Networking - Networking Basics - Discussion

Discussion Forum : Networking Basics - Networking Basics (Q.No. 7)
7.
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?
192.168.168.129-190
192.168.168.129-191
192.168.168.128-190
192.168.168.128-192
Answer: Option
Explanation:
256 - 192 = 64. 64 + 64 = 128. 128 + 64 = 192. The subnet is 128, the broadcast address is 191, and the valid host range is the numbers in between, or 129-190.
Discussion:
46 comments Page 3 of 5.

Mehulkumar Patel said:   8 years ago
I think answer should be B. Because 2nd subnet should be 128-192. I. E 128 is network address and 192 is broadcast address. Hence valid host range should be 129-191. So answer is B.

Iglooroy said:   8 years ago
The answer is right they said "usable" ip address, so the network is 192.168.168.128 to .191, 128 is network address and 191 is broadcast address.

Richard said:   8 years ago
256-192 = 64 so increment by 64.

Networ - Broadcast
64 - 127
128 - 191 > so the usable host address range is 127 - 190 (answer is A).
192 - 255.

Rohini said:   1 decade ago
Why the value should be between 129 and 191. It should be between 128-191. Also why the broadcast address is 191?

Robel said:   9 years ago
128 is called a network address and 191 is called broadcast address.

Thanks it is easy to understand.

Teddy said:   1 decade ago
@Richa.

Where did you get that 191?

As you said n/w add starts from 128 and end to 191 how?

Frank said:   9 years ago
@Ravi Sankar.

Why you call 64 as magic number and why not 192? Please help me.

Vini said:   1 decade ago
I don't know that how we can find the IP address so please explain in details?

Sumit sain said:   10 years ago
I really don't understand these answer correctly. Please explain it in brief.

Ritu said:   1 decade ago
Can anybody explain me this question in more elaborate way from the scratch?


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