Confused how it works?
Simple example. . .
Suppose you have a metro-E connection with a 100 Mbps Ethernet hand-off, but you've only contracted for 15 Mbps of bandwidth. (Why do this? It allows for easy bandwidth growth.)
What happens if you exceed your contracted bandwidth allowance? Well, your metro-E vendor might just discard all the excess traffic or they might charge a premium for excess usage. For either reason, we decide we don't want to violate our contracted bandwidth.
So to insure we don't violate our metro-E contract, we do:
policy-map ShapeExample
class class-default
shape average 15000000
interface fastethernet 0/0
service-policy output ShapeExample
What the above does, it "counts" the transmitted bits during some time interval (for this explanation, we'll use one second intervals, but actual time intervals are usually in the 25 to 40 millisecond range), and once the logical limit is reached, excess will be queued. If there are queued packets, they will be dequeued, again as to not exceed the logical bit limit for the time interval. (Although frames/packets are what are queued/dequeued, we'll use bit counts and ignore actual frames/packets, because the bit volume of the frames/packets is what is really considered.)
To start, during the first second we "count" 10,000,000 bits. These pass straight through.
For the next/second we "count" another 15,000,000 bits, which again, we allow to pass through, but more bits want to be transmitted. We queue all of those, which are another 45,000,000.
For the next/third second, since we have 45,000,000 bits queued, we transmit another 15,000,000 and transmit no more. During this third second, we also receive another 5,000,000 bits, which are also queued. Our queue now has 45,000,000 less 15,000,000 plus 5,000,000 or 35,000,000 bits.
For the next two seconds, we dequeue 15,000,000 bits each second. During those two seconds no additional traffic is seen by the shaper. So now we have 35,000,000 bits less 30,0000,000 leaving 5,000,000 bits in the queue.
Lastly, during the next second, we dequeue the final 5,000,000 bits but we also receive another 5,000,000. Since the combination is only 10,000,000, which is still below our 15,000,00 bits per second limit, we let them pass through too.
What the shaper has done, is insure our average transmission rate has not exceeded 15 Mbps, although the circuit physically transmits at 100 Mbps.
