Computer Networks MCQ - Computer Networks Section 3

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

Congestion window increases exponentially.

62 A client process P needs to make a TCP connection to a server process S. Consider the following situation: the server process S executes a socket ( ), a bind ( ) and a listen ( ) system call in that order, following which it is preempted. Subsequently, the client process P executes a socket ( ) system call followed by connect ( ) system call to connect to the server process S. The server process has not executed any accept ( ) system call. Which one of the following events could take place?

As there is no accept() at server side, the connect() at client could not get anything that can respond to its requests, so error occurs.

63 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?

The maximum packet lifetime = 64 seconds i.e. a sequence number increments after every 64 seconds.
So, minimum permissible rate = 1/64 = 0.015 per second

64 One of the header fields in an IP datagram is the Time-to-Live (TTL) field. Which of the following statements best explains the need for this field?

It can be used to prevent packet looping because when Time-to-Live field reaches to ‘0’ means when there is no any type of acknowledgement received, then TTL decrement continuously and when it reaches ‘0’, then it stop looping of packet.

65 Consider an instance of TCP’s Additive Increase Multiplicative Decrease (AI MD) algorithm where the windows 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 time-out occurs during the fifth transmission. Find the congestion window size at the end of the tenth transmission.

Window size for 1st transmission = 2 MSS
Window size for 2nd transmission = 4 MSS
Window size for 3rd transmission = 8 MSS
Threshold reached, increase linearly (according to AIMD)
Window size for 4th transmission = 9 MSS
Window size for 5th transmission = 10 MSS
Time out occurs, resend 5th with window size starts with as slow start.
Window size for 6th transmission = 2 MSS
Window size for 7th transmission = 4 MSS
Threshold reached, now increase linearly (according to AIMD)
Additive Increase: 5 MSS (since 8 MSS isn’t permissible anymore)
Window size for 8th transmission = 5 MSS
Window size for 9th transmission = 6 MSS
Window size for 10th transmission = 7 MSS