On Feb 20, 11:21 pm, query.c...@gmail.com wrote:
> Hi ,
>
> I am one of the members who is managing a Wide Area Network . The
> Backbone speed of
> this network is around 1.4 gbps and the links that connect to
> this backbone are of 10 mpbs
> and 100mpbs speed .
> I am managing a LAN which connects to the above WAN with a 10mbs
> link. Some other
> sites are connected with 100 mbps link . Now to gain better
> performance , some TCP
> tuning had to be done at the Host level. So , I was trying to
> calculate the size of the TCP
> window (Bandwidth Delay Product ) which is given by
>
> BDP = bandwidth * round trip time.
>
> My Doubt is regarding this formula . For bandwidth , which bandwidth
> I should consider, the
> speed of host's NIC (100 mbps), or the speed of the link (10 mbps )
> which connects us
> to the WAN or the speed of the Backbone (1.4 gbps).
>
> And for Round Trip Time , I communicate with different links with
> different RTT . e.g With
> one link the RTT is 26.678 ms and with another is 67.786 mps. So
> which RTT should I
> consider.
>
> Please somebody help me out in understanding the formula correctly.


The formula is correct. The bandwidth you need to use is the
*effective* bandwidth of the entire circuit. If that includes a
10Mbps link, it obviously cannot be *larger* than 10Mbps. It can be
smaller due to various forms of overhead (for example, bits dedicated
to framing, the IP and TCP overheads, other traffic sharing the link,
performance limitations of various routers along the way, the
performance of the hosts, etc.).

The RTT again is for the circuit in question. If you're talking to
two hosts, both with no link slower than 10Mbps, and one has a RTT of
25ms and the other 50, the BDP of the first is 31250 bytes (less
overhead) and the second is twice that. OTOH if you have a hundred
users trying to download files over the same link, you can reduce
those BDPs by a factor of 100.