The IP datagram for a TCP ACK message is 40 bytes long: it contains 20 bytes of TCP header and 20 bytes of IP header. Assume that this ACK is traversing an ATM network that uses AAAL5 to encapsulate IP packets. How many ATM packets will it take to carry the ACK? What if AAL3/4 is used instead?
Thanks for your time!
2007-01-13 04:59:14 · 1 answers · asked by Teddy 1 in Computers & Internet ➔ Computer Networking
First, this sounds like a certification test question. ATM is largely dead in networking today. And AAL3/4 truly are dead.
Second, ATM uses cells, not packets.
That said, the AAL5 overhead is 24 bytes on a 40 byte packet. Since 40+24=64 and 64 > 53, you end up with 2 ATM 53 byte cells to support that single IP packet.
AAL3/4 overhead is only 4 bytes, so the 40 byte packet fits in a single cell.
2007-01-15 11:35:42 · answer #1 · answered by tony1athome 5 · 0⤊ 0⤋