chunjihong 发表于 2018-7-17 11:43:17

Cisco EIGRP 路由度量值计算实验


  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 via 24.24.24.2, 01:09:05, FastEthernet2/0
  2.0.0.0/24 is subnetted, 1 subnets
  D    2.2.2.0 via 24.24.24.2, 00:26:13, FastEthernet2/0
  3.0.0.0/24 is subnetted, 1 subnets
  D    3.3.3.0 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 via 34.34.34.3, 01:09:06, FastEthernet3/0
   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 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
  *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
  *256=156160
  实验结束
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