1)+ddxpℓ+1(0)+γpℓ+1(2)−μ(x+ω)pℓ+1(1)+μ(ℓ+1)pℓ+1(0). Here, we used I{ℓ≤R−1}=1 in this range
of ℓ, and all functions in (5.23) are evaluated at x. After rearranging (5.23) and using (5.20), (3.3),
and (3.7) to evaluate pℓ(j) for j=0,1,2, we are led to(5.24)pℓ(3)−pℓ−1(3)−[pℓ+1(3)−pℓ(3)]=(x+γ)pℓ+1(2)−μ(x+ω)pℓ+1(1)+μ(ℓ+1)pℓ+1(0)−γpℓ(2)−[(x+γ)pℓ(2)−μ(x+ω)pℓ(1)+μℓpℓ(0)−γpℓ−1(2)]+12xF0′′+F0′+F1′′+xF1′+F1+ℓJ1′′+(ℓ+1)[xJ1′+J1]+2J2′+4ℓK2′,ℓ≤−1. We sum (5.24)
from ℓ=−m to ℓ=−1 (with m≥1) to obtain(5.25)p−m(3)−p−m−1(3)−[p0(3)−p−1(3)]+(x+γ)p−m(2)−μ(x+ω)p−m(1)−μmp−m(0)−γp−m−1(2)−(x+γ)p0(2)+μ(x+ω)p0(1)−γp−1(2)=m{12xF0′′+F0′+F1′′+xF1′+F1+2J2′−12(m+1)(J1′′+4K2′)−12(m−1)(xJ1′+J1)},m≥1. Note that
(5.25) remains true if m=0. Using (5.22) in (5.25) gives us(5.26)p−m(3)−p−m−1(3)+(x+γ)p−m(2)−μ(x+ω)p−m(1)−μmp−m(0)−γp−m−1(2)=−16F0′′′+12(F1′′−J1′′)−F2′+J2′−K2′+R{12xF0′′+F0′+F1′′+xF1′+F1+12(R+1)[xG1′+G1]+12RG1′′+G2′+RH2'}+m{12xF0′′+F0′+F1′′+xF1′+F1+2J2′−12(m+1)[J1′′+4K2′]−12(m−1)[xJ1′+J1]}.
Using our previous results for p−m(0), p−m(1), and p−m(2), we have(5.27)μmp−m(0)+μ(x+ω)p−m(1)+γp−m−1(2)−(x+γ)p−m(2)=μmF0+μ(x+ω)[F1−mJ1]−xF2+(mx−γ)J2+[(2m+1)γ−m2x]K2. Adding (5.26)
and (5.27), we obtain an explicit expression for p−m(3)−p−m−1(3) (in terms of F0, F1, J1, F2, J2, and K2) which is
quadratic in m. By summing from m=0 to m=n−1, we get(5.28)p0(3)−p−n(3)=−16nF0′′′+12n(F1′′−J1′′)−n(F2′−J2′+K2′)+nR{12xF0′′+F0′+F1′′+xF1′+F1+12(R+1)[xG1′+G1]+12RG1′′+G2′+RH2′}+12n(n−1)[12xF0′′+F0′+F1′′+xF1′+F1+2J2′]−16(n3−n)[J1′′+4K2′]−16(n3−3n2+2n)[xJ1′+J1]+n[μ(x+ω)F1−xF2]+12μn(n−1)[F0−(x+ω)J1]+[12(n2−n)x−nγ]J2+[n2γ−16(2n3−3n2+n)x]K2. This holds for
all n≥0. We write (5.28) as(5.29)p−n(3)(x)=F3(x)−nJ3(x)+n2K3(x)−n3L3(x), where F3(x)=p0(3)(x), and J3, K3, and L3 may be
identified from (5.28).We recall thatp−n(x)=ϕ(x,n/λ) for n≥0, and the coefficient of λ−3/2 in the
expansion (3.20) is(5.30)ϕ(3)(x,0)+nϕy(2)(x,0)+12n2ϕyy(1)(x,0)+16n3ϕyyy(0)(x,0). Comparing
(5.29) to (5.30) yields
(5.31)L3(x)=−16ϕyyy(0)(x,0),K3(x)=12ϕyy(1)(x,0),J3(x)=−ϕy(2)(x,0). In view of
(3.19), we have
(5.32)L3(x)=−16e−x2/2g0′′′(x), while (5.28)
shows that
(5.33)L3(x)=−16[J1′′(x)+xJ1′(x)+J1(x)]−13[2K2′+xK2(x)]. But, from
