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Question 3 [ 2 5 ] a . What is the typical size of the header for an IPv 4 datagram? [ 1 ] b

Question 3[25]
a. What is the typical size of the header for an IPv4 datagram? [1]
b. How is the 16-bit IP checksum field calculated? [2]
c. How is the TTL field used on the Internet? What purpose does it serve? [2]
d. Which field is used for IP congestion control? How does it work? [2]
e. Why is datagram fragmentation not an issue in IPV6? How is this achieved? [2]
f. A message of 2300 bytes is to be transmitted over 2 IP networks. The first has an MTU of 2300 bytes and the second network has an MTU of 1500 bytes. How many fragments are expected at the destination? Show the values of the following IP header fields and flags for each fragment on the two networks: Packet length, MF, Fragment Offset and Identification. [7]
Page 3 of 5
Degree and Diploma Examinations: June 2018 Introduction to Data Communication and Networking: CSI 21M2
g. What does ICMP stand for? Briefly explain the key areas of functionality for ICMP giving an example. [3]
h. Convert the MAC address CC-12-DE-4A-BD-88 into the corresponding binary digits. [3]
i. Given the following subnet masks determine the number of a) host bits available and b) the number hosts available. [3]
i.,255.255.240.0
ii.,255.255.255.0
iii. ,255.255.255.128
Question 4[25]
a. What is the role of the transport layer in the TCP/IP protocol stack? [2]
b. Describe the functionality provided by the Transmission Control Protocol (TCP)[2]
Consider the TCP connection mechanism.
c. What mechanism is used to set up a TCP connection? [1]
d. Why is the mechanism needed? [2]
e. Outline how the mechanism works. [3]
For each of the following pair of terms, define each term, and clarify the key difference(s) between the two terms. Be clear and concise.
f. Flow Control and Congestion Control [4]
g. What is the role of a window in TCP?[1]
The output on the next page shows a tcpdump trace (similar to Wireshark) of the network packets exchanged between two computers on a network. Use your knowledge of TCP and IP to answer the following questions.
h. What application-layer protocol is (likely) being used for this conversation? [1]
i. What is the IP address of the client? [1]
j. What is the Round-Trip Time (RTT) between the client and the server? [1]
k. How much application-layer payload is in the client's first data packet? [1]
I. What is the TCP Initial Sequence Number (ISN) proposed by the server? [1]
m. What is the Maximum Transmission Unit (MTU) size on this network? [1]
n. What is the Maximum Segment Size (MSS) used by the server's TCP?[1]
o. In total, how many bytes of application-layer data does the server send? [1]
p. Which end initiates the closing of this TCP connection: client or server? [1]
q. In total, how long did this conversation last? [1]
Page 4 of 5
Degree and Diploma Examinations: June 2018
Introduction to Data Communication and Networking: CSI 21M2
Time(s) SourceIP DestinationIP Size TCP SPort DPort SeqNum AckNum Window Flags
29.884192.168.1.9,136.159.5.1744 TCP 1035,80133227,0 win: 32768S
29.886,136.159.5.17,192.168.1.9,44 TCP 80,1035,3310607972,133228 win: 24820 SA
29.888192.168.1.9136.159.5.1740 TCP 1035,80133228,3310607973 win: 32768A
29.948192.168.1.9136.159.5.17418 TCP 1035,80133228,3310607973 win: 32768 PA
29.952136.159.5.17192.168.1.940 TCP 80,1035,3310607973,133606 win: 24820 A
29.955136.159.5.17192.168.1.9329 TCP 8010353310607973,133606 win: 24820 PA
29.959136.159.5.17192.168.1.91500 TCP 8010353310608262133606 win: 24820 A
29.960136.159.5.17192.168.1.91500 TCP 8010353310609722133606 win: 24820 PA
29.962192.168.1.9136.159.5.1740 TCP 1035801336063310609722 win: 31019 A
29.970136.159.5.17192.168.1.9,1500 TCP 80,10353310611182133606 win: 24820 A
29.972136.159.5.17192.168.1.91500 TCP 801035,3310612642133606 win: 24820 A
29.972192.168.1.9136.159.5.1740 TCP 1035,801336063310612642 win: 28099 A
29.973136.159.5.17,192.168.1.9,429 TCP 80,1035,3310614102133606 win: 24820 FA
29.974192.168.1.9136.159.5.1740 TCP 103580133606,3310614492 win: 26250A
30.072192.168.1.9136.159.5.1740 TCP 1035801336063310614492 win: 26250 A
30.136192.168.1.9136.159.5.1740 TCP 103580133606,3310614492 win: 31370 A
30.141192.168.1.9136.159.5.1740 TCP 1035801336063310614492 win: 0 F
End
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