i do have 2600 series router connected to both of my branch - HO connectivity with WIC 2T. bandwidth is 2 MB and i have set the 'bandwidth 2048' on interface config mode on both side. when i check for 'sh int serial 1/1' i got following results.
available Bandwidth is 1536 kb only. is it a real measurement since i set both side BW as 2048 kb. ? 1536 means, right now router transferring data over the only at 1.5 megabit/sec..? what mean by Output queue xx/xxxx/xxx..? i am not very deep into this things.
another problem is bandwidth availability. when i a copy 15 MB it is using only 40 % of my bandwith (MRTG graph display), so copying 15MB take more time. why it cannot use the entire bandwidth to copy the file. is there any command to use full bandwidth. .?
tough my questions are bit lengthy i these kinda basic doubts will clear others too.
The command " bandwidth 2048" is only for the routing protocol to understand. The routing protocol you configure will decide the best path ( OSPF,EIGRP) to take depending on the interface bandwidth command.
If you have the clock set in ur CSU/DSU as 2048 Mb, then you will surely get the bandwidth.
Available bandwidth=1536 means that BW available for data trafffic is 1536 Mbps. Rest is used for routing updates, keepalives traffic.
Go to the interface configuration where the 15 Mb passes through. Change the load-interval to 30 seconds. The load-interval refreshes the measured load under your 'show interface' output. the load is displayed from 0 to 255, where 0 is no data and 255 is 100% of linkusage. The load is calculated on the bandwidth parameter under you interface configuration. So if you've configured 2048 and you only have 1536 then 191 will be the maximum load of your link (1536).
When you are copying 15 Mb login in to your router and type 'show interface serial 0/0' or whatever interface you are using. You should see something like this;
reliablility 255/255, txload 6/255, rxload 76/255
Examine these values as they are the most accurate about the amount of Kb/s.
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