(5.21) and (3.21), we get K2(x)=(1/2)e−x2/2g0′′(x), and from (3.9), J1(x)=−e−x2/2g0′(x). Then, we can easily verify that (5.32) is consistent
with (5.33). Also, from (5.28), we find that(5.34)−2K3(x)=12xF0′′+F0′+xF1′+F1+F1′′+μF0−μ(x+ω)J1+xJ1′+J1+xJ2+2J2′+(x+2γ)K2=ddx[12xF0'+12F0+F1′+xF1+xJ1+J2]+μF0−μ(x+ω)J1+J2′+xJ2+(x+2γ)K2. We show that
this is the same as −ϕyy(1)(x,0). We recall that ϕ(1)(x,y)=ψ(1)(x,x+y) is given by
(5.5) and J2 is expressed in
terms of F1 in (3.38). From
(3.38) and (3.21), it follows that(5.35)F1′+xF1+J2=e−x2/2{12(x2−1)g0(x)+[γx+μ(x+ω)]g0(x)⁢−(R−1)γ[g0′(x)−xg0(x)]}=e−x2/2{12(x2−1)g0(x)+Λ′(x)Dxg0(x)}, where 𝒟x is the operator
in (5.13), with η replaced by x. The second equality in (5.35) follows from (3.40)
and (5.6). Using (3.9), we obtain(5.36)12xF0′+12F0+xJ1=12e−x2/2[(1−x2)g0(x)−xg0′(x)] which when
combined with (5.35) gives
(5.37)12xF0′+12F0+xJ1+F1'+xF1+J2=e−x2/2[Λ′(x)Dxg0(x)−12xg0′(x)]. We use (5.37)
in (5.34), also noting that
(5.38)J2(x)=−ϕy(1)(x,0)=−e−x2/2{16x(x2−3)g0′(x)+g1′(x)+Λ(x)[Dxg0′(x)+μg0′(x)]}, which follows
from (5.5). Then, (5.34) becomes
(5.39)−2K3(x)=e−x2/2{−Dxg0(x)+12(x2−1)g0′(x)−12xg0′′(x)+Λ′(x)[Dxg0′(x)+μg0′(x)]}+μF0−μ(x+ω)J1+J2′+xJ2+(x+2γ)K2. Here, we also
used Λ′′(x)=xΛ′(x)−1. Now, from (5.21) and (3.9), we obtain(5.40)μF0−μ(x+ω)J1+(x+2γ)K2=(x2+γ)F0′′+[x(x+2γ)+μ(x+ω)]F0′+[μ+μx(x+ω)+12(x2+1)(x+2γ)]F0=e−x2/2[μg0(x)+μ(x+ω)g0′(x)+(12x+γ)g0′′(x)]. Using (5.38) to
compute J2′+xJ2 and (5.40), we
get(5.41)−2K3(x)=e−x2/2{16x(x2−3)g0′′(x)+g1′′(x)+Λ(x)[Dxg0′′(x)+2μg0′′(x)]}. In view of
(5.5), the above is the same as 2K3(x)=ϕyy(1)(x,0)=ψηη(1)(ξ,ξ).We next examine the relationJ3(x)=−ϕy(2)(x,0)=−ψη(2)(ξ,ξ). We will use this to ultimately obtain g1(η) in (5.5), which
will complete the determination of the O(1/λ) correction
terms in (3.1) and (3.14). We first note from (5.28) and (5.29) that(5.42)J3(x)=−16F0′′′+12(F1′′−J1′′)−F2′+J2′−K2′+R{12xF0′′+F0′+F1′′+xF1′+F1+12(R+1)[xG1′+G1]+12RG1′′+G2′+RH2′}−12[12xF0′′+F0′+F1′′+xF1′+F1+2J2′]+16[J1′′+4K2′]−13[xJ1′+J1]+μ(x+ω)F1−xF2−12μ[F0−(x+ω)J1]−(12x+γ)J2−16xK2. We solve (5.42)
for the combination J3+F2′+xF2, that we rewrite as(5.43)J3+F2′+xF2=W(x)+Z(x)+RU(x), where
