On Indexed Data Broadcast

18 which we denote by (j) and which does not depend on t as desired. Finally we have: TUNE n X j=1 p j max t T(t; j) n X j=1 p j r(j) = n X j=1 p j r(h + 2(q ? 1)(log q d (j) + 1)) n X j=1 p j r(h + 2(q ? 1)(log q 2d j + 1)) n X j=1 p j r(h + 2q(log q d j + 1)) = 2qr(log q n X j=1 p p j + n X j=1 p j log q 1 p p j) + (h + 2q)r 4qr log q n X j=1 p p j + (h + 2q)r where the third inequality follows from the deenition of (j) and the last inequality follows from the convexity of the logarithm function. 2 Acknowledgement We would like to thank Badrinath, Ed Cooman and Peter Winkler for helpful discussions. We are also thankful to an anonymous referee for many valuable comments on an earlier version of this paper. 17 each of length at most r (buckets). Thus we have ACC 1 2 n X j=1 p j ((2d (j) =q + h)r + d (j) + 1) = 1 2 n X j=1 p j ((1 + 2r q)d (j) + hr + 1) 1 2 (1 + 2r q)(n X j=1 p j (d (j) + 1)) + hr + 1 2 = (1 + 2r q)ACC 0 + hr + 1 2 ; where the last equality is by Lemma 6. 2 3.2 Proof of Lemma 11 Let T(t; j), where 1 t N B and j 2 n], be the amount of time units spent by the client listening to the channel from the moment t (in a broadcast cycle) when the request for data item j is made to the time when data item j is received. Then TUNE = 1 N B N B X t=1 n X j=1 p j T(t; j) n X j=1 p j max t T(t; j): So it suuces to give an upper bound on max t T(t; j). Following Proposition 4, it is easy to see that T(t; j) is the number of time units spent on listening to (or equivalently, broadcasting) the nodes in V (t; j). In fact, since the number of time units needed to broadcast any node in the …