COMPUTER NETWORKS
CONE Lab - Link - Solutions to numerical exercises
Solution
to Problem 12 (by J.F. Kurose and K.W. Ross)
a), b), c) See Figure below.
d)
- Forwarding
table in A determines that the datagram should be routed to interface
111.111.111.002. - Host A uses
ARP to determine the LAN address for 111.111.111.002, namely
22-22-22-22-22. - The adapter
in A creates and Ethernet packet with Ethernet destination address
22-22-22-22-22-22. - The first
router receives the packet and extracts the datagram. The forwarding
table in this router indicates that the datagram is to be routed to 122.222.222.003. - The first
router then uses ARP to obtain the associated Ethernet address, namely
55-55-55-55-55-55. - The process continues until the packet has reached Host F.
e)
ARP in A must now determine the LAN address of 111.111.111.002. Host A
sends out
an ARP query packet within a broadcast Ethernet frame. The first router
receives the
query packet and sends to Host A an ARP response packet. This ARP response
packet is
carried by an Ethernet frame with Ethernet destination address 00-00-00-00-00-00.
Solution
to Problem 13 (by J.F. Kurose and K.W. Ross)
Wait for 51,200 bit times. For 10 Mbps, this wait is
51.2
x 103 bits : 10 x 106 bps
= 5.12 msec
For 100 Mbps, the wait is 512 µsec.
Solution
to Problem 14 (by J.F. Kurose and K.W. Ross)
At t = 0 A transmits. At t
= 576, A would finish transmitting. In
the worst case, B
begins transmitting at time t = 224. At time t = 224 + 225
= 449 B's first bit arrives at
A . Because 449 < 576, A aborts
before completing the transmission of the packet, as it
is supposed to do.
Thus A cannot finish transmitting before it
detects that B transmitted. This implies that
if
A does not detect the presence of a host, then
no other host begins transmitting while A
is transmitting.