# Should I Rest at the Travelator or Not at the Travelator?

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?

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

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

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.