Networking - Subnetting - Discussion
Discussion Forum : Subnetting - Subnetting (Q.No. 8)
8.
You have an interface on a router with the IP address of 192.168.192.10/29. Including the router interface, how many hosts can have IP addresses on the LAN attached to the router interface?
Answer: Option
Explanation:
A /29 (255.255.255.248), regardless of the class of address, has only 3 host bits. Six hosts is the maximum number of hosts on this LAN, including the router interface.
Discussion:
11 comments Page 1 of 2.
Anup kumar said:
3 years ago
It should be 7, if we consider the router interface address because in formula -2 means we are subtracting the network address and broadcast address, the network address is the router address so we have to add +1.
(2)
Fondo said:
3 years ago
255.255.255.248.
Take 256-248 this will give us 8hosts.
Then, take 8 - 2 (network ID and broadcast ID) you'll get 6 hosts.
Take 256-248 this will give us 8hosts.
Then, take 8 - 2 (network ID and broadcast ID) you'll get 6 hosts.
(2)
Manu Ratheesh said:
10 years ago
Here the host id is 3 bits. So the number of IP addresses is possible is 2^3 = 8. But the first IP is used as the network address and the last IP is used as the broadcast address.
So 8-2 = 6 host addresses are possible.
So 8-2 = 6 host addresses are possible.
(1)
David said:
8 years ago
The number of valid host is 6; but the number of host is 8. The formula 2^n - 2 is to get the number of valid host. If the question ask for valid hosts within the subnet, then the answer will automically be 6. But asking for just host, the answer should be 8. WORDING OF THE QUESTION IS CRITICAL IN DUBNETTING.
(1)
Ankit said:
1 decade ago
It should be 8 because 2^3=8. 3 bits are for host. 5 bits for netid. 2^5=32 subnets are there and 8 hosts.
Arpit said:
1 decade ago
For host, we calculate like, 2^3 - 2, so it will be 6.
Vinay said:
1 decade ago
So for LAN, 3 host bits are constant?
Ankur said:
10 years ago
192.168.192.10/29 this is class b address so network bits 29-24 = 5 and host bits 32-29 = 3.
So it it should be 8 because 2^3 = 8, 3 bits for host.
So it it should be 8 because 2^3 = 8, 3 bits for host.
Nigam singha said:
9 years ago
Wrong option write answer in 6 as 2 (pow 32 - 29) - 2 = 6.
Chaithanya katari said:
8 years ago
Simple, you need to know remaining host so ( 2^h)-2.
To find host 29-32=3.
2^3-2=8-2=6.
To find host 29-32=3.
2^3-2=8-2=6.
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