Queue-Limit Formula

From: Matt Bentley <mattdbentley_at_gmail.com>
Date: Fri, 8 Oct 2010 15:07:44 -0600

*Hi GS:*
*
*
*Question about queue-limits. Seeing a lot of output drops on a serial
interface - I believe due to the default queue-limits within the policy-map
applied to it. In times past, I've just picked a bigger queue-limit value
and "hope" that takes care of it.*
*
*
*RTR1#sh int s1/0*
*Serial1/0 is up, line protocol is up*
* Hardware is DSXPNM Serial*
* Description: to RTR2*
* Internet address is x.x.x.x/30*
* MTU 4470 bytes, BW 8000 Kbit/sec, DLY 200 usec,*
* reliability 255/255, txload 42/255, rxload 82/255*
* Encapsulation PPP, LCP Open*
* Open: IPCP, crc 16, loopback not set*
* Keepalive set (10 sec)*
* Last input 00:00:00, output 00:00:00, output hang never*
* Last clearing of "show interface" counters 1w3d*
* Input queue: 0/75/43/0 (size/max/drops/flushes); Total output drops:
209319 <<== Output Drops*
* Queueing strategy: Class-based queueing*
* Output queue: 0/1000/0 (size/max total/drops)*
* 5 minute input rate 2599000 bits/sec, 701 packets/sec*
* 5 minute output rate 1339000 bits/sec, 628 packets/sec*
*
*
*So, the 5 minute output average is 1339000 bits per second, or 167,375
bytes per second. At 628 packets per second, this means average packet size
of around 245 bytes on average.*
*
*
*So assuming the default queue-limit (6**4-packets) and default packet size
(245bytes), the class could hold 15,680 bytes, or 125,440 bits at any one
time before the queue-limit is breached and subsequent traffic is dropped.
 On a 8Mbps WAN interface, it would take about 15ms to send 125,4000 bits.*
*
*
*So, that makes the classes able to absorb a 15ms burst? **Is that correct?
*
*
*
*Is there a formula you use to determine what the queue-limit should be,
based on packet size, the bandwidth of the class, etc.,*
*
*
*I'm very confused. Thanks,*
*
*
*
*

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Received on Fri Oct 08 2010 - 15:07:44 ART

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