I recently went on some planes. Here is what I think about whenever I go on a plane.
A basic plane has N rows of M seats, divided by an aisle. For instance, maybe N=50 and M=6.
The standard routine by which people exit a plane looks like this. First, about two thirds of the people positioned next to the aisle stand up and prepare their items to leave. Then the first two or so walk off the plane (the people from seats x and w in the diagram). Then the two people in seats v and y get out, prepare their things, and walk off. Then u and z do that. Then the next couple of people in the original queue walk off, allowing the people in their row to climb out, get their stuff, and leave. And so on, for every row.
Let:
A be the time it takes to prepare one’s things to go
B be the time it takes to walk one seat forward in the plane (such that it takes the last person NB to walk from their seat to the front of the plane)
Then the time it takes for the usual procedure is:
time for one row to depart * number of rows + time for last person to walk off
= number of people in a row * time to take your stuff and step forward for the next person to get out * number of rows + number of rows * time to walk forward one row
= M(A+B) * N + BN
= AMN + BMN + BN
Here is an alternative method. Everyone in row Q stands and collects their things. They all walk off the plane. Row R stands and repeats the process. And so on.
Here is how long this procedure would take:
time for one column to depart * number of columns
= (time for a person to get their stuff + time for each person to move one seat forward, allowing the person behind them to start walking + time for the last person to walk all the way off) * number of columns
= (A + B(2N-1)) * M
= AM + 2BMN – BM
(In both cases I assumed that each person can only start walking forward after the person in front of them has moved one seat forward. So the last person in line takes B(N-1) time to start moving, and then BN time to get out. This is probably not quite right, but near enough.)
It might not be intuitively obvious, but in general AMN + BMN + BN is much bigger than AM + 2BMN – BM, if we assume it takes substantially longer to collect your bags than it does to walk a couple of steps forward. In fact, it is (A-B)M(N-1) + BN bigger, if we assume that I can do algebra.
For example if there are fifty rows (N = 50) of 6 seats (M = 6), and gathering your stuff takes ten seconds, and walking forward one seat takes one second, we have:
alternative method
= AM + 2BMN – BM
= 10*6 + 2*6*50*1 – 1*6
= 654 seconds
Intuitively, if a whole column can get their stuff together at once, that is a lot faster than everyone standing waiting while one person at the front gets their stuff. It’s bad if A gets multiplied by MN.
Each method is bottlenecked by something happening MN times – as many times as there are seats on the plane. In one case, we are bottlenecked by each person taking their stuff down one at a time and then taking a step forward. In the other, it is just the time it takes to walk forward one seat.
You can’t hope to be faster than about 2BMN: the time it would take for every person to walk off the plane in single file if they wait for the person ahead of them to move one seat forward before they start walking. So my proposed method is not much worse than theoretically optimal.
You might notice that the time to completely empty the plane isn’t the same as the time lost, because how bad it is for the plane to not be emptied yet depends on how many people there are still in it. If people leave fairly evenly throughout the disembarking process, the utilitarian cost is roughly time*total people/2. Intuitively, each person is there for half the time the plane is disembarking. This means the total time is proportional to the value lost, so we can ignore this factor. People don’t quite get out evenly throughout the process in these two procedures, but near enough.
Probably the algebra in this post is wrong in places, but I think the gist is correct.
So, the alternative method seems superior to me. Why isn’t it used?
I recently went on some planes. Here is what I think about whenever I go on a plane.
A basic plane has N rows of M seats, divided by an aisle. For instance, maybe N=50 and M=6.
Let:
A be the time it takes to prepare one’s things to go
B be the time it takes to walk one seat forward in the plane (such that it takes the last person NB to walk from their seat to the front of the plane)
Then the time it takes for the usual procedure is:
time for one row to depart * number of rows + time for last person to walk off
= number of people in a row * time to take your stuff and step forward for the next person to get out * number of rows + number of rows * time to walk forward one row
= M(A+B) * N + BN
= AMN + BMN + BN
Here is an alternative method. Everyone in row Q stands and collects their things. They all walk off the plane. Row R stands and repeats the process. And so on.
Here is how long this procedure would take:
time for one column to depart * number of columns
= (time for a person to get their stuff + time for each person to move one seat forward, allowing the person behind them to start walking + time for the last person to walk all the way off) * number of columns
= (A + B(2N-1)) * M
= AM + 2BMN – BM
(In both cases I assumed that each person can only start walking forward after the person in front of them has moved one seat forward. So the last person in line takes B(N-1) time to start moving, and then BN time to get out. This is probably not quite right, but near enough.)
It might not be intuitively obvious, but in general AMN + BMN + BN is much bigger than AM + 2BMN – BM, if we assume it takes substantially longer to collect your bags than it does to walk a couple of steps forward. In fact, it is (A-B)M(N-1) + BN bigger, if we assume that I can do algebra.
For example if there are fifty rows (N = 50) of 6 seats (M = 6), and gathering your stuff takes ten seconds, and walking forward one seat takes one second, we have:
usual method
= AMN + BMN + BN
= 10*6*50 + 1*6*50 + 1*50
= 3350 seconds
alternative method
= AM + 2BMN – BM
= 10*6 + 2*6*50*1 – 1*6
= 654 seconds
Intuitively, if a whole column can get their stuff together at once, that is a lot faster than everyone standing waiting while one person at the front gets their stuff. It’s bad if A gets multiplied by MN.
Each method is bottlenecked by something happening MN times – as many times as there are seats on the plane. In one case, we are bottlenecked by each person taking their stuff down one at a time and then taking a step forward. In the other, it is just the time it takes to walk forward one seat.
You can’t hope to be faster than about 2BMN: the time it would take for every person to walk off the plane in single file if they wait for the person ahead of them to move one seat forward before they start walking. So my proposed method is not much worse than theoretically optimal.
You might notice that the time to completely empty the plane isn’t the same as the time lost, because how bad it is for the plane to not be emptied yet depends on how many people there are still in it. If people leave fairly evenly throughout the disembarking process, the utilitarian cost is roughly time*total people/2. Intuitively, each person is there for half the time the plane is disembarking. This means the total time is proportional to the value lost, so we can ignore this factor. People don’t quite get out evenly throughout the process in these two procedures, but near enough.
Probably the algebra in this post is wrong in places, but I think the gist is correct.
So, the alternative method seems superior to me. Why isn’t it used?