第十三课：EIGRP中文必威体育精装版 思科认证网络助理工程师CCNA培训PPT资料方案策划.pptVIP

DUAL 示例 (续) Q Q (1) (1) (1) (2) (2) (a) A E D C B E EIGRP FD AD Topology (a) 3 (fd) via D 3 2 (Successor) via C 4 3 D EIGRP FD AD Topology (a) **ACTIVE** -1 (fd) via E (q) via C 5 3 (q) C EIGRP FD AD Topology (a) 3 (fd) via B 3 1 (Successor) via D via E 4 3 DUAL 示例 (续) R Q (1) (1) (1) (2) (2) (a) A E D C B E EIGRP FD AD Topology (a) **ACTIVE** -1 (fd) via D via C 4 3 (q) D EIGRP FD AD Topology (a) **ACTIVE** -1 (fd) via E (q) via C 5 3 C EIGRP FD AD Topology (a) 3 (fd) via B 3 1 (Successor) via D via E DUAL 示例 (续) R (1) (1) (1) (2) (2) (a) A E D C B E EIGRP FD AD Topology (a) 4 (fd) via C 4 3 (Successor) via D D EIGRP FD AD Topology (a) **ACTIVE** -1 (fd) via E (q) via C 5 3 C EIGRP FD AD Topology (a) 3 (fd) via B 3 1 (Successor) via D via E DUAL 示例 (续) R (1) (1) (1) (2) (2) (a) A E B E EIGRP FD AD Topology (a) 4 (fd) via C 4 3 (Successor) via D D EIGRP FD AD Topology (a) 5 (fd) via C 5 3 (Successor) via E 5 4 (Successor) C EIGRP FD AD Topology (a) 3 (fd) via B 3 1 (Successor) via D via E D C DUAL 示例 (续) (1) (1) (1) (2) (2) (a) A E D C B E EIGRP FD AD Topology (a) 4 (fd) via C 4 3 (Successor) via D D EIGRP FD AD Topology (a) 5 (fd) via C 5 3 (Successor) via E 5 4 (Successor) C EIGRP FD AD Topology (a) 3 (fd) via B 3 1 (Successor) via D via E DUAL 示例 (开始) (1) (1) (1) (1) (2) (2) (a) A E D C B E EIGRP FD AD Topology (a) 3 (fd) via D 3 2 (Successor) via C 4 3 D EIGRP FD AD Topology (a) 2 (fd) via B 2 1 (Successor) via C 5 3 C EIGRP FD AD Topology (a) 3