Ethereal-users: [Ethereal-users] sequence/acknowledgement number question

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Date: Thu, 11 Aug 2005 10:03:57 -0400
I have a question regarding how tcp sequence numbers are calculated and/or 
acknowledged.  I reviewed the RFC, Steven's, and other resources and what 
I read does not jive with what I see displayed or what happens in the read 
world.  What I read is the acknowledgement number is one greater than the 
sequence number.   This seems to only hold true if there is zero bytes of 
data sent.  In example one I see the acknowledgement number 3322332584 for 
the sequence number 3322332583:3322332583(0)

Example 1:
15:45:13.141102 IP (tos 0x0, ttl  60, id 22513, offset 0, flags [none], 
length:60) 10.10.10.241.37679 > 10.10.10.145.22: S 
3322332583:3322332583(0) win 65535 <mss 1460,nop,wscale 
4,nop,nop,timestamp[|tcp]>
15:45:13.141468 IP (tos 0x0, ttl  60, id 63343, offset 0, flags [none], 
length:60) 10.10.10.145.22 > 10.10.10.241.37679: S 
3524895916:3524895916(0) ack 3322332584 win 17376 <mss 1460,nop,wscale 
0,nop,nop,timestamp[|tcp]>

Now example two seems to blow the acknowledgement numbers as I read about 
them.  The sequence numbers are 3322333792:3322334032, I would expect the 
acknowledgement number to be 3322334033 not 3322334032.  Why is the ack 
3322334032 and not 3322334033.  I have been pulling what's left of my hair 
out. I have asked a Track 3 alumni and the two network dudes and they were 
equally perplexed as I was when I showed them this data.

Example 2:
15:45:14.356295 IP (tos 0x0, ttl  60, id 22570, offset 0, flags [none], 
length:292) 10.10.10.241.37679 > 10.10.10.145.22: P 
3322333792:3322334032(240) ack 354898565 win 32761 <nop,nop,timestamp 
1125902767 1122933152>
15:45:14.359383 IP (tos 0x0, ttl  60, id 63509, offset 0, flags [none], 
length:244) 10.10.10.145.22 > 10.10.10.241.37679: P 
3524898565:3524898757(192) ack 3322334032 win 17376 <nop,nop,timestamp 
1122933152 1125902767>

I would really appreciate if somebody could explain this too me.  This has 
been eating away at me and I am assuming it is something very simple.

Regards,
Jim


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