(5.44)W(x)=−12[xF1′+F1]+μ(x+ω)F1−12μF0+12μ(x+ω)J1−(12x+γ)J2−16xK2,Z(x)=−16F0′′′−14xF0′′−12F0′−13[J1′′+xJ1′+J1+K2′],U(x)=12xF0′′+F0′+F1′′+xF1′+F1+12(R+1)[xG1′+G1]+12RG1′′+G2′+RH2′. Next we note
that Z and U may be
integrated explicitly and we write(5.45)U(x)=ddxU(x),Z(x)=ddxZ(x), where
(5.46)U(x)=12xF0′+12F0+F1′+xF1+12(R+1)xG1+12RG1′+G2+RH2,(5.47)Z(x)=−16F0′′−14xF0′−14F0−13[J1′+xJ1+K2]. It follows that
(5.48)W(x)+Z′(x)+RU′(x)=−ϕy(2)(x,0)+ϕx(2)(x,0)+xϕ(2)(x,0)=ψξ(2)(x,x)+xψ(2)(x,x). The right side
of (5.48) was computed in (5.12).UsingF1(x)=ψ(1)(x,x), (5.38), and the identities F0(x)=e−x2/2g0(x), J1(x)=−e−x2/2g0′(x), and K2(x)=(1/2)e−x2/2g0′′(x), we evaluate W(x) in terms of g0(x), and then use (5.12). After some simplification, this
leads to(5.49)W(x)−ψξ(1)(x,x)−xψ(2)(x,x)=e−x2/2[γg1′(x)+μ(ω+x)g1(x)]+[∫−x∞e−u2/2du]ddx[γg1′(x)+μ(ω+x)g1(x)]+e−x2/2Λ(x)[γddx+μ(ω+x)]Dxg0(x)+[∫−∞xe−u2/2Λ(u)du]ddx{[γddx+μ(ω+x)]Dxg0(x)}−μ2e−x2/2[(ω+x)g0′(x)+g0(x)]−μ2∫−x∞e−u2/2duddx[(ω+x)g0′(x)+g0(x)]+γ6e−x2/2[(1−x2)g0′′(x)+(x3−3x)g0′(x)]+μ6e−x2/2{(1−x2)[(ω+x)g0′(x)+g0(x)]+(x3−3x)(ω+x)g0(x)}+x12e−x2/2[(x2−3)g0(x)−g0′′(x)]. Since this must
be equal to −[Z′+RU′], we try to write the right side of (5.49) as a
perfect derivative. To this end, we note that(5.50)ddx[e−x2/2((1−x2)g0−xg0′)]=e−x2/2x[(x2−3)g0−g0′′]. Adding Z′+RU′ to (5.49), we
rewrite that equation as(5.51)ddx{(∫−x∞e−u2/2du)[γg1'(x)+μ(ω+x)g1(x)]+(∫−∞xe−u2/2Λ(u)du)[γddx+μ(ω+x)]Dxg0(x)−12μ(∫−x∞e−u2/2du)[(ω+x)g0′(x)+g0(x)]+16e−x2/2(1−x2)[γg0′(x)+μ(ω+x)g0(x)]+112e−x2/2[(1−x2)g0(x)−xg0′(x)]+Z(x)+RU(x)}=0.
We next evaluate Z in terms of g0, and U in terms of g0 and g1, and then we integrate (5.51) (and thus explicitly
obtain g1). Since J1=−xF0−F0' and K2 is in (5.21),
we have(5.52)Z(x)=112[xF0′+(2x2−1)F0]=112e−x2/2[xg0′(x)+(x2−1)g0(x)]. We thus note
that Z(x) is canceled by
the bracketed term that precedes it in (5.51).Using (5.46), we explicitly calculate U(x), recalling that G1 is given by
(5.15), and G2+RH2=(G2+H2)+(1/2)(R−1)(2H2) can be computed
