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

Question

a. 114.34.2.8

b. 132.56.8.6

c. 208.34.54.12

d. 251.34.98.5

Answer 1

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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Comments

what answer 10.50.13.40 NET ID

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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

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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

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Thanks for the explanation. You're correct.

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You are assigned to use 132.56.8.6 IP address in two different networks , then how can you subnet it ? Do the calculation.

Answer 2

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.

Comments

Well written. Thank you for contributing.

Answer 3

199.108.61.2 and 235.73.47.8