Converting from bandwidth to traffic gives different results depending on operators position?

Question

This must be a stupid question, but nevertheless I find it curious:

Say I have a steady download of 128Kbps.

How much disk space is going to be consumed after a hour in Megabytes?

128 x 60 x 60 / 8 / 1024 = 56.25 MB

But

128 x 60 x 60 / 1000 /8 = 57.6 MB

So what is the correct way to calculate this?

Thanks!

Answer 1

Ok, I found out an official explanation from Symantec on the matter: http://seer.entsupport.symantec.com/docs/274171.htm

It seems the idea is to convert from bits to bytes as early as possible in calculation, and then the usual 1024 division comes in place.

I just hope it's a standard procedure, and not Symantec imposed one :).

Answer 2

In one calculation you're dividing by 1000, but in another you're dividing by 1024. There shouldn't be any surprise you get different numbers.

Officially, the International Electrotechnical Commission standards body has tried to push "kibibyte" as an alternative to "kilobyte" when you're talking about the 1024-based version. But if you use it, people will laugh at you.

Answer 3

Please remember that there is overhead in any transmission. There can be "dropped" packets etc. Also there is generally some upstream traffic as your PC acknoledges receipt of packets. Finally since packets can be received out of order, the packets themselves contain "extra" data to all the receiver to reconstruct the data in the proper order.