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atanis

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Connection speed

L.S,

How should I convert "kbps" into "K/sec"?
Internet connection monitoring software like NetMedic gives kbps and Netscape Communicator K/sec.
Some people say, divide kbps by 8, some by 9 or 10 depending on the part of the routing.

What is correct?

Adrie Tanis
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tstrob

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atanis

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Thanks for your reply.

If I understand it correctly, it means that my modem is working at a slightly higher speed.
Let's say, the above mentioned software shows  2.5 K/sec or 20.000kbps; what would be my modem speed (internal 28K8) or the speed along the line from my ISP?
Could you elaborate a bit on that?
atanis,

Let me try to explain better...

The speed given by your modem manufacturer is alway the speed along the line to your ISP. (This is the heigher value)
The Data your computer transfers is a little bit slower because of the overhead (Stopbit, and Parity)

Example:
Your modem has 28K8 that means: Devided by 8: Your modem transmitts 3600 Bytes per second to your ISP ( this includes overhead)
If you have set the connectionsettings to  8,N,1 (8 Databits, No Parity, 1 Stopbit), your modem has to transfer 9 Bits for 1 Byte your Computersoftware transmitts.
So the speed your software can transmitt with is only 3200 Bytes per second (400 Bytes per second overhead for stopbits)

Your software only can show you the slower value. To convert speeds from different softwareproducts (as asked in your question) you always have to devide/multiply by 8, because this are both values without overhead.

So, your software shows 2.5K/sec  = 20.000kbps is the data your software transmitts.
But over the line you transmit 22500kbps because your modem sets a stopbit after 8 Databits.

Hope this helps...
Don't forget about "compression".

If you send, say, 12000 8-bit bytes to the modem,
the modem may compress the 96000 bits,
so it may need to send as little as 32000 bits,
i.e., using a four-to-one compression ratio.
The modem groups that 32000 bits into 4000 bytes,
and sends one "start" bit, eight "data" bits, and one "stop" bit, for each byte.
So 40000 bits are sent, at, say, 33600 bits/second, thus taking just over one second.

So, your "effective" transfer-rate is
rate is 12000/(40000/33600),
i.e., 10080 (8-bit) bytes/second (9.8Kb/Sec.)
even though the modem indicates 3360 (10-bit) bytes/second.