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 1 of 2.

Umesh Prajapati said:   5 years ago
As I solved this question :

Nework id :- 192.168.19.24.
First vaild host :- 192.168.19.25.
Last vaild host - 192.168.19.30.
Broadcost IP address :- 192.168.19.31.
Subnet mask :- 255.255.255.248.

Please check it.
(4)

Harry said:   8 years ago
First host address 192.168.19.25 255.255.255.248 is assigned to the router as said in the question, after that next available address is (as given in options) 192.168.19.26 255.255.255.248 which would assign to the server.
(1)

Jahangir alam said:   7 years ago
I think here first usable host is 192.168.19.25. Because here 192.168.19.24 is the network address.
(1)

Bharti singh said:   7 years ago
The first host address should be :192.168.19.25.

Because up to 29 bits it is fixed part and the remaining part is 3 bits.
192.168.19.00011000 -> subnet block id.
.00011001-> this will be the first host for this subnet.
(1)

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.

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?

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?

Manu Ratheesh said:   8 years ago
I think answer is (D).

Solution:

Given CIDR is 192.168.19.24/29 which means the NID is 29 bits and HID is 3 bits.

So the subnet mask becomes 255.255.255.248.

Given that the router has the first available host address (i.e, SubnetID), which is given by:

(192.168.19.24) AND (255.255.255.248) = 192.168.19.24, which is the IP address of the router.

We know that 192.168.19.24 = 192.168.19.00011000.

Here the NID part is 192.168.19.00011 (29 bits) and host id part is remaining 3 bits.

In order to get the broadcast address (address of the server), set all host id bits to 1. Then it becomes 192.168.19.00011111.

i.e, 192.168.19.31 which is the IP address of the server.

Bijit Sarkar said:   8 years ago
I think it the range coming in 192.168.19.24 to 192.168.19.32.

So the 1st host coming is 192.168.19.25.

Please check it.


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