Need help understanding throughput

Please provide more detail about the file copy operation.

FTP or Windows copy ( read NETBIOS over TCP/IP)?

Size of file being copied, how long it take, etc, etc

Depending on the latency of your link you may be hitting window size problems (which will do this type of thing). If you have someone that is comfortable with Linux (to the point of building a custom built kernel)

is a pointer to a tool that with a Linux server on one end of your link and a machine (preferably the machine doing the file transfer) on the other end with a Java enabled web browser will tell you all sorts of interesting things about the link and more importantly make suggestions about buffer changes on the end hosts to get better performance. Its a marvelous tool, it helped us take a highish latency (17 msec) gig link from 35 megabits per second to

995 megabits per second (window scaling and increased tcp buffers in the kernel) on one of our research light paths.

Peter Van Epp / Operations and Technical Support Simon Fraser University, Burnaby, B.C. Canada

First question - how do you measure the throutput? Do you check the interface utilization on the router itself? Or 1200/1400/1800 Kbit/s is the speed, shown at your application? What application do you use? Is it FTP transfer? What is your end-point (where do you transfer to/from)? What Internet connection do they have? Do they limit transfer per-session?

Another aspect - FTP (and any other application) measure throutput at the Application Layer of OSI model (file size divided by time). But real transfer has some additional information - FTP adds TCP header (application information), then IP protocol adds IP header (IP addresses), then Frame Relay or HDLC at your T1 adds it's own information. And only then the frame gets into the wire. Lets say the frame is 1500 bytes. Out of these 1500 bytes "overhead" could be 60-70 bytes.

Then, any TCP connection has some extra packets in addition to normal data transfer (confirmations, retransmissions, out-of-sequence frames, etc.). Thus could be minimal, but counting all this together could give you that

25% of "missing bandwidth".

Good luck,

Mike

Add 'load-interval 30', to start 10-50 sessions simultaneously and then periodically check 'sh int', then see what you get.

Kind regards, iLya

Also TCP works by tentatively increasing the amount of data it tries to send until a packet goes missing, whereupon it backs off and starts increasing again. The missing packet may be dropped by a router somewhere on the path or it may be because an output queue overflowed in the host itself. Thus for a single TCP stream throughput against time is a sawtooth pattern; it can never fill the whole capacity of the network.

Sam

YOu have to be configured into MLPPP between the two interfaces (interface 1 with 24 channels and the second interface with the 14 channel) This creates a virual single interface that will then allow of dynamic use. Otherwise you only have 2 PtP

Michael Stokes, CTO stokesTechnologies, > I need some helping understanding why 25% of the available bandwidth I