Cisco EIGRP 路由度量值计算实验

chunjihong

　　1、如图搭建拓扑，做基本配置，确保直连网络连通性。
　　2、配置EIGRP
　　R1(config)#router eigrp 100
　　R1(config)#no au
　　R1(config)#network x.x.x.x x.x.x.x
　　3、配置完成EIGRP，测试全网的连通性。
　　4、EIGRP的度量值计算公式为：
256*{K1（10^7/带宽）+K2（10^7/带宽）/（256-负载）+K3（延迟）+K5/(可靠性+K4)}默认情况下，K1和K3是1，其他的K值都是0.　　所以通常情况下，度量值=256×（10^7/最小带宽+累积延时）
　　5、接口带宽和延迟通过“show interface”查看
　　R4#show int f 2/0
　　FastEthernet2/0 is up, line protocol is up
　　Hardware is AmdFE, address is cc03.1494.0020 (bia cc03.1494.0020)
　　Internet address is 24.24.24.4/24
　　MTU 1500 bytes,BW 100000 Kbit, DLY 100 usec,

　　>　　Encapsulation ARPA, loopback not set
　　Keepalive set (10 sec)
　　Full-duplex, 100Mb/s, 100BaseTX/FX
　　----------------------------------------------------------------------------------
　　R1#show int loopback 0
　　Loopback0 is up, line protocol is up
　　Hardware is Loopback
　　Internet address is 1.1.1.1/24
　　MTU 1514 bytes, BW 8000000 Kbit, DLY 5000 usec,

　　>　　Encapsulation LOOPBACK, loopback not set
　　---------------------------------------------------------------------------------
　　R2#show int f 0/0
　　FastEthernet0/0 is up, line protocol is up
　　Hardware is AmdFE, address is cc01.1494.0000 (bia cc01.1494.0000)
　　Internet address is 12.12.12.2/24
　　MTU 1500 bytes,BW 100000 Kbit, DLY 100 usec,

　　>　　Encapsulation ARPA, loopback not set
　　---------------------------------------------------------------------------------
　　R4#show ip rout
　　Codes: C - connected, S - static, R - RIP, M - mobile, B - BGP
　　D - EIGRP, EX - EIGRP external, O - OSPF, IA - OSPF inter area
　　N1 - OSPF NSSA external type 1, N2 - OSPF NSSA external type 2
　　E1 - OSPF external type 1, E2 - OSPF external type 2
　　i - IS-IS, su - IS-IS summary, L1 - IS-IS level-1, L2 - IS-IS level-2
　　ia - IS-IS inter area, * - candidate default, U - per-user static route
　　o - ODR, P - periodic downloaded static route
　　Gateway of last resort is not set
　　34.0.0.0/24 is subnetted, 1 subnets
　　C    34.34.34.0 is directly connected, FastEthernet3/0
　　/ 1.///0.0.0/24 is subnetted, 1 subnets
　　D    1.1.1.0 [90/158720] via 24.24.24.2, 01:09:05, FastEthernet2/0
　　2.0.0.0/24 is subnetted, 1 subnets
　　D    2.2.2.0 [90/156160] via 24.24.24.2, 00:26:13, FastEthernet2/0
　　3.0.0.0/24 is subnetted, 1 subnets
　　D    3.3.3.0 [90/156160] via 34.34.34.3, 01:08:54, FastEthernet3/0
　　4.0.0.0/24 is subnetted, 1 subnets
　　C    4.4.4.0 is directly connected, Loopback0
　　23.0.0.0/24 is subnetted, 1 subnets
　　D    23.23.23.0 [90/30720] via 34.34.34.3, 01:09:06, FastEthernet3/0
　　[90/30720] via 24.24.24.2, 01:09:06, FastEthernet2/0
　　24.0.0.0/24 is subnetted, 1 subnets
　　C    24.24.24.0 is directly connected, FastEthernet2/0
　　12.0.0.0/24 is subnetted, 1 subnets
　　D    12.12.12.0 [90/30720] via 24.24.24.2, 01:09:07, FastEthernet2/0
　　---------------------------------------------------------------------------------
　　根据以上信息，计算R4到达1.1.1.0网络的度量值
　　[10^7/100000+(100+100+5000)/10]*256=158720
　　100000是到达1.1.1.0网络的链路的最小带宽，lookback口为8000000；
　　100+100+5000是R4 F2/0,R2 F0/0, r1 lookback 0口的延迟之和
　　计算R4到达2.2.2.0网络的度量值
　　[10^7/100000+(100+5000)/10]*256=156160
　　实验结束