A network administrator is designing an ACL. The networks 192.168.1.0/25, 192.168.0.0/25, 192.168.0.128/25, 192.168.1.128/26, and 192.168.1.192/26 are affected by the ACL. Which wildcard mask, if any, is the most efficient to use when specifying all of these networks in a single ACL permit entry?
To efficiently specify all of the mentioned networks in a single ACL permit entry, we need to find a wildcard mask that matches all the networks.
The wildcard mask is used in conjunction with the network address to define the range of addresses to permit or deny in an ACL.
To find the most efficient wildcard mask, we need to identify the common bits in the network addresses. Let’s analyze the binary representation of the subnet masks for each network:
Network 192.168.1.0/25: subnet mask 255.255.255.128 or /25 (binary: 11111111.11111111.11111111.10000000).
Network 192.168.0.0/25: subnet mask 255.255.255.128 or /25 (binary: 11111111.11111111.11111111.10000000).
Network 192.168.0.128/25: subnet mask 255.255.255.128 or /25 (binary: 11111111.11111111.11111111.10000000).
Network 192.168.1.128/26: subnet mask 255.255.255.192 or /26 (binary: 11111111.11111111.11111111.11000000).
Network 192.168.1.192/26: subnet mask 255.255.255.192 or /26 (binary: 11111111.11111111.11111111.11000000).
By comparing the binary representation, we can see that the first 26 bits are common among all the network addresses. Therefore, the most efficient wildcard mask to cover all these networks in a single ACL permit entry would be 0.0.0.63, which has the binary representation 00000000.00000000.00000000.00111111.
Using this wildcard mask, we can construct the ACL permit entry as follows:
permit ip any 192.168.1.0 0.0.0.63.
This entry will effectively permit traffic to all the mentioned networks while denying traffic to any other addresses.
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