Max Crew or Min Crew?

Some players have previously wondered if using minimum crew is better than using maximum crew. In this article, I will use a simple scenario as the analogy to explain.

Imagine you have 10 minutes and 6 handling crew to handling as many planes as possible. There will only be a single plane type, and this plane can take 2 handling crew at best. 

When using 1 crew, it takes 10 minutes to handle; when using 2 crew, it takes 5 minutes to handle. Remember the handling time is inversely proportional to the number of handling crew; the more crew you assign, the lesser the time - and the resultant handling time is the handling time with 1 crew divided by the number of crew assigned.

Firstly, let's see how many planes we can handle without using ramp agent. Using 1 crew for all planes, we can handle 

(6 / 1) * (10 / 10) = 6

planes within the timeframe. Using 2 crew, that would be 

(6 / 2) * (10 / 5) = 6

planes. So there isn't a difference in the case of no ramp agent.

Now let's see the capacity when using ramp agent. For 1 crew, the capacity is 

(6 / (1+1)) * (10 / 10) = 3 

planes. For 2 crew, the calculation would be 

(6 / (2+1)) * (10 / 5) = 4

planes. So using more crew allows one extra plane to be handled within the 10 minutes given.

Here is an informal proof for showing that maximum crew is better than minimum crew. The calculation we just did can be simplified to a formula. The number of planes handled in a period, N, can be calculated by 

N = (T / C') * (H / h)

where T is the total number of crew, C' is the final number of crew needed, H is the handling time with 1 crew, and h is the final handling time. 

Since h is inversely proportional to C, the number of crew (excluding ramp agent), their relationship can be expressed as 

h / H = 1 / C

and be re-arranged to 

h = H / C

So the expression becomes 

N = (T / C') * (H / (H / C))

Cancelling out H gives 

N = (T / C') * C

and re-arranging gives 

N = T * C / C'

So we end up with 

N = T * C / C'

Since T is constant, let it divide N to give the number of planes handled per period per crew. This quantity will be called Q. The expression now becomes 

Q = C / C'

Ideally we want Q to be as high as possible to handle as many planes with as little time and crew as possible.

In the first case without ramp agent, C' would become C since there's no need for one extra crew per plane. The expression therefore becomes

Q = C / C 

and hence

Q = 1

for any number of crew. The number of crew therefore doesn't matter when the ramp agent is not used.

In the second case of hiring a ramp agent, C' would become C + 1. The expression becomes 

Q = C / (C + 1)

By turning this into a function, we can visualise how efficient the handling becomes.

Q(C) = C / (C + 1)

The graph for the function is shown in the figure in this section of the article.

As the number of crew (x-axis) increases, the efficiency (y-axis) also increases. It is therefore better to use more crew - the maximum number of crew - when the ramp agent is used.

Notice the asymptote y = 1 or Q = 1 - the curve is closer to that value but never reaches it especially at low values (C < 20). This means that theoretically more planes can be handled without the ramp agent present. However, not automating the services induces delays from the player. These delays can be caused by human reaction time or idle time from handling other planes or doing other tasks outside of the game. These delays therefore result in Q < 1.

Last but not least, remember that this is a simplification, and here the total number of crew and the plane type is assumed to be constant. Nonetheless this calculation can be principally applied to any plane and any airport in the game.

So to conclude the original question, it is more efficient to use maximum crew than minimum crew. Another conclusion that can be drawn from this is that ramp agent may allow more planes to be handled more quickly and less crew-intensively, though the efficiency difference is not significant because of human delays.