Test: Data Link Layer & MAC-Sublayer- 2 - Computer Science Engineering (CSE) MCQ

07-01 - 12 - 03 - 04 - 05. The ethernet has more than one destinations but not all. So it is a multicast address not a broadcast address.

Where n is maximum sequence number selective repeat

CRC guarantees that all burst error of length equal to the degree of the polynomials are detected and also burst errors affecting an odd number of bits are detected.

In Go back-N maximum window size 2^N- 1
And range of sequence member 0 to 2^N- 1
So, 0 to 63.

It n data packets sent we need less than n data packet acknowledgments.

In token ring N number of system choose 1 monitor. The responsibility of this monitor is
(i) Lost of Token,
(ii) Garbage frame,
(iii) Orphan frame.
If a station is disabled the token generated by it is removed from the token rina,

The station who sending data strips the data frame from the token ring network.

Single parity can detect single bit error, while block sum check can more than one bit error and CRC can detect all errors present in data.
So. sinale Daritv < Block sum check < CRC.

With selective repeat ARQ of 7 bits. Maximum window size = 2^7-1 - 26 = 64 Maximum sequence number = (64 -1) = 63.

Two dimensional parity check consist parity bits for each data unit as well as an entire data unit of parity bits.

Reference polynomial x⁴ + r³ + 1 =11001
Datasequence = 1010101010
On dividing 1010101010 by 11001,
We get CRC (i.e. remainder) = 0010
On appending this CRC on data sequence, we get the actual message transmitted i.e. 10101010100010.

First we find the Transmission Delay

= Packet size / Bandwidth

= 64 bytes / 1.536 Mbps

= (64 × 8 bits) / (1.536 × 10⁶ bits per sec)

= 333.33 μsec

Then we find the Propagation Delay

For 1 km, propagation delay = 6 μsec

For 3000 km, propagation delay = 3000 x 6 μsec = 18000 μsec

Value of a is calculated by  :

= T_p / T_t

= 18000 μsec / 333.33 μsec

= 54

Bits required in sequence number field

= ⌈log₂(1+2 x a)⌉

= ⌈log₂(1 + 2 x 54)⌉

= ⌈log₂(109)⌉

= ⌈6.76⌉

= 7 bits

Hence, option 2) is correct.

Bits for sequence no. = 2^{no of bits} > Window size
=>    W = 110
2⁷ = 128 > 110
Hence 7 bits used for sequence number.

Data rate = 1 Mbps
T_p = 270 msec
F = 125 bytes
= 125 x 8 bits
= 1000 bits

Four bit sequence number
For Go-back-N, W= 16-1 = 15