Download Solution to Assignment #3

Computer Communications Network (COMP312)
Solution to Assignment Three
(On or before 3:30pm on 9 April, 2004)
1. At least three ports need to be disabled---one for each loop. For example,
LAN segment
Switch connection
LAN switch
2.
At the end of iteration 0
to bridge 0
(0, 0, 0)
to LAN e
to LAN d
(4, 4, 0)
(3, 3, 0)
5
(5, 5, 0)
(5, 5, 0)
to LAN g
to LAN f
At the end of iteration 1
to bridge 0
(0, 0, 0)
to LAN e
to LAN d
(5, 0, 1)
(3, 0, 1)
5
(5, 0, 1)
(5, 0, 1)
to LAN g
to LAN f
At the end of iteration 2: Same as that for the end of iteration1.
1
3.
a) Both B2 and B5 can choose either root bridge. For example, B5 chooses the original one and
B2 chooses the wrong one:
B1
A
B5
B
B1
C
B5
B1
D
B2
B7
K
B7
B1
E
F
B1
B1
B1 B1
B1
G
B1
H
Root bridge
B6
B4
B4
I
J
B4
Root ports
Designated
bridges
b) When B5 chooses the wrong root tree (B2), the tree is not partitioned. Otherwise, LANs A
and C are partitioned from the rest of the network.
4. No. The subnet mask can be 255.255.254.0 or 255.255.252.0. Reasons:
 From the highest-order byte of the IP address, we know that this is a class B address.
Therefore the first two bytes represent the network address.
 From the broadcast address, it is easy to see that the last byte must be host ID.
The remaining question is: which bits in the 3rd byte are used for subnet? By comparing the
broadcast address (00001011) and the IP address (00001010), it is possible that either the leftmost
6 bits or the leftmost 7 bits are used for subnet.
5. XY and YX are OK, because they perceive each other on the same network. YZ and ZY
are also OK for the same reason. XZ is OK but ZX is not OK, because Z thought that X is on
a different network and there is no default router.
6.
a. Host 2 will reassemble the fragments.
b. Router 2 will reassemble the fragments.
c. Host 2 will reassemble the fragments.
In the regular IP forwarding, the destinations are supposed to reassemble the fragments, because
fragments generally travel different paths. Thus host 2 is responsible for the reassembly in cases
(a) and (c). However, in case (b) the destination is really router 2. Thus router 2 is responsible for
the reassembly.
7.
a. Unreliable broadcast. It would be very costly to ensure every broadcast datagram is received
correctly.
b. Yes, B will receive the datagram because the IP module will be able to match the destination
IP address with its own. On the other hand, all other hosts, although received the datagram
from the data link module, will discard the datagram after checking with their IP addresses.
2
c
i. R receives the datagram and does not forward it.
ii. R receives the datagram and performs a data link broadcast on 142.252.2.0.
iii. R uses the normal IP unicast routing to forward the datagram (the host part is ignored in
the forwarding).
iv. R uses the normal IP unicast routing to forward the datagram (the host part is ignored in
the forwarding).
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