02-09-2011 07:14 AM - edited 03-06-2019 03:26 PM
what would be the best Route Summarization for this network which cover all 4 network
192.168.1.100/30
192.168.1.104/30
192.168.1.108/30
192.168.1.112/30
pls mention the method and one more request could any body tell me the summarization method so i learn do to summarization
if i break last octat number in binary
100 - 1100100
104 - 1101000
108 - 1101100
112 - 1110000
only first two bit from left side match so as per my calculation it summary route would be 192.168.1.100/26 but i don't know i did right or wrong so pls also tell me how to know my summarization is right or not
as per my knowledge summary address 192.168.1.100/26 has block size of 64 so it starts from
192.168.1.0
192.168.1.64
192.168.1.128 so my 4 networks comes in between 64-128
pls pls teach me summarization
Solved! Go to Solution.
02-09-2011 11:02 AM
Hi,
192.168.1.100/30
192.168.1.104/30
192.168.1.108/30
192.168.1.112/30
you select the interesting octet which is last octet here and convert to binary like you did
100 - 01100100
104 - 01101000
108 - 01101100
112 - 01110000
beginning from left how many bits have you in common? The first 3 so the mask is indeed /27 ( /24 + 3 common bits) and the summary octet is
01100000 which is equal to decimal 96
So summary is 192.168.1.96/27 which gives a range from 192.168.1.97 to 192.168.1.126 so it is indeed a valid summary but it encompasses a lot more subnets than the ones given.
So if we look at the 3 first numbers we have 4 bits in common so summary is 192.168.1.96/28 which gives a range of 192.168.1.97 to 192.168.1.110
and we're left with the last subnet not summarized.
Regards.
Alain.
02-09-2011 11:02 AM
Hi,
192.168.1.100/30
192.168.1.104/30
192.168.1.108/30
192.168.1.112/30
you select the interesting octet which is last octet here and convert to binary like you did
100 - 01100100
104 - 01101000
108 - 01101100
112 - 01110000
beginning from left how many bits have you in common? The first 3 so the mask is indeed /27 ( /24 + 3 common bits) and the summary octet is
01100000 which is equal to decimal 96
So summary is 192.168.1.96/27 which gives a range from 192.168.1.97 to 192.168.1.126 so it is indeed a valid summary but it encompasses a lot more subnets than the ones given.
So if we look at the 3 first numbers we have 4 bits in common so summary is 192.168.1.96/28 which gives a range of 192.168.1.97 to 192.168.1.110
and we're left with the last subnet not summarized.
Regards.
Alain.
02-09-2011 07:33 PM
hi Cadetalain
what you above told i got it.just pls pls tell me little bit more handone on summarization
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