from (3.30) and (3.32). After some cancellation of terms, we obtain(5.53)U(x)=[12γ(R+1)(x+γ)+μ(ω+x)]F0(x)+12γF0′(x)−γF1(x)=e−x2/2[{12γ[Rx+(R+1)γ]+μ(ω+x)}g0(x)+12γg0′(x)]−γe−x2/2{16x(x2−3)g0(x)+g1(x)+Λ(x)Dxg0(x)}. Using (5.52)
and (5.53), we integrate (5.51), subject to the condition that the solution
decays exponentially as x→+∞. Hence,(5.54)(∫−x∞e−u2/2du)[γg1′(x)+μ(ω+x)g1(x)−12μ(ω+x)g0′(x)−12μg0(x)]+(∫−∞xe−u2/2Λ(u)du)[γddx+μ(ω+x)]Dxg0(x)+16e−x2/2(1−x2)[γg0′(x)+μ(ω+x)g0(x)]+12Re−x2/2{γ[Rx+(R+1)γ]g0(x)+2μ(ω+x)g0(x)+γg0′(x)}−Rγe−x2/2[16x(x2−3)g0(x)+g1(x)+Λ(x)Dxg0(x)]=0. What remains is
a linear first-order ordinary differential equation for g1, which is readily solved by multiplying by the
integrating factor:(5.55)exp⁡[(μ/(2γ))(ω+x)2](∫−x∞e−u2/2du)R+1. We introduce
the notation
(5.56)E(x)=∫−x∞e−u2/2du and note that
(5.57)E′(x)=e−x2/2,E(∞)=2π,E(x)∼e−x2/2−x,x⟶−∞. Then, we have
(5.58)∫exp⁡[(μ/(2γ))(ω+x)2][E(x)]R{γg1′(x)+μ(ω+x)g1(x)−Rγe−x2/2E(x)g1(x)}dx=γexp⁡[(μ/2γ)(ω+x)2][E(x)]Rg1(x). From (3.43), we
have
(5.59)g0(x)=A0exp⁡[−μ2γ(ω+x)2][E(x)]R and thus
(5.60)γg0′(x)+u(ω+x)g0(x)=A0Rγe−x2/2exp⁡[−μ2γ(ω+x)2][E(x)]R−1. With (5.60), we
have
(5.61)∫e−x2/2exp⁡[(μ/(2γ))(ω+x)2][E(x)]R+1{(1−x2)[γg0′(x)+μ(ω+x)g0(x)]−Rγ(x3−3x)g0(x)}dx=A0Rγ∫e−x2/2{(1−x2)e−x2/2[E(x)]2−x3−3xE(x)}dx=−A0Rγe−x2/2(1−x2)E(x),∫e−x2/2exp⁡[(μ/(2γ))(ω+x)2][E(x)]R+1{γg0′(x)+μ(ω+x)g0(x)+Rγxg0(x)}dx=−A0Rγe−x2/2E(x).   Furthermore,
(5.62)∫exp⁡[(μ/(2γ))(ω+x)2][E(x)]R{R(ω+x)e−x2/2E(x)g0(x)−(ω+x)g0′(x)−g0(x)}dx=−A0∫[μγ(ω+x)2−1]dx=A0[μ3γ(ω+x)3−ω−x],(5.63)∫e−x2/2exp⁡[(μ/2γ)(ω+x)2][E(x)]R+1g0(x)dx=A0log⁡[E(x)],(5.64)∫exp⁡[(μ/(2γ))(ω+x)2][E(x)]R+1{(∫−∞xe−u2/2Λ(u)du)[γddx+μ(ω+x)]Dxg0(x)⁢−Rγe−x2/2Λ(x)Dxg0(x)}dx=γ(∫−∞xe−u2/2Λ(u)du)exp⁡[μ2γ(ω+x)2][Dxg0(x)][E(x)]−R−1⁢+(R+1)γ∫{e−x2/2∫−∞xe−u2/2Λ(u)du[E(x)]R+2⁢−e−x2/2Λ(x)[E(x)]R+1}exp⁡[μ2γ(ω+x)2]Dxg0(x)dx. To obtain
(5.64), we used exp[(μ/(2γ))(ω+x)2][γ(d/dx)+μ(ω+x)]F(x)=γ(d/dx){exp[(μ/(2γ))(ω+x)2]F(x)} and integrated
by parts. Combining (5.58) and (5.61)–(5.64),
we integrate (5.54) to get(5.65)exp⁡[μ2γ(ω+x)2][E(x)]−Rg1(x)=A1+A0μ2γ[ω+x−μ3γ(ω+x)3]+A0R6e−x2/2(1−x2)E(x)+A02R2e−x2/2E(x)−A02R(R+1)γlog⁡[E(x)]−∫−∞xe−u2/2Λ(u)du[E(x)]R+1exp⁡[μ2γ(ω+x)2]Dxg0(x)+(R+1)∫x∞{e−v2/2∫−∞ve−u2/2Λ(u)du[E(v)]R+2−e−v2/2Λ(v)[E(v)]R+1}⋅exp⁡[μ2γ(ω+v)2]Dvg0(v)dv. Here, A1 is a constant
