I think this will be complicated by your topology because you are seeing split horizon in effect.
The all-links conmand shows all routes received whether they are FS or not.
The issue is from R5's perspective it has three equal cost paths to 192.168.100.0/24 assuming youi have left the default delays on the interfaces and all interfaces are fast ethernet.
So R5 receives a route from R2 to 192.168.100.0/24 and it uses R2 as one of it's paths. R5 also receives the same route advertisements from R3 and R4 but it cannot advertise them to R2 because it is using R2 as a next hop for that network.
So split horizon stops it advertsing that route.
Note this should also apply to R3 and R4.
In addition you say you want R2 to use R5 -> R3 etc. but even if we didn't face split horizon R5 would actually use R3 and R4 unless you manipulated the metrics there as well.
So it may not be the best topology to use to for influencing metrics although i would have to sit down and work it all out to make sure.
In terms of working out the delays EIGRP sums all the delays of the outgoing interfaces ie.
R2 -> R1 -> R7 -> SW1
1) SW1 advertise the route with the delay included in the EIGRP route advertisement (the delay here is the delay to the SVI)
2) R7 receives the route advertisement and adds the delay of the interface used to connect to SW1 to the delay in the EIGRP route advertisement to work out the locally calculated distance.
3) R7 then advertises the route to R1 with the sum of the delays ie. the delay of it's interface that connects to SW1 + the delay advertised by SW1
3) R1 then adds the delay of it's interface to the delay value received in the EIGRP route advertisement to work out it's locally calculated distance. It then sends a route advertisement to R2 which again contains a delay value with is the sum of delays from R1 -> R7 -> SW1
4) R2 adds it's own interface delay to that delay value to calculate the distance.
So you need to work out the delays like that.
For a route to be a FS it's advertised distance must be less than the FD so you need R5 to advertise a distance that is less than the FD on R2.
Note here we are assuming the FD on R2 is via the direct path to R1 -> R7 -> SW1
But you also need to make sure that once R2 has calculated the distance via R5 it is greater than it's current FD via R1 or it will choose R5 as the path ie
R2's FD must be less than R5s advertised distance but greater than R2's locally calculated distance via R5.
Like i say though with your toplogy things may not work out correctly due to other things but you can always experiment and you will probably learn a fair bit.
Hopefully that makes some sense.
I'll check back tomorrow if you need any futher clarification and nobody else responds.
Finally EIGRP terminology is much misused in terms of FD, FS and how these things are worked out etc so if Peter Paluch sees this he may well need to correct some of what i have said if i have made any glaring errors.
Peter, if you do see this please feel free to join in.
Jon
