Refer to the exhibit. EIGRP has been configured on all routers in the network. The command metric weights 0 0 1 0 0 has been added to the EIGRP process so that only the delay metric is used in the path calculations.
Which router will R1 select as the successor and feasible successor for Network A?
Select the best response.
A.
R4 becomes the successor for Network A and will be placed in the routing table. R2 becomes the feasible successor for Network A.
B.
R4 becomes the successor for Network A and will be included in the routing table. No feasible successor will be selected as the advertised distance from R2 is higher than the feasible distance.
C.
R2 becomes the successor and will be placed in the routing table. R4 becomes the feasible successor for Network A.
D.
R2 becomes the successor and will be placed in the routing table. No feasible successor will be selected as the reported distance from R4 is lower than the feasible distance.
Somebody could explain me the answer?, Thanks.
well..as we get from the exhibit from R1 to reach Network A we have two paths and according to EIGRP we will choose the path with lowest metric cost
*R1-R4-R5=20
R1-R2-R3-R5=30
SO we will choose first one as primary path which call it in EIGRP Successor and one of Eigrp features is to maintain high reliability using through using this feature DUAL is to find back up path incase of failure of primary under condition FD(S)>AD(FS)
FD=fessiable distance=it is path from source to destination (R1 TO R5)
AD=Advertise distance=it is path from my neighbor to destination (R2 TO R5) and From (R4 TO R5)
so any clairfication you are more than welcome
Can some one please explain me why Answer “A” is not correct.
As Per eigrp rule, a path can be feasible successor only and only if advertise distance from the nieghbor is less than the current feasble distance.
from path 1 feasible distance ::R1-R4-R5= 20,
advertise diatance from path 2 (from R2) is :: R2-R3-R5= 25, which is not less than feasible distance(20). so there can not be a feasible successor.