The Last Buer First Serve Priority Policy is
Stable for Stochastic Re{entrant Linesy
Sunil Kumar and P.R. Kumarz
Abstract
In this paper, the stability of the Last Buer First Serve (LBFS) buer priority
policy, in the sense of positive Harris recurrence, is shown under fairly general assumptions on the interarrival and service time distributions, whenever the nominal load at
each machine is less than one. This is done by rst describing the uid limits, and then
showing that the uid limits empty in nite time from all suitable initial conditions,
which Dai [1] has recently shown to be sucient for positive Harris recurrence.
1 Introduction
The Last Buer First Serve (LBFS) buer priority policy is a well known scheduling policy
for re{entrant lines. In this policy, the priority order of buers contending for service at a
machine is the reverse of the order in which they are visited by a part. That is, a buer
receives priority over all contending buers which are upstream of it. The stability of this
policy was rst studied by Lu and Kumar [2], for the case when the interarrival and service
times are deterministic and bursty. They showed that the condition that the nominal load
at each machine be strictly less than one, which is obviously necessary for stability, is also
sucient for stability.
Recently, Dai [1] has proposed a method of establishing positive Harris recurrence, which
is considered \stability" in this paper, for queueing networks using the uid limit models. He
Please address all correspondence to the rst author.
The research reported here has been supported in part by the National Science Foundation Grant Nos.
ECS{90{25007 and ECS{92{16487, and the U. S. Army Research Oce under Contract No. DAAL{03{88{
K{0046.
z Department of Electrical and Computer Engineering, and the Coordinated Science Laboratory, University of Illinois, 1308 West Main Street, Urbana, IL 61801.
y
1
has shown that if the uid limits empty in nite time from every suitable initial condition,
and stay empty, then the underlying stochastic queueing network is stable. He has also
illustrated the use of this technique by proving the stability of the First Buer First Serve
(FBFS) policy. This paper analyzes the uid limits by identifying a sequence of bottlenecks,
and then invokes Dai's results, to establish the stability of the LBFS policy. It also identies
techniques for analyzing the uid limits through the integral equations identied in Dai [1],
which may be useful in further work in the area.
The network we consider consists of S machines f1; 2; : : : ; S g. Parts arrive at buer b1,
located at machine (1) 2 f1; : : : ; S g, with the interarrival times being an i.i.d. sequence
f(n)g. Upon completing service, they proceed to buer b2 located at machine (2) 2
f1; : : : ; S g. Let bL at machine (L) be the last buer visited. The sequence f(1); : : : ; (L)g
is the route of the part. Since one may have (i) = (j ) for some pairs i and j with i 6= j , we
say that the system is a re-entrant line. Let us suppose that parts in bi require i.i.d. service
times fi (n)g, with mean 1=i = mi < 1, from machine (i). Let us also suppose that
E [(1)] = = 1= < 1, and that f(n)g and fi(n)g for i = 1; 2; :::; L are mutually
independent. We assume that the interarrival times are unbounded. That is for any x > 0,
Prob ((1) > x) > 0. Finally, we assume that1 for some integer n and some function p(x) 0
R
on IR+ with 01 p(x)dx > 0,
"
Prob a X
n
i=1
# Z
(i) b b
a
p(x)dx; for any 0 a b:
For each machine 2 f1; 2; : : :; S g, dene
0
= @
X
f1kL; (k)=g
1
mk A :
We call the nominal load for machine .
If buers bj and bk share the same machine, i.e., (j ) = (k), and j < k, then priority is
given to bk . The priority discipline is assumed to be preemptive resume. Therefore if k < L,
bk can be worked on by (k) only if bj = 0 for all j = k + 1; : : : ; L; with (j ) = (k). Note
1
We are grateful to Sean Meyn for pointing out the need for this additional assumption.
2
that bL is never preempted. We also assume that the service discipline is non{idling, that
is, a machine cannot be idle when any of the buers located at that machine is non{empty.
This is the LBFS discipline.
The main result of this paper is the following theorem.
Theorem Under the assumptions stated above, the re{entrant line operating under the
LBFS policy is positive Harris recurrent if
< 1; for all 2 f1; 2; : : : ; S g:
(1)
The rest of the paper is devoted to proving this theorem.
2 The Proof of the Theorem
The main steps of the proof are as follows:
Describe the uid limits of the stochastic system.
Show that the uid limits empty in nite time from every possible initial condition,
and stay empty. This is done as follows:
{ Show that the last buer bL empties at some nite time tL.
