Patulong po kong pano sagutin tong tanong na to?

[amalinao]

1. What is the summary for these range of addresses 172.32.48.0/24 - 172.32.50.0/24? (Indicate the subnet mask for this summary in the answer)
2.What is the first valid host on the subnet that the node 192.168.22.68 255.255.255.240 belongs to? (No need to indicate the subnet mask in the answer)

 

[uchija_zero]

try ko lang po
1. What is the summary for these range of addresses 172.32.48.0/24 - 172.32.50.0/24? (Indicate the subnet mask for this summary in the answer)
netmask 255.255.255.0
172.32.48.0 network address
172.32.48.1 -first host
172.32.48.254 - last host
172.32.48.255 - broadcast

172.32.50.0 network
172.32.50.1 first host
172.32.50.254 last host
172.32.50.255 broadcast

2.What is the first valid host on the subnet that the node 192.168.22.68 255.255.255.240 belongs to? (No need to indicate the subnet mask in the answer)
/28 class C
192.168.22.64 - network address
192.268.22.65 - first host
192.168.22.78 - last host
192.168.22.79 - broadcast

 

[pinch]

Hope this helps. Try to use this po.
http://adminsub.net/

 

[r a Z e]

Last year pa pala post ni ts... Malamang nacheckan na yung assignment niya

 

[uchija_zero]

Last year pa pala post ni ts... Malamang nacheckan na yung assignment niya

hehehe kahapon ko lang nakita nung may senisearch ako

 

[imba_kan]

Last year pa pala post ni ts... Malamang nacheckan na yung assignment niya

malay mo lods bumagsak at least balikan nya tong thread nya hahahah

 

[CHOKNOX GSM]

1. The range of addresses you provided is from 172.32.48.0 to 172.32.50.255, with a subnet mask of /24 indicating that the first 24 bits of the IP address are used to identify the network portion and the remaining 8 bits are used for host addresses. This means that there are three networks in this range: 172.32.48.0/24, 172.32.49.0/24, and 172.32.50.0/24. Each network can accommodate up to 256 hosts, excluding the network and broadcast addresses.


2. To determine the first valid host on the subnet that the node 192.168.22.68/28 belongs to, we need to perform a bitwise AND operation between the IP address and the subnet mask to obtain the network address.

The subnet mask 255.255.255.240 in binary is 11111111.11111111.11111111.11110000. Performing the bitwise AND operation:

11000000.10101000.00010110.01000100 (192.168.22.68)

& 11111111.11111111.11111111.11110000 (255.255.255.240) ----------------------------------- 11000000.10101000.00010110.01000000 (192.168.22.64)

Therefore, the network address for this subnet is 192.168.22.64.

The first valid host on this subnet is obtained by adding 1 to the network address, which gives us 192.168.22.65. So, the first valid host on the subnet that the node 192.168.22.68/28 belongs to is 192.168.22.65.