Cisco IOS Min. Memory for Routers

[kenneth.rogers]

I assume this is the best place to post regarding IOS.  We are beginning the process of upgrading our IOS images on 1841 and 2811 series remote routers.  With the new 12.4.24.T2 code in advance enterprise the min. memory required is 256MB.  Our routers contain 239MB as ordered.  The code has loaded successfully on several test routers.  My concern is there any long term effects of being below Cisco's recommended memory for the code.

Since we are below the requirement, what commands and statistics would you all suggest watching closely to ensure the stability of the router with its internal processes.  The heaviest "load" we could put on the routers, config wise, will be with crypto maps, ZBF and WAAS

I look forward to your thoughts.

thank you

[Leo Laohoo]

Can you post a "sh ver" of both models of the routers please?

[kenneth.rogers]

Leo.....please see attached .txt for the sh ver.

thanks

[Leo Laohoo · Accepted solution]

Cisco 2811 (revision 53.51) with 247808K/14336K bytes of memory.

247 + 14 = 261mb

Cisco 1841 (revision 5.0) with 235520K/26624K bytes of memory.

235 + 26 = 261mb

[kenneth.rogers]

Thanks so much Leo for the info.  I guess I was looking at the NVRAM stats instead of the memory.  Learn something new everyday to look for!

thanks again

[Leo Laohoo]

Glad to be of assistance.

[JORGE RODRIGUEZ]

kenneth,   to add to leo's post ..

For future reference you can always look at  2800 ISR series specs  table.. found at http://www.cisco.com/en/US/prod/collateral/routers/ps5854/prod_qas0900aecd80169bd6.html

2800 series routers at least comes from factory with default memory listed in above link  ,  for the 2811 series default is 256MB,  and  maximun  RAM capacity of 768MB   , 2811 have two slots for adding more mem which may be a combination of ONE 256MB and ONE 512 to equal 768MB Max MEM.

In any given router you may  retrieve absolute memory size the device have by simple calc ,   from your 2811 sh ver output  Cisco 2811 (revision 53.51) with 247808K/14336K bytes of memory   [ simply add both figures   247808 + 14336  =  262144  devided by 1024 = 256  ]  which will give you 256MB .

Regards