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 UWA week 37 (2nd semester, week 7) ↓
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helpful
5:06am Sun 10th Sep, Christopher M.

"Minn Minn Khant Maw" <23*7*0*3@s*u*e*t*u*a*e*u*a*> wrote:
> Since it is mentioned that the data transfer rates are measured in Bytes/second, 2000Bytes with a speed of 3,000,000Bps will take 0.0006666.... This is practically 0 in the program. So if my understanding is correct, this data write should only take 20usecs, which is just to acquire the DATABUS.
It is not meaningful to say "This is practically 0 in the program". If something starts at, say, t=10 and takes 0.0006666, then you can't say it finishes at t=10. Moreover, if we're only using integers to represent time, then we have to say that it has finished at/by t=11.

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