Networking - Subnetting - Discussion

Discussion Forum : Subnetting - Subnetting (Q.No. 11)
11.
You need to configure a server that is on the subnet 192.168.19.24/29. The router has the first available host address. Which of the following should you assign to the server?
192.168.19.0 255.255.255.0
192.168.19.33 255.255.255.240
192.168.19.26 255.255.255.248
192.168.19.31 255.255.255.248
Answer: Option
Explanation:
A /29 is 255.255.255.248, which is a block size of 8 in the fourth octet. The subnets are 0, 8, 16, 24, 32, 40, etc. 192.168.19.24 is the 24 subnet, and since 32 is the next subnet, the broadcast address for the 24 subnet is 31. 192.168.19.26 is the only correct answer.
Discussion:
14 comments Page 2 of 2.

Unknown said:   9 years ago
Guys if I'm correct you count up to 256 but I see some where people counted to 255 maybe?

Nick said:   9 years ago
I think D is the right answer.

192.168.19.24/29 --> 11000000.10110000.00010011.00011000 with the first 5 bits of the 4th octet defining the sub net 24 and last 3 bits defining the host number. So if the next sub net is 32--> 11000000.10110000.00010011.00100000 then the broadcast address for sub net 24 would be 11000000.10110000.00010011.00011111.

Am I missing something?

Nani said:   1 decade ago
How the block size is 8 in the fourth octet?

Bhadana said:   1 decade ago
Subnet id is 192.168.19.24/29 or mask is 255.255.255.248.

First host of the subnet is 11000000.10110000.00010011.00011001 that is 192.168.19.25/29.


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