26 Mar
2011
26 Mar
'11
5:21 p.m.
Hello,
On sab, mar 26, 2011 at 04:31:42 +0100, Marek Lindner wrote:
Looking at the calculus above, I'm proposing to use TTL = ~135
Bye!
--
Antonio Quartulli
..each of us alone is worth nothing..
Ernesto "Che" Guevara
Hi,
I think it is wrong: TQ is not decremented by hop_penalty each time, but it is
multiplied by (TQ_MAX-hop_penalty/TQ_MAX), then the number of hops is
bigger than what you said. With hop_penalty = 10 we can probably reach
~135 hops (raw calculus 255*(TQ_MAX-hop_penalty/TQ_MAX)^135) =~ 0)
In these days I was wondering whether the TTL
field in the OGM packet is
really useful or not...Due to hop_penalty an OGM will be discarded as
soon as the TQ will reach 0 (which is a sort of TTL mechanism itself).
At this point why is the TTL field needed?
uhm.., you probably are right. As long as there is a hop penalty it also
functions as a TTL. Keep in mind that it is possible to set the hop penalty to
zero. With the current default hop penalty of 10 the maximum number of hops is
limited to 25. Someone could probably say that a node far at
least TTL hops will never
be reached by a unicast packet, then it is meaningless to let it know
about me. But then it could be possible to recalibrate the TTL such that
it has to be equal to the maximum length in number of hops of the longest
path the OGM can traverse.
Not quite sure what you are proposing. A TTL of TQ_MAX / hop penalty ?
26 Mar
26 Mar
5:45 p.m.
New subject: [B.A.T.M.A.N.] TTL yes, TTL no
I think it is wrong: TQ is not decremented by
hop_penalty each time, but it is
multiplied by (TQ_MAX-hop_penalty/TQ_MAX), then the number of hops is
bigger than what you said. With hop_penalty = 10 we can probably reach
~135 hops (raw calculus 255*(TQ_MAX-hop_penalty/TQ_MAX)^135) =~ 0)
I agree with the principle. However, you need to take rounding into
effect. The kernel does integer arithmetic and TQ is always an integer
value. I also got it wrong in my last post. I used the wrong round
function when calculating the values. Using rounddown() i get:
255
245
235
225
216
207
198
190
182
174
167
160
153
147
141
135
129
123
118
113
108
103
98
94
90
86
82
78
74
71
68
65
62
59
56
53
50
48
46
44
42
40
38
36
34
32
30
28
26
24
23
22
21
20
19
18
17
16
15
14
13
12
11
10
9
8
7
6
5
4
3
2
1
0
After 73 hops, TQ will be zero.
Looking at the calculus above, I'm proposing to
use TTL = ~135
TTL of 73 would be more accurate i think.
Which leads me to a new question. At WBM we talked about moving the
HNAs out of the OGMs. If we had a really big mesh, with more than 73
hops from side to side, do we get into a situation where we have HNAs
from more than 73 hops away, but no idea how to route to them since
OGMs got dropped when TQ reached 0?
Andrew
6:15 p.m.
New subject: [B.A.T.M.A.N.] TTL yes, TTL no
Hi,
On Sat, Mar 26, 2011 at 06:45:48PM +0100, Andrew Lunn wrote:
Which leads me to a new question. At WBM we talked
about moving the
HNAs out of the OGMs. If we had a really big mesh, with more than 73
hops from side to side, do we get into a situation where we have HNAs
from more than 73 hops away, but no idea how to route to them since
OGMs got dropped when TQ reached 0?
I have another question: why do we use a HOP_PENALTY instead of an
additional HOP_COUNT field? In this way the OGM is spread as long as
possible. Is there any problem in this concept I'm not seeing?
Bye!
--
Antonio Quartulli
..each of us alone is worth nothing..
Ernesto "Che" Guevara
4449
days inactive
4449
days old
b.a.t.m.a.n@lists.open-mesh.org
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Andrew Lunn -
Antonio Quartulli