Contribution Strategy

I want to achieve maximum contribution per day in an alliance: what schedule should I fly, or what variables are the most important?

This guide covers:

• constructing objective functions and performing optimisation with derivatives
• arrive at a optimal pure-contribution strategy

Experimental

This guide is a preview and not finalised!

Advanced

Optimisation

First, list out all independent and dependent variables we should consider:

We know from collecting data that, if we decrease the \(\text{CI}\downarrow\):

• the contribution \(\$_{\text{C}}\uparrow\)
• but at the same time, the aircraft speed \(v'\downarrow\) and also, flight time \(T'\uparrow\)

We also found that increasing distance \(d\uparrow\) will cause the contribution \(\$_{\text{C}}\uparrow\) and flight time \(T'\uparrow\).

Now what happens to the contribution per hour \(\frac{\$_{\text{C}}}{T'}\)? Well, it'll always increase, if we assume that contribution (numerator) grows faster than the flight time (denominator).

But how do we test if our assumption is true? We need calculus.

Also, if I want to increase my contribution per hour, should I prefer increasing the distance, or decreasing the CI? Hmm... they both decrease the flight time and increase the contribution in different ways... We need calculus.

Objective Function \(J\)

When we have multiple equations, we should try to combine them into one equation, which gives us a singular output. Why? It is easier to perform differentiation on it, which will be key to understanding how one variable affects the other.

Now, our goal is to make one equation, that outputs the contribution per hour given distance \(d\), \(\text{CI}\) and \(v\). We call this the objective function \(J\). It's really useful because we can try to calculate \(\frac{\partial J}{\partial \text{CI}}=0\) to find the optimal CI (more on this later).

But let's deviate a bit before proceeding. As route planners, CI is something we set after we know the route and target flight time \(T'\). It'd be more useful to know contribution as a function of \(d\), \(T'\) and \(v\) instead. To do this, we need three formulae:

• relationship between speed, distance and time
• relationship between new speed, original speed and CI
• the contribution formula.

Let's first combine the first two, then work our way to the third.

Exercise 7: Suppose I have an aircraft with speed \(v\) flying a distance \(d\). What CI do I need to set to get a flight time of \(T'\)? Write an equation for it and the interval for which it is valid.

When CI is applied, the new speed is

\[ \begin{equation} v' = v(0.0035\text{CI} + 0.3) \end{equation} \]

Speed \(v\) is defined as the distance flown \(d\) over flight time \(T\):

\[ \begin{equation} v = \frac{d}{T} \end{equation} \]

Rearranging Eq 1 for CI and substituting Eq 2 for \(v'\):

\[ \begin{align*} \text{CI} &= \frac{\frac{v'}{v}-0.3}{0.0035} \\ &= \frac{2000d}{7vT'}-\frac{600}{7} \end{align*} \]

This formula is only valid when \(0 \le \text{CI} \le 200\), simplifying:

\[ \begin{align*} 0 &\le \frac{2000d}{7vT'}-\frac{600}{7} \le 200 \\ 0 &\le \frac{2000d}{vT'}-600 \le 1400 \\ 600 &\le \frac{2000d}{vT'} \le 2000 \\ 0.3vT' &\le d \le vT' \\ \end{align*} \]

Again, our goal is to express the objective function \(J\) with inputs \(d\), \(v\), \(T'\). So far, we have done so for \(\text{CI}\) and it's time to do the same for \(J\).

Exercise 8: Find \(J(d, v, T')\) by combining the contribution formula and CI.

The contribution formula is:

\[ \begin{equation} \$_\text{C} = \min\left( k_\text{gm}kd\left(1 + \frac{2(200 - \text{CI})}{200}\right),152 \right) \end{equation} \]

\[ \begin{equation} \begin{align*} k_\text{gm} &= \begin{cases} 1.5 & \text{if realism} \\ 1 & \text{if easy} \end{cases} \\ k &= \begin{cases} 0.0064 & \text{if } d < 6000 \\ 0.0032 & \text{if } 6000 < d < 10000 \\ 0.0048 & \text{if } d > 10000 \\ \end{cases} \end{align*} \end{equation} \]

We can get rid \(\text{CI}\) using the formula we found earlier:

\[ \begin{equation} \text{CI} = \frac{2000d}{7vT'}-\frac{600}{7}, d \in [0.3vT', vT'] \end{equation} \]