{ At tL, identify an upstream bottleneck which caused bL to empty, and show that
all the buers downstream of that bottleneck (including bL) remain empty for all
t > tL.
{ Iterate the argument, at each time identifying a new upstream bottleneck and a
new nite time, and showing that all buers downstream of it remain empty after
that time.
{ Eventually, the iterations terminate, showing that the uid limit is completely
empty at some nite time and remains so thereafter.
3
First we introduce the notation and some preliminaries from Dai [1]. Let Qk (t) denote
the queue length, and vk (t) the residual service time at buer bk at time t. Let u(t) denote
the residual interarrival time at time t. Let E u(t) := maxfn : u + (1) + + (n 1) tg.
Let Dk (t) denote the number of departures from buer bk up to time t. Then the total
number of arrivals to buer bk is given by:
A (t) = Q (0) + E u(t);
1
1
Ak (t) = Qk (0) + Dk (t); k = 2; : : : ; L:
1
Immediately, we have
Qk (t) = Ak (t) Dk (t):
Let Tk(t) be the amount of time in [0; t] that (k) spends on bk , and let Bk (t) =
PL
fj :1j k; and (j )=(k)g Tj (t) be the amount of time spent on buers bk and those with
higher priority by machine (k). Let Ik (t) = t Bk (t) denote the time not spent on
fbj : k j L; and (j ) = (k)g by (k). Because of the non{idling assumption
and the LBFS priority, we have
20
Z 1 6B
4@
1 3
Qj (t)C
A ^ 175 dIk (t) = 0:
X
and (j)=(k)g
The state of the system is given by X (t) := (Q1(t); : : :; QL(t); u(t); v1(t); : : :; vL(t)). For
x = (q1; :::; qL; u; v1; :::; vL) we dene jxj := PLk=1(qk mk + u + vk ). In the sequel, we shall
denote explicit dependence on an initial condition x by adding a superscript x to the variable
of interest. For any function f , let
fx (t) := jx1j f x(jxjt); for t 0 and x > 0;
denote its scaled version.
For simplicity, let us make the additional assumption that all the service and interarrival
time distributions are exponential. This allows us to completely ignore the eects of u and
vk . The state x(t) then consists only of the queue lengths (Q1(t); : : :; QL(t)). This will be
relaxed later.
Next we state the result from Dai [1].
0
fj : kj L
4
Lemma 1 For every sequence fxng with jxnj ! 1, there is a subsequence fxnl g such that
as l ! 1,
D;
A;
Q;
T;
I);
(E xnl ; D xnl ; Axnl ; Q xnl ; Txnl ; Ixnl ) ) (E;
(2)
where \)" denotes uniform convergence on compact sets. Furthermore, the limit processes,
called the uid limits, satisfy:
E (t) = t;
D k (t) = k T(t);
A (t) = q + E (t);
Ak (t) = qk + D k (t); k 2;
Q k (t) = Ak (t) D k (t);
X
Ik(t) = t
1
(3)
(4)
(5)
1
(6)
1
Tj (t);
fj :kj L and (j )=(k)g
0 Tk (t) Ij (t) t for all j > k with (j ) = (k);
0 Tk (t1) Tk (t2) and 0 Ik (t1) Ik(t2) if t1 t2;
0
Z 1B
@
X
1
Q j (t) ^ 1C
A dIk (t) = 0;
(7)
(8)
(9)
(10)
(11)
fj :kj L and (j )=(k)g
jIk(t) Ik (s)j jt sj; jTk(t) Tk (s)j jt sj; and Q k (t) is also Lipschitz; (12)
0
and
L
X
k=1
m k qk 1 :
(13)
Proof. Recall that all distributions are exponential, and u; fvkg are identically zero under
the simplifying assumption. Equation (3) follows from Lemma 4.2 of Dai [1], and (4) follows
from the proof of Corollary 4.2. Equations (5{7) are the queue length variants of equations
(4.18{4.19) of Theorem 4.1 of [1]. Equations (8{11) are obtained trivially from the fact that
these hold for each sample path of the stochastic process, because of the preemptive LBFS
policy. Equation (12) follows from (8,9,10). Equation (13) is a consequence of the scaling
by 1=jxj.
2
Having described a set of necessary conditions (3{13) for the uid limits, we now show
through a sequence of lemmas that these limits empty in nite time, and stay empty, from
5
every initial condition q satisfying (13).
Lemma 2 (i) Q L(tL) = 0 for some nite tL > 0 which depends only on and fk g.
