Juniper Networks G10 CMTS manual Ideal HFC Plant Configuration Issues

Ideal HFC Plant Configuration Issues

! The cable modem belongs to a downstream or upstream interface on which a traffic	•
scheduling policy is assigned. Packets that exceed the maximum sustained traffic rate	•
(MSTR) are dropped or shaped, depending on the traffic scheduling policy configuration.	•
Issue the show cable policy traffic-schedulingcommand to display configured traffic	•
scheduling policies.	•
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! The cable modem belongs to a downstream or upstream interface on which a	•
congestion control policy is assigned, such as random early detection (RED), leading to	•
dropped packets. Issue the show cable policy congestion-controlcommand to display	•
configured congestion control policies.	•
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! Congestion exists in the upstream.	•
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You can compute the approximate upstream channel utilization by monitoring the	•
ifInOctets object in the DOCS-IF-MIB (RFC 2670). The ifInOctets object contains the total	•
number of octets received on an interface, including data packets as well as MAC layer	•
packets, and includes the length of the MAC header. However, this object does not	•
account for the PHY layer overhead—preamble, FEC, and guard time—which consumes	•
a certain percent of the available raw channel bandwidth. The following procedure	•
explains how to compute the approximate upstream channel utilization using an SNMP	•
MIB browser:	•
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1. Set the SNMP polling time to a value large enough to capture a statistically	•
significant amount of upstream traffic. In this example, assume the polling time is	•
60 seconds.	•
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2. Browse the ifInOctets object for the interface that corresponds to the upstream	•
interface you are measuring. Wait for the value of the object to change and record	•
this value. Assume the value is 33,019,041 octets.	•
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3. Wait 60 seconds for the value of the object to change and record this value. Assume	•
the value is 65,903,162 octets.	•
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4. Subtract the value of the object measured in step 2 from the value measured in	•
step 3 to obtain the number of octets received by the CMTS on this upstream	•
interface over the polling time: (65,903,162–33,019,041=32,884,121 octets).	•
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5. Multiply the value computed in step 4 by eight (to convert to bits), then divide by	•
the polling time to compute the upstream channel bandwidth (without the PHY	•
overhead): [(32,884,121 octets * 8) / 60 sec]=4,384,549 bps.	•
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6. Compute the maximum available raw bandwidth by multiplying the symbol rate of	•
the channel by the number of bits/symbol. Assume the symbol rate is 2560	•
Ksym/sec, and the modulation is QPSK (2 bits/symbol), which yields a bandwidth of	•
5,120,000 bps.	•
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7. Computing the PHY overhead, and hence the channel efficiency, is a non-trivial	•
exercise because it is dependent on the mix of transmissions that use a particular	•
interval usage code, packet sizes, and the modulation profile and mini-slot size of	•
the channel. Using practical values for these variables, assume a channel efficiency	•
of 92 percent (be aware that the channel efficiency can be lower depending on the	•
assumptions made). Derating the maximum available raw bandwidth of the	•
channel by 92 percent yields 4,710,400 bps (5,120,000 bps*0.92).	•
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Troubleshooting 153