Subject: Rule of Thumb for Ad Hoc Channel Capacity

From: Chip Elliott <[email-masked]>
Date: Wed, 21 Nov 2001 15:44:08 -0500

Hi MANET scientists,

This is in regards the results documented in the
fairly recent paper by Gupta, Gray, & Kumar:
"An Experimental Scaling Law for Ad Hoc Networks",
and also recent FCS tests in New Jersey.

I had been surprised by the numbers on the Y axis
of the graphs at the end of the paper, mainly because
I was reading the graphs wrong. They give the per-node
throughput (as clearly stated in their captions), and thus
must be multiplied by the # of nodes to give the useful
channel capacity, assuming little or no spatial reuse.

A few BBN friends and I reverse-engineered the numbers
this way, and believe they're consistent with a simple rule
of thumb for estimating channel capacity in ad hoc wireless
networks. This rule of thumb is based on results from the
SURAN days of packet radio, ie, in the 1980s:

* To estimate the available local channel capacity in a
    multihop packet radio network with 802.11 like channel
    access, divide the "raw" channel bandwidth by 4.

For instance, a 2 Mbps 802.11 radio network will, if everything
is implemented right, give about 500 Kbps total bandwidth
as seen for UDP traffic flows through the network.

The rationale is as follows. Consider a multi-hop scenario
shown below, where each of the letters indicates an ad hoc
radio node.

       A ---> B ---> C ---> D

When A transmits to B, everything in some "radius" around A must
be silent for the duration - it can neither transmit nor receive. Ditto
when B sends to C, and when C sends to D.

Now let's follow a packet from A to D. First A transmits. Then B
transmits (and A must be silent). Then C transmits, and B is
forbidden to receive. Finally, when C's transmission is complete,
A may finally send the next packet to B. Thus a given traffic
flow can only use 1/3 of the raw channel bandwidth.

In most cases, this won't be the only traffic in the network.
In fact, there will be contention for the channel. Since 802.11 is
at first cut somewhat like Ethernet, one can guestimate about 70%
utilization of the channel when there's contention.

Taken all together, this means that maximal good throughput
in some RF region will likely be about 1/4 of the available
channel bandwidth.

Any comments on this (overly?) simple rule of thumb? It is not
a careful approximation, and certainly will be wrong for a number
of cases, but I personally think it's an OK going-in position for
estimating channel capacity. But comments & better thoughts would be
much appreciated.

Cheers,

-- Chip

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From: Yu-Ching Hsu <[email-masked]>
Date: Thu, 22 Nov 2001 15:16:30 +0800 (CST)

Hi,
  In the A---> B---> C---> D example, it sounds reasonable to divide the
"raw" channel bandwidth by 4. However, this example considers
one-dimensioned configuration. When considering 2-dimensioned cases, while
C sending packets to D, A can receives/sends packets from/to other nodes
except B. Then it would be too rough to divide the bandwidth by 4.

Yu-Ching

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From: Lichun Bao <[email-masked]>
Date: Tue, 27 Nov 2001 12:05:25 -0800 (PST)

I think the rule is an over-simplified case of channel capacity
estimation. The factors determining the capacity available to individual
nodes include:

1. Network load, both overall and per node;
2. Contention level (number of neighbors having packet to transmit);
3. The MAC protocol (backoff manner in case 802.11);
4. Network topology;
5. Packet size.

Lichun