Question of EIGRP Calculation [7:72091]

2003-07-10 Thread dovelet
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




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RE: Question of EIGRP Calculation [7:72091]

2003-07-10 Thread hebn9999
hi,
  unlike ospf, eigrp process doesn't know the toppology of the whole
network.
   from R1's perspective,there is only one path to reach 192.168.1.0 with
nexthop is R2's s0.
   the successor is actually the nexthop instead of route.

  hebn


- Original Message -
From:"dovelet" 
To:[EMAIL PROTECTED]
Subject:Question of EIGRP Calculation [7:72091]
Date:Thu, 10 Jul 2003 15:30:30 +0800
 >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
__

===




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