[Cst-1b] FEC/ARQ protocol

[Hugh Nimmo-Smith]

Sorry for slow reply.

Assumptions... erm... CRC identifies all errors... maybe delay is constant
or something.

Not really sure...

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 21:07
> To: Hugh Nimmo-Smith
> Cc: cst-1b@srcf.ucam.org
> Subject: Re: [Cst-1b] FEC/ARQ protocol
>
>
> Dear Hugh
>
> Thanks
>
> What assumptions have you made ?
>
> The assumption I made : that CRC correct all errors
>
> Love
> Phebe
>
> On Sun, 30 Apr 2000, Hugh Nimmo-Smith wrote:
>
> > 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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> > >
> >
> >
>
>
>
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