full utilization of bandwidth and its physical significance

Question

I was reading the following extract:

A LFN is a network that offers a high bandwidth but also a very high delay. An example could be a satellite connection. These connections offer a high bandwidth but the delay is also quite high since you have to send your signal 22000 miles up to the satellite and another 22000 miles down to reach the receiver. You can expect a round trip time anywhere between about 500-1000 ms.

The problem here is that when the sender sends some data, it has to be wait a very long time for an acknowledgment of the receiver before it can send the next data. During the time we are waiting, nothing happens so we don’t utilize the full bandwidth of our link.

Given that we don't utilize full bandwidth of our link when the sender has to wait for a long time for an acknowledgement, what do we actually mean when we say this?

Does full utilization of bandwidth means that the sender needs to pump the data continuously? Also, let me know if the data transfer rate of the sender bears any relation to the bandwidth of the link?

From what I have perceived, data transfer rate of the sender is independent of bandwidth of the link as long as it is less than the latter. Am I correct?

Answer 1

Suppose this scenario:

The sender and the receiver share a 1 Mbps bandwith media.

If the sender needs to send a file of 2 Mbit, in an ideal world it will take 2 seconds using the full bandwith.

But in the real world we have two problems:

• The 2 Mbit file must be divided in chunks because the sending and receiving buffers are limited and because it gives you a more detailed control of errors.

• The media has a propagation delay from end to end.

Let's make each chunk of 25000 bits long, giving us a total of 800 chunks.

Let's suppose the media has a propagation delay of 10 milliseconds.

Start-Stop protocol:

Protocols using a start-stop technique send each chunk and stops until an acknowledge returns, in that case and supossing zero errors, to send each chunk of 2500 bits takes 20 milliseconds and give us a bandwith use of 125 Kbps. In that fashion it will take 16 seconds to transmit the file.

Sliding Window protocol:

A sliding window technique allows to send chunks one after another and receive acknowledgments in an asynchronous way.

Ssupposing a window of 10 then the sender put 10 chunks on the media. It takes 100 milliseconds and before sending the next one, the acknowledge for the first chunk must have arrived. If everything is right and there are zero errors, it will be 110 milliseconds for 25000 bits giving us a use of 227 Kbps ( more efficient that the first case). It will take 8,8 seconds to send the file.

A bigger chunk (5000 bits) can give us 455 Kbps and a transfer time of 4,4 seconds.

All this supossing zero errors.

The bandwith is a limit but the data transfer is a result of the protocol used, the propagation delay and the errors in the media.

Answer 2

You might want to give RFC 2488 a look. It addresses your question and offers recommendations to alleviate the effects on satellite links.

https://www.rfc-editor.org/rfc/rfc2488 Enhancing TCP Over Satellite Channels using Standard Mechanisms