Download test 1 - answer

EKT355 ADVANCED COMPUTER NETWORK
TEST I
DATE
: 29 MARCH 2013
DURATION : 1 HOUR
NAME
:
MATRIC NO
:
Instructions:
1. This is a close book test. No books and notes are allowed to be referred to during the
test.
2. Answer ALL questions. Please write your answer in the column provided.
3. Duration : 1.0 Hour.
4. Number of pages (including this page) : 6
TOTAL MARKS / 40
1
1. One of the services provided by the Link Layer is the reliable delivery service. Compare
the Link Layer reliable delivery service and the Transport Layer reliable delivery service.
(Explain in terms of how the reliable delivery service can be achieved in Link Layer as compared
to in Transport Layer, the advantage and disadvantage of performing reliable delivery service in
Link Layer instead of in Transport Layer.)
[4 marks]
A:
Similar to transport layer reliable delivery service such as in TCP protocol, a link layer
delivery service can be achieved with ACKs and retransmissions. A link layer reliable
delivery service is often used for links that are prone to high error rates, such as wireless
link.
Disadvantage(s):
– created unnecessary overhead for low bit error links, such as fiber.
Advantage(s):
+ Link layer retransmissions avoid having to retransmit over the entire path. Transport
and application layer retransmissions must be sent over the entire path, thus incurring
higher latency and wasting bandwidth.
+ Link layer retransmissions can hide non-congestion losses from higher layer protocols.
For example, wireless link layer retransmissions can prevent error losses from being
misinterpreted by TCP as congestion losses.
2. In CRC error detection method, for the 5-bit generator, G = 10011 and the data, D =
1010101010. [Note: G = 10011 is a CRC-5-ITU standard].
a) Calculate the value of R.
[3 marks]
A: R = 0100
b) Explain how this method can be used to detect any single bit error in data D.
[2 marks]
A: If we divide K by G, then the reminder is not zero. In general, if G contains at least
two 1’s, then a single bit error can always be detected.
2
3. Suppose the information content of a packet is the bit pattern 1010 0111 0101 1001 and
an even parity scheme is being used.
a) Determine the value of the Parity bit field if a single bit parity scheme is used.
[1 mark]
A: 1
b) Determine the value of the Parity bits field if a 2-dimensional bit parity scheme is
used.
[4 marks]
A:
10100
0 1 1 1 1 Parity bits
01010
10010
0 0 0 1 1 Parity byte
Transmitted bits: 10100 01111 01010 10010 00011
4. a) List 4 desirable characteristics to be considered when designing a multiple access
protocol (MAC protocol) for a broadcast channel in Link Layer.
[4 marks]
A: 1. When 1 node wants to transmit, it can send at rate R.
2. When M nodes want to transmit, each can send at average rate R/M.
3. Fully decentralized (no synch, no admin/master node).
4. Simple (inexpensive to implement).
b) Based on answer in 4 (a), determine which of these characteristics are employed in the
following MAC protocols.
[4 marks]
MAC Protocol
TDM
Characteristic(s)
2, 4
Slotted Aloha
1, 2, 4
CSMA
1, 3, 4
Token Passing
1, 2, 3, 4
3
5. Suppose 4 active nodes; A, B, C and D are competing for access to a channel using
slotted ALOHA. Assume each node has an infinite number of packets to send. Each node
attempts to transmit in each slot with probability p. The first slot is numbered slot 1, the
second slot is numbered slot 2, and so on.
a) If p is a number between 0 and 1 and
If p(A) = probability that A succeeds in a slot
p(A) = p(A transmits and B does not and C does not and D does not)
= p(A transmits) p(B does not transmit) p(C does not transmit) p(D does
not transmit)
= p(1 – p)(1 – p)(1 – p) = p(1 – p)3
Calculate the probability that node A succeeds for the first time in slot 5.
[2 marks]
A: Hence, p(A succeeds for first time in slot 5)
= (1 – p(A))4 p(A) = (1 – p(1 – p)3)4 p(1 – p)3
b) Calculate the probability that some node (either A, B, C or D) succeeds in slot 4.
[3 marks]
A:
p(A succeeds in slot 4) = p(1-p)3
p(B succeeds in slot 4) = p(1-p)3
p(C succeeds in slot 4) = p(1-p)3
p(D succeeds in slot 4) = p(1-p)3
p(either A or B or C or D succeeds in slot 4) = 4p(1-p)3
(because these events are mutually exclusive)
6.
IP: 192.168.1.1
MAC: AA-AA-AA-AA-AA-AA
IP: 192.168.2.1
MAC: BB-BB-BB-BB-BB-BB
IP: 192.168.1.2
MAC: 11-11-11-11-11-11
Host A
Host B
IP: 192.168.2.2
MAC: 22-22-22-22-22-22
Host D
Host C
Figure 1: Network Topology
4
Consider the network shown in Figure 1. Assume ARP Tables at the hosts and router are
empty initially, and that routing data tables are already set up in the hosts and the router.
Also assume that each host knows the IP addresses of all other hosts. If Host A sends an
IP datagram to Host B:
a) Determine the IP address being resolved in the initial ARP request and the resolved
MAC address.
A: IP address being resolved is 192.168.2.1
Resolved MAC address 11-11-11-11-11-11
b) Determine who sends the ARP request and who sends the ARP reply.
A: Host A sends the ARP request and the router sends the ARP reply.
c) For each network on the end-to-end path between the hosts, what are the destination
IP address and source IP address in the packet header? And what are destination
MAC address and source MAC address?
A:
The destination IP address is 192.168.2.1 and source IP address is 192.168.1.1 on
both networks.
On the first network (Host A network), the destination MAC address is 11-11-11-1111-11 and the source MAC address is AA-AA-AA-AA-AA-AA.
On the second network (Host B network), the destination MAC address is BB-BBBB-BB-BB-BB the source MAC address is 22-22-22-22-22-22.
[10 marks]
7. A PPP framing protocol uses flags with bit stuffing for transmission. The original data to
be transmitted contains:
0001111111001111101000111111111000011111
Given the FLAG byte: 01111110 (used to mark the start and end of a frame)
Perform the bit stuffing and determine the resulting frame.
[3 marks]
A:
011111100001111101100111110010001111101111000011111001111110
-oooOooo5