Should I Rest at the Travelator or Not at the Travelator?
$begingroup$
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?
mathematics optimization real
$endgroup$
add a comment |
$begingroup$
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?
mathematics optimization real
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1
$begingroup$
for the example of purely walking without travelators, ain't the resting time be 500/9 =55.56s ?
$endgroup$
– Kryesec
2 hours ago
$begingroup$
@Kryesec whoops, fixed, thanks!
$endgroup$
– athin
6 mins ago
add a comment |
$begingroup$
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?
mathematics optimization real
$endgroup$
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?
mathematics optimization real
mathematics optimization real
edited 6 mins ago
athin
asked 4 hours ago
athinathin
7,71022473
7,71022473
1
$begingroup$
for the example of purely walking without travelators, ain't the resting time be 500/9 =55.56s ?
$endgroup$
– Kryesec
2 hours ago
$begingroup$
@Kryesec whoops, fixed, thanks!
$endgroup$
– athin
6 mins ago
add a comment |
1
$begingroup$
for the example of purely walking without travelators, ain't the resting time be 500/9 =55.56s ?
$endgroup$
– Kryesec
2 hours ago
$begingroup$
@Kryesec whoops, fixed, thanks!
$endgroup$
– athin
6 mins ago
1
1
$begingroup$
for the example of purely walking without travelators, ain't the resting time be 500/9 =55.56s ?
$endgroup$
– Kryesec
2 hours ago
$begingroup$
for the example of purely walking without travelators, ain't the resting time be 500/9 =55.56s ?
$endgroup$
– Kryesec
2 hours ago
$begingroup$
@Kryesec whoops, fixed, thanks!
$endgroup$
– athin
6 mins ago
$begingroup$
@Kryesec whoops, fixed, thanks!
$endgroup$
– athin
6 mins ago
add a comment |
2 Answers
2
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$begingroup$
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 $V_mfrac{100-P}{P}k$ meters of walking. Now the caveats: you gain an additional $max(kV_T, K-kV_m)$ meters standing on the travelator, and $min(frac{V_T}{V_m}(V_mfrac{100-P}{P}k-(N-K)),0)$ meters walking on the travelator. So you can solve for that :P
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
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 $V_mfrac{100-P}{P}k$ meters of walking. Now the caveats: you gain an additional $max(kV_T, K-kV_m)$ meters standing on the travelator, and $min(frac{V_T}{V_m}(V_mfrac{100-P}{P}k-(N-K)),0)$ meters walking on the travelator. So you can solve for that :P
$endgroup$
add a comment |
$begingroup$
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 $V_mfrac{100-P}{P}k$ meters of walking. Now the caveats: you gain an additional $max(kV_T, K-kV_m)$ meters standing on the travelator, and $min(frac{V_T}{V_m}(V_mfrac{100-P}{P}k-(N-K)),0)$ meters walking on the travelator. So you can solve for that :P
$endgroup$
add a comment |
$begingroup$
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 $V_mfrac{100-P}{P}k$ meters of walking. Now the caveats: you gain an additional $max(kV_T, K-kV_m)$ meters standing on the travelator, and $min(frac{V_T}{V_m}(V_mfrac{100-P}{P}k-(N-K)),0)$ meters walking on the travelator. So you can solve for that :P
$endgroup$
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 $V_mfrac{100-P}{P}k$ meters of walking. Now the caveats: you gain an additional $max(kV_T, K-kV_m)$ meters standing on the travelator, and $min(frac{V_T}{V_m}(V_mfrac{100-P}{P}k-(N-K)),0)$ meters walking on the travelator. So you can solve for that :P
answered 3 hours ago
phenomistphenomist
8,0823052
8,0823052
add a comment |
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
answered 31 mins ago
NautilusNautilus
3,694523
3,694523
add a comment |
add a comment |
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1
$begingroup$
for the example of purely walking without travelators, ain't the resting time be 500/9 =55.56s ?
$endgroup$
– Kryesec
2 hours ago
$begingroup$
@Kryesec whoops, fixed, thanks!
$endgroup$
– athin
6 mins ago