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.
Arunkumar said:
9 years ago
Option C is correct.
The subnet is 255.255.254.0 for10.16.3.0 because the CIDER value is 23.
The subnet is 255.255.254.0 for10.16.3.0 because the CIDER value is 23.
(1)
Arunkumar said:
9 years ago
Hi @Davies.
The answer :10.255.255.255.
The answer :10.255.255.255.
John said:
9 years ago
Good and enough explanation thanks to all.
Nana said:
8 years ago
Buy I thought the up address is class A. Am I right?
Colly said:
8 years ago
What is the broadcast for 10.1.1.1.8? Please, explain me.
Nia said:
8 years ago
I think first address should be 10.16.2.0 instead of 10.16.2.1.
because 2^(32-23) = 2^9,
3 : 0 0 0 0 0 0 1 1,
65 : 0 1 0 0 0 0 0 1.
9 means 1 bit from 3rd octet and rest of them from 4th octet,
0 0 0 0 0 0 1 [0. 0 0 0 0 0 0 0 0] = 2.0,
0 0 0 0 0 0 1[1. 1 1 1 1 1 1 1 1 ] = 3.255.
because 2^(32-23) = 2^9,
3 : 0 0 0 0 0 0 1 1,
65 : 0 1 0 0 0 0 0 1.
9 means 1 bit from 3rd octet and rest of them from 4th octet,
0 0 0 0 0 0 1 [0. 0 0 0 0 0 0 0 0] = 2.0,
0 0 0 0 0 0 1[1. 1 1 1 1 1 1 1 1 ] = 3.255.
(1)
Shailendra kumar yadav said:
8 years ago
The lowest host address in the subnet is 10.16.2.1 255.255.254.0.
It is incorrect because the lowest host address in the subnet is 10.16.3.254 255.255.254.0.
It is incorrect because the lowest host address in the subnet is 10.16.3.254 255.255.254.0.
Monty said:
8 years ago
Why not option A? Please explain.
Sourav chettr said:
8 years ago
Why option A is not correct?
Usama said:
7 years ago
B) is the correct answer.
The correct explanation is;
First, understand the question, Describe the IP 10.16.3.65/23
let's start from the beginning 11111111.11111111.11111110.00000000/23 which mean we are in the third octet.
How many zeroes we have in the third octet? One so 2^1=2 each network.
Now the answer will be:
First =10.16.0.0 (to) 10.16.1.255
10.16.0.0 this is N/W
10.16.0.1 first IP
10.16.1.254 last IP
10.16.1.255 Broadcast
Second= 10.16.2.0 (to) 10.16.3.255
10.16.2.0 N/W
10.16.2.1 First IP
10.16.3.254 last IP
10.16.3.255 Broadcast
And the question asking IP of 10.16.3.65 which is between the second N/W the first or lowest host in second N/W is 10.16.2.1 and last 10.16.3.254 the 255 is broadcast.
Note: 10.16.3.65/23 lock to the 23 which mean 8+8+7+0. (11111111+11111111+11111110+00000000) How many Zeros we have in the third octet? One So you need to use this formula to get the IP each N/W 2^(number of zeros) in the octet where we are.
The correct explanation is;
First, understand the question, Describe the IP 10.16.3.65/23
let's start from the beginning 11111111.11111111.11111110.00000000/23 which mean we are in the third octet.
How many zeroes we have in the third octet? One so 2^1=2 each network.
Now the answer will be:
First =10.16.0.0 (to) 10.16.1.255
10.16.0.0 this is N/W
10.16.0.1 first IP
10.16.1.254 last IP
10.16.1.255 Broadcast
Second= 10.16.2.0 (to) 10.16.3.255
10.16.2.0 N/W
10.16.2.1 First IP
10.16.3.254 last IP
10.16.3.255 Broadcast
And the question asking IP of 10.16.3.65 which is between the second N/W the first or lowest host in second N/W is 10.16.2.1 and last 10.16.3.254 the 255 is broadcast.
Note: 10.16.3.65/23 lock to the 23 which mean 8+8+7+0. (11111111+11111111+11111110+00000000) How many Zeros we have in the third octet? One So you need to use this formula to get the IP each N/W 2^(number of zeros) in the octet where we are.
(8)
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