0 votes 0 votes Computer Networks subnetting network-addressing computer-networks + – Çșȇ ʛấẗẻ asked Jan 18, 2017 Çșȇ ʛấẗẻ 664 views answer comment Share Follow See 1 comment See all 1 1 comment reply rajatmyname commented Aug 9, 2017 reply Follow Share From where have you taken this question? 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes Class B Total Networks : First two octets (16 bits) and first two bits(10) are fixed. Therefore, remaining 14 bits are required for network. => 214 = 16384 Total Hosts : Each network has 216 - 2 = 65536 -2 = 65534 Therefore, correct option is (c) targate2018 answered Jan 18, 2017 targate2018 comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes In class B we have 16 bits for NID and 16 bits for HOST ID . Out of the 16 bits of NID we have 2 bits fixed that is network prefix. So, we are left with 14 bits . So , total networks = 2^14. We have 16 bits in HOST ID part . So, 2^16 -2 hosts per network are possible . dragonball answered Jun 27, 2017 • edited Jun 27, 2017 by dragonball dragonball comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes no. of hosts = 2^16-2 =65534 since one address is reserved for dba and the other for network id no. of networks=2^14=16384 since the first two bits are fixed popo040 answered Jun 29, 2017 popo040 comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes Range of Class B considering 1st octet is $128$ to $191$, Then number of networks = $64 \times 2^8$ (because default subnet mask of class B is 255.255.0.0) = $16384$ Number of hosts per network = $2^{16} = 65536$ Mk Utkarsh answered Sep 9, 2018 Mk Utkarsh comment Share Follow See all 0 reply Please log in or register to add a comment.