642 views

4 Answers

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)

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 .

edited by
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
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 = $2^{16} = 65536$

Related questions

2 votes
2 votes
3 answers
1
aditi19 asked Oct 23, 2018
3,482 views
If Direct Broadcast Address of subnet is 201.15.16.31. Which of the following will be subnet mask ?1. 255.255.255.2402. 255.255.255.1923. 255.255.255.1984. NOTAExplain yo...
1 votes
1 votes
2 answers
2
Çșȇ ʛấẗẻ asked Dec 25, 2016
1,262 views
An Internet Service provider (ISP) has the following chunk of CIDR- based IP addresses available 252.224.128.0/22. Now it gives its quarter of addreeses to X, Calculate ...
1 votes
1 votes
1 answer
4
admin asked Mar 31, 2020
1,840 views
There is a need to create a network that has $5$ subnets, each with at least $16$ hosts. Which one is used as classful subnet mask ?$255.255.255.192$$255.255.255.248$$255...