## TCP

 Question 1

Consider a long-lived TCP session with an end-to-end bandwidth of 1 Gbps (= 109 bits/second). The session starts with a sequence number of 1234. The minimum time (in seconds, rounded to the closest integer) before this sequence number can be used again is _________.

 A 33 B 34 C 35 D 36
Computer-Networks       TCP       GATE 2018       Video-Explanation
Question 1 Explanation:
In TCP, Sequence number field is 32 bit, which means 232 sequence number per byte are possible. Whatever be the starting sequence number the possible number will be 232bytes
The process of using all the sequence number and repeating a previously used sequence number.
The time taken to wrap around is called wrap around time:
Minimum Time = Wrap around time = Total number of bits in sequence number / Bandwidth = 232 * 8 / 109 = 34.35 == 34 (closest integer)
 Question 2

Identify the correct sequence in which the following packets are transmitted on the network by a host when a browser requests a webpage from a remote server, assuming that the host has just been restarted.

 A HTTP GET request, DNS query, TCP SYN B DNS query, HTTP GET request, TCP SYN C DNS query, TCP SYN, HTTP GET request D TCP SYN, DNS query, HTTP GET request
Computer-Networks       TCP       GATE 2016 [Set-2]       Video-Explanation
Question 2 Explanation:
When a browser requests a web page from a remote server then that requests (URL address) will be mapped to IP address using DNS query, then TCP synchronization takes place after that HTTP verify whether it is existed in the web server or not.
 Question 3

Suppose two hosts use a TCP connection to transfer a large file. Which of the following statements is/are False with respect to the TCP connection?

1. If the sequence number of a segment is m, then the sequence number of the subsequent segment is always m+1.
2. If the estimated round trip time at any given point of time is t sec, the value of the retransmission timeout is always set to greater than or equal to t sec.
3. The size of the advertised window never changes during the course of the TCP connection.
4. The number of unacknowledged bytes at the sender is always less than or equal to the advertised window.
 A III only B I and III only C I and IV only D II and IV only
Computer-Networks       TCP       GATE 2015 [Set-1]
Question 3 Explanation:
I. False.
If the sequence no. of the segment is m, then the sequence number of the subsequent segment depends on the current segment size.
II. True.
If the estimated RTT at any given point of time is t second, then the value of the re-transmission timeout is always set to greater than or equal to t sec.
III. False.
The size of the advertised window may change during the course of the TCP connection depending on the processing capability at the receiver's side and the network traffic.
IV. True.
The number of unacknowledged bytes at the sender is always less than or equal to the advertised window, because the sender never sends no. of bytes greater than advertised window.
 Question 4

Assume that the bandwidth for a TCP connection is 1048560 bits/sec. Let α be the value of RTT in milliseconds (rounded off to the nearest integer) after which the TCP window scale option is needed. Let β be the maximum possible window size the window scale option. Then the values of α and β are

 A 63 milliseconds, 65535×214 B 63 milliseconds, 65535×216 C 500 milliseconds, 65535×214 D 500 milliseconds, 65535×216
Data-Structures       TCP       GATE 2015 [Set-2]
Question 4 Explanation:
TCP header sequence number field consist 16 bits. The maximum number of sequence numbers possible = 216 = 65,535.
The wrap around time for given link = 1048560 * α. The TCP window scale option is an option to increase the receive window size. TCP allows scaling of windows when wrap around time > 65,535.
==> 1048560 * α > 65,535*8 bits
==> α = 0.5 sec = 500 mss
Scaling is done by specifying a one byte shift count in the header options field. The true receiver window size is left shifted by the value in shift count. A maximum value of 14 may be used for the shift count value. Therefore maximum window size with scaling option is 65535 × 214.
 Question 5

Consider the following statements.

I. TCP connections are full duplex.
II. TCP has no option for selective acknowledgment.
III. TCP connections are message streams.
 A Only I is correct B Only I and III are correct C Only II and III are correct D All of I, II and III are correct
Computer-Networks       TCP       GATE 2015 [Set-3]
Question 5 Explanation:
In TCP, as sender and receiver can send segments at the same time, It is FULL-DUPLEX. TCP can use selective ACK and TCP uses byte streams that is every byte is send using TCP is numbered.
There are 5 questions to complete.

Register Now