Is there a formula describing the relationship between bit error rate, frame size, and normalised effective throughput?
to clarify, normalised effective throughput is the same as channel utilisation.
so far I have "throughput = 1/(1+2a)" where "a = tprop*R/L" (tprop=propagation time or delay,R = bit rate), however I am unsure how bit error rate fits into this – does bit error rate have much to do with bit rate.
Any help/pointers/where to get help would be appreciated. are there any other forums that I should post this on?
EDIT:
Just to give some more background info:
basically this is for a project(topic: analysis of optimal frame size and bit error rate relationship on lossy links), where we have to write a MATLAB GUI (graphical user interface) and the inputs are bit error range (ie max and min bit error rates) and frame size range (max and min frame/packet sizes). We have to calculate the normalised effective throughput of the link from those inputs, and make a 3d graph.(x:BER, y:FSR, Z:throughput)
@clabacchio i was talking about general parameters – L stands for number of bits in a frame.
@kellenjb we haven't been given any info on how errors are handled, so i suppose we just make an assumption. (and i deleted the other posts, sry about that)
Best Answer
Yes, you will be able to arrive at a formula which will at best be as useful as the accuracy of the assumptions that you make, and which will have a statistical component due to effects of stat variation of noise and resync and ... on the results.
You'd have to define things rather more tightly before you could get a useful formula.
eg the data getting through is D_sent x (1-BER)
(BER fractional rate eg 1 in 10^6 = 0.000001).
BUT ANY error either needs a resend or has forward correction overhead.
If any error occur at all then you need a resend and
throughput drops to D_sent - (packets with error(s) in them per second = pweitps) x (packet data content length )
And you also have the back channel overhead which may be part of your total bandwidth budget or may not.
To convert BER to pweitps you either assume every single error destroys a packet (as above), or you need to know statistical grouping of errors.
If each error bit loses one packet and if net error free data rate is D and if packet length is P then you lose BER x P bits/second due to errors so rate simplistically reduces to ~= (D - BER x P)/ D of previously.
If you lose synchronisation due to errors and must resynchronise that is a rate reduction and you need to know how resync varies wih error rate and types of error hit.
And much more.
SO the answer is yes, a formula can be derived which is as useful as the assumptions that you make and which will hav a ststaistical component due to effects of stat variation of noise and resync and ... on results.