New feature! Upgrade and increase expert visibility of your issues with Priority Questions.
Given address 18.104.22.168/27, what is the network and broadcast address?
consider only the 4th octet, where the subnet mask (27bits) boundary occurs.
step1. how many host bits? number of host bits = 5
step2. what is the starting address for the address block containing given address?
possible starting addresses in the 4th octet are: 0, 32, 64, 96, 128, 160,....
Ans: network address: 22.214.171.124 (first address)
broadcast address: 126.96.36.199 (last address)
Add your voice to the tech community where 5M+ people just like you are talking about what matters.
Join the community of 500,000 technology professionals and ask your questions.