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asked in Computer Networks by Active (2.1k points) | 117 views
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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)

answered by Active (1.8k points)
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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 .

answered by Active (2.1k points)
edited by
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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
answered by (81 points)
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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 = $2^{16} = 65536$

answered by Boss (23.7k points)

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