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A packet of 10 Kb is to be downloaded from a web server.
Find the time needed to download this packet using:
a) A dial up telephone connection at 28 Kbps
b) A cable modem at 28 Mbps.

Question Posted / fifi

Answers were Sorted based on User's Feedback



A packet of 10 Kb is to be downloaded from a web server. Find the time needed to download this pac..

Answer / david tilley

The straight maths answer is: Ti = Fs/Sp
Where:
Ti...Time
Fs..FileSize
Sp..Speed

a) Ti = 10,000 / 28,000
= 357ms
b) Ti = 10,000/ 28,000,000
= 357uS

However.... In practice transfer times would be longer than
this depending on the overhead added to handle error
checking/recovery.

Is This Answer Correct ?    30 Yes 6 No

A packet of 10 Kb is to be downloaded from a web server. Find the time needed to download this pac..

Answer / vaibhav

the correct answer is below

10kb=10*1024bytes=10240 bytes
28Kilobitsps=28/8KiloBytesps=3.5KBps=3.5*1024Bytesps=3584Bps

time=10240/3584=2.85 s

do same with 28.8Mbps=28.8*1024*1024.

Is This Answer Correct ?    7 Yes 5 No

A packet of 10 Kb is to be downloaded from a web server. Find the time needed to download this pac..

Answer / john

I agree this is the correct one.

10kb=10*1024bytes=10240 bytes
28Kilobitsps=28/8KiloBytesps=3.5KBps=3.5*1024Bytesps=3584Bps

time=10240/3584=2.85 s

do same with 28.8Mbps=28.8*1024*1024.

Is This Answer Correct ?    2 Yes 3 No

A packet of 10 Kb is to be downloaded from a web server. Find the time needed to download this pac..

Answer / anil kumar. k. k

A cable modem at 28 Mbps.

Is This Answer Correct ?    2 Yes 6 No

A packet of 10 Kb is to be downloaded from a web server. Find the time needed to download this pac..

Answer / vijayan. g

A dial up telephone connection at 28 Kbps

Is This Answer Correct ?    3 Yes 8 No

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