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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.
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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 |
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 |
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 |
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