Re: bandwith measurement

[Brian Candler <B.Candler at pobox.com>]

On Tue, Jul 03, 2001 at 06:36:54PM +0400, Zeimm Auladin wrote:
> I installed pathchar and you need to type ./pathchar <ipaddress> or the 
> <hostname> you want to measure the bandwidth to.
> 
> I tested it and from a 100Mb to a 100Mb port on the same Layer 4 switch I 
> got a 23Mb/s bottleneck. Is it because the algorithm is for measuring links 
> between routers? It's been fairly accurate on the Internet bandwith however.

Pathchar's algorithm is flawed, especially for links which involve layer 2
hops (i.e. switches) since they introduce additional latency which it
doesn't expect. It may be useful for networks consisting of serial links
though.

There is a method I use for testing the bandwidth of links, just using ping.
It is best to do it on the link when there is as little traffic as possible,
although it does work on a loaded link.

(1) ping -c100 -s100 <far-end-IP-address>

This sends a hundred 128-byte pings. Looks at the stats and choose the
LOWEST round-trip time. Let's call this T1 milliseconds. If the link is
loaded then you will see a big variation in round trip times, but once in a
while your packet will sneak through first which is why you choose the
lowest RTT.

(2) ping -c1400 -s100 <far-end-IP-address>

This sends a hundred 1428-byte pings. Look at the stats and again choose the
LOWEST round-trip time. Let's call this T2 milliseconds.

Now, (T2-T1) is the round-trip time for 1300 bytes, and this subtraction
removes both framing overhead and distance-related latency (e.g. satellite
links).

So, assuming the bandwidth in both directions is the same, the bandwidth in
bits per second is:

           1300 x 8
         --------------
         (T2-T1)/1000/2

Example using my DSL link at home:

T2 = 86.0ms
T1 = 25.1ms

Bandwidth = (1300*8)/((86-25.1)/1000/2) = 341,500 bits per second.

(It's actually asymmetric, 1Mbps downstream and 256K upstream)

Regards,

Brian.

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