NIELIT 2017 DEC Scientist B - Section B: 36

Question

If the number of networks and number of hosts in class $B$ are $2^m, (2^n-2)$ respectively. Then the relation between $m,n$ is

• A. $3m=2n$
• B. $7m=8n$
• C. $8m=7n$
• D. $2m=3n$

Comments

For class B network, no. of networks = 2^14

No of host per network = 2^16 -2

so, m = 14, n = 16

for relation, m/n = 14/16 = 7/8

m/n = 7/8 => 8m = 7n

Answer 1

answer is (c)

FOR CLASS B no of network=2^14

No of host PER NETWORK=2^16 -2

so 14*8=16*7

Answer 2

In class B addresses, the leading bits in the network id part of the IP address are $10$  i.e. 2 bits are reserved.

We know that $16$ bits each are reserved for network and host id part in the  $32$-bit IP address.

So, the number of networks possible in class B would be:  $2^{14}$

And the number of hosts would be:  $2^{16}-2$

So, $m=14$  and  $n=16$.

$\therefore$  Option C satisfies the relation,

Answer 3

Total number of networks in class-B network is 214
Total number of hosts per network is 216-2
Substitute 8m and 7n
⇒ 14*8=16*7
112=112

Answer 4

number of networks= 2^m = 2^14   [m=14 (=7*2)]

 

and number of hosts in class B =(2^n) −2 = (2^16)-2  [n=16 (=8*2)]

 

Now to equal m and n, multiply m by 8 and n by 7.

So, 8m=7n

 

Answer: C