| Justin2112 2002-09-19, 4:15 pm |
| I just like . . .figured this out this afternoon (with help from people here) - began work on it a while ago,
but I've come up with something that has worked on every range - so here ya go
Justin2112’s Network IP address summarization tool
1) The first network address in the range must be divisible by 2.
2) The networks that you are summarizing must be consecutive.
Example ( 192.71.84.0, 192.72.85.0, 192.71.86.0, 192.71.87.0 )
For this example, assume a standard class C mask.
3) For the IP address portion of summarized address, use the first address in
range you are summarizing: 192.71.84.0
4) For the mask, use this formula:
Break the range into binary: ( we know the first two octects match, so I will start with the third octect for this example )
84= 01010100
85= 01010101
86= 01010110
87= 01010111
Now just look for the common 1’s and 0’s. ( I have bolded them ). Now used the matched bits to create the subnet. In this case, the first 6 bits match. To create you mask change the first six bits to a value of 1, so you’ll have:
11111100 change that back into dotted decimal format, for this case its 252. Your subnet mask then becomes 255.255.252.0
Your summary address then becomes: 192.71.84.0 255.255.252.0 which represents the above 4 networks.
For addresses you have summarized that have the standard bit masks, 8,16 and 32 you can then use the following formula to check your work.
Take your mask that you made ( in this case 252 ) and subtract that number from 256. Here we get 256-252=4. 4 is the number of networks summarized!
5) Starting with VLSM address, ( non standard bit masks can be summarized ).
Let’s use this example: 130.10.32.32 255.255.255.240 and
130.10.32.48 255.255.255.240
This meets the first criteria ( the network number is divisible by two ), in this case the network number is the 32 in the fourth octect. Even though the numbers in the fourth octect are not consecutive, the networks ARE! So again we break it down into binary. Again, I’ll just use the fourth octect because the first three match.
32= 00100000
48= 00110000
So for this we will use the 224 mask because 11100000 is 224, and I’ll place it in the fourth octect this time because that is where the network number is, so I get:
Summary address = 130.10.32.32 255.255.255.224
Thanks to everyone here !
PEACE!@
Justin |