AD vs FD with variance clarification

JohnnyBiggles

Please verify:

Path 1:

[R1] ======= [R2] ======= [R3] ======= [R4]
.............20...................25.....................20.................


Path 2:

[R1] ======= [R5] ======= [R6] ======= [R4]
.............10....................10....................10................

Path 3:

[R1] ======= [R7] ======= [R8] ======= [R4]
...............15..................10....................5..................



FD

[1-2-3-4] = 20 + 25 + 20 = 65 {AD of 45 > FD of successor 30 = not feasible successor(?)}
[1-5-6-4] = 10 + 10 + 10 = 30 {Lowest FD, so Best path = successor}
[1-7-8-4] = 15 + 10 + 5 = 30 {AD of 15 < FD of successor 30 = Feasible successor(?)}

AD
[1-2-3-4] = 25 + 20 = 45
[1-5-6-4] = 10 + 10 = 20
[1-7-8-4] = 10 + 5 = 15

So, if variance is set to 2 and the FD of the successor is 30, then 30 x 2 = 60. Therefore any FS with an FD less than or equal to 60 is a path used for load balancing?

(So final paths with variance 2 are:
[1-5-6-4]
[1-7-8-4])

Is this correct?? (I hope what I displayed makes sense)

FloOz

I'm a bit tired so bear with my answer. Basically the variance command will only load balance on another path as long as that paths AD is lower then the current best paths FD. So yes you are correct with this statement {AD of 15 < FD of successor 30 = Feasible successor}

networker050184

Kind of hard to follow along without a diagram, but it seems your math is correct.

theodoxa

Path 1 has an AD > the FD of Paths 2 and 3, so it would not be a Feasible Successor and would not be used for Load Balancing.