Jon Marshall Fri, 07/06/2007 - 08:46
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Hi


Well you could do


192.168.6.40 0.0.0.7

host 192.168.40.49


although this is dependant on how 192.168.6.0 -> 39 hsa been subnetted (if indeed it has ).


HTH


Jon

dgahm Fri, 07/06/2007 - 11:10
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Sparky,

Here is the last octet binary for your range of numbers.


00101000 40

00101001 41

00101010 42

00101011 43

00101100 44

00101101 45

00101110 46

00101111 47

00110000 48

00110001 49


Note that only the 128, 64, and 32 bit value positions do not change, and are always 001 (32). If you use a mask of 0.0.0.31 to "don't care" the changing bits, you get a range of 32-63 for your summary. So a single summarization exceeds your range.


If you move two more bits right with the must match position (00101 or 40) the remaining 3 "don't care" bits mask 0.0.0.7 give you a range of 40-47. This is closer, but you still need 48 and 49.


The 48-49 range can be matched with a 0.0.0.1 mask.


So, 2 statements are required:


192.168.6.40 0.0.0.7

192.168.6.48 0.0.0.1


Please rate helpful posts.


Dave



Jon Marshall Fri, 07/06/2007 - 11:46
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Hi Dave


Thanks for catching that. My mistake with the .48 address.



Deserves rating


Jon

Jon Marshall Fri, 07/06/2007 - 11:46
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Hi Dave


Thanks for catching that. My mistake with the .48 address.


Apologies to sparky.


Deserves rating - only it won't let me rate it. Any ideas why - does Sparky have to rate it first ?


Jon

dgahm Fri, 07/06/2007 - 13:11
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Jon,

Fortunately perfection is not a requirement for posting on NetPro -- if it was, none of us would be here.


Strange on the rating, never had a problem with it.


Dave

Jon Marshall Fri, 07/06/2007 - 22:37
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Dave


Thanks for this. I stil can't rate this post which i don't understand so i have rated your multicast post in IP telephony which worked !!


Jon

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