I’m working through questions from computer networking top down approach and was hoping to get some clarity on the end-to-end delay calculation. At this stage we are not considering processing, transmission or propagation delay.
N = link
L = packet length
R = transmission rate
The end-to-end Delay is N*L/R
(N
links in series, using store-and-forward between the links).
The author asks to generalize the above formula for sending P packets over N links.
The solution I derived is (N*L/R)*P
.
Is this a reasonable solution?
PS. This is not home working.
Best Answer
So, this assumes that all packets are the same length and all links are the same bandwidth.
I think a mistake you are making is that your formula looks at the end-to-end delay for a single packet, then multiplies it by the number of packets. In this case, the first packet would be sent end-to-end, then the second packet would be send end-to-end etc.
In reality, many packets can be in transit within the network at any one time. If the only delays we are considering are serialisation, then the following would happen:
So really, you need to look at the time taken to send all packets across the first link, one-by-one, then look at the time taken for the last packet to traverse the remaining links.
So if you consider (L/R) as the time taken to transport one packet across one link.
Then consider there are P of these delays to transport all packets over the first link, one-by-one
And there are (N-1) of these delays to transport the last packet over the remaining links
Then the formula for this would be:
Note: this formula is very simplistic and assumes there are no propagation delays, node processing delays or queuing delays etc.