Is it possible to solve this without guessing?

atran · 2013-12-31 03:53:19

Hi;

A, B, C can walk at 5 km/hr. They have a car that can accomodate any two of them which travels at 50 km/hr. Can they reach a point 62km away in less than 3 hrs?

How can I solve this without guessing?
Thanks for help.

Last edited by atran (2013-12-31 04:26:33)

bobbym · 2013-12-31 05:40:58

Hi;

Happy New Year soon.

What is wrong with guessing and then refining?
Guesstimating is done all the time in numerical work. An ansatz is a useful thing.

So far I wish I was getting that answer by any means!

atran · 2013-12-31 09:46:49

Hi bobbym; Happy New Year to you too!

bobbym · 2013-12-31 09:48:37

I have hidden your answer because I would like to continue to work on it myself and see if I can solve it.

Also, I would like to prove that is the smallest possible answer.

Here is an obvious point that I overlooked in my solution, one person must remain in the car!

gAr · 2013-12-31 18:00:02

Hi,

Last edited by gAr (2013-12-31 18:38:40)

gAr · 2013-12-31 19:32:59

bobbym · 2013-12-31 22:48:29

Hi gAr;

gAr · 2014-01-01 15:55:10

Hi bobbym,

bobbym · 2014-01-01 22:56:34

Hi gAr;

I will post my solution so you can check. I might be making some mistake in interpreting the problem.

gAr · 2014-01-01 23:18:48

Okay.

bobbym · 2014-01-01 23:21:03

It will take some time to make it readable.

gAr · 2014-01-01 23:23:38

That's fine.

bobbym · 2014-01-01 23:27:41

I worked it in the form of a program so that I could try different f's easily.

(b is the position of b in km, c is the position of c in km, car is the position of car in km, time is the amount of time used so far in hrs ) a always drives the car. Everyone starts at 0 and tries to get to 62.

b = 0;
c = 0;
car = 0;
time = 0;

Drop b, f of the way, meanwhile c is walking.

f = 1/2;
d = f*(62);
time = time + d/50;
b = d;
c = c + 5 (d/50);
car = d;

Output {b, c, car, time} = {31, 3.1, 31, .62}

I go back to get c, in the meantime b is walking.

Solve 50 t + 5 t = Abs[car - c] for t.

t = .62

b = (b + 5 t)
c = b;
car = b;
time = time + t

Output
{b, c, car, time} = {36.63, 36.63,36.63,1.74}

I take b home ( the 62km point), meanwhile c is walking.

time = time + (62 - b)/50;
b = 62;
car = 62;
c = c + 5 ((62 - b)/50);

Output {b, c, car, time} = {62, 36.63, 62, 2.25}

Go back and get c.

Solve 50 t + 5 t = car - c for t
c = (c + 5 t)
car = c;

Take c home ( both b and c are at the 62km mark with a)

time = time + (62 - c)/50

( both b and c are at the 62km mark with a)

Output time = 2.71570

gAr · 2014-01-02 00:24:05

Thanks, I did not think of going to and fro, maybe we can minimize it some more, let me see.

bobbym · 2014-01-02 00:37:08

Hmmm, it suggests that the minimum with this idea is 651/275.

Feels like there is something wrong?!

gAr · 2014-01-02 00:46:37

Can't say for now, working it out!

gAr · 2014-01-02 18:50:03

Wait, I was solving the problem for everybody to reach the destination at an exact same time.

You solved for one of them to reach at the earliest?

I used your idea dropping midway and going back to pick the other person. I did it as an infinite series,
going to and fro.

d=62 
a=0
b=d/2                       # one person is dropped here
a+=(b-a)/10                 # other person here when the car goes to b
st=a                        # keeps a checkpoint of every iteration
car=b                       # car's location
dist=d/2                    # total distance traveled by the car
while (d-b)>0.000000000001: # loop till it converges
    a+=(b-st)/11            # the lagging person meets the car at this point
    dist += (b-st)*10/11    # car travels this distance to meet at 'a'
    b+=(b-st)/11            # in the meantime, the other person also travels the same distance
    car=a                   # car goes back
    a=b                     # 'b' is always located after a
    b+=(d-b)/2              # 'b' is moved by half of the remaining distance
    a+=(b-car)/10           # car takes the person at 'a' to 'b'
    dist+=(b-car)           # add the distance and loop
    st=a
    car=b

print dist/50  # 2.95692307692302 hours, the same answer!

