Consider L=1000, A=1, B=999, C=1000, M=1001. Here M>C but unfortunely there's no feasible plan.
I have followed this post from the beginning.I made calculations to see if I could estimate the time for the walkers to reach the other end.My inital estimate was 4 hours I started by sending M to pick up the slowest walker and to drop him at a point where the time to do the same for the others would keep him from crossing the finish line.Rate of walkers .1 K per minute,.133, .166. Rate of M 2K per minute.It takes M to meet a 57 minutes.They approach each other at 2.1 K per minute. The trip back to the 90 mile mark takes 47 minute. The pick up point was at mile 5.7. Picking up the others took a total of 247 minutes. I can see that the drop offs should have been closer to the end.Final back and forth movements ofM to get walkers in correct positions would take little time.
To the second question: Go here http://www.mathhelpforum.com/math-he...tml#post660726
Integer example (2 walkers) where 1 walker is driven backward.
L = 1080 (X to Y), M = 100, A = 20, B = 80
M and B meet after 6 hours.
M drives B backward for 2 hours.
B then walks to Y : 10 hours ; total = 18 hours
After the 1st 8 hours, A has walked 160.
M is now 120 from A: takes 1 hour to meet A.
Distance from Y now 900: takes 9 hours ; total = 18 hours.
to have ALL walkers at fastest walker speed, and the Motorcycle is useless.
That's the only way to reach destination at same time. Agree?
Remark on case where a walker is driven backward:
it is possible to get a solution where a walker is driven back BEHIND (West of) starting point X.
The wording of your problem does not prevent that.
Say a=100, b=99, c=98 and m=99.9, L=100
what do you say? m first meets a and drive him back to somewhere between b and c, then pick up c to make everybody arrive the same time at destination. Calculations should be similar to that done here http://www.mathhelpforum.com/math-he...tml#post660891 and a feasible plan exists.