Given a station with 4 interfaces: Interface name Address and netmask Ethernet0

Question

Given a station with 4 interfaces:

Interface name          Address and netmask

Ethernet0               64.81.209.250/24

Ethernet1               64.81.212.1/22

Ethernet2               64.81.208.80/26

Ethernet3               64.81.211.1/23

And a routing table of

Network         netmask           nexthop

64.81.208.128   255.255.255.128   64.81.215.250      

64.81.208.0     255.255.255.192   64.81.208.126

defaultroute                      64.81.210.254

For questions 9 and 10 you have a network 128.64.144.0/20 and you want

to subnet it.

9. If you wanted to subnet 128.64.144.0/20 into 4 networks of 1024

addresses (each supporting 1022 hosts) what would the the 4 sub-network

addresses be? I'm looking for 4 subnets specified as XX.XX.XX.XX/YY.

10. If you wanted to further subnet ONE of the 4 networks above into networks

of 64 address (supporting 62 hosts) what would the new netmask be? How

many sub-networks would you create?

Explanation / Answer

9. Given subnet 128.64.144.0/20
We need to subnet the above into 4 networks. Each 1024 hosts (1022 valid).
Inorder to have 4 hosts, we need to borrow 2 bits so that we can have 2^2 = 4 networks.
So, new subnetmask will be /22. There are 10 bits for host portion remaining (32-22 = 10). So, there can be 2^10 = 1024 hosts. Outof all the available hosts in the network, one is for network address and one is reserved for broadcast address. So, total valid hosts are 1024-2 = 1022 addresses.

10. The above network was subnetted into 4 networks

Binary notation
10000000.01000000.10010000.00000000
11111111.11111111.11110000.00000000 (/20)

4 subnetworks are

128.64.144.0/22
128.64.148.0/22
128.64.152.0/22
128.64.156.0/22

First subnet is 128.64.144.0/22

We need to subnet the above network with 64 hosts each. So, we need 6 bits in host portion (2^6 = 64). There are 4 bits still remaining in the host portion, with which we could make 2^4 = 16 subnets.

New subnet mask will be 128.4.144.0/28

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