RE: TCP Expected Throughput
<Hakim.Badis <at> lri.fr>
2005-09-02 21:17:17 GMT
> If you have a MANET, expected throughput will be considerably less than
> the channel capacity. How much less will depend on several factors.
> Here are two:
> 1. Retransmissions for forwarding. Suppose you have A B C D and A is
> sending a message consisting of two packets to D via B and C. First, A
> sends the first packet to B. Then B sends it to C. During this time,
> A will sense B's carrier and not send the second packet. Then, C sends
> to D. Even if A does not sense C's signal, any attempt to send the
> second packet would fail, due to a data collision at B. Hence, A will
> not be able to send the second packet for about three times the nominal
> transmission time for the first packet. Also, while this is going on,
> other nodes in the vicinity won't be able to use the channel, either.
> Depending on traffic patterns, node density and node configuration, this
> could easily cut your effective channel capacity by two thirds.
> 2. Retransmission for errors. In a MANET, the effective range is the
> result of a stochastic balance between the likelyhood of success
> transmittion and the probability of selecting a node during a search.
> Because of fine-grained variations in signal strength due to fading, a
> packet may be lost at any distance. However, as distance increases, the
> probability of having to retransmit a randomly dropped packet increases.
> If the maximum number of MAC retries is fairly high, the protocol is
> likely to stick with a poor link, compensating by retransmitting
> packets, rather than switching to a better path. If this is the case,
> you could lose a significant fraction of your capacity due to
> retransmission of dropped packets.
> In short, the throughput is what you get, depending on a number of
> factors. I would say that 500 Kb/s is not a surprising figure, though
> it might be possible to do a bit better.
If you see the work of Gupta and kumar, as the number of nodes per unit
area increases, the throughput per source-to-destination (SD) pair
decreases approximately like 1/sqrt(n). This is the best performance
achievable even allowing for optimal scheduling, routing, and relaying of
packets in the networks and is a somewhat pessimistic result. The number
of hops in a typical route is of order sqrt(n). So, 500Kb/s is conform to
the model. But this is not a rule, throughput depends on many factors:
interference, buffer size, traffic patterns, node density , etc. and it is
is difficult to determine the realy throughput.
> John P. Mullen, Ph.D.
> (505) 646-2958
> jomullen <at> nmsu.edu
> From: manet-bounces <at> ietf.org [mailto:manet-bounces <at> ietf.org] On Behalf
> Of Hasan Holandi
> Sent: Friday, September 02, 2005 1:51 PM
> To: manet <at> ietf.org
> Subject: [manet] TCP Expected Throughput
> Dear All,
> I have a very trivial (but important to me) question:
> Basically, I would like to know when we have a 4 hop ad hoc network with
> a 2 Mb/s channel, what is the expected throughput?? Will that be also
> 2Mb/s (which of course in reality would be much less due to MAC
> contention) or it would be 500 Kb/s?? In other words if i have file size
> of 16Mb, theoretically will it take 8 sec or 32 sec??
> I would appreciate if can clarify this issue with your response.
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