From Graydon Snider comes a dilemma of delivery:
A restaurant at the center of Riddler City is testing an airborne drone delivery service against their existing fleet of scooters. The restaurant is at the center of a large Manhattan-like array of square city blocks, which the scooter must follow.
Both vehicles travel at the same speed, which means drones can make more deliveries per unit time. Assume that (1) Riddler City is circular in shape, as you may recall (2) deliveries are made to random locations throughout the city and (3) the city is much, much larger than its individual blocks.
In a given amount of time, what is the expected ratio between the number of deliveries a drone can make to the number of deliveries a scooter can make?
Extra credit: In addition to traveling parallel to the city blocks, suppose scooters can also move diagonally from one corner of a block to the opposite corner of the block. Now, what is the new expected ratio between the number of deliveries a drone can make and the number of deliveries a scooter can make?
A Scooter Path is the sum of its horizontal and vertical components. Note that there are many possibilities to traverse a path with combinations of horizontal and vertical roads, but the total distance for each of horizontal and vertical remains the same for a specific destination.
Using polar coordinates, the sum of all Scooter Paths can be represented as:
$$8\cdot\int\limits_{r = 0}^{1} \int\limits_{\theta = 0}^{\frac{\pi}{4}} \left(r\sin \theta + r\cos \theta\right) \cdot d\theta \cdot dr$$
$$= 8\cdot\int\limits_{r = 0}^{1} \left(-r\cos \theta + r\sin \theta \bigg]_{0}^{\frac{\pi}{4}}\right) \cdot dr$$
$$= 8\cdot\int\limits_{r = 0}^{1} \left(\left(-\dfrac{1}{\sqrt{2}} + \dfrac{1}{\sqrt{2}}\right) - (-r + 0)\right) \cdot dr$$
$$= 8\cdot\int\limits_{r = 0}^{1} r \cdot dr = 8\cdot\dfrac{r^2}{2}\bigg]_{0}^{1} = 4$$
A Drone Path is simply the shortest distance to the destination.
The sum of all Drone Paths can be represented as
$$8\cdot\int\limits_{r = 0}^{1} \int\limits_{\theta = 0}^{\frac{\pi}{4}} r \cdot d\theta \cdot dr$$
$$= 8\cdot\int\limits_{r = 0}^{1} r\theta \Big]_{0}^{\frac{\pi}{4}} \cdot dr$$
$$= 8\cdot\int\limits_{r = 0}^{1} \dfrac{\pi r}{4} \cdot dr = 8\cdot\dfrac{\pi r^2}{8}\bigg]_{0}^{1} = \pi$$
Since the number of deliveries is proportional to the inverse of distance, the expected ratio between the number of deliveries a drone can make to the number of deliveries a scooter can make is:
$$\dfrac{\dfrac{1}{\pi}}{\dfrac{1}{4}} = \dfrac{4}{\pi} \approx 1.2732$$
In other words, a drone can make about 27% more deliveries.
A Scooter Path is the sum of its diagonal and horizontal components for this $\dfrac{1}{8}$ of the city. Again, there are many possibilities to traverse a path with combinations of diagonal and horizontal roads, but the total distance for each of diagonal and horizontal remains the same for a specific destination.
The diagonal distance is $\sqrt{2}y$. The remaining horizontal distance is $x-y$.
The sum of all Scooter Paths can be represented as:
$$8\cdot\int\limits_{r = 0}^{1} \int\limits_{\theta = 0}^{\frac{\pi}{4}} \left(\sqrt{2}r\sin \theta + r(\cos \theta-\sin \theta)\right) \cdot d\theta \cdot dr$$
$$8\cdot\int\limits_{r = 0}^{1} \int\limits_{\theta = 0}^{\frac{\pi}{4}} r\left((\sqrt{2}-1)\sin \theta + \cos \theta\right) \cdot d\theta \cdot dr$$
$$= 8\cdot\int\limits_{r = 0}^{1} r\left(-(\sqrt{2}-1)\cos \theta + \sin \theta \bigg]_{0}^{\frac{\pi}{4}}\right) \cdot dr$$
$$= 8\cdot\int\limits_{r = 0}^{1} r\left(\left(-(\sqrt{2}-1)\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right) - \left(-(\sqrt{2}-1)1 + 0\right)\right) \cdot dr$$
$$= 8\cdot\int\limits_{r = 0}^{1} r(2\sqrt{2}-2) \cdot dr = 8\cdot(2\sqrt{2}-2)\dfrac{r^2}{2}\bigg]_{0}^{1} = 8\sqrt{2}-8$$
The expected ratio between the number of deliveries a drone can make to the number of deliveries a scooter can make is now:
$$\dfrac{\dfrac{1}{\pi}}{\dfrac{1}{8\sqrt{2}-8}} = \dfrac{8\sqrt{2}-8}{\pi} \approx 1.0548$$
In other words, a drone can now only make about 5.5% more deliveries.
