Computer Networks MCQ - Computer Networks Section 3

1 In a data link protocol, the frame delimiter flag is given by 0111. Assuming that bit stuffing is employed, the transmitter sends the data sequence 01110110 as

Delimiter flag: 0111
Data sequence: 01110110
So, for a flag of 4 bits we will compare data sequence with pattern of 3 bits i.e. 011. Data sequence : 0 1 1 1 0 1 1 0
The underlined bits are found matched, hence, 0 is stuffed.
Thus resulting in the data sequence as 0110101100 which is option (d).

2 A bit-stuffing based framing protocol uses an 8- bit delimiter pattern of 01111110. If the output bit-string after stuffing is 01111100101, then the input bit-string is

The given 8-bit delimiter pattern is 01111110.
The output bit-string after stuffing is 01111100101.
The above bold underline bit is stuffed bit.
So input bit-string must be 0111110101.

3 A computer on a 10 Mbps network is regulated by a token bucket. The token bucket is filled at a rate of 2 Mbps. It is initially filled to capacity with 16 Megabits. What is the maximum duration for which the computer can transmit at the full 10 Mbps?

Let duration is t sec.
= 16 + 2t = 10 * t
= t = 2 sec

4 A host is connected to a Department network which is par t of a University network. The University network, in turn, is part of the Internet . The largest network in which the Ethernet address of the host is unique is

Ethernet address is basically the MAC address, which is supposed to be unique to a NIC. Thus it is unique over the Internet.

5 A router has two full-duplex Ethernet interfaces each operating at 100 Mb/s. Ethernet frames are at least 84 bytes long (including the Preamble and the Inter-Packet-Gap).
The maximum packet processing time at the router for wirespeed forwarding to be possible is (in micro-seconds)

Minimum transmission time
= (84 bytes) / (2*100 bps)
= (84 * 8 bits) / (2*100 bps)
= 3.36 seconds