NB100 - Printable Version +- How2pass.com Forums (https://www.how2pass.com/forum) +-- Forum: CCNP (https://www.how2pass.com/forum/forum-6.html) +--- Forum: CCNP ENARSI 300-410 Forum (https://www.how2pass.com/forum/forum-15.html) +--- Thread: NB100 (/thread-337.html) |
NB100 - yostin - 03-25-2010 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. Re: NB100 - joshuad31 - 03-25-2010 Hello, 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. Joshua Re: NB100 - kostea - 07-07-2010 Hello. 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. kos |