Question

Break 78.56.128.0/17 into four equal size subnetworks. They are (in dotted decimal notation):

Answer 1

We have Address as 78.56.128.0/17 .

/17 means that total number of addresses in the given network is 32-17= 15 i.e.2^15 Addresses

Now we want to divide it into 4 equal size subnetworks. So number of addresses each subnetwork will get = 2^15 / 4 = 2^13

As one subnet has 2^13 addresses so netmask for each subnet is / 19

As Initially we have 17 bits So we get subnet mask as 255.255.128.0

We have /19 so net mask will be 255.255.224.0. It means that last 13 bits are of host .

Lets divide we get

Subnetwork No.	Subnetwork address	Netmask	Range of addresses
1.	78.56.128.0/19	255.255.224.0	78.56.128.0 - 78.56.159.255
2.	78.56.160.0/19	255.255.224.0	78.56.160.0 - 78.56.191.255
3.	78.56.192.0/19	255.255.224.0	78.56.192.0 - 78.56.223.255
4.	78.56.224.0/19	255.255.224.0	78.56.224.0 - 78.56.255.255

This is how we can divide the given IP into 4 equal size subnets where each have 2^13 addresses .

Thank You

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