For the given network , 10.0.0.0/25 , what is the first and last address (network and broadcast)?
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A. B. C. D.Answer - A.
This can be determined using the CIDR calculator.
Reference the below link for the CIDR calculator.
http://www.subnet-calculator.com/cidr.phpThe network 10.0.0.0/25 is a Class A private network with a subnet mask of 255.255.255.128. This network has 128 IP addresses ranging from 10.0.0.0 to 10.0.0.127.
To find the first and last address (network and broadcast), we need to apply the following steps:
Step 1: Identify the network address The network address is the first IP address in the subnet range. To calculate this, we need to perform a bitwise AND operation between the IP address and the subnet mask.
In this case, the subnet mask is 255.255.255.128, which in binary is:
11111111.11111111.11111111.10000000
Performing a bitwise AND operation between the subnet mask and the IP address 10.0.0.0 gives:
00001010.00000000.00000000.00000000 (10.0.0.0)
Therefore, the network address for the given network is 10.0.0.0.
Step 2: Identify the broadcast address The broadcast address is the last IP address in the subnet range. To calculate this, we need to invert the subnet mask and perform a bitwise OR operation with the network address.
Inverting the subnet mask gives:
00000000.00000000.00000000.01111111
Performing a bitwise OR operation between the inverted subnet mask and the network address 10.0.0.0 gives:
00001010.00000000.00000000.01111111 (10.0.0.127)
Therefore, the broadcast address for the given network is 10.0.0.127.
The correct answer is A. 10.0.0.0 and 10.0.0.127, where the first IP address is the network address and the last IP address is the broadcast address.