Refer to the exhibit. A network administrator created an IPv6 ACL to block the Telnet traffic from the 2001:DB8:CAFE:10::/64 network to the 2001:DB8:CAFE:30::/64 network. What is a command the administrator could use to allow only a single host 2001:DB8:CAFE:10::A/64 to telnet to the 2001:DB8:CAFE:30::/64 network?

IT Questions BankCategory: CCNA SecurityRefer to the exhibit. A network administrator created an IPv6 ACL to block the Telnet traffic from the 2001:DB8:CAFE:10::/64 network to the 2001:DB8:CAFE:30::/64 network. What is a command the administrator could use to allow only a single host 2001:DB8:CAFE:10::A/64 to telnet to the 2001:DB8:CAFE:30::/64 network?

Question:

Refer to the exhibit. A network administrator created an IPv6 ACL to block the Telnet traffic from the 2001:DB8:CAFE:10::/64 network to the 2001:DB8:CAFE:30::/64 network. What is a command the administrator could use to allow only a single host 2001:DB8:CAFE:10::A/64 to telnet to the 2001:DB8:CAFE:30::/64 network?

  • permit tcp 2001:DB8:CAFE:10::A/64 2001:DB8:CAFE:30::/64 eq 23
  • permit tcp host 2001:DB8:CAFE:10::A 2001:DB8:CAFE:30::/64 eq 23 sequence 5
  • permit tcp 2001:DB8:CAFE:10::A/64 eq 23 2001:DB8:CAFE:30::/64
  • permit tcp host 2001:DB8:CAFE:10::A eq 23 2001:DB8:CAFE:30::/64

Explanation: When an IPv6 ACE is created and is to be processed before an existing ACE is processed, the next command entered must use the sequence argument with a number lower than the existing ACE. This allows an entry to be placed before an existing entry, as the default sequence numbers are commonly numbered by increments of 10. Thus, using a sequence number of 5 on an ACE will place it in front of a prior existing entry with a sequence number of 10.

Exam with this question: CCNP Enterprise: Advanced Routing (Version 8.0) – Infrastructure Security and Management Exam
Exam with this question: Modules 8 - 10: ACLs and Firewalls Group Exam Answers

Subscribe
Notify of
guest

0 Comments
Inline Feedbacks
View all comments
0
Would love your thoughts, please comment.x
()
x