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asked in Computer Networks by (197 points)
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208.10.128.192/26
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is it only possible answer???

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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

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answered by (21 points)
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Can u please elaborate, how u came to this conclusion!!!
0
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

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