Test: Communication & Congestion Control - Computer Science Engineering (CSE) MCQ

Total bit per character = 8 bit data + 2 stop bit +1 start bit (#) = 11 bits
no of characters = 2400/11 = 218.18
Since it is asked for transmitted characters we take floor and answer is 218.

A) Because RS 232 requires a start before each byte transmission for synchronization..

In serial communication in beginning '0' is padded as start bit and one or two '1's are padded as stop bit.
and those bits are for synchronize receiver

The baud rate is the rate at which information is transferred in a communication channel. Serial ports use two-level (binary) signaling, so the data rate in bits per second is equal to the symbol rate in bauds.

"9600 baud" means that the serial port is capable of transferring a maximum of 9600 bits per second."
So, transmission rate here = 9600 bps
An eight bit data (which is a char) requires 1 start bit, 2 stop bits and 1 parity bit = 12 bits.
So, number of characters transmitted per second = 9600 / 12 = 800

T1: 1 char. = ( 8 + 2 + 1 + 1) = 12 bit
Transfer Rate = 1200/12 = 100 char/sec.
T2: Transfer character in bits = 24 + 240 = 264 bits
In 264 = 30 character
Then 1200 = ?
264/30 = 1200/X
X = 136.3 character / sec.
so correct option is (C)

Answer is B
Here we can spent at max 5 packets per Time unit 5000 /1000.
So Whatever which is not sent is backlog.
So First Time Unit => 6 ,
Backlog in First time unit => 6-5 => 1 This one gets added to next Time units load
Second time unit => 9 + 1 (One from Previous Time Unit)
Backlog in Second time Unit = 10-5 => 5 (This one gets added to next Time Units load.)
Total Backlog this way = 1+5+3+5+2+0+0+0+0+1+0+5+7+7+10+8+9+6+10+10=89
Avg Backlog=89/20=4.45
The average number of backlogged of packets per time unit during the given period is 4.45 , (Option B) .

Ans should be (B)
Current Sender window = min (Congestion Window, Advertised Window)= min(4KB, 6KB)= 4KB

increase is exponential in the Slow Start Phase.
answer = option D

The Answer is correct , but method of solving is wrong .
At
t=1, =>2mss
t=2, =>4mss
t=3, =>8mss
t=4, =>9mss (after threshold additive increase)
t=5, =>10mss (fails)
Threshold will be reduced by n/2 i.e. 10/2 = 5.
t=6, =>2mss
t=7 =>4mss
t=8, =>5mss
t=9, =>6mss
t=10, =>7mss.
So at the end of 10th transmission congestion window size will be 8 mss.