Hello, im getting ready to take my 642-642 Qos test and i dont understand a couple of anwsers of pass4sure regarding the same subject
wich is calculatintg Tc


if Tc = Bc / Cir


Which of the following police commands would be required to enable single-rate policing at
approximately 128 kbps, with the Bc set to cause Tc = 10ms? (Note that a comment line shows
where the police command would be added to the configuration.)


a. police cir 128000 Bc 1280 conform-action transmit exceed-action transmit
violate-action discard
b. police cir percent 8 conform-action transmit exceed-action transmit violate-action
discard
c. police cir 128000 Tc 10 conform-action transmit exceed-action transmit violateaction
discard
d. police cir percent 8 Bc 10 ms conform-action transmit exceed-action transmit
violate-action discard

the right accordign to p4s is b

0.010(10ms) = Bc / 128000
Bc = 128000 * 0.010 , wich is not 8% of 128000

In here is the same calculation i dont understand

what command would list the

calculated Tc value, and what would that value be?


! point 2
shape average 256000 5120


a. show policy-map, Tc = 125 ms
b. show policy-map, Tc = 20 ms
c. show policy-map, Tc = 10 ms

the chosen by p4s is c, i though it was b

Tc = 5120 / 256000

5120 / 256000 = 0.02 (20 ms)


Im probably wrong but i´d like to understand why

Thanks in advace, cesar

I'm coming up with the same answer as you cesar - 0.02 (20 ms) but the book is stating the answer is c (10ms). I understand why.

I meant ..." I don't understand why". Sorry in case there was any confusion from my previous answer.

Hello Cesar.

I find two mistakes about this question:

1) As you mention, Tc must be 20ms.

2) The command "show policy-map" only shows the policy-map configuration. The command "show policy-map interface" shows statistics. It is in the last command where you can find the Tc value (refer to the Cisco Command Reference or the QoS exam Self-Study Guide examples).

Best regards.