LinkBack URL
About LinkBacksYa...really then, all it is a "2 walker" solution, with 3rd walker made to fit in:Originally Posted by godelproof
L / (time-for-2-walkers) .
No.>EDIT: Did your program search for cases where M may carry a person backwards?
Ya...really then, all it is a "2 walker" solution, with 3rd walker made to fit in:
L / (time-for-2-walkers) .
No.>EDIT: Did your program search for cases where M may carry a person backwards?
see the example in #9. there you must carry B backwards. I suspect even in this case my formular gives correct answer... are you able to modify your program to include this cases, too? If so, can you still "see" my fomular "fits" perfectly what your modified program does?
Sorry for my claim. It is wrong...
Consider L=1000, A=1, B=999, C=1000, M=1001. Here M>C but unfortunely there's no feasible plan.
Me, too! I didn't expect the general problem to turn out to be so complicated. Why nobody responses? I guess most people just browse the first few posts of the thread and think it's just a matter of computation of a specific problem? Anyway this thread is getting too long... Perhaps it's better to post the generalised problem to some other subforum?
Hello.
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.
bjh
Now to the first question, the answer is NO! See #21 and #22.Two questions:
1) Given any configuration, if a feasible plan exists, so does a feasible plan that takes least time
. Is
always equal to
in #10, upon substituting
2) Given any configuration
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.
Huh? If Motorcycle speed < fastest walker speed, then ONLY possible case is
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.
I don't agree
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.
After M drives A back, M then drops A off, then M goes forward to pick up C.
But A will also go forward : both leave together going forward, and M will never
be able to catch up, since speed slower....
  |
.