that will be fixed by normalization.We thus writeg1 as(5.66)g1(x)=exp⁡[−μ2γ(ω+x)2][E(x)]R⋅{A1+N(x)+A0[M(x)+Re−x2/2(1−x2)6E(x)]}, where
(5.67)M(x)=μ2γ[ω+x−μ3γ(ω+x)3]+R2e−x2/22E(x)−R(R+1)γ2log⁡[E(x)],(5.68)N(x)=−∫−∞xe−u2/2Λ(u)du[E(x)]R+1exp⁡[μ2γ(ω+x)2]Dxg0(x)+(R+1)∫x∞e−v2/2[E(v)]R{∫−∞ve−u2/2Λ(u)du[E(v)]2−Λ(v)E(v)}⋅exp⁡[μ2γ(ω+v)2]Dvg0(v)dv, where 𝒟vg0(v)=γg0′′(v)+μ(ω+v)g0′(v)+μg0(v) is as in
(5.13).We next determineA1 by
normalization and then obtain correction terms to the blocking probabilities B1 and B2. This requires that we evaluate the integrals ∫−∞∞ϕ(1)(x,0)dx and ∫−∞∞∫0∞ϕ(1)(x,y)dydx=∫−∞∞∫ξ∞ψ(1)(ξ,η)dηdξ. From (5.5), we have(5.69)ϕ(1)(x,0)=e−x2/2[16(x3−3x)g0(x)+g1(x)+Λ(x)Dxg0(x)], and from
(3.43), we calculate 𝒟xg0 and obtain(5.70)exp⁡[(μ/(2γ))(ω+v)2][E(v)]RDvg0(v)=A0Rγe−v2/2E(v){(R−1)e−v2/2E(v)−[v+μγ(ω+v)]}.
Consider the contribution to ∫−∞∞ϕ(1)(x,0)dx that comes from
the term Λ(x)𝒟xg0(x) in (5.69) and
the part of g1 that is
proportional to N(x) (cf. (5.66)). We obtain(5.71)∫−∞∞e−x2/2{Λ(x)Dxg0(x)+exp⁡[−μ2γ(ω+x)2][E(x)]RN(x)}dx=∫−∞∞e−x2/2{[Λ(x)−∫−∞xe−u2/2Λ(u)duE(x)]Dxg0(x)+(R+1)exp⁡[−μ2γ(ω+x)2][E(x)]R∫x∞e−v2/2[E(v)]R+2Dvg0(v)⋅[∫−∞ve−u2/2Λ(u)du−E(v)Λ(v)]exp⁡[μ2γ(ω+v)2]dv}dx=μγ∫−∞∞(ω+x)[E(x)]R+1exp⁡[−μ2γ(ω+x)2]⋅∫x∞e−v2/2[E(v)]R+2[∫−∞ve−u2/2Λ(u)du−E(v)Λ(v)]exp⁡[μ2γ(ω+v)2]Dvg0(v)dv⁢dx. Here, we wrote
(5.72)e−x2/2(R+1)[E(x)]R=ddx[E(x)]R+1 and integrated  by
parts. We also have
(5.73)∫−∞∞e−x2/2(x3−3x)g0(x)dx=−∫−∞∞e−x2/2(1−x2)g0'(x)dx=A0∫−∞∞(1−x2)e−x2/2exp⁡[−μ2γ(ω+x)2][E(x)]R[μγ(ω+x)−Re−x2/2E(x)]dx. Using (5.71)
and (5.73), we integrate (5.69) and get
(5.74)∫−∞∞ϕ(1)(x,0)dx=∫−∞∞exp⁡[−μ2γ(ω+x)2][E(x)]R⋅(e−x2/2[A1+A0μ6γ(ω+x)(1−x2)+A0M(x)]⁢+A0μR(ω+x)E(x)∫x∞[∫−∞ve−u2/2Λ(u)du−Λ(v)E(v)]⁢⋅e−v2[E(v)]3{(R−1)e−v2/2E(v)−[v+μγ(ω+v)]}dv)dx. Here, we also
used (5.70) to eliminate 𝒟xg0 from the
expression.Next, we consider(5.75)∫−∞∞∫0∞ϕ(1)(x,y)dy⁢dx=∫−∞∞∫x∞ψ(1)(x,η)dη⁢dx=∫−∞∞∫x∞e−x2/2[16(x3−3x)g0(η)+Λ(x)Dηg0(η)+g1(η)]dη⁢dx. Integration by
parts shows that
(5.76)∫−∞∞e−x2/2(x3−3x)[∫x∞g0(η)dη]dx=∫−∞∞(1−x2)e−x2/2g0(x)dx,∫−∞∞e−x2/2Λ(x)[∫x∞Dηg0(η)dη]dx=∫−∞∞[∫−∞xe−u2/2Λ(u)du]Dxg0(x)dx,∫−∞∞e−x2/2[∫x∞g1(η)dη]dx=∫−∞∞E(x)g1(x)dx.
