Task 8.1

Hi,

Any body can do a deep dive into how 8.1 SG is being derived out?

Comments

  • Bc= CIR/tc = 5000000/0.032= 160000

     

    Be=(Pir -Cir )/Tc= (7500000 - 5000000)/ 0.032 = 80000

     

    that's why you need :" shape average 5000000 160000 80000"

    Now ...Can someone tell me why the bandwidth is set to 45000 in the interface serial 0/0?

  • I believe its because of the DS3/T3 link which is 44.736 Mbit/s or 45000 Kbps

     

    HTH,

  • Yes,you are right I totally missed it.

    Thanks!

  • If the bandwidth command was left off the interface, wouldnt the shaping still occur?  OR is it because since the default bandwidth of 1544kbit a speed greater than that can never be achieved.

     

    Thanks.

  • You need to set the bandwidth to 45000 for your bandwidth reservations, if you do not set the bandwidth and stick with the default you will not have enough bandwidth to allocate for your qos classes.

    The bandwidth has nothing to do with the speed. It is used to signal routing protocols and QoS  of the correct bandwidth of a link

    HTH

    Good luck!

  •  

    Hi Qqabdal,

    This is correct, just like what case is with minCIR and CBWFQ/LLQ. I found the following by Scott Morris as useful info:

    "Each policy can only use what is available to it.

     "bandwidth" command on the interface will obviously limit the top policy.

     Each
    lower policy gets limitations from any listed values, which include the
    "bandwidth" or "priority" guarantee as well as the "shape" or "police"
    limiters.  The lower of any values (worst case scenario) is what is used
    to calculate information for any sub-policies."

    QOS- CBWFQ/LLQ- Nested Policy-map configuration with bandwidth percent

     

    Regards,

    Bassam

  • I was pretty sure that I applied the shaping without having the bandwidth applied to the interface.  I will check again.

  • Bc= CIR/tc = 5000000/0.032= 160000

     

    Be=(Pir -Cir )/Tc= (7500000 - 5000000)/ 0.032 = 80000

     

    that's why you need :" shape average 5000000 160000 80000"

    Now ...Can someone tell me why the bandwidth is set to 45000 in the interface serial 0/0?

    you got it wrong. BC = 5000000*0.032 = 160000

    like wise for the second equation. you need to multiply not divide. 

  • Would it be ok for that task to set the bandwidth to 5000 ? That would be enough to guarantee bandwidth to our queues, and it would more closely reflect the actual bandwidth available to us.

    Would that be ok during the exam ?

     

    Thanks.

  • Hello,

     

    I thought that the way to calculate Be = AR-CIR/Tc. Where in this question, AR will be 45M???

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