(ii) Let bk be the buer satisfying
k := maxfi < L j
X
fj :ij L
and (j)=(i)g
mj mLg:
(14)
If no such i exists, take k := 0. Then, the buers bk+1; :::; bL are empty at time tL.
Proof. (i) Consider PLk Q k (t). Suppose Q L(s) > 0 for all s 2 (0; t]. Then IL(t) = 0
=1
because of (11), and the fact that IL(0) = 0. So by (8), TL(t) = t. This means that
L
L
X
X
Q k (t) = qk (L )t;
k=1
k=1
by (4{7). By (1) this means that Q L(t) = 0 for some t PLk=1 qk =(L ). From (13),
tL K=(L ), where K depends only on and fk g.
I; T are Lipschitz. If k = L 1, there is nothing to prove as we already
(ii) Recall that Q;
know that Q L(tL) = 0. So we consider the case when k L 2. The proof is by induction.
We note that from (4,6,7), for Q L(tL) = 0, we must have
D L (tL) D L (tL ) L; for all 2 [0; ];
1
1
0
(15)
for suitably small 0 > 0. Now if Q L 1(tL) > 0, then by (4,7), Q L 1(s) > 0, for all
s 2 [tL ; tL] for suitable > 0. By (11), since (k +j ) 6= (L) for j = 1; :::; L k 1, we have
IL 1(tL) IL 1(s) = 0 for all s 2 [tL ; tL]. By (4,8), this implies that D L 1 (tL) D L 1(s) >
L(tL s), since bL 1 satises the condition mL 1 < mL, because k < L 1. This contradicts
(15). So we have Q L 1(tL) = 0, and D L 1 (tL) D L 1 (tL ) L ; for all 2 [0; 0].
Now assume that for some j > k, and all i = j + 1; :::; L, we have Q i(tL) = 0 and
D i (tL) D i (tL ) L ; for all 2 [0; 0] for some 0 > 0 suciently small. Then, we
claim that these two relations also hold for bj . Suppose not. Then either Q j (tL) > 0, or, for
every 0 > 0, there is a < 0 such that D j (tL) D j (tL ) > L. Now if Q j (tL) > 0, then
6
by (4,7), Q j (s) > 0, for all s 2 [tL ; tL] for suitably small > 0. But by the induction
hypothesis, D l(tL) D l (tL ) L for all l > j , < 0 ^ . From (4), this implies that
Tl(tL) Tl(tL ) mlL for all l > j . So
X
fl:j<l<L
and (l)=(j)g
2
h
i
Tl(tL) Tl(tL ) 64
X
fl:j<l<L and (l)=(j )g
Thus, from (8{11), we have
Tj (tL) Tj (tL
2
) 641
X
fl:j<l<L
and (l)=(j)g
3
ml75 L:
3
mlL75 :
So
2
3
X
D j (tL) D j (tL ) j 641
mlL 75 :
fl j<l<L and l j g
hP
i
P
Rewriting fl j<l<L and l j g ml as fl jl<L and l j g ml mj , we have
"
#
L
D j (tL) D j (tL ) L + (1 )j (16)
j
. But by (14), j > L . Thus, we conclude that in
where j := P
m
l
fl jl<L and l j g
either case, for every > 0, there is a < such that D j (tL) D j (tL ) > L. But by
the induction assumption, D j (tL) D j (tL ) L; for all 2 [0; ]. By (6,7), this
implies that Qj (tL) > 0, which is a contradiction. So Qj (tL) = 0 and D j (tL) D j (tL ) L; for all 2 [tL ; tL].
2
:
:
( )= ( )
:
( )= ( )
( )= ( )
1
(
:
( )= ( )
)
0
0
+1
+1
0
+1
0
Lemma 3 Let bk and tL be as in the previous lemma. Then
Q L( + tL) = = Q k ( + tL) = 0; for all 0:
+1
Proof. If k = 0, dene D k (t) := t. Otherwise D k (t) is as in (4). Suppose Q L(t) > 0 for
some t > tL. We know Q L(tL) = 0. So, by (6,7), D L 1 (t) D L 1 (tL) > L(t tL). But
Q L 1(tL) = 0. So D L 2 (t) D L 2(tL) > L (t tL), and so on. Thus, we can conclude that
7
D j (t) D j (tL) > L(t tL), for all j = k; :::; L 1. If k = 0, we immediately obtain a
contradiction. If k > 0, by (4, 8{11), this implies that
2
3
h
i
X
X
Tl(t) Tl(tL) > 64
mj 75 L(t tL):
fl kj L and l j g
fl kj L and l j g
But the right hand side above is at least as large as (t tL), by (14). This is a contradiction.
