[6923] in cryptography@c2.net mail archive
Average packet size (math)
daemon@ATHENA.MIT.EDU (Ron Rivest)
Sun Apr 16 23:10:38 2000
Date: Sun, 16 Apr 2000 22:50:18 -0400 (EDT)
Message-Id: <200004170250.WAA26855@ibis.lcs.mit.edu>
From: Ron Rivest <rivest@theory.lcs.mit.edu>
To: smb@research.att.com
Cc: cryptography@c2.net, mrobshaw@supanet.com
Steve --
To make the argument clearer (since I received an inquiry
about it):
(total work) = (setup cost per packet)*(total number of packets)
+ (encryption cost per byte)(total number of bytes)
for any data stream. Thus:
(work/byte) = (setup cost per packet)*(total number of packets)/(total number of bytes)
+ (encryption cost per byte)
= (setup cost per packet) / (average number of bytes per packet)
+ (encryption cost per byte)
by which we see that it really is the *average* packet size that matters, not
the minimum, mode, or median...
Cheers,
Ron
