Exam 2 of Computer Networks (ICE 1230)
2008.4.1
Answer in English and Total 15 points
1. In our rdt protocols, why did we need to introduce sequence
numbers? Also, why did we need to introduce timers? (1 point)
Sequence numbers are required for a receiver to find out whether an arriving packet
contains new data or is a retransmission.
To handle losses in the channel. If the ACK for a transmitted packet is not received
within the duration of the timer for the packet, the packet (or its ACK or NACK) is
assumed to have been lost. Hence, the packet is retransmitted.
2. Answer true or false to the following questions and briefly
justify your answer:
a. With the SR protocol, it is possible for the sender to receive an
ACK for a packet that falls outside of its current window.
True. Suppose the sender has a window size of 3 and sends packets 1, 2, 3 at t0 .
At t1 (t1 > t0) the receiver ACKS 1, 2, 3. At t2 (t2 > t1) the sender times out and
resends 1, 2, 3. At t3 the receiver receives the duplicates and re-acknowledges 1, 2,
3. At t4 the sender receives the ACKs that the receiver sent at t1 and advances its
window to 4, 5, 6. At t5 the sender receives the ACKs 1, 2, 3 the receiver sent at
t2 .These ACKs are outside its window.
b. With GBN, it is possible for the sender to receive an ACK for a
packet that falls outside of its current window.
True. By essentially the same scenario as in (a).
c. The alternating-bit protocol is the same as the SR protocol
with a sender and receiver window size of 1.
True.
d. The alternating-bit protocol is the same as the GBN protocol
with a sender and receiver window size of 1.
True. Note that with a window size of 1, SR, GBN, and the alternating bit protocol
are functionally equivalent. The window size of 1 precludes the possibility of out1
of-order packets (within the window). A cumulative ACK is just an ordinary ACK
in this situation, since it can only refer to the single packet within the window.
(2 points)
3. Consider the following plot of TCP window size as a function of
time.
Assuming TCP Reno is the protocol experiencing the behavior
shown above, answer the following questions. In all cases, you
should provide a short discussion justifying your answer. (2 points)
a. Identify the intervals of time when TCP slow start is operating.
TCP slowstart is operating in the intervals [1,6] and [23,26]
b. Identify the intervals of time when TCP congestion avoidance
is operating.
TCP congestion advoidance is operating in the intervals [6,16] and [17,22]
c. After the 16th transmission round, is segment loss detected by
a triple duplicate ACK or by a timeout?
After the 16th transmission round, packet loss is recognized by a triple duplicate
ACK. If there was a timeout, the congestion window size would have dropped to
1.
d. After the 22nd transmission round, is segment loss detected by
a triple duplicate ACK or by a timeout?
After the 22nd transmission round, segment loss is detected due to timeout, and
2
hence the congestion window size is set to 1.
e. What is the initial value of Threshold at the first transmission
round?
The threshold is initially 32, since it is at this window size that slowtart stops and
congestion avoidance begins.
f. What is the value of Threshold at the 18th transmission round?
The threshold is set to half the value of the congestion window when packet loss
is detected. When loss is detected during transmission round 16, the congestion
windows size is 42. Hence the threshold is 21 during the 18th transmission round.
g. During what transmission round is the 70th segment sent?
During the 1st transmission round, packet 1 is sent; packet 2-3 are sent in the 2nd
transmission round; packets 4-7 are sent in the 3rd transmission round; packets 815 are sent in the 4th transmission round; packets15-31 are sent in the 5th
transmission round; packets 32-63 are sent in the 6th transmission round; packets
64 – 96 are sent in the 7th transmission round. Thus packet 70 is sent in the 7th
transmission round.
h. Assuming a packet loss is detected after the 26th round by the
receipt of a triple duplicated ACK, what will be the values of
the congestion control window size and of Threshold?
The congestion window and threshold will be set to half the current value of the
congestion window (8) when the loss occurred. Thus the new values of the
threshold and window will be 4.
4. Do the routers in both datagram network and virtual-circuit
networks use forwarding tables? If so, describe the forwarding
tables for both classes of networks. (1 point)
Yes, both use forwarding tables. For a VC forwarding table, the columns are :
Incoming Interface, Incoming VC Number, Outgoing Interface, Outgoing VC
Number. For a datagram forwarding table, the columns are: Destination Address,
Outgoing Interface.
5. Consider a datagram network using 8-bit host addresses.
Suppose a router uses longest prefix matching and has the
following forwarding table:
3
Prefix Match
Interface
1
0
11
1
111
2
otherwise
3
For each of the four interfaces, give the associated range of
destination host addresses and the number of addresses in the
range. (1 Point)
Destination Address Range
10000000
through (64 addresses)
10111111
11000000
through(32 addresses)
11011111
11100000
through (32 addresses)
11111111
00000000
through (128 addresses)
Link Interface
0
1
2
3
01111111
6. We saw that there is no network-layer protocol that can be used
to identify the hosts participating in a multicast group. Given this,
how can multicast applications learn the identities of the hosts that
are participating in a multicast group?
(1 point)
The protocol must be built at the application layer. For example, an application may
periodically multicast its identity to all other group members in an application-layer
message.
7. In these days, the Internet needs to include the wireless
communications network in addition to the wired networks. What is
a way to extend the congestion control of TCP in this case? Why?
4
(2 point)
Packet losses occur from network congestion and wireless communication.
Since the wired Internet does not consider the packet loss due to errors in
wireless communication, the congestion control of TCP should be revised. A
way is …
8. The most obvious technique for achieving broadcast is a flooding
approach. What is it? What is a fatal flaw? What could be remedies
(at least two)?
The endless multiplication of broadcast packets result in broadcast storm.
Sequence-number-controlled flooding and reverse path forwarding could be
remedies.
(2 point)
9. Describe the overall steps of Link State Routing algorithm. And
show the technical issues (at least two) and their solutions of its
implementation as Open Shortest Path First.
(2 points)
1.
discover its neighbors and learn their network address.
2. measure the delay or cost to each of its neighbors.
3. construct a packet telling all it has just learned.
4. send this packet to all other routers.
5. compute the shortest path to every other router.
For example, OSPF has an architecture consisting of areas where the link
state packets are flooding inside. Sequence number is with age number in
order to solve …
5