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+4 votes

What is the maximum number of characters (7 bits + parity) that can be transmitted in a second on a $19.2$ Kbps line. This asynchronous transmission requires $1$ start bit and $1$ stop bit.

  1. $192$
  2. $240$
  3. $1920$
  4. $1966$
in Computer Networks by Boss (30.8k points) | 3k views

2 Answers

+10 votes
Best answer
bandwidth is 19.2Kbps

I think

19.2*10^3 bits -is transfer per second

19200 bit is transfer per second

Now they say 1 character is made 7 bits + 1 parity = 8 bits

Since it is asynchronous whenver you are sending it would require 1 start and stop bit

So 10 bit now is required to send 1 character

Number of character that can be send is 19200/10 = 1920 character
by Loyal (9.9k points)
selected by
Why value of kilo is taken as 1000 instead of 1024 in bandwidth conversion.

If we take 1024 then the number of character is (19.2*1024)/(7+1+1+1) = 1966.08 = 1966

This answer is given in option D.

A Kilobit per second is a unit used to measure data transfer rates and is based on "Decimal multiples of bits". The symbol for Kilobit per second is Kbps or kb/s or kbit/s. There are 8 Kilobits per second in a Kilobyte per second.

Kilobyte (KB or K) = 1024 bytes (2 ^ 10) "binary kilobyte"
kilobyte (kB) = 1000 bytes (10 ^ 3) "decimal kilobyte"

Kilobit (Kb) = 1024 bits (2 ^ 10) "binary kilobit"
kilobit (kb) = 1000 bits (10 ^ 3) "decimal kilobit"

KBps = Kilobytes (1024 bytes) per second (binary)
kBps = kilobytes (1000 bytes) per second (decimal)

Kbps = Kilobits (1024 bits) per second (binary)
kbps = kilobits (1000 bits) per second (decimal)

bps = bits per second

in simple words ,

K- Binary system

k- Decimal system


So according to you K (capital) =1024 and k (small) =1000 for kilo.

Then in above question, bandwidth is given as 19.2Kbps then it should be equivalent to 19.2 * 1024 bps but in above solution value of 'K' is 1000. This is what contradicting.

Please confirm (means still need help) which is correct.

In Networking jargon, K/k represents 1000 bits only, not 1024 bits.
0 votes
Answer(C):In actual question paper it is given as 19.2kbps so we have to take 10^3 and given solution is 1920.

if we take 19.2Kbps then none of option is correct only we can select option (D) that too approx.

Also keep one thing in mind whenever we do data transfer we take kbps and whenever we have to calculate file size in memory we take Kbps.
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