1Mb (1 Mega Bits) = 1024 Kb (Kilo Bits) = 1024 * 1024 Bits = (1024 * 1024)/8 bytes = 131072 bytes = 131072 / 1024 KB (Kilo Bytes) = 128 KB (Kilo Bytes)

So
10 Mbps = 128 Kilobytes/second

The calculation above starts with 1Mbit.
So, 10 Mbps would be 10 * 128 Kilobytes/second.

maeb3

Ya..

1 Mbps = 128 Kilobytes/second
10 Mbps = 10 * 128 Kilobytes/second

1.192092896      MBps
1220.703125      KBps
1250000      Bps

10000000      bps
10000      Kbps
10      Mbps

First, you must figure out the number of bits per second.  10Mbps is 10,000,000 bits per second.  Then, to convert the bits to Bytes, you have to divide by 8, because there are 8 bits in a Byte.  Then, it's a little different since TECHNICALLY the notation is by factors of 1024 which is just really a factor of 2 (2^10).  Anyway, 1 Kilobyte = 1024 bits, so you have to divide by 1024 to get the Kilobytes which comes out to 1,220.7 KB/sec.

Hi when calculatins speed you do not use 1 MB=1024kb is should be 1mb=1000kb. only in storage you use the 1024kb=1mb:

1 Mbps =1000 Kbps =0.125 MB/sec = 125 KB/sec
so 10Mbps=125KB/sec *10

Also try this link

http://www.mediaroad.com/products/speedcheck/free_tools/

Jeez guys just give him the answer and dont confuse him:

Answer is: 1250 KB/sec

or

10 Mbps
10000 Kbps
1.25 MB/sec

>>1027 Bytes per KiloByte

Typo.  1024 Bytes per KiloByte.

pseudocyber is CORRECT!

I forgotten about the concept :'(
Realized I am wrong.


hongjun

Thanks Hongjun. :)

You deserved it :)

Sorry if this confuses the issue further:

10Mbps = 10,000 kbps. It takes 8 bits to code one byte (character), so one might expect 10,000 kbps to be equivalent to 1250 kBytes per second. Things are not that simple.

As pseudocyber points out, when used in measuring communication speeds in bits per second, the prefix kilo- means 1000. In computer file sizes in bytes, the prefix kilo- is commonly taken to mean 1024, so you cannot translate from kbps to kBytes per second just by dividing by 8.

The maximum amount of real user data that can be accommodated in one data packet is 1460 bytes. Then there are 20 bytes of TCP overhead, plus 20 bytes of IP overhead, plus 18 bytes of MAC overhead, making 1518 bytes to be transmitted to carry 1460 bytes, a 4% overhead. Then between each packet there will be an inter-packet gap of indeterminate size.

So at a true transmission rate of 10 Mbps, the apparent user data rate will be significantly below 1250 kBytes per sec, because both the above effects will reduce the equivalent rate in kBytes per sec. The maximum sustained data download rate you can expect is 10,000 * 1000/1024 * 1460/1518 * 1/8 = 1174 kBytes/sec, and not 1275 as a simple division by 8 might suggest.

"and not 1275 as a simple division by 8 might suggest"

sorry  - should be "and not 1250 as a simple division by 8 might suggest"

yikes!!!! and i thought i was going to get a quick small answer saying (****)==(****)

thanks all  :)

@mnamiri

But my answer is not correct!
I made a mistake.

http:#15342405 is correct.

Anyway, I will request for a reopen of this question so you can accept the correct comment.

>>(Max theoretical througput) and 10Mbps = 1220.703125 KBps (Kilobytes per second).

Max theoretical throughput is my way of saying - including protocol overhead and interframe gap.

https://www.experts-exchange.com/questions/21639913/Reopen-pls.html

@mnamiri

This question has been reopened. I admit my answer was incorrect and am sorry to mislead you. http:#15342405 is the correct answer. Please accept pseudocyber's comment.


hongjun

Thanks all. :)

:)