Transport Layer

Question 1

The transport layer protocols used for real time multimedia, file transfer, DNS and email, respectively are:

• A. TCP, UDP, UDP and TCP
• B. UDP, TCP, TCP and UDP
• C. UDP, TCP, UDP and TCP
• D. TCP, UDP, TCP and UDP

Answer: (C)

TCP (Transmission Control Protocol) and UDP(User Datagram Protocol) are two main transport layer protocols. TCP is connection oriented and UDP is connectionless, this makes TCP more reliable than UDP. But UDP is stateless (less overhead), that makes UDP is suitable for purposes where error checking and correction is less important than timely delivery. For real time multimedia, timely delivery is more important than correctness. --> UDP For file transfer, correctness is necessary. --> TCP DNS, timely delivery is more important --> UDP Email again same as file transfer --> TCP

Question 2

The transport layer protocols used for real time multimedia, file transfer, DNS and email, respectively are:

• A. TCP, UDP, UDP and TCP
• B. UDP, TCP, TCP and UDP
• C. UDP, TCP, UDP and TCP
• D. TCP, UDP, TCP and UDP

Answer: (C)

TCP (Transmission Control Protocol) and UDP(User Datagram Protocol) are two main transport layer protocols. TCP is connection oriented and UDP is connectionless, this makes TCP more reliable than UDP. But UDP is stateless (less overhead), that makes UDP is suitable for purposes where error checking and correction is less important than timely delivery. For real time multimedia, timely delivery is more important than correctness. --> UDP For file transfer, correctness is necessary. --> TCP DNS, timely delivery is more important --> UDP Email again same as file transfer --> TCP

Question 3

Which of the following transport layer protocols is used to support electronic mail?

• A. SMTP
• B. IP
• C. TCP
• D. UDP

Answer: (C)

E-mail uses SMTP as application layer protocol. TCP and UDP are two transport layer protocols. SMTP uses TCP as transport layer protocol as TCP is reliable.

Question 4

Consider an instance of TCP’s Additive Increase Multiplicative Decrease (AIMD) algorithm where the window size at the start of the slow start phase is 2 MSS and the threshold at the start of the first transmission is 8 MSS. Assume that a timeout occurs during the fifth transmission. Find the congestion window size at the end of the tenth transmission.

• A. 8 MSS
• B. 14 MSS
• C. 7 MSS
• D. 12 MSS

Answer: (C)

See question 2 of http://www.geeksforgeeks.org/computer-networks-set-2/

Question 5

A layer-4 firewall (a device that can look at all protocol headers up to the transport layer) CANNOT

• A. block entire HTTP traffic during 9:00PM and 5 :0OAM
  
• B. block all ICMP traffic
  
• C. stop incoming traffic from a specific IP address but allow outgoing traffic to the same IP address
  
• D. block TCP traffic from a specific user on a multi-user system during 9:00PM and 5:00AM
  

Answer: (D)

A.Can Block entire HTTP traffic by blocking TCP port 80 and it is possible because it is L4 firewall. 
D) As it is L4 firewall can not block packets based on user identity because it is the responsibility of Application layer

Question 6

While opening a TCP connection, the initial sequence number is to be derived using a time-of-day(ToD) clock that keeps running even when the host is down. The low order 32 bits of the counter of the ToD clock is to be used for the initial sequence numbers. The clock counter increments once per millisecond. The maximum packet lifetime is given to be 64s. Which one of the choices given below is closest to the minimum permissible rate at which sequence numbers used for packets of a connection can increase?

• A. 0.015/s
• B. 0.064/s
• C. 0.135/s
• D. 0.327/s

Answer: (A)

The maximum packet lifetime is given to be 64 seconds in the question. Thus, a sequence number increments after every 64 seconds.
So, minimum permissible rate = 1 / 64 = 0.015 per second

Thus, option (A) is the answer.

Please comment below if you find anything wrong in the above post.

Question 7

Which of the following system calls results in the sending of SYN packets?

• A. socket
• B. bind
• C. listen
• D. connect

Answer: (D)

socket() creates a new socket of a certain socket type, identified by an integer number, and allocates system resources to it. bind() is typically used on the server side, and associates a socket with a socket address structure, i.e. a specified local port number and IP address. listen() is used on the server side, and causes a bound TCP socket to enter listening state. connect() is used on the client side, and assigns a free local port number to a socket. In case of a TCP socket, it causes an attempt to establish a new TCP connection. When connect() is called by client, following three way handshake happens to establish the connection in TCP. 1) The client requests a connection by sending a SYN (synchronize) message to the server. 2) The server acknowledges this request by sending SYN-ACK back to the client. 3) The client responds with an ACK, and the connection is established. Sources: Berkeley sockets, TCP Connection Establishment and Termination

Question 8

In the slow start phase of the TCP congestion control algorithm, the size of the congestion window

• A. does not increase
• B. increases linearly
• C. increases quadratically
• D. increases exponentially

Answer: (D)

See Question 2 of http://www.geeksforgeeks.org/computer-networks-set-7/

Question 9

Which one of the following uses UDP as the transport protocol?

• A. HTTP
• B. Telnet
• C. DNS
• D. SMTP

Answer: (C)

UDP is a stateless ,connectionless and unreliable protocol.
HTTP needs connection to be established and thus, uses TCP.
Telnet is a byte stream protocol which again needs connection establishment ,thus uses TCP.
DNS needs request and response ,it needs a protocol in which a server can answer the small queries of large number of users. As UDP is fast and stateless it is the most suitable protocol and thus, it is used in DNS querying .
SMTP needs reliability and thus, uses TCP.

Question 10

Let the size of congestion window of a TCP connection be 32 KB when a timeout occurs. The round trip time of the connection is 100 msec and the maximum segment size used is 2 KB. The time taken (in msec) by the TCP connection to get back to 32 KB congestion window is _________.

• A. 1100 to 1300
• B. 800 to 1000
• C. 1400 to 1600
• D. 1500 to 1700

Answer: (A)

Current size of congestion window in terms of number of segments
                            = (Size in Bytes)/(Maximum Segment Size)
                            = 32KB / 2KB 
                            = 16 MSS
          
When timeout occurs, in TCP's Slow Start algorithm, threshold is 
reduced to half which is 16KB or 8MSS. Also, slow start phase begins 
where congestion window is increased twice. 
So from 1MSS to 8 MSS window size will grow exponentially. 
Congestion window becomes 2MSS after one RTT and becomes 4MSS after
2 RTTs and 8MSS after 3 RTTs.  At 8MSS, threshold is reached and
congestion avoidance phase begins.  In congestion avoidance phase,
window is increased linearly. So to cover from 8MSS to 16MSS, it needs
8 RTTs

Together, 11RTTs are needed (3 in slow start phase and 8 in congestion
avoidance phase).