From: Carolyn Camarda (ccamarda@xxxxxxxxxxxxx)
Date: Mon Feb 18 2002 - 12:32:27 GMT-3
In the book on page 712 Caslow tasks you to create an ACL to permit /24 from
172.16.17.0 through 172.16.151.0
His solution is:
acl 1 d 172.16.0.0 0.0.15.255
acl 1 d 172.16.16.0 0.0.0.255
acl 1 p 172.16.0.0 0.0.127.255
acl 1 p 172.16.128.0 0.0.15.255
acl 1 p 172.16.144.0 0.0.7.255
acl 1 p 172.16.152.0 0.0.0.255
Why the last statement permitting the .152 network. Through .151 should be
to .151.255 which should be handled by acl entry 5. What am I missing
here?
This archive was generated by hypermail 2.1.4 : Thu Jun 20 2002 - 13:46:26 GMT-3