194 12. Protocols for All-Optical Networks
Proof.
Let A be the delay range of the round, b-~rther let us denote the worm
traversing path j by
wj,
and its delay by 6j. Clearly, there are BA ways to
choose a wavelength and a delay for Wl. In the following we show that for
any delay 51 of worm
wi,
there are at least L~.__!1 ways to assign a delay 5i+1
to worm
Wi+l
such that
wi+l
blocks
wi.
According to the construction of the type-1 structure,
wi+l
starts/b~-~J +
1 levels after
wi.
Hence, if 6i+t <_ 5i + [~_!j then
wi+x
is at least one level
ahead of
wi
during the routing. On the other hand, if 5i+1 _> 6i - /b--~J
then
wi+l
is at most L - 1 levels ahead of
wi
during the routing. Since
l{6i - L~-~J,...,& + LL~J} n [A]I > [~J + 1 > ~ for A _> L, the
number of ways to assign delays to
Wi+l
such that
wi
is blocked by
wi+l
is
at least L-1
2
Thus altogether there are at least
BA(b_~)i
ways to choose delays and
wavelengths for the worms such that the worms traversing the first i paths
are discarded. Hence this happens with a probability of at least
BA(h~-!)i _ (L_I~ i
(BA)i+I \ 2BA ]
D
Consider now the situation that it takes t + 1 rounds to route the worms
traversing the first t + 1 paths in a type-1 structure. This could happen, e.g.,
if in round i only
wt-i+2
is able to reach its destination, and the worms
wl,... ,wt-i+l
are discarded. According to the lemma above, for L > 2 the
probability of such an event is at least
'( L-i
H (B(A i +
L))t-i+2 = H \ 2BF,5~. ~ L) ] ' (12.4)
i=1 /=1
where A~ > 1 is the delay range for round i. Clearly, the number of time
steps necessary for the t rounds is at least
12(~ti=l (Ai + D + L)).
Given a
fixed A =
~ti_ 1 Ai,
the product in (12.4) gets minimal if Ai + L = ( t - i +
1)(A + t. L) / (~-~1) for all i E { 1,..., t}. This is shown in the following lemma.
Lernma 12.4.6.
Consider xl,...,xn E R+ with y = ~in=lxi. Then, for
n n+l
every a E
[O,y], YIi=l(Xi q-or) ~
gets maximal if xi + a = i(y + n.a)/( 2 )
for all i E
{1,...,n}.
Proof.
For n = 1, the lemma is trivially true. We will show by complete
induction on n that the assumption above is also true for n > 1. Suppose
= n ( 2 )) is the
that we have already shown that
f(y,n) 1-Ii=l(i(y + na)/n+l i
maximal value the product of the
(xi + (~)i
can reach. Then we want to find
the
Xn+l + a
for which
f(y -
Xn+l, n)- (Xn+l + a) n+l gets maximal. Clearly,
f(y -
Xn+l,
n) .
(Xn+l +
(9/) nq-1 ~---
[,~+I~
q'- or)n+1
(Y -
Xn+l + '. (Xn+l 9 g(n) , 02.5)