Huy
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How could you do that in an exam with no calculators!Originally posted by cyrax83
That 9600 bps question had me a bit stumped too.. but i worked it out, i put 3.4 seconds because :
4KB = 4*1024 Bytes = 4096 Bytes
4096 Bytes = 4096 * 8 Bits = 32768 Bits
Since modem is 9600 bits per second
32768 / 9600 = approx 3.4 seconds
I wrote 4.3s in the exam, but my way of working it out was:
9600 bits per sec / 8 bits per byte = 1200 kilobytes per sec
1200 / 1024 bytes = 1.171875 kilobytes per sec
4 KB / 1.171875 kilobytes per sec
= 3.4133333333333333333333333333333
that's how i usually work things out, anyway,
for example:
56K = 56,700 bps
56,700 bps / 8 = 7087.5 Kbps
7087.5 Kbps / 1024 = 6.92138671875 KBps (Kbytes)
Theoretical bandwidth for 56K is about 7KB/s.
I should have started off by knowing this, (slipped my mind),
then dividing 56K by 9.6K
(about 6)
Dialup downloads at 7KB/s (tops)
meaning for a 9.6K modem, it'll be about 1/6th of 56K,
7/6 is roughly 1.1KB/s
4KB file, going at a max of 1.1KB/s
4 / 1.1 = ~3.6
Which is close enough to 3.4 secs, moreso than 4.3 seconds
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