CIDR to the inverse

gouki2005

ok i found this question in other forum and i dont know how to get the answer please help me here

which two subnetworks would be included in the summarized address of 172.31.80.0/20?

A 172.31.17.4/30
B 172.31.51.16/30
C 172.31.64.0/18
D 172.31.80.0/22
E 172.31.92.0/22
F 172.31.192.0/18

the correct answer are D,E

ok how to get that answer??

Akiii

gouki2005 wrote: »

ok i found this question in other forum and i dont know how to get the answer please help me here

which two subnetworks would be included in the summarized address of 172.31.80.0/20?

A 172.31.17.4/30
B 172.31.51.16/30
C 172.31.64.0/18
D 172.31.80.0/22
E 172.31.92.0/22
F 172.31.192.0/18

the correct answer are D,E

ok how to get that answer??

/20 = block of 16 (256-240)

172.31.0.0, 172.31.16.0, 172.31.32.0, 172.31.48.0, 172.31.64.0, 172.31.80.0

So basically wich adresses can be found in the intervallum of 172.31.80.0 - 172.31.96.0

And the answer is D, E

Hope you get it

billyr

172 . 31 . 80 . 0 /20
11111111.11111111.11110000.0000
255 . 255 . 240 . 0


Have a look at it this way it may make a bit more sense. If you notice 20 bits have been set to 1 and are reserved for the network or shared in common with the summarized addresses whichever you prefer.

Once you have this all you are looking for is the increment or block size. To determine this as Akiii mentions you can deduct 240 from 256 to give you a block size of 16.
I prefer to get my block size by looking at the value of the last bit that you have turned on or set to 1. This would be the 16 bit in the third octet.

gouki2005

yeah thx i used subnetting made easy method

20-24 = 4 so 2^4 = 16

172.31.0.0
172.31.16.0
172.31.32.0
172.31.48.0
172.31.64.0
172.31.80.0
172.31.96.0

so yeah i thoung it was a convert the address to binary question