3
$\begingroup$

This is a real-life puzzle encountered by one of my friends, again.

After landing and arriving at the airport, I want to pick my baggage up quickly. The problem is that the baggage claim is $500$ m away. Luckily, along the way there are many travelators. Let's say, the total length of travelators is $250$ m (hence, the total length of non-travelators is also $250$ m). The travelator speed is $1$ m/s. I can also walk in $1$ m/s but I need to reserve $10\%$ of my walking time to take a rest.

How long and where should I take the rest to quickly arrive at the baggage claim? Should I rest at the travelator or not at the travelator?

Note: Assume you are going to walk for $700$ s, you need to take a rest for $70$ s and can only truly walk for $630$ s. Let's say there is no travelator in the statement, then you need a total of $\frac{5000}{9}=555.55...$ s ($55.55...$ s for rest and $500$ s for walking) to arrive at the baggage claim.

Bonus: Can you generalize the solution for $N$ m away, $K$ m part of travelators, $V_T$ m/s of travelator speed, $V_m$ m/s of my speed, and $P \%$ of rest?

$\endgroup$
1
  • 1
    $\begingroup$ for the example of purely walking without travelators, ain't the resting time be 500/9 =55.56s ? $\endgroup$
    – Kryesec
    Feb 7 '19 at 5:18
3
$\begingroup$

You should

rest at the travelator.

The intuition behind this decision:

You want to rest at the travelator to maximize the time that you are on a travelator. This decreases the effective distance that you have to walk.

Calculation

For the initial problem, suppose you rest $k$ seconds on the travelator, giving you $9k$ meters of walking distance. You gain an additional $k$ meters standing on the travelator, plus $9k-250$ meters of "walking distance on the travelator". So $19k-250=500$ or $k = \frac{750}{19}$, which means it will take you $\frac{7500}{19} \approx 394.7$ seconds. (Compare that with $375+37.5=412.5$ seconds if you walked the entire way and stopped at the end.)

Generalization calculation

It works similarly. If you rest $k$ seconds on the travelator, you get $\frac{100-P}{P}k$ seconds of walking, which covers $V_m\frac{100-P}{P}k$ meters of walking. Now the caveats: you gain an additional $\max(kV_T, K-kV_m)$ meters standing on the travelator, and $\min(\frac{V_T}{V_m}(V_m\frac{100-P}{P}k-(N-K)),0)$ meters walking on the travelator. So you can solve for that :P

$\endgroup$
1
$\begingroup$

Let's say you're resting for $x$ seconds while not on a travelator and $y$ seconds on one. You have to spend $250$ seconds to walk on the floor and $(250-y)/2$ to walk on the travelator.

$9x+9y = (250-y)/2+y$
$18x+17y = 250$

To keep $x+y+250+(250-y)/2$ at a minimum, we have to minimize $x+y/2 -> 2x+y$. Since $18x+17y = 250 = (2x+y)*9+8y$, maximizing $y$ by resting only on a travelator is the best option.

Generalization:

Let's say you're resting for $x$ seconds while not on a travelator and $y$ seconds on one. You have to spend $(N-K)/V_m$ seconds to walk on the floor and $(K-y*V_T)/(V_m+V_T)$ to walk on the travelator.

$100x+100y = P*[x+y+(N-K)/V_m+(K-y*V_T)/(V_m+V_T)]$
$(100/p-1)(x+y) = (N-K)/V_m+(K-y*V_T)/(V_m+V_T)$

To minimize the right-hand side, we have to minimize $(K-y*V_T)/(V_m+V_T)$ by maximizing $y$ by means of resting only on a travelator.

$\endgroup$

This site is temporarily in read only mode and not accepting new answers.

Not the answer you're looking for? Browse other questions tagged .