jskfan
asked on
EIGRP Best Path Selection
EIGRP Best Path Selection
Assuming R1 has 2 paths to reach route 11.11.11.11 behind R4
R1--->R2--->R3----R4
R1---->R5--->R6----R4
The bandwidth : R1 to R2 :1.544kbps R2 to R3: 128Kbps R3 to R4 : 1.544kbps
The bandwidth : R1 to R5 :256Kbps R5 to R6: 256Kbps R6 to R4 : 256 Kbps
Which path R1 will choose to reach R4 ?
Thank you
Assuming R1 has 2 paths to reach route 11.11.11.11 behind R4
R1--->R2--->R3----R4
R1---->R5--->R6----R4
The bandwidth : R1 to R2 :1.544kbps R2 to R3: 128Kbps R3 to R4 : 1.544kbps
The bandwidth : R1 to R5 :256Kbps R5 to R6: 256Kbps R6 to R4 : 256 Kbps
Which path R1 will choose to reach R4 ?
Thank you
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The path is calculated through the topology and the throughput in the first line R1-R2-R3-R4 is the 128kb connection.
so the total K value of that road is higher then the R2-R5-R5-R4 road.
so the total K value of that road is higher then the R2-R5-R5-R4 road.
ASKER
by the way 1.544kbps I meant 1.544Mbps
So the Throughput via : R1--->R2--->R3----R4 is higher even with one link that has 128 Kbps
So why would it take the path : R1---->R5--->R6----R4 instead of R1--->R2--->R3----R4
So the Throughput via : R1--->R2--->R3----R4 is higher even with one link that has 128 Kbps
So why would it take the path : R1---->R5--->R6----R4 instead of R1--->R2--->R3----R4
ASKER
Thanks
ASKER
if so then Why ?