Networking - Subnetting - Discussion
Discussion Forum : Subnetting - Subnetting (Q.No. 1)
1.
Your router has the following IP address on Ethernet0: 172.16.2.1/23. Which of the following can be valid host IDs on the LAN interface attached to the router?
- 172.16.1.100
- 172.16.1.198
- 172.16.2.255
- 172.16.3.0
Answer: Option
Explanation:
The router's IP address on the E0 interface is 172.16.2.1/23, which is 255.255.254.0. This makes the third octet a block size of 2. The router's interface is in the 2.0 subnet, and the broadcast address is 3.255 because the next subnet is 4.0. The valid host range is 2.1 through 3.254. The router is using the first valid host address in the range.
Discussion:
36 comments Page 4 of 4.
Shipukangra@gmail.com said:
9 years ago
How can be 172.168.3.0 is a host id? It is a network id.
Salim said:
1 decade ago
How many network will divide by this subnet mask ?
How you know the number of host to each subnet will be from 2.1 to 3.254 ?
How you know the number of host to each subnet will be from 2.1 to 3.254 ?
Akshay said:
9 years ago
How you guys calculate? The router's IP address on the E0 interface is 172.16.2.1/23, which is 255.255.254.0. How? Please tell me. I am new in subnets.
Jot Singh said:
9 years ago
Guys What I think is.
Block size is 2(256-254:FROM LAST NON-ZERO OCTET IN SUBNET MASK).
So the subnets would be divided by a factor of 2.
Starting from
172.16.0.0-172.16.1.255(including a network address and broadcast address).
As we are considering E0 starting from _._.2.1.
The subnet it comes under is as 127.16.2.0(being n/w id) and the flow is as follows:
172.16.2.0 NID.
172.16.2.1 FIRST VALID HOST.
172.16.2.255.
172.16.3.0 *VALID ADDRESS(as it is not a n/w id).
172.16.3.1.
172.16.3.254 LAST VALID HOST.
172.16.3.255. BROAD CAST.
According to which it should be none of these networks.
Hope it will help you.
Block size is 2(256-254:FROM LAST NON-ZERO OCTET IN SUBNET MASK).
So the subnets would be divided by a factor of 2.
Starting from
172.16.0.0-172.16.1.255(including a network address and broadcast address).
As we are considering E0 starting from _._.2.1.
The subnet it comes under is as 127.16.2.0(being n/w id) and the flow is as follows:
172.16.2.0 NID.
172.16.2.1 FIRST VALID HOST.
172.16.2.255.
172.16.3.0 *VALID ADDRESS(as it is not a n/w id).
172.16.3.1.
172.16.3.254 LAST VALID HOST.
172.16.3.255. BROAD CAST.
According to which it should be none of these networks.
Hope it will help you.
Nirmal said:
9 years ago
@Nikhil.
Because 7 bits are opened in the third octet.
Because 7 bits are opened in the third octet.
Nikhil said:
10 years ago
How did we get the subnet mask as 255.255.254.0?
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