Now simplifying our objective function \(J = \$_\text{C} / T'\):

\[ \begin{align*} J(d, v, T') &= \min\left(\frac{k_\text{gm}kd}{T'}\left(3 - \frac{\text{CI}}{100}\right), \frac{152}{T'} \right) \\ &= \min\left(\frac{k_\text{gm}kd}{T'}\left(\frac{27}{7} - \frac{20d}{7vT'}\right), \frac{152}{T'} \right) \\ &= \min\left(\frac{27k_\text{gm}kd}{7T'} - \frac{20k_\text{gm}kd^2}{7vT'^2}, \frac{152}{T'} \right) \end{align*} \]

For the remainder the guide, I will assume easy mode, with \(k_{\text{gm}}=1\).

Great! Now we \(J\) as a single multivariable function with 3 variables.

We want to understand the relationships between variables, and plotting is always a great way to understand things. Unfortunately, we cannot quite visualise everything at once because we don't have 4 dimensions!

But if we freeze any one variable constant, \(J\) now has 2 independent variables ("inputs") and one dependent variable ("output"). For instance, when we have three inputs \((d, v, T')\), we choose to freeze \(d=6000\) as a constant, then \((v, T')\) are the two remaining inputs. Now, set \(v\) as the x-axis, \(T'\) as the y-axis - and compute \(J\) for every single combination and colour it:

This is called a contour plot. Here, we can observe "bands" of the same colour - imagine walking on a hill: walking along this path does not change your elevation. In this case, the region of the darkest blue all have the same value: the best contribution per hour.

Exercise 9: If you are familiar with Python, can you reproduce the graph above? (hint: you might want to use np.minimum for CI bounds-checking, np.where for the piecewise \(k\) and matplotlib for plotting.)

I've written the ci and cont functions separately. Try changing ci to cont!

import matplotlib.pyplot as plt
import numpy as np

def ci(d, v, tp):
    ci_ok = (d >= 0.3*v*tp) & (d <= v*tp)
    ci = 2000*d / (7*v*tp) - 600 / 7
    return np.where(ci_ok, ci, np.nan)

def cont(d, v, tp):
    k = np.where(d < 6000, 0.0064, np.where(d < 10000, 0.0032, 0.0048))
    cont = np.minimum(k*d*(3-ci(d, v, tp)/100), 152)
    return cont/tp

dconst, vconst, tpconst = 6000-1e-9, 1049*1.1*1.5, 12
ds = np.linspace(0, 20000, 1000)
vs = np.linspace(300, 1800, 1000)
tps = np.linspace(0, 16, 1000)

fig = plt.figure(figsize=(16, 7))
gs = fig.add_gridspec(1, 3)

vm, tpm = np.meshgrid(vs, tps)
ax0 = fig.add_subplot(gs[0, 0])
ct0 = ax0.contourf(vm, tpm, cont(dconst, vm, tpm), levels=20, cmap='turbo_r')
cb0 = fig.colorbar(ct0, ax=ax0, location="bottom")

dm, tpm = np.meshgrid(ds, tps)
ax1 = fig.add_subplot(gs[0, 1])
ct1 = ax1.contourf(dm, tpm, cont(dm, vconst, tpm), levels=20, cmap='turbo_r')
cb1 = fig.colorbar(ct1, ax=ax1, location="bottom")

dm, vm = np.meshgrid(ds, vs)
ax2 = fig.add_subplot(gs[0, 2])
ct2 = ax2.contourf(dm, vm, cont(dm, vm, tpconst), levels=20, cmap='turbo_r')
cb2 = fig.colorbar(ct2, ax=ax2, location="bottom")

plt.show()

You can view the middle plot on Desmos 3D as well.

Concept Check: Just by looking at the contour plot, how do we achieve the highest contribution per hour?

The highest contribution per hour are regions with the darkest blue:

• Left Plot, constant \(d\):
  • faster original aircraft speed \(v\)
  • shorter target flight time \(T'\). but if it is too short, it'll get worse
• Middle Plot, constant \(v\):
  • choose distances less than 6000km
  • along some linear line, they all have equally good contribution per hour
• Right plot, constant \(T\) for a 12hr departure schedule and \(v=1200\):
  • when \(d<6000\), greater distances perform better.
  • when \(d>10000\), greater distances perform poorer.

