networking

Question

Which of the following is the valid host range for the subnet on which the IP address 192.168.168.188 255.255.255.192 resides?

A. 192.168.168.129-190 B. 192.168.168.129-191 C. 192.168.168.128-190 D. 192.168.168.128-192

Comments

options A is correct,

host can range from 192.168.168.129 to 192.168.168.190.

Answer 1

Logically ANDing the IP address 192.168.168.188  with subnet mask 255.255.255.192 , we get 192.168.168.128 as the subnet id.

For subnet mask 255.255.255.192, #valid_hosts = 2⁶ - 2 = 62

Since it is class C address, we'll concentrate only on last octet i.e. 128

128 = 10000000

Host bits of all 0's and all 1's will not be used for any host as they belong to subnet id and broadcast address respectively. So the first address and last address of the  subnet cannot be assigned to any host.

10000000 and 10111111 will be excluded.

$\therefore$ Last octet of first host = 10000001 and Last octet of last host = 10111110

So valid range of IP addresses that can be assigned to hosts is 192.168.168.129 to 192.168.168.190

Comments

Last address should be ........190 (A) should be correct.

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@just_bhavana

.10 is subnet id, therefore

host will range from .10000001 to .10111110 (i.e .129 to .190)

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Yes, thanks. Edited.

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@just_bhavana

sorry, last host will be .10111110 (.190)