# Variance question

Can someone please expain why the answer is 3 on this one?
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InstructorCommented:
First, router A will only have 3 routes to the data center.

BEH, DEH and FEH

BCH would not have been advertised to A because it has a cost of 30 while BEH is only 20. The same is true for FGH since FEH has a lower cost.

To determine the number of unequal cost paths, first find the BEST path. In this case DEH at 30.

Then multiply that by the variance which is 2. 30*2=60.

Any paths of 60 or better will be used.

DEH is already in the routing table at a cost of 30
BEH is 60
FEH is 40
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Commented:
Paths must also meet the feasibility condition for them to be considered by variance i.e. their advertised distance must be less than the current FD. They must be successor's or feasible successors.
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Commented:
A-D-E-H main route
A-B-C-H =70 ( this route comply to the FD condition), but ( A-D-E-H) x variance=60 > A-B-C-H=70, so this route will not be added to the routing table.
Correct answer should be A ( 1 route ).
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Commented:
sorry :D , instead of ( A-D-E-H) x variance=60 > A-B-C-H=70, ( A-D-E-H) x variance=60 < A-B-C-H=70
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