Cisco CCNA (200-301)Network FundamentalsMedium
A network engineer needs to configure a Cisco IOS access control list (ACL) statement that matches any host within the 172.16.16.0/22 network. Which wildcard mask should be used in the ACL entry?
- A0.0.0.255
- B0.0.7.255
- C0.0.3.255
- D0.0.1.255
Show answer & explanationAnswer & explanation
Correct answer: C. 0.0.3.255
A /22 subnet mask is 255.255.252.0. The wildcard mask is calculated by subtracting each octet from 255: 255-255=0, 255-255=0, 255-252=3, 255-0=255, giving 0.0.3.255. This matches all addresses from 172.16.16.0 through 172.16.19.255.
Why the other options are wrong
- A. 0.0.0.255 corresponds to a /24 network, not /22.
- B. 0.0.7.255 corresponds to a /21 network.
- D. 0.0.1.255 corresponds to a /23 network.
Wildcard Mask Calculation
A wildcard mask is the inverse of a subnet mask, used in ACLs and OSPF to specify which bits must match exactly.
- Wildcard = 255.255.255.255 minus subnet mask
- 0 bits mean must match, 1 bits mean don't care
- /22 = 255.255.252.0 subnet -> 0.0.3.255 wildcard
Memory trick: Wildcard is the subnet mask's mirror image (255 minus each octet).