From (5.68) and (5.70), we obtain(5.77)∫−∞∞{E(x)exp⁡[−μ2γ(ω+x)2][E(x)]RN(x)+(∫−∞xe−u2/2Λ(u)du)Dxg0(x)}dx=A0R(R+1)γ∫−∞∞exp⁡[−μ2γ(ω+x)2][E(x)]R+1⋅∫x∞[∫−∞ve−u2/2Λ(u)duE(v)−Λ(v)]e−v2[E(v)]2⋅{(R−1)e−v2/2E(v)−[v+μγ(ω+v)]}dvdx. With (5.66), (5.68), and (5.75)–(5.77), we obtain
(5.78)∫−∞∞∫0∞ϕ(1)(x,y)dy⁢dx=∫−∞∞exp⁡[−μ2γ(ω+x)2][E(x)]R+1⋅[A1+A0M(x)+16A0(R+1)e−x2/2(1−x2)E(x)+A0R(R+1)γ⋅∫x∞[∫−∞ve−u2/2Λ(u)duE(v)−Λ(v)]e−v2[E(v)]2{(R−1)e−v2/2E(v)−[v+μγ(ω+v)]}dv]dx. Using ∫Z∞e−u2/2du=π/2Erfc(Z/2) and (5.6), we
can show that(5.79)∫−∞ve−u2/2Λ(u)du−E(v)Λ(v)=π2∫0∞[Erfc(−v2)e−t2/2evt−Erfc(t−v2)]dtt, which helps in
the numerical evaluation of (5.78).Finally, we calculate the blocking probabilities.
Using (3.1), (3.3), and (3.5), we obtain
(5.80)B1=1λ∫−∞∞[pR(0)(x)+1λpR(1)(x)+O(λ−1)]dx=1λ∫−∞∞{F0(x)+1λ[F1(x)−R(x+γ)F0(x)−RF0'(x)]+O(λ−1)}dx=A0λ∫−∞∞e−x2/2[1−Rλ(x+γ)]exp⁡[−μ2γ(ω+x)2][E(x)]Rdx+1λ∫−∞∞ϕ(1)(x,0)dx+O(λ−3/2), where the last
integral is given by (5.74) in terms of A0 and A1. To obtain (5.80), we used the scaling (2.9) and
(2.10) in (2.4), and approximated the sum by an integral. Since the integrand
has exponentially small tails, the finite limits in (2.4) may be replaced by
infinite ones, with an error, that is, o(λ−N) for all N.Similarly, we obtainB2 as(5.81)B2=1λ∫−∞∞∑ℓ=0R[pℓ(0)(x)+1λpℓ(1)(x)+O(λ−1)]dx=1λ∫−∞∞∑ℓ=0R{F0(x)+1λF1(x)−ℓλ[(x+γ)F0(x)+F0′(x)]+O(λ−1)}dx=A0λ(R+1)∫−∞∞e−x2/2[1−R2λ(x+γ)]exp⁡[−μ2γ(ω+x)2][E(x)]Rdx+R+1λ∫−∞∞ϕ(1)(x,0)dx+O(λ−3/2).
Finally, we determine A1 from the
normalization (2.3). Again, using (2.9), (2.10), (3.1), and (3.12), we obtain(5.82)∫−∞∞∫0∞ϕ(0)(x,y)dydx+121λ∫−∞∞ϕ(0)(x,0)dx+1λ∫−∞∞∑ℓ=1Rpℓ(0)(x)dx+1λ∫−∞∞∫0∞ϕ(1)(x,y)dydx+O(λ−1)=1. Here, the
second integral in (5.82) comes from the Euler-MacLaurin approximation as we
go from a discrete sum to an integral over y=0. Note that the expansion on the ℓ scale, for 1≤ℓ≤R, leads to the third term in (5.82). For ℓ≤0, the expansion on the y scale contains
that on the ℓ scale. The
leading term in (5.82) regains (3.45) and determines A0. The O(1/λ) terms lead to(5.83)A0(R+12)∫−∞∞e−x2/2exp⁡[−μ2γ(ω+x)2][E(x)]Rdx+∫−∞∞∫0∞ϕ(1)(x,y)dydx=0. In view of
(5.78), (5.83) may be viewed as a linear equation for A1, and thus all the correction terms are now known
fully.To summarize the calculations in this section, we have
determinedg1(x) in
(5.66)–(5.68), with A1 computed from
(5.78) and (5.83). In terms of g1, the second term in the expansion on the (x,y) (or (ξ,η)=(x,x+y)) scale is
given by (5.5). Then, F1(x)=ϕ(1)(x,0)=ψ(1)(ξ,ξ) and the second
terms on the ℓ scale are given
by (3.5) for 0≤ℓ≤R, and by (3.10) for ℓ<0 (with F0(x) in (3.21) and
(3.43)).


