From: daniel (danyccie@rediffmail.com)
Date: Sun Apr 01 2007 - 22:41:45 ART
Thank you for your reply. I find an issue with this summerisation.10.10.103.0/24 - 10.10.13.110.10.104.0/23 - 10.10.13.1These two covers 103,104 and 105
10.10.100.0/22 - 10.10.13.2 10.10.106.0/24 - 10.10.13.2These two covers 100,101,102,103 and 106There is an overlap of 103 which will create a loop in my scinario and goof up things.So can you give me a little more accurate solution.Thanks,Dany.On Mon, 2 Apr 2007 06:15:42 +1000 \"CCIE 19999\" wrote Dany, I think, if you do the following summarization, you should be able to squeeze the 6 routes into 4. 10.10.103.0/24 10.10.104.0/24 & 10.10.105.0/24 all going out through 10.10.13.1 When converting the third octet into binary we get the following: 0110 0111 - 103 0110 1000 - 104 0110 1001 - 105 Consider only last two and the difference is only in the last bit, and we can summarize these 3 routes into two as: 10.10.103.0/24 - 10.10.13.1 and the summarised 2nd and third as: 10.10.104.0/23 - 10.10.13.1 By applying the same technique you can summarise the 10.10.101.0/24 & 10.10.102.0/24 into 10.10.100.0/22 - 10.10.13.2 and 10.10.106.0/24 - 10.10.13.2 Hope this helps. RS ----!
- Original Message ----- From: \"daniel \" To: Cc: Sent: Monday, April 02, 2007 3:34 AM Subject: Re :Re: Re :Re: Static Route > OK. I understand.No let me change the scinario a little bit.How can I > squeeze the following static routes in to 4. > router#(config) ip route 10.10.103.0 255.255.255.0 > 10.10.13.1router#(config) ip route 10.10.104.0 255.255.255.0 > 10.10.13.1router#(config) ip route 10.10.105.0 255.255.255.0 10.10.13.1 > > router#(config) ip route 10.10.101.0 255.255.255.0 > 10.10.12.1router#(config) ip route 10.10.102.0 255.255.255.0 > 10.10.12.1router#(config) ip route 10.10.106.0 255.255.255.0 > 10.10.12.1Regards,Dany.On Mon, 2 Apr 2007 00:30:55 +1000 \"Radioactive > Frog\" wrote XOR the 3rd ocatate and put tha tin the ACL Similar to below > and I believe that you can squeeze your routes/ACL\'s in 3 lines. 128 64 > 32 16 8 4 2 1 103= 01100111 104= 01101000 105= > 01101001 ------------------------ = 0000111 = 4+2+1 = &!
gt; 5 ----------------------- 10.10.103.0 255.255.5.0 > >!
; Hi, &g
t; > > Thank you for your reply. My problem is I\'m allowed to use only 4 > > access-lists to acheive the same. Please tell me how to do this. > > Thanks, > > Dany. > > > > On Sun, 1 Apr 2007 19:28:13 > +1000 \"Radioactive Frog\" wrote Hi Daniel, It\'s > preety much very > simple. Assuming you\'ve cisco router. Otherwise every > router have > their different interface to configure. 1) Go to config t > ro! > uter#config t router#(config) ip route 10.10.103.0 255.255.255.0 > > 10.10.13.1 router#(config) ip route 10.10.104.0 255.255.255.0 > 10.10.13.1router#(config) ip route > 10.10.105.0 255.255.255.0 > 10.10.13.1 2) router#config t router#(config) ip > route 200.200.101.0 > 255.255.255.0 10.10.12.1 router#(config) ip route > 200.200.102.0 > 255.255.255.0 10.10.12.1 router#(config) ip route > 200.200.106.0 > 255.255.255.0 10.10.12.1 HTH, Frog > > Regards, > Dany > ____________!
___________________________________________________________ > Subscription information may be found at: > http://www.groupstudy.com/list/CCIELab.html > > Regards, > Dany > > _______________________________________________________________________ > Subscription information may be found at: > http://www.groupstudy.com/list/CCIELab.html _______________________________________________________________________ Subscription information may be found at: http://www.groupstudy.com/list/CCIELab.html
Regards,
Dany
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