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I was curious if some one could help explain why there are 3 paths.

If metric is >60  only 2 paths meet that requirement
If metric is >=60 then 4 paths meet that requirement.



You must view the question this way.
Step one: Chart out the FD and AD of all routes

route FD AD
ADEH 30 20
ABCH 70 30
ABEH 60 20
AFGH 60 40
AFEH 40 20

Which ones now meet the feasibility condition (search wikipedia if you are not familiar with this) of AD Feasible Successor < FD Successor.

There are only two routes with AD less than 30.  These are the two routes with the AD of 20.  The variance is set at 30 x 2 = 60.  How many of these routes that meet the feasibility condition also satisfy this requirement.  Two.  Thus total routes are 1 Successor 2 Feasible successor.



I  don't think that ABEH will be used for load balancing. It must have a value less than 60 but it has 60. So, i think the correct answer is 2.