How'd they get that answer?

wolfdoggwolfdogg Member Posts: 2 ■□□□□□□□□□
in CCNA & CCENT
Hi all.. just got a wild hair that I'd like to study for a certification, so literally started trying to learn stuff yesterday. I'm green as green can be. So.. I figured out how to determine the subnet from given ip and mask and have been getting those all right at subnettingquestions.com.

I'm sure I'll have a bunch of "huh?" questions, so I thought I'd start my thread. :)

First one..
What is the last valid host on the subnetwork 172.23.91.128/26?

I determined the subnet mask is 255.255.255.192

So.. I go through the Cisco math and get my m# (magic number) of 64, which completes my subnet of: 172.23.91.64

The answer is: 172.23.91.190

Why is 190 the last valid host? I'm confused. Thx!
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  • behbeh Member Posts: 10 ■□□□□□□□□□
    So /26 means a block size of 2^(32-26)=64


    If you add the blocksize (64) to the network address (172.23.91.12icon_cool.gif you'll get the network address of the NEXT subnet.

    So 172.23.91.192 is the network address of the next subnet. That means that 172.23.91.191 is the broadcast address of the network you're looking at. This then means that 172.23.91.190 is the last valid host.

    EDIT: Let me know if I should clarify something.
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  • wolfdoggwolfdogg Member Posts: 2 ■□□□□□□□□□
    Perfect! Thx beh! Makes complete sense.. and you just taught me another "bit" about the broadcast address. Thx!!
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