From: Tim Chan (timanji@yahoo.com)
Date: Fri Jul 14 2006 - 02:00:21 ART
The formula is:
Bc = CIR * Tc
CIR = Bc / Tc
Tc = Bc / CIR
Tc in this case is 0.032
spycharlies@hotmail.com wrote: Hey Guys, i came across this question in one of the workbooks(IEWB), i totally have no idea on how the Bc was calculated.
R1,R2 & R3's serial interface connects to the frame cloud
R1 supports maximum transmission rate of 896kbps
R2 -> max. of 640kbps
R3 -> max. of 1536kpbs
Task
-----
Configure Frame Relay traffic shaping on these devices to allow
the highest possible average rate without any circuit becoming overwhelmed.
R1 & R3 shld be able to transmit a max of 3584 bytes per shaping interval
on the circuit btw them, while R2 & R3 shld be able to transmit a max of 2560 bytes.
Here is the solution
---------------------
R1:
map-class frame-relay 896Kbps
frame-relay cir 896000
frame-relay bc 28672
!
interface Serial0/0
frame-relay class 896Kbps
frame-relay traffic-shaping
R2:
---- map-class frame-relay 640Kbps frame-relay cir 640000 frame-relay bc 20480 ! interface Serial0/0 frame-relay class 640Kbps frame-relay traffic-shapingR3: ---- map-class frame-relay 640Kbps frame-relay cir 640000 frame-relay bc 20480 ! map-class frame-relay 896Kbps frame-relay cir 896000 frame-relay bc 28672 ! interface Serial1/0 frame-relay traffic-shaping frame-relay interface-dlci 301 class 896Kbps frame-relay interface-dlci 302 class 640Kbps
Any hints on how the Bc was calculated????
thanks in advance
cheers uyota
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