I'm being quite ambiguous here: we will see later how to know "when it's too short" or how to calculate the "linear line".

Now, we want to know what is the equation that gives me the darkest blue region?

To answer it, imagine yourself blindfolded, at the red region (bottom of the hill) and wanting to reach the "peak". The strategy would be, climb whatever direction with the steepest slope. If we follow it long enough, we will find ourselves confused because there are no more slopes to climb - that's the peak!

The direction with the steepest ascent is called the gradient \(\nabla J\). It is a vector that tells me which direction to go:

\[ \nabla J=\left[\begin{array}{c} \frac{\partial J}{\partial d} \\ \frac{\partial J}{\partial v} \\ \frac{\partial J}{\partial T'} \end{array}\right] \]

When \(\nabla J=\left[\begin{array}{c}0 \\ 0 \\ 0\end{array}\right]\), it means we should stop moving because it holds the position of the plateau - the critical point.

The \(\frac{\partial J}{\partial d}\), \(\frac{\partial J}{\partial v}\) and \(\frac{\partial J}{\partial T'}\) are the partial derivatives of \(J\). Setting them all equal to zero gives us the critical point.

Partial Derivatives

Finding partial derivatives of \(J\) is effectively the same as the ordinary derivatives, but you freeze all but one variable fixed.

Case 1: Optimal \(T'\)

To find the optimal time, we need to freeze \(d\) and \(v\) to "collapse" it to a single-variable function. We can do so by taking the leftmost plot (which already froze \(d=6000\)) then freezing \(v=1730.85\) (taking a slice with grey line):

The green line here indicates the optimal time that maximises \(J\): our goal is to find this value.

The plot on the right represents \(J(d=6000, v=1730.85, T')\), a single variable function.

Remember earlier in our concept check, we said "shorter flight time will yield higher contribution per hour, but if it is too short, it'll get worse"? Indeed, it increases up until some point \(T'_\text{optimal}\) then drops dramatically: this point can be found with partial derivatives.

Example: For an A388 with speed mod (\(v=1730.85\)) and \(d=6000\), find the optimal target flight time \(T'_\text{optimal}\) that maximises \(J\). Verify your answer on the plot.

We would like to evaluate the location at which \(\frac{\partial J}{\partial T'}=0\).

Recall \(J\):

\[ J = \min\left(\frac{27kd}{7}T'^{-1} - \frac{20kd^2}{7v}T'^{-2}, 152T'^{-1} \right) \]

Treat \(d, v, k\) here as numbers and differentiate with respect to \(T'\), using \(\frac{d}{dT'}(T'^n) = nT'^{n-1}\):

\[ \frac{\partial J}{\partial T'} = \min\left(-\frac{27kd}{7T'^2} + \frac{40kd^2}{7vT'^3}, -\frac{152}{T'^2} \right) \]

Setting it to 0 yields:

\[ \begin{align*} \frac{27kd}{7T'^2} &= \frac{40kd^2}{7vT'^3} \\ 27 &= \frac{40d}{vT'} \\ T' &= \frac{40d}{27v} = \frac{40 \cdot 6000}{27 \cdot 1730.85} = 5.135 \end{align*} \]

This gives us the time at which the contribution is maximised¹. It indeed matches with the green vertical line! We can also evaluate \(J\) at this point:

\[ \begin{align*} J(T'_\text{optimal}) &= \min\left(\frac{27kd}{7}\cdot\frac{27v}{40d} - \frac{20kd^2}{7v}\cdot\frac{27^2v^2}{40^2d^2}, 152\cdot\frac{27v}{40d} \right) \\ &= \min\left(\frac{729}{560}kv, \frac{102.6v}{d} \right) \\ &= \min\left(\frac{729 \cdot 0.0064 \cdot 1730.85}{560}, 29.588\right) = 14.420 \end{align*} \]

Indeed, this also matches the plot!

We now reach a very useful equation:

\[ T'_{\text{optimal}} = \frac{40d}{27v} \\ \]

This gives us the optimal* time, given some distance \(d\) and original aircraft speed \(v\). The contribution per hour at this time is also:

\[ J(T'_\text{optimal}) = \min\left(\frac{729}{560}kv, \frac{102.6v}{d} \right) \]

Important note on optimality

By optimal, I mean locally optimal.

