Thread: Motorcycle and Walking Rates and Times

a,b and c leave on foot at the same time for at speed 6km/h, 8km/h and 10km/h respectively. =120km. At the same time they leave , a motocycle leaves to pick them up. Speed of the motorcycle is 120km/h. It can only take one person at a time. However, it can stop at any time so as to let the person on it to get off and catch another person to pick up. Now a,b and c want to arrive at simultaneously, and as quickly as possilbe. How long will it take them, with the help of the motorcycle? How far do they walk and travel on the motorcycle respectively?

X[A,B,C].....................................120.......... .............................[M]Y
Distance of straight path X to Y is 120.
A, B and C are at X, and motorcycle M is at Y; speeds are 6,8,10,120 respectively.
All leave at same time, A,B,C going East, M going West.
M can pick up ONLY one person at a time....and so on...
Is that what you mean?

If so, what work have YOU done on this?
Is this homework, or from a puzzle site?
Can you answer this:
M meets A, then drives A to a point 36 from Y: how long does that take?

Originally Posted by Wilmer

X[A,B,C].....................................120.......... .............................[M]Y
Distance of straight path X to Y is 120.
A, B and C are at X, and motorcycle M is at Y; speeds are 6,8,10,120 respectively.
All leave at same time, A,B,C going East, M going West.
M can pick up ONLY one person at a time....and so on...
Is that what you mean?

If so, what work have YOU done on this?
Is this homework, or from a puzzle site?
Can you answer this:
M meets A, then drives A to a point 36 from Y: how long does that take?

Thanks! the question is as you understand it. I heard it from a friend. A (B or C) either walks or travels on M. So the speed is either 6 (8 or 10) or 120. Suppose A, B and C have traveled on M t1, t2 and t3 hours respectively when they arrive at Y, that they spend the same amount of time in the journey means t1+(120-120t1)/6=t2+(120-120t2)/8=t3+(120-120t3)/10=T (these equations assume, implicitly, correctly or incorrectly, that it is never optimal for M to take one person on it and drive BACKWARDS some distance towards X). During the journey M keeps moving at 120. So it travels 120T distance on the jouney. I have no idea how to proceed from here on...

I find that for this type, always easier to set up a simpler case; you can then "see"
what's involved; have a look at this one:

X[A,B].....................................120.......... ..............................[C]Y
Distance of straight path X to Y is 120.
A, B are at X, and motorcycle C is at Y; speeds are 5,6,10 respectively.

Code:

1: C meets A in 120 / (5 + 10) = 8 hours; B has travelled 8 * 6 = 48
X......40.........[C,A,8h]................80.......................Y
X.........48...........[B,8h]................72....................Y

2: C drives A for 6 hours, to 20 from Y; B goes another 6 * 6 = 36
X......................100........................[C,A,14h]...20...Y
X....................84....................[B,14h].......36........Y

3: C is now 16 from B; turns around and meets B in 1 hour; A travels 5
X.......................105...........................[A,15h]..15..Y
X......................90...................[C,B,15h]......30......Y

4: C drives B for 3 hours; A travels 15; both arrive at Y at same time
X..............................120..........................[A,18h]Y
X..............................120........................[C,B,18h]Y

That's all you get from me!
Should be easier now for you to set this up your own way, and bring in the 3rd walker.

Originally Posted by Wilmer

I find that for this type, always easier to set up a simpler case; you can then "see"
what's involved; have a look at this one:

X[A,B].....................................120.......... ..............................[C]Y
Distance of straight path X to Y is 120.
A, B are at X, and motorcycle C is at Y; speeds are 5,6,10 respectively.

Code:

1: C meets A in 120 / (5 + 10) = 8 hours; B has travelled 8 * 6 = 48
X......40.........[C,A,8h]................80.......................Y
X.........48...........[B,8h]................72....................Y
 
2: C drives A for 6 hours, to 20 from Y; B goes another 6 * 6 = 36
X......................100........................[C,A,14h]...20...Y
X....................84....................[B,14h].......36........Y
 
3: C is now 16 from B; turns around and meets B in 1 hour; A travels 5
X.......................105...........................[A,15h]..15..Y
X......................90...................[C,B,15h]......30......Y
 
4: C drives B for 3 hours; A travels 15; both arrive at Y at same time
X..............................120..........................[A,18h]Y
X..............................120........................[C,B,18h]Y

That's all you get from me!
Should be easier now for you to set this up your own way, and bring in the 3rd walker.

Thanks... It's one of the ways they can do it. But how do you know it is the QUICKEST WAY they can do it? Remember they not only need to arrive at Y at the same time, but also as quickly as possilbe. I think when there're 3 people, all the complications come from finding the QUICKEST WAY to arrive at the same time to Y.

interestingly, after some purely algebraic calculations, i discover that in case of 2 walkers, as long as they arrive at Y at the same time, it makes no difference whether the the motorcylce picks up a first or b first --- it all takes exactly the same amount of time!! i don't know if it holds true for 3 or more walkers (it likely does!). Anyway i find this result remarkable. It seems that under the contraint that all walkers arrive at Y simultaneously, time is a conserved quantity! What is the intuitive or deeper reason underlying this result?

i now have a better idea than going into the details! The insight is this: motorcycle m starts out at Y and finally comes back to Y, so its displacement is zero! What does this mean? it means that no matter how it zigzags among all the walkers, when all's done, it has equal amount of time going forward and going backward! If we accept that it is never optimal for m to carry a person backwards, then m is going backward only when it takes no one on it. Similarly, if we also accept that it is never optimal for m to going forward without carrying a person on it, then m is going forward only when it is loaded.

