# Motorcycle and Walking Rates and Times

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• Jun 16th 2011, 11:16 AM
Wilmer
Re: Motorcycle and Walking Rates and Times
Quote:

Originally Posted by godelproof
> No need for that! Solution to the problem exists as long as M>C...

Ya...really then, all it is a "2 walker" solution, with 3rd walker made to fit in:
L / (time-for-2-walkers) .

Quote:

>EDIT: Did your program search for cases where M may carry a person backwards?
No.
• Jun 16th 2011, 05:58 PM
godelproof
Re: Motorcycle and Walking Rates and Times
Quote:

Originally Posted by Wilmer
>EDIT: Did your program search for cases where M may carry a person backwards?

No.

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?
• Jun 17th 2011, 10:36 PM
godelproof
Re: Motorcycle and Walking Rates and Times
Quote:

Originally Posted by Wilmer
No need for that! Solution to the problem exists as long as M>C...

Ya...really then, all it is a "2 walker" solution, with 3rd walker made to fit in:
L / (time-for-2-walkers) .

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.

Quote:

Originally Posted by Wilmer
Lots of math superstars at this site (does not include me!), and find it sort of strange
nobody else has responded...since I find this a very interesting "puzzle".

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?
• Jun 18th 2011, 03:01 AM
bjhopper
Re: Motorcycle and Walking Rates and Times
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
• Jun 18th 2011, 04:23 AM
godelproof
Re: Motorcycle and Walking Rates and Times
Quote:

Originally Posted by godelproof
Two questions:

1) Given any configuration \$\displaystyle ({a}_{1},{a}_{2},...,{a}_{n},m,L)\$, if a feasible plan exists, so does a feasible plan that takes least time \$\displaystyle {T}_{least} \$. Is \$\displaystyle {T}_{least}\$ always equal to \$\displaystyle {T}_{min}\$ in #10, upon substituting \$\displaystyle ({a}_{1},{a}_{2},...,{a}_{n},m,L)\$?

2) Given any configuration \$\displaystyle ({a}_{1},{a}_{2},...,{a}_{n},m,L)\$, how can we decide if a feasible plan exists at all?

Now to the first question, the answer is NO! See #21 and #22.

To the second question: Go here http://www.mathhelpforum.com/math-he...tml#post660726
• Jun 18th 2011, 05:28 AM
Wilmer
Re: Motorcycle and Walking Rates and Times
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.
• Jun 18th 2011, 06:25 AM
godelproof
Re: Motorcycle and Walking Rates and Times
Quote:

Originally Posted by Wilmer
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.

Nice. And my formular gives: (1080/20+1080/80)/(1/2+20/80+80/20)=270/19\$\displaystyle \approx\$14.21, which is obviously impossible(Crying).
• Jun 18th 2011, 06:37 AM
godelproof
Re: Motorcycle and Walking Rates and Times
Quote:

Originally Posted by bjhopper
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

Thanks~ So how can this example further help us?
• Jun 18th 2011, 06:56 AM
bjhopper
Re: Motorcycle and Walking Rates and Times
godelproof you didn't answer my question.

bjh
• Jun 18th 2011, 07:12 AM
godelproof
Re: Motorcycle and Walking Rates and Times
Quote:

Originally Posted by bjhopper
godelproof you didn't answer my question.

bjh

• Jun 18th 2011, 09:45 AM
Wilmer
Re: Motorcycle and Walking Rates and Times
Quote:

Originally Posted by godelproof
....it may exist even if M<C. But that's contingent.

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.
• Jun 18th 2011, 03:26 PM
bjhopper
Re: Motorcycle and Walking Rates and Times
My question. How many hours have you estimated are required to complete the task. Just give your numbers without lengthy comments.Thanks.

bjh
• Jun 18th 2011, 05:11 PM
godelproof
Re: Motorcycle and Walking Rates and Times
a1=.1 a2=.133 a3=.166 m=2
What is L? L=2.1*57=1197/10?
• Jun 18th 2011, 06:44 PM
godelproof
Re: Motorcycle and Walking Rates and Times
Quote:

Originally Posted by Wilmer
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?

I don't agree(Smirk)
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.
• Jun 18th 2011, 09:03 PM
Wilmer
Re: Motorcycle and Walking Rates and Times
Quote:

Originally Posted by godelproof
Say a=100, b=99, c=98 and m=99.9, L=100
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.

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....
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