If we had just arbitrarily chosen \(d=6000\) and \(v=1000\), we cannot guarantee that the contribution is the absolute max!

If we can prove \(d\) and \(v\) is indeed optimal, using this equation will give us globally optimal contribution. We do that by also evaluating partial derivatives with respect to \(v\) and \(d\).

I can now find all optimal times \(T'_{\text{optimal}}\) at all speeds \(v\) and plot them with a green line:

Notice that this line is the "ridge" of our mountain that maximises \(J\)! If we also make regularly spaced cuts along the speed \(v\), we can clearly see that the green dots correspond to the locally optimal \(T'_{\text{optimal}}\) in that cut.

Case 2: Optimal \(d\)

coming soon™ ;)

Case 3: Optimal \(v\)

coming soon™ ;)

Interesting Properties

Exercise 10: What is the optimal CI to achieve maximum contribution per hour? Assume we are flying distances less than 6000km. (Tip: Notice that the objective function is different because we now take CI as the input.)

Recall the contribution formula is:

\[ \$_\text{C} = \min\left( kd\left(1 + \frac{2(200 - \text{CI})}{200}\right),152 \right) \]

The objective function \(J\) should take in \(\text{CI}\) and \(T'\) only. Define it to be \(\$_\text{C}/T'\) and substitute \(d=v'T'=T'v(0.0035\text{CI}+0.3)\):

\[ \begin{align*} J(\text{CI}, T') &= \min\left( kv(0.0035\text{CI}+0.3)\left(3 - \frac{\text{CI}}{100}\right),\frac{152}{T'} \right) \\ &= \min\left( kv(0.0105\text{CI} + 0.000035\text{CI}^2 + 0.9 - 0.003\text{CI}),\frac{152}{T'} \right) \\ &= \min\left( kv(\underbrace{-0.000035\text{CI}^2 + 0.0075\text{CI} + 0.9}_{\text{a quadratic function!}}),\frac{152}{T'} \right) \end{align*} \]

Now, to find the global maxima, we need \(\frac{\partial J}{\partial \text{CI}}\) and the boundary points. First, the partial derivative of \(J\) with respect to \(\text{CI}\) (note that \(k\) is a piecewise function returning a constant so \(\frac{\partial k}{\partial \text{CI}} = 0\)):

\[ \frac{\partial J}{\partial \text{CI}} = \begin{cases} kv \left(-0.00007\text{CI} + 0.0075\right) & \text{if } J < 152 \\ 0 & \text{otherwise} \\ \end{cases} \]

For a fixed target flight time \(T'\), this is a decreasing function (as \(\frac{\partial^2 J}{\partial \text{CI}^2} < 0\)), meaning that contribution grows with CI initially, but slowly plateaus off and starts decreasing. This is the local maxima and can be found by solving for \(\frac{\partial J}{\partial \text{CI}} = 0\):

\[ \begin{align*} kv(-0.00007\text{CI}+0.0075) &= 0 \\ \text{CI} &= \frac{-0.0075}{-0.00007} = 107.143 \end{align*} \]

Interestingly, a CI of 107 is guaranteed to be the local optimum, regardless of aircraft speed! In the particular case where \(d<6000\), it is the global optima because \(k\) is the largest. But we need to be more careful for \(d>6000\): it is possible to get a better contribution if we had chosen \(d=6000\).

Mixed Strategies

Exercise 11: Reducing CI obviously increases contribution, but decreases profit. Some players like to choose a hybrid contribution-profit strategy: say \(k_c = 0.4\) means 40% profit, 60% contribution. Formulate an objective function and comment on the optimal strategy.

coming soon™ ;)

Summary

• identify all IVs, DVs and Combine equations into a single objective function \(J\).
• contour plots are helpful to visualise combinations of variables.
• gradient \(\nabla J\) is the direction of steepest ascent, solving for zero corresponds to critical points
• partial derivatives freeze all but one variable constant.
• if \(f(x, y)\) and \(x(t), y(t)\), \(\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}\), forming a tree.
• evaluate the nature of critical points with Hessian and check boundaries for global maxima.

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1. Technically, we should also check that \(\frac{\partial^2 J}{\partial T'^2}<0\) to verify that it is indeed concaving downwards, and is thus the maximum. ↩