Bits bytes transfer rates??? :S

bps(n0*x) + lin(n1*x) + num(n2*x) + [..] + relative weight of an africaan finch(nn*x)=oo   (-1)^k x^(2k+1)  oo
sum - sum (ni)^(2k+1)-k=0      (2k+1)!     i=0= true speed of transfer

does this help?

No....

oldgreyguy is yanking your chain ... posting in the Lounge for actual technical solutions is bound to get you a light-hearted or whimsical response.

"oldgreyguy is yanking your chain"...I realised that, although I don't know where to put this question since it doesn't really fit into much of a category...Any suggestions?? :S

Sounds like Networking would be appropriate.  You can post a note in Community Support to move this.  When in doubt, Miscellaneous seems to be a catch-all.

I am not a chainyanker... but I am glad UniPerson understood the jest


..... hmmmm chainyanker... is that word allowed in the lounge?

It's a Merkin saying ;-)

Try the Homework topic area. You're sure to get loads of help there.

Define "characters"...

i'm going to guess about 6 and 2/3rds seconds

>>My usual rule of thumb is to divide the bit-rate by 10 to guesstimate the byte-rate

Actually if the bit rate applies to a modem the 10 "by rule of thumb" is actually correct when you consider the stop bits per byte. If you assume a PPP connection which can have frame sizes greater than 1000 bytes and add a couple of bytes on for the compressed header (Van Jacobson "on") then anything around 8.4 to 8.5 seconds is going to be correct.

Depends, if it's a direct serial connection, no modems involved, figure 10 bits per char for async, 8 bits for sync, but add a few bytes of overhead.

If it's going thru modems, many modems will try to do a bit of compression, if it's text, it can easily be compressed by 30-50% so times will vary....

Thanks modulo :)

----------------------

"Try the Homework topic area. You're sure to get loads of help there."....thanks for your sarcasm. I'm actually preparing for an exam and I don't know how to do this question on the practise paper.

Right, so if I'm using a synchronous connection the connection is faster since I don't take into account the overhead like the ACK/NACK's. A calculation like the following would be good enough then:

7x1,000=7,000 bits
7,000/1,200=5.833...seconds?

Although if I'm using asynchronous and there's an overhead of 10 bytes it'd be something like:

7x1,000=7,000 bits + (10x8) = 7,080
7,080/1,200=5.9...seconds?

Somehow that doesn't seem too right... :S Anybody?

Youre right, its not right.

Sychronous doesnt imply ack/nack.   Acks are a separate error correcttion protocol, a whole different layer.

Synchonous usually means the data is sent in blocks, usually with a few preceding SYNC bytes.  So theres no start or stop bits on each byte, but a few sync bytes on each block.

Async by itself doesnt imply any ac/nacks either, but it does mean at least one start and stop bit around each byte.  Sometimes 1.42, 1.5, or 2 stop bits.  

> I'm actually preparing for an exam and I don't know how to do this question on the practise paper

Sarcasm aside, studying for exams *is* homework....

Thanks all! :-) Sorry for the delay been a bit busy lately...

daydreams,

>>Please do not accept my answer if you plan to grade it less than an A


they must have missed that...  ;-)


that's ok - i was the first one to take a guess, which was right in the middle of the too accepted answers and i got nothing....  :-)

>..and i got nothing....  :-)

I would rather get nothing than get a B!! That's just me. I answered this before it got moved from the "Lounge Resource Channel", that's why I said that about the A grade.