[mrtg] MRTG and bandwidth delay product
Lyle Giese
lyle at lcrcomputer.net
Wed Dec 31 14:49:29 CET 2008
In order to meet the criteria for their formula IMHO, you need a
scenerio where you have time for at least three TCP packets to leave the
sender before you get the ACK for the first packet back. The higher the
bandwidth on the link, the sooner you hit critical mass.
So I would think you need to work back from this:
Size of TCP packet/speed of line * 3 = time to send three TCP packets.
(You will need to know the size of the TCP packets of course. A sniffer
will help here.)
(speed of line is the speed of the slowest throttle point and is 4kB/s
in your example.)
The 'time to send three TCP packets' would need to be at least less than
1/2 the RTT, IMHO for this to kick in. In other words, if the 'time to
send three TCP packets' is less than 1/2 the RTT, the three packets
could leave the sender before the sender has a chance to receive the
first packet ack from the receiver. Only then would waiting for packet
ack start to slow down the FTP protocol.
This gap between the sending of a packet and the receiption of the ack
is where this limitation kicks in. Now in the old days, Zmodem protocol
was well defined and you knew how many packets would be sent before the
sender would stop sending and wait for an ack. I have not delved into
the TCP packet protocol at this level to determine that parameter. I am
using the number three to help make this explanation understandable.
Plus in your setup, you seem to be limiting FTP to 4kB/s, but are
measuring the bit/bye output of the NIC on the FTP server. The 4kB/s
limitation is set before the overhead bits of the TCP packet is added
while you are measuring the combined bit/byte output of the entire
packet via the NIC.
Lyle
nangineni praneeth wrote:
> Oh!i get it now....Thanks for your explanation.But is there any way
> for me to show that the network throughput is being affected by latency...
>
> Regards
> Venkat
> --- On *Wed, 12/31/08, Lyle Giese /<lyle at lcrcomputer.net>/* wrote:
>
> From: Lyle Giese <lyle at lcrcomputer.net>
> Subject: Re: [mrtg] MRTG and bandwidth delay product
> To: nv_2030 at yahoo.com
> Cc: mrtg at lists.oetiker.ch
> Date: Wednesday, December 31, 2008, 9:34 AM
>
> nangineni praneeth wrote:
>> Thanks for your reply.I studied in various books and web articles
>> that the throughput of a link is limited by Round trip time.
>> Actually the product which i was speaking about earlier is also
>> referred to as TCP receive window size which is the maximum
>> amount of received data, in bytes, that can be buffered at one
>> time on the receiving side of a connection. The sending host can
>> send only that amount of data before waiting for an acknowledgment .
>>
>> The link below explains it in much more detail
>> http://www.speedguide.net/faq_in_q.php?category=89&qid=185
>>
>> I would be glad if anyone could show me a way which would let me
>> prove that by limiting the ftp server to send data at constant
>> rates i am able to operate the network efficiently
>>
>> For eg: I am limiting the server to send traffic at 4kBps hence
>> the effective throughput is xB/s and the effecting throughput of
>> the network is yB/s when the server is sending the traffic at
>> 8kB/s and so on.....
>>
>> Regards
>> Venkat
>>
>>
>> --- On *Wed, 12/31/08, Lyle Giese /<lyle at lcrcomputer.net>/* wrote:
>>
>> From: Lyle Giese <lyle at lcrcomputer.net>
>> Subject: Re: [mrtg] MRTG and bandwidth delay product
>> To: nv_2030 at yahoo.com
>> Cc: mrtg at lists.oetiker.ch
>> Date: Wednesday, December 31, 2008, 8:37 AM
>>
>> nangineni praneeth wrote:
>>> Hello everyone....
>>>
>>> I am working in a lab environment...This is how configured
>>> my network.....
>>>
>>> I designed a enterprise campus network which operates at
>>> 100mbps...The enteprise edge router in the enterprise campus
>>> network connects to the branch office router using a frame
>>> relay connection. The frame relay PVC operates at
>>> 8kB/s...The clients in the branch office connect to the
>>> branch office router with links operating at 100mbps...I
>>> have setup a filezilla ftp server in the enterprise campus
>>> network and limited the speed at which it can send the data
>>> to the clients at 4kB/s....
>>>
>>> So i am calcuting the rtt with MRTG and monitoring network
>>> with MRTG....MRTG calculated the max round trip time to
>>> 340ms..between the client in the branch office and the
>>> server in the enterprise campus network.....I wanted to
>>> theoritically prove that the link cannot carry more than x
>>> bytes of data or in other words because of certain latency
>>> the effective bandwidth of the link is limited to only
>>> xbytes/second...I used bandwidth delay product to do this......
>>>
>>> BDP=0.340*8kBps=2.72kB/s
>>>
>>> So this is saying that the effective throughput of the link
>>> with a latency of 340 ms is 2.72kB/s......But when i am
>>> using MRTG to graph bandwidth the max bandwidth it was
>>> showing was 4kBps which is what we should expect...
>>>
>>> My question is how can i theoritically prove that the
>>> effective throughput of the link is xbytes per second ..I
>>> think i am misunderstanding the bandwidth delay
>>> product.....Can anyone correct me if i am wrong....
>>>
>>> I wish you all a good new year ahead..
>>> Regards
>>> Venkat
>>>
>> Delay has a heavy effect on a stop-start transfer protocol
>> like Xmodem or Ymodem. But for streaming protocols(FTP is a
>> streaming protocol), delay has a minimal effect on throughput.
>>
>> Limiting FTP to 4k over an 8k line, you are going to get
>> close to the 4k max in practice, because bandwidth is not a
>> limiting factor. FTP is a streaming protocol and depends on
>> the TCP protocol to ensure that what is being sent is what
>> will be received and there is no acknowledgement sent back by
>> the sender for each packet(as Xmodem or Ymodem did over
>> dialup) unless there is a transmission error. Since there is
>> no acks required by the sender, there is no delay in waiting
>> for those acks.
>>
>> So I am wondering why you think transmission delay would
>> introduce a reduction in speed, since the sender is not
>> waiting for any responses on a per packet basis from the
>> receiver? I could see that in a protocol like Xmodem, Ymodem
>> or Zmodem where per packet acks needed to be returned to the
>> sender.
>>
>> Lyle Giese
>> LCR Computer Services, Inc.
>>
>>
>>
>>
> Ok, at the TCP packet level is where the acknowledgements are
> done. Now in their article you have to have a enough throughput
> so that the acks being sent back are not getting back fast enough,
> so that the sender has to wait for acknowledgements. You need to
> calculate how long it takes to send a TCP packet over your line.
> You have to enough bandwidth to send three or more packets before
> the acknowledgement for the first packet is received by the sender.
>
> I don't think 8kilobytes/second is fast enough to meet that
> criteria, in order for their formula to engage.
>
> Lyle
>
>
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