OK: same actors and general script as in Godelproof's torture!
Speeds: A=1, B=2, C=4 and M=5.
ALL pick-up and drop-off points are at INTEGER hours.
What is SHORTEST X to Y distance?
Well, I find a rather simple way to work your problem out
But first let me ask you and everybody a couple of questions, too:
1) A=1, B=2, C=4 and M=5, XY=L as in your problem. Then the minimum time is $\displaystyle {T}_{least}=\alpha$$\displaystyle L$, for some constant $\displaystyle \alpha$. What is the exact value of $\displaystyle \alpha$?
2) A=10, B=30, C=40, D=50, M=75, L=whatever you want. Is there a feasible plan? Yes or no?
Here's how to figure out $\displaystyle \alpha$:
Spoiler:
As for integer solutions Wilmer required, I appologize for my earlier claim. There's no simple way around it. One has to delve into the dizzy details. Better to write a program to seach and check for it. Too much load to calculate it by hand! However, we still must have $\displaystyle \alpha=\frac{7}{17}$, as calculated in the spoiler.
WHOA!! We're not looking at the same problem!
MY FAULT; my original problem:
> OK: same actors and general script as in Godelproof's torture!
> Speeds: A=1, B=2, C=4 and M=5.
> ALL pick-up and drop-off points are at INTEGER hours.
> What is SHORTEST X to Y distance?
I FORGOT to mention: A, B, C and M are ALL at X ; SORRY!
Now try again
Well... then this is not very interesting... M picks up A and carries him somewhere before C, then turn back and on its way to B, carries C some distance backwards, then catches B and carry him forwards to catch A, when C just catches up with A, too. Does your plan work in the same way?
Well, posting my solution in case I forget!!
> Speeds: A=1, B=2, C=4 and M=5.
> ALL pick-up and drop-off points are at INTEGER hours.
> What is SHORTEST X to Y distance?
ANSWER: 72 miles (36 hours)
M carries A for 9 hours; A continues to Y, takes 27 hours.
M returns, picks up C after 1 hour (hour#10), keeps going back to X.
(the 9 hours was in order to make it possible for M to pick up C in 1 hour)
C now goes to Y: takes 18 hours (72 / 4 = 18)
In meantime, B goes to Y, taking 36 hours (72 / 2 = 36)
Only works for this specific case.
Main "trick" is have M in position to pick up C only 1 hour after dropping off A.
In case useful to you, I found ONLY 1 case with 4 walkers, keeping L < 10000.
L = 6912; a = 12, b = 24, c = 32, d = 33 and m = 96 ; time = 198.
(ALL pick=ups and drop-offs being at INTEGER hours)
1: M meets A at hour# 64 ; takes A East for 54 hours; A takes another 80 hours
2: M meets B at hour#144; takes B East for 30 hours; B takes another 24 hours
3: M meets C at hour#180; takes C East for 9 hours; C takes another 9 hours
4: M meets D at hour#192; takes D to Y in 6 hours
Your "cute" formula will show above as correct.