1 votes 1 votes Computer Networks ip ip-addressing computer-networks network-addressing + – Vaishnavi01 asked Sep 12, 2018 • edited Mar 11, 2019 by Naveen Kumar 3 Vaishnavi01 617 views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply Magma commented Sep 12, 2018 reply Follow Share 208.10.128.192/26 0 votes 0 votes BASANT KUMAR commented Sep 23, 2018 reply Follow Share is it only possible answer??? 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes 208.10.128.64/27 208.10.128.96/27 208.10.128.128/27 208.10.128.160/27 208.10.128.192/27 208.10.128.224/27 These are the possible answers pawangupta answered Sep 28, 2018 • reshown Sep 30, 2018 by pawangupta pawangupta comment Share Follow See all 2 Comments See all 2 2 Comments reply Vaishnavi01 commented Sep 28, 2018 reply Follow Share Can u please elaborate, how u came to this conclusion!!! 0 votes 0 votes pawangupta commented Sep 30, 2018 reply Follow Share ECE = 208.10.128.0/26 last octet is 00000000/26 Only first two bits are in network id other are bits for host since mec require 30 address so first 3 bits can be considered as id part other as host part Ece is given 00 as id part in last octet so we can't use 00 as id part of mec except all can be use that is 01, 10, 11 can be use as id part of mec and with each id part of mec 0 or 1 can be used since we are using first 3 bits of last octet so finally required address are 010/27 64/27 011/27 96/27 100/27 128/27 101/27. 160/27 110/27. 192/27 111/27. 224/27 ( last five bits are definitely 0 in all cases) The reason I have not considered 00/26 as id part of mec becoz if ECE further has divide its network it will consider as 001/27, 000/27 so these two address will not be considered otherwise all are accepted I hope I cleared ur doubt and not mess up with paragraph type of giving solution 1 votes 1 votes Please log in or register to add a comment.