So Q L(tL + ) = 0, for all > 0.
If k = L 1, there is nothing else to prove. So assume that k < L 1. Assume for
induction that Q L(tL + ) = = Q j (tL + ) = 0, for all > 0, for some j > k. For this,
by (6,7), we require that
D l (t) D l (t ) L;
(17)
for all l = j; :::; L, all t > tL and 0 < < t tL. Now if for some t > tL, Q j (t) > 0. Then
arguing exactly as we did while obtaining equation (16), we get D j (t) D j (t ) > L ;
which is a contradiction to (17). So the proof is complete by induction.
2
:
( )= ( )
:
( )= ( )
+1
Proof of the Theorem In the preceeding lemmas we have identied a time tL at which
the buers bk+1; : : : ; bL are all empty, and remain empty thereafter. If we now show that
there exists a time tk > tL, which depends only on and fi g, such that bk empties and
remains empty thereafter, then we are done by iteration, since there are only L buers. Now
the existence of such a tk follows analogously to Lemma 2 (i). Suppose Q k (s) > 0, for all
s 2 (tL; tL + ]. Then we have, from (8{11),
Tk (tL + ) Tk (tL) = X
h
i
Ti(tL + ) Ti(tL) :
(18)
and (i)=(k)g
But from Lemma 3, we can conclude that D i(tL + ) D i(tL) = D k (tL + ) D k (tL), for all
i = k + 1; :::; L. Using (4) and (18), we get
D k (tL + ) D k (tL) = P
:
fi:kiL and (i)=(k)g mi
Thus we conclude that
0
1
L
L
X
X
1
A :
0 Q l(tL + ) = Q l(tL) + @ P
m
i
l=1
l=1
fi:kiL and (i)=(k)g
fi:k<iL
8
Since the second term on the right hand side above decreases linearly as long as Q k stays
positive, we conclude that there is a time tk at which Q k (tk ) = 0. The fact that tk depends
only on and fi g follows from
L
L
X
X
Q l(tL) < tL + ql:
l
l
since the right hand side above is no larger than K 1 + L by (13) and Lemma 2 (i).
At time tk we next identify a k0 such that
X
X
k0 = maxfj < k j
mi mng:
=1
=1
(
fi:j iL and (i)=(j )g
)
and (n)=(k)g
This is the bottleneck which caused bk to empty. Then we repeat the arguments of Lemmas
3 and 4 to show that bk0+1; : : : ; bk are empty at tk and remain empty thereafter. Iterating
the argument proves the fact the uid limits remain empty after a nite time.
We note that we can relax the assumption that residual service times are zero, i.e., the
exponential assumption, since it is easy to see that when all downstream buers are empty,
a buer emptying at time tk in the exponential case, would empty at time tk + vk in general,
and P vk < 1. Also see Theorem 5.2 of Chen [3].
Now we apply Theorem 4.3 of Dai [1] to conclude the queueing network is positive Harris
recurrent.
2
fn:knL
3 Conclusions
We have shown that the LBFS policy is stable whenever the arrival rate is within capacity,
i.e., (1) is satised. This provides an useful extension of the stability of bursty deterministic
models, established in [2], to the stochastic setting, and an interesting application of the
results of [1], which proves their power. It also exhibits techniques of deducing the behavior
of the uid limits from the integral equations which they satisfy, and hence may be useful
in future work. Finally, we note that Dai and Weiss [4] have simultaneously and independently obtained a proof of the stability of LBFS. We also refer the reader to [4] for other
stability/instability results for some examples of systems.
9
References
[1] J. G. Dai, \On positive Harris recurrence of multiclass queueing networks: A unied
approach via uid limit models," tech. rep., Georgia Institute of Technology, 1993. To
appear in Annals of Applied Probability.
[2] S. H. Lu and P. R. Kumar, \Distributed scheduling based on due dates and buer priorities," IEEE Transactions on Automatic Control, vol. AC-36, pp. 1406{1416, December
1991.
[3] H. Chen, \Fluid approximations and stability of multiclass queueing networks I: Work
conserving disciplines," tech. rep., University of British Columbia, 1993.
[4] J. Dai and G. Weiss, \Stability and instability of uid models for certain re-entrant lines."
Preprint, February 1994.
10