Keep in mind how ACL's work (its binary). So, 55 in binary is 01100111. Effectively, the wildcard you typed would mean that in the fourth octet the host address must have 0's in the same positions as 01100111. As you can see, a 4th octet of 00000001 matches that wildcard, but this is address .1, which is not what he is trying to do.
Also, it wont work. The wildcard mask must be a multiple of two minus 1 (for example wildcard = .1, .3, .7, .15, .31 ... ,.127 etc)
We are pleased to announce availability of Beta software for 16.6.3.
16.6.3 will be the second rebuild on the 16.6 release train targeted
towards Catalyst 9500/9400/9300/3850/3650 switching platforms. We are
looking for early feedback from customers befor...
Introduction Featured Speakers Luis Espejel is the Telecommunications
Manager of IENova, an Oil & Gas company. Currently he works with Cisco
IOS® and Cisco IOS XE platforms, and NX to some extent. He has also
worked as a Senior Engineer with the Routing P...
In this session you can learn more about Layer 3 multicast and the best
practices to identify possible threats and take security measures. It
provides an overview of basic multicast, the best security practices for
use of this technology, and recommendati...