Galvanizing Growth

This challenge on growth involves algebra, calculus, and some combinatorics.

Consider a pandemic outbreak. It begins with patient zero in a random country. For this challenge, only consider the countries with the 50 largest populations, rounded to four significant digits. Now, there are two ways the pandemic can grow. It can either expand within a country, or expand by travel to another country. There are then several methods to stop an outbreak: cutting travel, quarantining, and social distancing. The impact of a pandemic can be calculated by the net cost of these method, plus the cost of the sickness. Now, the more seriously we try to do any of these methods, the more costly and inefficient it will become. We express this with the first equation below for quarantine and social distancing, the second for travel restrictions. Now, let the expansion of a virus within a country be dictated by the third equation below, and the expansion between any two, the fourth. The cost of each case is then given by the fifth equation. The challenge is to find a general formula, given a segment of a countries net cases curve, to determine how much a country should enact social distancing, and how strictly it should quarantine victims. And another, complementary function, given curve segments for two countries, that will give the restriction on travel that the first should impose on the latter. The curves represent the data from an incubation period, x, ago, and this is randomly generated each pandemic be the sixth equation. They recover after a recovery time, y, given by the seventh. In the below equations, c_n represents money spent on efforts in country n, i_n cost of cases in country n, p_n is population, and r is random number 0 ≤ r < 1, different each time. a_nx gives the number of interactions between people, with a_n1 within a country, a_n2 between two. s_n is the number of cases in country n, and g_n is the number of new cases in that country from that country. g_ab is the number of new cases spread from b to a. Good luck! A python simulation of this has been published for you to test your strategy.

a_{n1}=125\ e^\frac{-c_n}{100}\\[8pt] a_{n2}=5\ e^\frac{-c}{5}\\[16pt] g_n=\left\lfloor a_{n1}s_n{\left({3-2\sqrt2\operatorname{erfc^{-1}}{(r)}\over200}\right)}\left(1-\frac{s_n}{p_n}\right)\right\rfloor\\[16pt] g_{a_b}=\left\lfloor a_{b2}s_b{\left({3-2\sqrt2\operatorname{erfc^{-1}}{(r)}\over200}\right)}\left(1-\frac{s_a}{p_a}\right)\right\rfloor\\[16pt] i_n=s_n+10\left(s_n-\frac{p_n}5+|s_n-\frac{p_n}5|\right)\\[16pt] x=5-3\sqrt2\operatorname{erfc^{-1}}{(2r)}\\[8pt] y=11-5\sqrt2\operatorname{erfc^{-1}}{(2r)}