## 6. Numerical Studies

We test the numerical accuracy of our asymptotic
expansions, focusing on the blocking probabilitiesB1 and B2. The numerical results are obtained by solving the
linear system (2.1) with the normalization (2.3). We simply omitted the
equation with n1=n2=0 in (2.1) and
replaced it by (2.3), thus obtaining an inhomogeneous problem with a unique
solution.We solved (2.1) by two different methods. First, we
simply used the program MAPLE to solve the linear system numerically. We also
tried an iteration method of the form p(n1,n2;M+1)=p(n1,n2;M)+T/MAX⋅Lp(n1,n2;M), where Lp=0 is the basic
equation (2.1). Starting from some initial guess p(n1,n2;0) and iterating
up to M=MAX−1 correspond to
solving approximately for the transient solution for this model, from time t=0 to t=T. We verified that choosing T sufficiently
large leads to the same results as the MAPLE solution of (2.1).Since the asymptotic results are expansions in powers
of1/λ with
coefficients expressed in terms of (γ,μ,ω), we input the five parameters (C,R,γ,μ,ω), calculate λ from (2.8),
then ν and κ from (2.7), and
solve (2.1) numerically. In Table 1, we have R=2, γ=1, μ=1, and ω=1, and we compare the exact (numerical) results for B1 and B2 with the one-
and two-term asymptotic approximations. We give various values of C and also
tabulate the corresponding values of λ, as computed from (2.8). We see that the one-term
approximations always overestimate the true values, while the two-term
approximations underestimate them. The two-term approximations are more
accurate especially for the second blocking probability B2 and for larger
values of C. In Table 2, we have ω=0, and in Table 3, ω=−1, with the other parameter values unchanged. With
decreasing ω (which
corresponds to increasing the total load (cf. (2.7)), we get similar results,
but the overall asymptotics (both one- and two-term) are getting somewhat
worse. Also note that the two-term approximations may sometimes lead to
negative answers, and this is explained in what follows.Table 1
R=2, γ=1, μ=1, ω=1.
C(λ)B1B2Exactasy-1asy-2Exactasy-1asy-25(2.25).09741.3504<0.6293>1.441810(5.13).09906.2320.01921.4881.6961.428915(7.90).09384.1869.04879.4194.5609.387420(10.6).08874.1612.05848.3756.4837.354725(13.3).08432.1440.06203.3442.4321.329130(16).08051.1314.06315.3201.3943.308540(21.3).07432.1138.06262.2849.3416.277350(26.5).06948.1019.06087.2598.3059.254360(31.8).06556.09320.05886.2407.2796.236470(37).06229.08638.05688.2254.2591.2221Table 2
R=2, γ=1, μ=1, ω=0.
C(λ)B1B2Exactasy-1asy-2Exactasy-1asy-25(1.5).03572.2801<0.4381.8404.108210(4).04771.1715<0.3293.5146.240115(6.5).04921.1345<0.2804.4037.234720(9).04860.1143.01515.2500.3431.221025(11.5).04743.1011.02352.2284.3035.208030(14).04613.09170.02791.2120.2751.196640(19).04361.07871.03171.1882.2361.178350(24).04140.07004.03283.1714.2101.164360(29).03948.06371.03292.1586.1911.153270(34).03781.05884.03258.1484.1765.1442Table 3
R=2, γ=1, μ=1, ω=−1.
C(λ)B1B2Exactasy-1asy-2Exactasy-1asy-25(1).00990.1911<0.2574.5735<010(3.11).01820.1082<0.1856.3248.0769815(5.34).02083.08270<0.1566.2481.103520(7.61).02170.06926<0.1392.2077.106325(9.92).02189.06068<0.1270.1820.104230(12.2).02179.05462.00421.1178.1638.100840(16.9).02123.04645.00999.1045.1393.0937450(21.6).02053.04106.01257.09516.1232.0875560(26.4).01984.03719.01382.08806.1115.0823370(31.2).01919.03422.01443.08242.1026.07791We next consider a different purely numerical approach
to estimating the coefficients in the expansions of theBj. We choose some C0, and for C=C0−1, C0, and C0+1, we equate(6.1)B1=T1λ+T2λ+T3λ3/2,B2=S1λ+S2λ+S3λ3/2. Note that Bi=Bi(C) and λ=λ(C), for fixed values of (R,γ,μ,ω). Thus, (6.1) may be viewed as 3×3 systems of
linear equations for the Ti and Si, respectively. This allows us to numerically estimate
the first three coefficients in the asymptotic series. In Table 4, we consider C0 in the range of
5 to 70, and give the Ti and Si, fixing (R,γ,μ,ω)=(2,1,1,1). We see that the sequence of T1 and S1 does appear to
converge as C0→∞; the convergence of T2 and S2 is slower, and
that of T3 and S3 is even slower.