Instead of keeping track of time, keep track of the distance traveled by the car.

'b' and 'a' are two points to keep track of the persons, they don't indicate the identity.

And I believe the lowest attainable answer for one of them to reach the earliest under the constraints is still 2.937777 hours.

The earlier LPP holds good no matter where the persons are dropped or picked.
All it indicates is the time for the distance traveled by walking and by car, it doesn't matter where on the path they are in the car or not.

Last edited by gAr (2014-01-02 19:13:47)

bobbym · 2014-01-02 22:53:01

Hi gAr;

I said it did not feel right. My mistake appears to be right here:

Go back and get c.

Solve 50 t + 5 t  =  car - c for t
c = (c + 5 t)
car = c;

Take c home ( both b and c are at the 62km mark with a)

time = time + (62 - c)/50

( both b and c are at the 62km mark with a)

Although the car makes two trips I only added to time for one of them! That is why it is below the correct minimum.

gAr · 2014-01-02 23:02:57

I too was thinking you might have missed adding,
What's your answer now?

bobbym · 2014-01-02 23:14:04

I stared at my work for hours yesterday and did not see the error. As soon as you posted post #17 I found the error in 5 minutes or less. I just got disgusted and deleted the whole notebook. I am sorry.

gAr · 2014-01-02 23:21:10

Okay, never mind.

I too worked for hours to set that code right for that idea. Only after seeing the output it struck me that it's not an issue whether they walk at a stretch or not!

Anyway, thanks for your idea, we are now doubly sure about the answer.

bobbym · 2014-01-02 23:27:09

How did you get the idea for the linear programming solution?

Save that for a little later, I need to get some sleep. See you then.

gAr · 2014-01-02 23:35:41

There were constraints, so I think it was natural.

We wanted to make the car's time dependent on A and B's times. Rest is easy!

Okay, see you later.

Last edited by gAr (2014-01-02 23:43:33)

bobbym · 2014-01-03 04:49:25

Hi gAr;

Thanks, I did not think about using it.

gAr · 2014-01-03 04:57:03

You're welcome.

Math Is Fun Forum

#1 2013-12-31 03:53:19

Is it possible to solve this without guessing?

#2 2013-12-31 05:40:58

Re: Is it possible to solve this without guessing?

#3 2013-12-31 09:46:49

Re: Is it possible to solve this without guessing?

#4 2013-12-31 09:48:37

Re: Is it possible to solve this without guessing?

#5 2013-12-31 18:00:02

Re: Is it possible to solve this without guessing?

#6 2013-12-31 19:32:59

Re: Is it possible to solve this without guessing?

#7 2013-12-31 22:48:29

Re: Is it possible to solve this without guessing?

#8 2014-01-01 15:55:10

Re: Is it possible to solve this without guessing?

#9 2014-01-01 22:56:34

Re: Is it possible to solve this without guessing?

#10 2014-01-01 23:18:48

Re: Is it possible to solve this without guessing?

#11 2014-01-01 23:21:03

Re: Is it possible to solve this without guessing?

#12 2014-01-01 23:23:38

Re: Is it possible to solve this without guessing?

#13 2014-01-01 23:27:41

Re: Is it possible to solve this without guessing?

#14 2014-01-02 00:24:05

Re: Is it possible to solve this without guessing?

#15 2014-01-02 00:37:08

Re: Is it possible to solve this without guessing?

#16 2014-01-02 00:46:37

Re: Is it possible to solve this without guessing?

#17 2014-01-02 18:50:03

Re: Is it possible to solve this without guessing?

#18 2014-01-02 22:53:01

Re: Is it possible to solve this without guessing?

#19 2014-01-02 23:02:57

Re: Is it possible to solve this without guessing?

#20 2014-01-02 23:14:04

Re: Is it possible to solve this without guessing?

#21 2014-01-02 23:21:10

Re: Is it possible to solve this without guessing?

#22 2014-01-02 23:27:09

Re: Is it possible to solve this without guessing?

#23 2014-01-02 23:35:41

Re: Is it possible to solve this without guessing?

#24 2014-01-03 04:49:25

Re: Is it possible to solve this without guessing?

#25 2014-01-03 04:57:03

Re: Is it possible to solve this without guessing?

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