Data Communication and Computer Networks MCQ - Functions of OSI and TCP/IP Layers

66 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.

In a TCP connection
If the sequence number of a segments is m. Then it is not necessary the subsequent segment is m + 1, because it depend upon no of packet sent is current segment.
Advertise window can not be fixed it may change on changing the capacity of receiver window.

67 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 with window scale option. Then the values of α and β are

(1) we have bandwidth delay product
= (1048560/8) B * α = 64 K
α = (64 K * 8 )/1048560
= 0.5s = 500 milliseconds.
(2) If window scaling happens, a 14 bit shift count is used in TCP header. So, the maximum possible window size gets increased from 2¹⁶ – 1 to (2¹⁶ – 1) * 2¹⁴ or from 65535 to 65535 * 2¹⁴

68 Host A sends a UDP datagram containing 8880 bytes of user data to host B over an Ethernet LAN. Ethernet frames may carry data up to 1500 bytes (i.e. MTU = 1500 bytes). Size of UDP header is 8 bytes and size of IP header is 20 bytes. There is no option field in IP header. How may total number of I P fragment s will be transmitted and what will be the contents of offset field in the last fragment?

UDP data = 8880 bytes
UDP header = 8 bytes
IP Header = 20 bytes
Total Size(excluding IP Header) = 8888 bytes.
Number of fragments = 8888/1480 = 7
Offset of last segment = (1480 * 6)/8 = 1110

69 Consider the following statements.
I. TCP connections are full duplex.
II. TCP has no option for selective acknowledgment
III. TCP connections are message streams.

In TCP, as sender and receiver can send segments at the same time, It is FULL-DUPLEX.