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 _________.
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)
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.
HTTP GET request, DNS query, TCP SYN
DNS query, HTTP GET request, TCP SYN
DNS query, TCP SYN, HTTP GET request
TCP SYN, DNS query, HTTP GET request
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.
I and III only
I and IV only
II and IV only
If the sequence no. of the segment is m, then the sequence number of the subsequent segment depends on the current segment size.
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.
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.
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.
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
63 milliseconds, 65535×214
63 milliseconds, 65535×216
500 milliseconds, 65535×214
500 milliseconds, 65535×216
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.
Consider the following statements.
I. TCP connections are full duplex.
II. TCP has no option for selective acknowledgment.
III. TCP connections are message streams.
Only I is correct
Only I and III are correct
Only II and III are correct
All of I, II and III are correct