Hi all, I have a question about EIGRP calculation and hope someone can help me. The network diagram is as follow:
[R2]-(s0)---------------(s0)-[R1] (e0) ---------------- (e0) (e0) [R3] [R4] (s0) (s0) | | | | | | (s0) (s0) [R5] [R6] (e0) (e0) ----------------- (192.168.1.0) In this network, all routers are running eigrp. The eigrp parameters are as follow: R1 (s0): delay:20,000us; bandwidth:1,536kbit R2 (e0): delay:1,000us; bandwidth:10,000kbit R3 (s0): delay:35,000us; bandwidth:2,048kbit R4 (s0): delay:20,000us; bandwidth: 1,536kbit R5 (e0): delay:1,000us; bandwidth:10,000kbit R6 (e0): delay:1,000us; bandwidth:10,000kbit >From R1 to the network 192.168.1.0, it should have 2 paths: Path1: R1->R2->R3->R5 Total delay: 57,000 (20,000+1,000+35,000+1,000) Min. Bandwidth: 1536kbit EIGRP Metric: 3,125,760 Advertised Distance: 2,196,992 Path2: R1->R2->R4->R6 Total delay: 420,000 (20,000+1,000+20,000+1,000) Min. Bandwidth: 1536kbit EIGRP Metric: 2,741,760 Advertised Distance: 2,229,760 As the metrics of Path2 is smallest, I think it should go through Path2 (R1->R2->R4->R6). However, I found that the router choice Path1 (R1->R2->R3->R5). It make sense! Because if a packet is in R2 and it wants to go to 192.168.1.x network, it should not go through R2->R4->R6 (metric: 2,229760), it should go through R2->R3->R5 (mtrics:2,196,992). My question is: from the books, it mentioned that the smallest EIGRP metric will becomes the FD and the route should be the Successor. In this example, the result is not true. Please help to correct me. Regards, Dovelet Message Posted at: http://www.groupstudy.com/form/read.php?f=7&i=72091&t=72091 -------------------------------------------------- FAQ, list archives, and subscription info: http://www.groupstudy.com/list/cisco.html Report misconduct and Nondisclosure violations to [EMAIL PROTECTED]
