[Fwd: Re: [Cst-1b] FEC/ARQ protocol]

[Nathan Dimmock]

Hi Garan

Archives of this list are available at
http://www.srcf.ucam.org/pipermail/cst-1b/

All the best,

Nathan

-------- Original Message --------
Subject: Re: [Cst-1b] FEC/ARQ protocol
Date: Sun, 30 Apr 2000 11:00:01 +0100
From: "Hugh Nimmo-Smith" <hrfn2@cam.ac.uk>
To: "Phebe Mann" <pm258@hermes.cam.ac.uk>,<cst-1b@srcf.ucam.org>

This is a solution (I think):

delay = round trip time = lamda

data rate without FEC = d1 = 10^6
data rate with FEC = d2 = 5^6

error rate without FEC = e1 = 10^-4
error rate with FEC = e2 = 10^-5

probability of a bit being OK without FEC = 1 - e1
probability of packet OK without FEC = p1 = (1 - e1)^1000

probability of a bit being OK with FEC = 1 - e2
probability of packet OK with FEC = p2 = (1 - e2)^1000

average number of times you have to send packet without FEC = 1/p1
average number of times you have to send packet with FEC = 1/p2

time to send a single (possibly with error) packet without FEC = time to
send 1000 bits + link delay = 1000/d1 + lamda
time to send a single (possibly with error) packet with FEC = 1000/d2 +
lamda

so average time to send a packet successfully without errors...

without FEC =  time1 = (1000/d1 + lamda) * 1/p1
with FEC = time2 = (1000/d2 + lamda) * 1/p2

you then say what happens when the two times are equal:

time1 = time2

and solve for lamda where the two systems give the same throughput (or
whatever)

(lamda + 1000/d1) / p1 = (lamda + 1000/d2)/ p2

lamda * p2 + 1000*p2/d1 = lamda * p1 + 1000*p1/d2

lamda * (p2 - p1) = 1000 (p2/d1 - p1/d2)

etc... (if my algebra skills haven't all vanished!)

and then say that if link delay is less than lamda then it is better to
not
use FEC otherwise FEC is worthwhile.

Hope this helps a bit

Hugh

> -----Original Message-----
> From: cst-1b-admin@srcf.ucam.org [mailto:cst-1b-admin@srcf.ucam.org]On
> Behalf Of Phebe Mann
> Sent: 30 April 2000 02:51
> To: cst-1b@srcf.ucam.org
> Subject: [Cst-1b] FEC/ARQ protocol
>
>
> Dear collegues
>
> Please take a look and let me know whether it's correct or not.
>
> 99/6/3 (final part)
>
> Information is transferred via a long, error-prone communication link. The
> link has a data rate of 10 Mbps and a constant delay. The bit error rate
> on the link is 1 bit in 10^4. A forward error correcting coder is
> available which can act in the following settings :-
>
> A simple ARQ protocol is used over the link. Packets have 32-bit CRCs. You
> may assume that the undetected error rate is less than 1 in 10^20, that
> is, effectively zero.
>
> Information is sent in 1000-bit packets, with a window of one packet. At
> what link delay would it be beneficial to use the FEC coder ?
>
> My Answer
>
> No FEC, with ARQ
>
> t1 = (1+10/10) (2 x Td + (1/10^4))
>
> (is 11/10 the right ratio to be here ? can I take any other ratio ?)
>
> With FEC, assume CRC correct all errors
> (what other assumptions do I also have to make ??)
>
> t2 = 2 x (Td + (1/10^4)
>
> m/k = 1/2
>
> For FEC to be beneficial
>
> t2 < t1
> 2 Td + (2/10^4) < (11/10) (2 x Td + (1/10^4))
> 2((11/10) - 1) Td > (2 - 11/110)(1/10^4)
> Td > ((20 - 11)/(1/5)) (1/10^4)
> Td > ((5x (9/10) )/ 10^4
> Td > (9/2) x 10^-4 s
> Td > 4.5 x 10^-4
>
> (is that the right approach ? if not, why )
>
>
> Thanks
>
> Phebe
>
>
>
>
>
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