Anyone can explain how following solution would work for burst traffic of 7.5M and queue length of 20 packets as asked by section 7.4
No idea. I have the same question. I did this because of that 20 packet and max burst to 7.5 Mbps comment. I'm not sure if this will even work:
access-list 174 permit ip 18.104.22.168 0.0.0.127 any!class-map match-all XYZ321 match access-group 174!policy-map XYZ321 class XYZ321 police cir 2500000 pir 7500000 conform-action transmit exceed-action transmit violate-action drop bandwidth 2500 queue-limit 20!map-class frame-relay FRTS service-policy output XYZ321
Isn't this policy applied to the interface which was configured with FRTS in previous task? If yes, limit 20 packets and burst up to 7.5M are controlled by map-class/FRTS.
So remaining part would be only about guarantee 2.5M in case of congestion.
This is my understanding of the task/solution.
That is how I see it. The lesson of this task is... Beware of the wording of each task and factor in previous task objectives if necessary.