Question

Wild Mask Example 220.78.168.0-A 220.78.175.0-B Web server IP add is 221.79.176.1 Wild Mask/mask A B And XOR 11011100 010011101010100000000000 Network 11011100 0100111010101111 00000000 11011100 010011101010100000000000220.78.168.0 000000000000000000000011100000000 0.0.7.255 255.255.248.0 What would be the implication for the following ACL? Access-list 101 permit tcp 220.78.168.0 0.0.7.255 host 221.79.176.1 eq 80 Questions: How many subnets can access the web server? How many hosts can access the web server?

Answer 1

Given that the Network addresses of

Network A and Network B are 220.78.168.0 and 220.78.175.0 respectively.

The web server IP address is 221.79.176.1

Configured command of ACL as Access-list 101 permit tcp 220.78.168.0 0.0.7.255 host 221.79.176.1 eq 80

The implications that we can draw from this is Network A i.e IP address 220.78.168.0 with wildcard mask as 0.0.7.255 means 220.78.168.0/21 because subnet mask is 255.255.248.0 (inverse of wildcard mask).

All the hosts in the given subnet can be permitted to access the port 80 of the webserver having IP 221.79.176.1. All tcp traffic with port 80 will be accepted by the server from the subnetwork mentioned in the command.

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Q1) As IP address is given as 220.78.168.0/21 so the number of subnets will be 2^5 ( 21 = 16 + 5).

So, 32 subnets are there in the given network.

But only the subnet 22 can access the webserver as per the command given above.

[Subnet range is 220.78.168.0 - 220.78.175.255]

Q2) As 32-21 = 11 (as the network address is 220.78.168.0/21 ) so the total number of hosts that can access the webserver will be 2^11 = 2046 hosts.

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