Test: Network & Its Security - Computer Science Engineering (CSE) MCQ

Bit rate is half the baud rate in Manchester encoding as bits are transferred only during a positive transition of the clock.

THE AMBIGUOUS QUESTION WITHOUT ANY STANDARD MENTION !!
BOTH A and B is correct , it is just the convention which determine the correct answer.

as per IEEE B is correct
as per G E Thomas A is correct.
as we ususally follow IEEE thus B is correct

Hence answer d

Answer: C
The start and stop bits are used to synchronize the serial receivers.

Answer will be option D.)
The bit stuffing is done after every two '11' (as flag is 0111) to differentiate the data part from the flag- there must not be
"111" in the data so after every 11 a '0' is added. The receiver also knows this and so it decodes every "110" as "11".
Therefore option D. is the answer.

Answer is C).
Question says TCP message is of size = 2100 B, hence header is encapsulated with the frame and send to IP layer, and IP
layer when receives this message thinks it as complete data to be send .
To be send across the first network, It is fragmented into 2 payloads of sizes 1200 and 900 .
No need to add IP Header size with 2100B, as fragmentation acts on data/Payload and not the packet.
Now, since, IP thinks 2100B as the data so, it will directly fragment it and hence, no need to add any header.
Payload / Fragment 1: 1200 B {1200 / 8 = 150, Except last fragment all the fragments should be divisible by 8}
Payload 2: 900 B
At the second network, 1200 B is fragmented into 400 B,400 B and 400 B .
Similarly, 900 B is fragmented into 400, 400 and 100 B.
Since, 2100 B is fragmented into 6 data payloads and Header is attached to all the payloads when they will be forwared to
DLL layer. So, Total Overhead = 6 * 20 = 120 B
Hence, DLL layer will receive total Data = 2100 + 120 = 2220 B and DLL will treat it as a complete data to be sent and
DLL header will be encapsulated in DLL frame.

for 50% utilization
tt/(tt+2tp)>=1/2
2tt>=tt+2tp

tt>=2tp
L/B>=2*tp
L>=2*tp*B
so here L=2*20*10^-3*4*10^3
=160 bits
so ans is d

Since there is no buffer.. and constraint given is there should not be any data lost, and no wastage of capacity as well..
Since data should not be lost, we calculate for the extreme case when all sources are on-time (that is transmitting)..
10 Mbps * n-station 100 Mbps
n-station = 10..
In the next part of the question it is given that the link is provided with large buffer and we are asked to find out large no. of stations..
for that we'll calculate expected value of bandwidth usage (if more data comes we store in buffer and due to expectation,
the buffer will be emptied soon):
E = 1/3 * 10 + 1/3 * 10 +...n-station times 100 Mbps [ total time is (1+2) = 3 then on time is 1 so 1/3 of BW]
=> 1/3 * 10 * n-station 100 Mbps
=> n-station = 30
so, option (A)