01-17-2014, 08:55 AM
(This post was last modified: 01-17-2014, 09:16 AM by zdormanjones.)
I think Mick's answer of 4 is correct, but the reasoning behind it is flawed in a way that might be helpful to address. One thing to remember about routing decisions is that a router doesn't choose the complete path it will send a packet along; it can only choose the next hop.
In light of that, ADEBCH would not be a path here. Let's say a packet gets forwarded from A to D and then from D to E. The focus is now on where E will send the packet. E has a route to H with a metric of 10. Its route through B has a metric of 40. With a variance of 2, E would not send the packet to B. It would forward it directly to H.
AFGH would also not be a path. Let's say a packet gets forwarded from A to F. The best path from there goes through E, and even with the variance command, it would not load balance between E and G because the advertised distance from G is not less than the distance through E. It doesn't meet the feasibility condition.
Now for a few words on what the answer may be.
A lot hinges on how the variance command works. We can all agree that to qualify for unequal load balancing, a route needs to meet the feasibility condition, and it's well established that the FC is a solid less than.
But Cisco documents I've looked through are vague on the exact variance algorithm. Testing on actual equipment shows that if the variance-adjusted metric is equal to the successor's metric, load balancing does occur.
Unfortunately, the answer to this question depends completely on how the question's author understands operation of the variance command.
I set up a rather elaborate test of this scenario which I documented in a different thread (search for the QID and you'll find it), and I found that traffic gets load balanced across 4 paths:
ADEH
AFEH
ABEH
ABCH
I think the correct answer is 4, but what's the Cisco answer? Either 2 or 4, and there's no way for us to tell. I really hope I don't get a question like this on the test.
In light of that, ADEBCH would not be a path here. Let's say a packet gets forwarded from A to D and then from D to E. The focus is now on where E will send the packet. E has a route to H with a metric of 10. Its route through B has a metric of 40. With a variance of 2, E would not send the packet to B. It would forward it directly to H.
AFGH would also not be a path. Let's say a packet gets forwarded from A to F. The best path from there goes through E, and even with the variance command, it would not load balance between E and G because the advertised distance from G is not less than the distance through E. It doesn't meet the feasibility condition.
Now for a few words on what the answer may be.
A lot hinges on how the variance command works. We can all agree that to qualify for unequal load balancing, a route needs to meet the feasibility condition, and it's well established that the FC is a solid less than.
But Cisco documents I've looked through are vague on the exact variance algorithm. Testing on actual equipment shows that if the variance-adjusted metric is equal to the successor's metric, load balancing does occur.
Unfortunately, the answer to this question depends completely on how the question's author understands operation of the variance command.
I set up a rather elaborate test of this scenario which I documented in a different thread (search for the QID and you'll find it), and I found that traffic gets load balanced across 4 paths:
ADEH
AFEH
ABEH
ABCH
I think the correct answer is 4, but what's the Cisco answer? Either 2 or 4, and there's no way for us to tell. I really hope I don't get a question like this on the test.