So half of the time m is loaded and half of the time empty. this is the crucial observation!

Now we can consider a more general problem. Let there be n walkers, with speeds a1,a2,...,an. Motorcycle has speed m and distance between X an Y is L. Suppose the total time taken is T hours, and each walker spends ti hours on the motorcycle, i=1,2,...,n. Then according to the above analysis, we can formulate the following n+1 linear equations (attachment), with n+1 unknowns, which are independent of picking orders and all the complicated details! it's also consistent with #6.

but again... #7 relies on two assumptions, i.e., in optimal solutions, m never carries a person backwards and it never go forward unloaded... anyone can give a proof of this?

Somebody once said, that in search of a proof we are forced to face the problem more honestly... Now using the notations in #7, consider the case of 2 walkers, if

L=30, a1=1, a2=25, m=30

then, in order to bring a1 and a2 to Y simultaneously, IT IS NECESSARY FOR THE MOTORCYCLE TO CARRY a2 BACKWARDS FOR SOME DISTANCE ! So here we can no longer use the equations in #7...

In order to avoid carrying people backwards or going unloaded forwards, and to legitimately use the equations in #7, m or L needs to be large enough. Why is that? Well, let's just suppose for the moment it is so. Hence if m and L are large enough, there is a way to bring all to Y at the same time in which the Motrocycle either carries people forward or goes back unloaded. According to #7 then, all such ways entail the same amount of time T. But how can we show T is also the smallest compared to all the possible ways that can bring walkers simultaneously to Y?

Well, nobody thinks it worth the effort... I thought about the problem for hours though... and here's my proof for #8 (Given the analysis in #9 we only consider the case where m is large enough)

Walkers are A@21, B@28, C@36, with motocyclist M@84.
Distance = 1680.

Shortest time?

Originally Posted by Wilmer

Walkers are A@21, B@28, C@36, with motocyclist M@84.
Distance = 1680.

Shortest time?

To answer the question in my way, the only thing I have to assume, without being able to prove, is that there IS a plan, in which the motorcycle either goes backward unloaded or goes forward loaded (call it the "regular plan"), AND which brings the three people to Y simultaneously (so it's "feasible"). This is also the assumption made in #9. If so, then according to #10, the regular feasible plan (RFP) entails the shortest time among all the feasible plans, which is given by:


= , where L=1680, and M are 21, 28, 36 and 84. So we have = =44.


Furthermore, according to #10 the time that is on the motorcycle forward is given by , which upon substitutions gives 12, 8 and 2 for the three people respectively.

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Now what troubles me is this: the above argument gives the right answer, IF an RFP exists. But what happens when all feasible plans are nonregular, as given in the example in #9? Why is this important? Well, because we don't know a priori, when using the equations in #10 to solve for the problem, that an RFP exists. It's just side effect of a nonconstructive solution Besides, I also want to be able to tackle the nonregular case as the one in #9

Any thoughts or suggestions are welcome!

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EDIT：Consider the minimizing problem in #10. There letting alone is not enough to guarantee that the motorcycle doesn't carry person i backwards! Because when we finish solving the equations it is possible for us to find . Since for all RFPs we have and for all i, there will be no RFP if for all i but for some i. Check it for #9 we have: , . And indeed #9 has no RFP. So to sum up: solve equations in #10 (or equivilently, in #7). If for some i, then no RFP exists.

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 , how can we decide if a feasible plan exists at all?

Originally Posted by godelproof

= , where L=1680, and M are 21, 28, 36 and 84. So we have = =44.

Very nice!
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".

I wrote a program to find integer cases for the "3 walkers case"
(that's how I got my example I posted).
I can now "see" that your formula above "fits" perfectly what my program does, but
I don't think I'd ever have been able to "condense" it as nicely as you did.

Keeping 0 < A < B < C < M < 100, I found minimum distance to be 1512 (time 66).

In my posted example, breakdown is:
A: @21: 672-32, @84: 1008-12 ; 1680-44
B: @28: 1008-36, @84: 672-8 ; 1680-44
C: @36: 1512-42, @84: 168-2 ; 1680-44

I have only one thought:
what happens if C (faster of the 3) is not picked up by M, because his speed
is such that he covers the distance at EXACTLY same time as A and B arrive?

Thanks for the puzzle; hope someone steps in to help further...

Originally Posted by Wilmer

I have only one thought:
what happens if C (faster of the 3) is not picked up by M, because his speed
is such that he covers the distance at EXACTLY same time as A and B arrive?

No need for that! Solution to the problem exists as long as M>C (it may exist even if M<C. But that's contingent)

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EDIT: Did your program search for cases where M may carry a person backwards?