Cisco EIGRP 路由度量值计算实验

sdsk

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,
reliability 255/255, txload 1/255, rxload 1/255
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,
reliability 255/255, txload 1/255, rxload 1/255
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,
reliability 255/255, txload 1/255, rxload 1/255
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

dryu999

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