### **Correction** to today's discussion

by Katerina Argyraki -
Number of replies: 0

We discussed the scenario where one end-system sends to the other one packet of size L, then a second packet of size 0.1L over three links, of rates 2R, R and 4R.

Somebody (I think it was Noémie or Inès) asked whether it would be correct to consider this scenario equivalent to the situation where we have 11 packets of size 0.1L.

I said it would be OK for computing the transfer time, but actually it would NOT be OK.

If we have 11 packets of size 0.1L, then transfer time is (according to the formula):

0.1L/2R + 1.1L/R + 0.1L/4R + prop. delays.

If we have 1 packet of size L and 1 of size 0.1L, then transfer time is (as we saw in class):

L/2R + L/R + L/4R + 0.1L/4R + prop. delays.

So, the outcome is quite different.

The reason is that one big packet introduces longer delays than many small packets, because each store-and-forward packet switch must receive all the bits of this one big packet before it starts transmitting them.

Sorry for the confusion.