Networking - Subnetting - Discussion
Discussion Forum : Subnetting - Subnetting (Q.No. 2)
2.
Which two statements describe the IP address 10.16.3.65/23?
- The subnet address is 10.16.3.0 255.255.254.0.
- The lowest host address in the subnet is 10.16.2.1 255.255.254.0.
- The last valid host address in the subnet is 10.16.2.254 255.255.254.0.
- The broadcast address of the subnet is 10.16.3.255 255.255.254.0.
Answer: Option
Explanation:
The mask 255.255.254.0 (/23) used with a Class A address means that there are 15 subnet bits and 9 host bits. The block size in the third octet is 2 (256 - 254). So this makes the subnets in the interesting octet 0, 2, 4, 6, etc., all the way to 254. The host 10.16.3.65 is in the 2.0 subnet. The next subnet is 4.0, so the broadcast address for the 2.0 subnet is 3.255. The valid host addresses are 2.1 through 3.254.
Discussion:
37 comments Page 3 of 4.
Ali said:
9 years ago
After calculating how did we come to know that "the possible networks will be 10.16.2.1 to 10.16.3.254"?
Sia said:
9 years ago
Can anyone know the answer for Abdulrazig Mohammed's question?
Milhem said:
9 years ago
Why option 4 is wrong? Explain me.
Abdulrazig mohammed said:
9 years ago
What is the broadcast for 10.1.1.1/8?
Priyanka said:
9 years ago
Why option 3 is wrong?
(The last valid host address in the subnet is 10.16.2.254, 255.255.254.0).
I think option 3 is wrong because 10.16.3.254 is the last. Since in 3rd octet its 3 not 2. Therefore, The last valid host address in the subnet is 10.16.3.254.
(The last valid host address in the subnet is 10.16.2.254, 255.255.254.0).
I think option 3 is wrong because 10.16.3.254 is the last. Since in 3rd octet its 3 not 2. Therefore, The last valid host address in the subnet is 10.16.3.254.
Priyanka said:
9 years ago
Why it is started from 10.16.2.0? Why not from 10.16.1.0 or from 10.16.3.0 please tell.
Mainodin said:
9 years ago
Why option 3 is wrong ?
Rohit said:
10 years ago
How come 10.16.2.254 255.255.254.0 is not a valid address keeping in mind that valid address for hosts range from 10.16.2.1 to 10.16.3.254?
Vishwajeet said:
1 decade ago
Then the possible networks will be 10.16.2.1 to 10.16.3.254, because this is the one subnet. So if we write 10.16.4.0 then this the another network. That's why we take the IP of 10.16.2.254 and 10.16.3.1. For more explanation see my previous post.
Vishwajeet mishra said:
1 decade ago
Yes, The answer (2 & 4) is a correct option, because now we are working in CIDR Value of (/23), which means we will work on only those bits which have the both combination's of 0's and 1's. So here this happens on third octat of subnet mask.
CIDR Value = /23. So the subnet in binary is 255.255.254.0 => 11111111.11111111.11111110.00000000.
So if we see the third octat then we will see that here is the combination of 0's and 1's. Now, let's find out the value of 1's in third octat. 2^7= 128.
And the total no. Of hosts in a subnetwork= 2^1=2. So here the subnetwork will be 0, 2, 4, 6, 8 and so on because the block size = 2.
So every subnet will made after the gap of 2. How to find the block size = we have to subtract the octat third from 256.
So here we subtract 256-254= 2. It defines the interval of the subnetwork in a subnetted network.
CIDR Value = /23. So the subnet in binary is 255.255.254.0 => 11111111.11111111.11111110.00000000.
So if we see the third octat then we will see that here is the combination of 0's and 1's. Now, let's find out the value of 1's in third octat. 2^7= 128.
And the total no. Of hosts in a subnetwork= 2^1=2. So here the subnetwork will be 0, 2, 4, 6, 8 and so on because the block size = 2.
So every subnet will made after the gap of 2. How to find the block size = we have to subtract the octat third from 256.
So here we subtract 256-254= 2. It defines the interval of the subnetwork in a subnetted network.
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