# Find the netid and the hostid of the following IP addresses:

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a. 114.34.2.8

b. 132.56.8.6

c. 208.34.54.12

d. 251.34.98.5

+1 vote
by
selected by (user.guest)

a. Class is A → netid: 114 and hostid: 34.2.8

b. Class is B → netid: 132.56 and hostid: 8.6

c. Class is C → netid: 208.34.54 and hostid: 12

d. Class is E → The address is not divided into netid and hostid.

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by
by (user.guest)
10.50.13.40 is a class A address type. Class A address (first octet) range from 1 - 127. So any address that takes the format of 1.x.x.x to 126.x.x.x is a Class A type.

In class A, netid is the first octet, while the rest is the hostid.
netid = 10
hostid = 50.13.40
by
10 net id
50.13.40 host id

class a - 1 net id and 3 host id
class b - 2 net id and 2 host id
class c- 3 net id and 1 host id
by (user.guest)
Thanks for the explanation. You're correct.
by
You are assigned to use 132.56.8.6 IP address in two different networks , then how can you subnet it ? Do the calculation.
+1 vote
by
a.  Net id : 114.0.0.0 , Host id : 0.34.2.8

b.  Net id : 132.56.0.0 , Host id : 0.0.8.6

c.  Net id :  208.34.54.0 , Host id : 0.0.0.12

d. This address is reserved for research purpose.
by (user.guest)
Well written. Thank you for contributing.
by
199.108.61.2 and 235.73.47.8

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