With the following subnets, identify all of those with overlaps: (Select all that apply)
1) 192.168.1.253/30
2) 192.168.1.245/29
3) 192.168.1.122/30
4) 192.168.1.113/28
5) 192.168.1.125/30
Since first 3 octets of all the addresses given are the same, let's consider the last octet
Last octet in binary is
1. 11111101/30
2. 11110101/29
3. 01111010/30
4. 01110001/28
5. 01111101/30
If we observe, ip addresses 3 and 5 are subnets of ip address 4 because the first four bits (Total 28 bits in the ip address) are the same for all three.
So, Subnet 3 overlaps with subnet 4
Subnet 5 Overlaps with subnet 4
With the following subnets, identify all of those with overlaps: (Select all that apply) 1) 192.168.1.253/30...
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