Linux and Latency

Guenter Geiger geiger at
Sat Aug 21 14:58:30 CEST 1999

reine at writes:
 > Here are my guesses:
 > The 2 in (2<<16) refers to number of fragments.
 > The 14 refters to size of fragments. 14 bits is 16384. 
 > (correct me if im wrong).

 this is exactely right. 
 The  parameter is 
 fragmentnumber<<16 + log2(fragmentsize)

 > It seems like if I stay on a total of 32768 (=2*16384)
 > I can use a number of variations like svar=(4<<16) | (13);

 This is funny behaviour and will get you rather high latency
 (assuming that you can't use the GETOSPACE calls either)

 BTW, what driver are you using ?


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