> Why a TCP window of about 26K is needed ?
> Can you show me your calculation...
> Thank you very much.
You must be able to transmit for a full round-trip time without exhausting
the window.
384kbps * 560ms = 215kbits = 25900 bytes
(1) 560ms is the roundtrip time I have pretty consistently found for the
router-geostationary orbit-router (and back) round trip
(2) k = 1000 here.
Hans Kruse, Associate Professor and Director
McClure School of Communication Systems Management, Ohio University
9 S. College Street
Athens, OH 45701
614-593-4891 voice, 614-593-4889 fax, [email protected]
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