The asymptotic results in Sections 3 and 5 show that for these parameter valuesTable 4
R=2, γ=1, μ=1, ω=1.
C0T1T2T3S1S2S310.4875−.7674.38821.556−1.227.467015.4993−.8271.46351.565−1.269.519520.5056−.8649.52071.568−1.291.553325.5096−.8922.56791.571−1.306.579030.5123−.9131.60781.572−1.317.598640.5158−.9431.67231.573−1.329.626050.5180−.9650.72561.575−1.340.653060.5195−.9806.76171.575−1.346.667870.5203−.9899.79481.575−1.340.6508(6.2)B1∼.52574λ+−1.0924λ,B2∼1.5772λ+−1.3717λ(ω=1). This is in good
agreement with Table 4. The data in Table 4 also give a rough estimate of the
third (O(λ−3/2)) terms in the
expansions of the blocking probabilities. These can be computed analytically by
continuing our expansions further, but the calculations are too foreboding.In Tables5 and 6, we again give
the Ti and Si for C0 between 5 and
70, but now with ω=0 and ω=−1, respectively. For these values, our asymptotic
analysis predicts thatTable 5
R=2, γ=1, μ=1, ω=0.
C0T1T2T3S1S2S310.2951−.5575.3160.9891−.8412.361015.3145−.6442.41351.006−.9203.450020.3230−.6911.47831.014−.9621.507725.3277−.7214.52671.018−.9880.549030.3308−.7433.56561.021−1.006.582140.3345−.7727.62491.023−1.029.628450.3366−.7920.66951.025−1.043.661060.3379−.8058.70481.026−1.053.684970.3390−.8186.74091.027−1.069.7309Table 6
R=2, γ=1, μ=1, ω=−1.
C0T1T2T3S1S2S310.1360−.2716.1556.5215−.4576.204115.1585−.3618.2467.5435−.5460.293420.1688−.4138.3122.5533−.5957.356025.1745−.4476.3622.5588−.6279.403430.1781−.4715.4018.5621−.6500.440140.1824−.5034.4622.5660−.6797.496450.1847−.5238.5069.5682−.6987.538160.1862−.5381.5418.5695−.7115.569770.1871−.5478.5677.5701−.7175.5855(6.3)B1∼.34312λ+−.89300λ,B2∼1.0293λ+−1.0983λ(ω=0),B1∼.19119λ+−.61754λ,B2∼.57358λ+−.77272λ(ω=−1). Again this is
in good agreement with the apparent limiting values of T1, T2, S1, and S2 as C0→∞. The data in Tables 4–6 show that the expansions do
indeed appear to be in powers of 1/λ, and that we correctly computed the leading two
terms. Note that in each case, the second coefficient (T2 and S2) is negative,
while the first and third ones are positive. This is consistent with the fact
that in Tables 1–3 the leading terms always overestimate the exact answer,
while the two-term approximations underestimate it. As we decrease ω, the ratio ∣T2/T1∣ increase, as do ∣S2/S1∣ (though these
are always larger). Hence, we expect that decreasing ω leads to
further cancellation between the first and second terms in the asymptotic
series, and this again is in agreement with the data in Tables 1–3. It also
explains why the two-term asymptotic approximations to B1 sometimes lead
to negative answers, for moderate C values.To summarize, we have shown that the asymptotic
approximations are reasonably accurate, though certainly not excellent, and
that there is merit to computing theO(1/λ) correction
terms unless C is quite small.
For small C, however, the one-term approximations may be
superior, as the two-term approximations may lead to negative answers. The
accuracy of the asymptotic approximations presumably increases as C increases
further, and the two-term approximations are presumably better than the
one-term approximations. However, limitations of the available computing
facilities have so far prevented the evaluation of the exact numerical results
for larger values of C.



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