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Appendix K
CDM-760 Advanced High-Speed Trunking Modem
Step 1 – The sum of all queues' minimum bandwidth (20000 kbps + total available bandwidth) =
30000 kbps. Once the sum of all minimums are met, there is a remainder of 10000 kbps overage
available for use.
The overage bandwidth is distributed to all non-served maximum bandwidth queues. In this
example, only Q1, Q2, and Q3 still have unmet maximum bandwidth.
Since Q4's minimum and maximum bandwidths are the same, Q4 is therefore excluded from the
remaining overage distribution.
The remaining total weights are (9+8+7) = 24.
Step 2 – Calculate each bandwidth share, based on total weights (24) and bandwidth availability
(10000 kbps) after the minimum is served (20000 kbps):
•
Q1 = 10000 * (9/24) = 3750 kbps
•
Q2 = 10000 * (8/24) = 3334 kbps
•
Q3 = 10000 * (7/24) = 2916 kbps
Step 3 – After the initial round of quantum distribution, Q1 required only 2000 kbps, so the
remaining 1750 kbps is therefore returned to the allocation pool for distribution to the
remaining queues.
Step 4 – Calculate each bandwidth share, based on total remained weights (15) and remain
bandwidth (1750 kbps):
•
Q3 = 1750 * (8/15) = 934 kbps
•
Q1 = 17500 * (7/15) = 816 kbps
Step 5 – After the second round, Q2 can be assigned a maximum of 4000 kbps, and the
remaining kbps is again returned to the allocation pool.
Step 6 – In the third round, only one queue remains and its maximum is not yet met. That queue
therefore takes the remaining quantum. Since there is no more leftover bandwidth, the
calculation loop stops here.
Step 7 – Each queue's share after all rounds:
•
Q1 = 2000 kbps
•
Q2 = 4000 kbps
•
Q3 = 4692 kbps
•
Q4 = 0 kbps
Note that the per-queue quantum is less than the configured minimum bandwidth.
K–9
Revision 4
MN-CDM760
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