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Math Help - Special "Motorcyclist-walkers" case.

  1. #1
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    Special "Motorcyclist-walkers" case.

    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?
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    Re: Special "Motorcyclist-walkers" case.

    Hi, Wilmer! Why do we find this motorcyclist's little problem to be TORTUROUSLY addictive???!

    This special case is more involved than your previous one. Let's wait to see what others have to say
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    Re: Special "Motorcyclist-walkers" case.

    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 {T}_{least}=\alpha L, for some constant \alpha. What is the exact value of \alpha?

    2) A=10, B=30, C=40, D=50, M=75, L=whatever you want. Is there a feasible plan? Yes or no?
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    Re: Special "Motorcyclist-walkers" case.

    Quote Originally Posted by godelproof View Post
    1) A=1, B=2, C=4 and M=5, XY=L as in your problem. Then the minimum time is {T}_{least}=\alpha L, for some constant \alpha. What is the exact value of \alpha?
    Minimum time = L / B where L = (2M - B)(M + C)
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    Re: Special "Motorcyclist-walkers" case.

    Quote Originally Posted by godelproof View Post
    2) A=10, B=30, C=40, D=50, M=75, L=whatever you want. Is there a feasible plan? Yes or no?
    Where is M; at X or Y?
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    Re: Special "Motorcyclist-walkers" case.

    Quote Originally Posted by Wilmer View Post
    Where is M; at X or Y?
    Sorry for that! M's at Y of course!
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    Re: Special "Motorcyclist-walkers" case.

    Quote Originally Posted by Wilmer View Post
    Minimum time = L / B where L = (2M - B)(M + C)
    No. L is given as distance between X and Y. What is \alpha?
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    Re: Special "Motorcyclist-walkers" case.

    Here's how to figure out \alpha:

    Spoiler:
    Consider only 2 walkers with A=1, B=2, M=5. Then by forumla here http://www.mathhelpforum.com/math-he...es-182882.html in #10, {T}_{least}=(\frac{L}{M-A}+\frac{L}{M-B})/(1/2+\frac{A}{M-A}+\frac{B}{M-B})=\frac{7}{17}L . When you add in the third walker C=4, {T}_{least} won't change because all M has to do with C is to drive her backwards some distance a couple of times on his way to pick up A or B, which wastes no time. So \alpha=\frac{7}{17}.


    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 \alpha=\frac{7}{17}, as calculated in the spoiler.
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    Re: Special "Motorcyclist-walkers" case.

    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
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    Re: Special "Motorcyclist-walkers" case.

    Quote Originally Posted by Wilmer View Post
    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?
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    Re: Special "Motorcyclist-walkers" case.

    Quote Originally Posted by godelproof View Post
    ...Does your plan work in the same way?
    No.
    QUIT asking questions! Just solve it. Remember: ALL in INTEGERS.

    HINT: L is < 100
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    Re: Special "Motorcyclist-walkers" case.

    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.
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    Re: Special "Motorcyclist-walkers" case.

    Quote Originally Posted by godelproof View Post
    2) A=10, B=30, C=40, D=50, M=75, L=whatever you want. Is there a feasible plan? Yes or no?
    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.
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