Algorithm Design

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Epilogue: Algorithms That Run Forever
Now, if we want to maintain nonnegative slacks over time, then we need
to worry about counterbalancing events that cause Slack(p) to decrease. Let’s
return to the description of a single time step and think about how such
decreases can occur.
One step under sped-up input/output queueing:
Packets ~rrive on input li~ks snd are placed in input buffers
The switch moves a matching
At most one packet departs from each output buffer
The switch moves a matching
Consider a given packet p that is unprocessed at the beginning of a time
step. In the arrival phase of the step, IC(p) could increase by 1 if the arriving
packet is placed in the input buffer ahead of p. This would cause Slack(p)
to decrease by 1. In the departure phase of the step, OC(p) could decrease
by !, since a packet with an earlier time to leave will no longer be in the
output buffer. This too would cause Slack(p) to decrease by 1. So, in summary,
Slack(p) can potentially decrease by 1 in each of the arrival and departure
phases. Consequently, we will be able to maintain property (i) if we can
guarantee that Slack(p) increases by at least 1 each time the switch mov~s
a matching. How can we do this?
If the matching to be moved includes a packet in I[p] that is ahead of p, then
IC(p) will decrease and hence Slack(p) will increase. If the matching includes
a packet destined for O[p] with an earlier time to leave than p, then OC(p) and
Slack(p) will increase. So the only problem is if neither of these things happens.
Figure E.B gives a schematic picture of such a situation. Suppose that packet
x is moved out of I[p] even though it is farther back in order, and packet y
is moved to O[p] even though it has a later timeto leave. In this situation, it
seems that buffers I[p] and O[p] have both been treated "unfairly": It would
have been better for I[p] to send a packet like p that Was farther forward, and
it would have been better for O[p] to receive a packet like p that had an earlier
time to leave. Taken together, the two buffers form something reminiscent of
an instability from the Stable Matching Problem.
In fact, we can make this precise, and it provides the key to finishing the
algorithm. Suppose we say that output buffer 0 prefers input buffer I to I’
if the earliest time to leave among packets of type (/, O) is smaller than the
earliest time to leave among packets of type (I’, 0). (In other words, buffer I
is more in need of sending something to buffer 0.) Further, we say that input
buffer I prefers output buffer 0 to output buffer O’ if the forwardmost packet
of type (I, O) comes ahead of the forwardmost packet of type (/, 0’) in the
ordering of I. We construct a preference list for each buffer from these rules;
I[p] (front)
Epilogue: Algorithms That Run Forever
Ix
O[p]
Figure E.3 Choosing a matching to move.
It would be unfair to move~
and ybut not movep.
J
and if there are no packets at all of type (I, 0), then I and 0 are placed at the
end of each other’s preference lists, with ties broken arbitrarily. Finally, we
determine a stable matching M with respect to these preference lists, and the
switch moves this matching M.
~ Analyzing the Algorithm
The following fact establishes that choosing a stable matching will indeed yield
an algorithm with the performance guarantee that we want.
(E.2) Suppose the switch always moves a stable matching M with respect
to the preference lists defined above. (And for each type (I, O) contained in
M, we select the packet of this type with the earliest time to leave). Then, for
all unprocessed packets p, the value Slack(p) increases by at least 1 when the
matching M is moved.
Proof. Consider any unprocessed packet p. Following the discussion above,
suppose that no packet ahead of p in I[p] is moved as part of the matching
M, and no packet destined for O[p] with an earlier time to leave is moved as
part of M. So, in particular, the pair (![p], O[p]) is not inM; suppose that pairs
(F, O[p]) and (![p], 0’) belong to M.
Nowp has an earlier time to leave than any packet of type (/’, O[p]), and it
comes ahead of every packet of type (![p], 0’) in the ordering of I[p]. It follows
that I[p] prefers O[p] to 0’, and O[p] prefers I[p] to I’. Hence the pair (![p], O[p])
forms an instability, which contradicts our assumption that M is stable. ,
Thus, by moving a stable matching in every step, the switch maintains
the property Slack(p) > 0 for all packets p; hence, by (E.!), we have shown
the following.
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