■■□□□□□□□□ Posts: 45Member ■■□□□□□□□□
Heres an example. Address 192.168.144.0/20 so whats the range of host address that would be forwarded according to this summary?

/20 is 255.255.240.0

Third octet is block size of 16. And the starting summary is 144, the next block of 16 is 160. so the network summary range is 144-159 in the third octet. I am trying to understand why the range ends in 159 and not 160 if the next block is 160? I am guessing because we do not use the last address we just -1. which equals 159. Can someone explain please. Thanks!

• ■■□□□□□□□□ Posts: 38Member ■■□□□□□□□□
alu408 wrote: »
I am trying to understand why the range ends in 159 and not 160 if the next block is 160? I am guessing because we do not use the last address we just -1. which equals 159. Can someone explain please. Thanks!

It's because 160 is the beginning of the next block of 16 (192.168.160.0 - 192.168.195.255 /20).
192.168.144.0 to 192.168.159.255 is actually a block of 16 because you have to count from 144 to 159, inclusive.
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• Posts: 160Member
alu408 wrote: »
I am trying to understand why the range ends in 159 and not 160 if the next block is 160? I am guessing because we do not use the last address we just -1. which equals 159. Can someone explain please. Thanks!

Since it starts from "0" as simple as it can be. Since everything can only be in the powers of 2 start counting from 0 and everything starts from even number and ends with odd.
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• Posts: 303Member
I will give you a nudge in the correct direction, if you take the third octet of your subnet mask of 11110000, now right out the values of the bits that are 0, which would be 1 + 2 + 4 + 8 = 15 . You are confusing the math behind this binary logic, what Majestic said above is absolutely correct.