Assume that you have been assigned the 192.168.1.0/24 network. You need to creat
ID: 3821530 • Letter: A
Question
Assume that you have been assigned the 192.168.1.0/24 network. You need to create 6 subnets. a. How many bits do you need to barrow from the host portion for the subnet address? b. What is the number of subnets that can be created with the number of bits in step a? c. How many bits can be used to create the host space? d. What is the number of host addresses available per subnet? e. What prefix/length notation would you use? f. What is the subnet mask in dotted decimal? g. What is the subnet and broadcast address of the first and last subnet?Explanation / Answer
1) We need 3 bits more in host area of subnet mask, thus our mask becomes 255.255.255.224.
2) 8 subnets can be created with 3 bits. From 0-255, with a gap of 32 in between.
eg 1st subnet will be from 192.168.1.0 to 192.168.1.31,
2nd subnet from 192.168.1.32 to 192.168.63... and so on.
3) In last octate of subnet mast, 3 bits are used for subnetting and rest 5 are used to create host space.
4) Host available per subnet is 32-2=30. (As first and last are reserved by Network ID and Broadcast ID)
5)192.168.1.0/27
6) 11111111.11111111.11111111.11100000
7)
1st subnet is 192.168.1.0 to 192.168.1.31, so 192.168.1.0 is the Subnet ID address and last address 192.168.1.31 is the broadcast address.
last i.e 8th subnet is 192.168.1.224 to 192.168.1.255, so 192.168.1.224 is subnet ID address and 192.168.1.255 is Broadcast address.