Imagine a highly infectious, people-borne plague for which there is no immunization or ready cure, e.g. leprosy or small pox in the 1800s, or bubonic plague in the 1500s assuming that the carrier was fleas on people (there is a good argument that people-fleas were the carrier, not rat-fleas). We’ll call these plagues zombie invasions to highlight understanding that there is no way to cure these diseases or protect from them aside from quarantining the infected or killing them. Classical leprosy was treated by quarantine.
I propose to model the progress of these plagues to know how to survive one, if it should arise. I will follow a recent paper out of Cornell that highlighted a fact, perhaps forgotten in the 21 century, that population density makes a tremendous difference in the rate of plague-spread. In medieval Europe plagues spread fastest in the cities because a city dweller interacted with far more people per day. I’ll attempt to simplify the mathematics of that paper without losing any of the key insights. As often happens when I try this, I’ve found a new insight.
Assume that the density of zombies per square mile is Z, and the density of susceptible people is S in the same units, susceptible population per square mile. We define a bite transmission likelihood, ß so that dS/dt = -ßSZ. The total rate of susceptibles becoming zombies is proportional to the product of the density of zombies and of susceptibles. Assume, for now, that the plague moves fast enough that we can ignore natural death, immunity, or the birth rate of new susceptibles. I’ll relax this assumption at the end of the essay.
The rate of zombie increase will be less than the rate of susceptible population decrease because some zombies will be killed or rounded up. Classically, zombies are killed by shot-gun fire to the head, by flame-throwers, or removed to leper colonies. However zombies are removed, the process requires people. We can say that, dR/dt = kSZ where R is the density per square mile of removed zombies, and k is the rate factor for killing or quarantining them. From the above, dZ/dt = (ß-k) SZ.
We now have three, non-linear, indefinite differential equations. As a first step to solving them, we set the derivates to zero and calculate the end result of the plague: what happens at t –> ∞. Using just equation 1 and setting dS/dt= 0 we see that, since ß≠0, the end result is SZ =0. Thus, there are only two possible end-outcomes: either S=0 and we’ve all become zombies or Z=0, and all the zombies are all dead or rounded up. Zombie plagues can never end in mixed live-and-let-live situations. Worse yet, rounded up zombies are dangerous.
If you start with a small fraction of infected people Z0/S0 <<1, the equations above suggest that the outcome depends entirely on k/ß. If zombies are killed/ rounded up faster than they infect/bite, all is well. Otherwise, all is zombies. A situation like this is shown in the diagram below for a population of 200 and k/ß = .6
Fig. 1, Dynamics of a normal plague (light lines) and a zombie apocalypse (dark) for 199 uninfected and 1 infected. The S and R populations are shown in blue and black respectively. Zombie and infected populations, Z and I , are shown in red; k/ß = 0.6 and τ = tNß. With zombies, the S population disappears. With normal infection, the infected die and some S survive.
Sorry to say, things get worse for higher initial ratios, Z0/S0 >> 0. For these cases, you can kill zombies faster than they infect you, and the last susceptible person will still be infected before the last zombie is killed. To analyze this, we create a new parameter P = Z + (1 – k/ß)S and note that dP/dt = 0 for all S and Z; the path of possible outcomes will always be along a path of constant P. We already know that, for any zombies to survive, S∞ = 0. We now use algebra to show that the final concentration of zombies will be Z∞ = Z0 + (1-k/ß)S0. Free zombies survive so long as the following ratio is non zero: Z0/S0 + 1- k/ß. If Z0/S0 = 1, a situation that could arise if a small army of zombies breaks out of quarantine, you’ll need a high kill ratio, k/ß > 2 or the zombies take over. It’s seen to be harder to stop a zombie outbreak than to stop the original plague. This is a strong motivation to kill any infected people you’ve rounded up, a moral dilemma that appears some plague literature.
Figure 1, from the Cornell paper, gives a sense of the time necessary to reach the final state of S=0 or Z=0. For k/ß of .6, we see that it takes is a dimensionless time τ of 25 or to reach this final, steady state of all zombies. Here, τ= tNß and N is the total population; it takes more real time to reach τ= 25 if N is high than if N is low. We find that the best course in a zombie invasion is to head for the country hoping to find a place where N is vanishingly low, or (better yet) where Z0 is zero. This was the main conclusion of the Cornell paper.
Figure 1 also shows the progress of a more normal disease, one where a significant fraction of the infected die on their own or develop a natural immunity and recover. As before, S is the density of the susceptible, R is the density of the removed + recovered, but here I is the density of those Infected by non-zombie disease. The time-scales are the same, but the outcome is different. As before, τ = 25 but now the infected are entirely killed off or isolated, I =0 though ß > k. Some non-infected, susceptible individuals survive as well.
From this observation, I now add a new conclusion, not from the Cornell paper. It seems clear that more immune people will be in the cities. I’ve also noted that τ = 25 will be reached faster in the cities, where N is large, than in the country where N is small. I conclude that, while you will be worse off in the city at the beginning of a plague, you’re likely better off there at the end. You may need to get through an intermediate zombie zone, and you will want to get the infected to bury their own, but my new insight is that you’ll want to return to the city at the end of the plague and look for the immune remnant. This is a typical zombie story-line; it should be the winning strategy if a plague strikes too. Good luck.
Robert Buxbaum, April 21, 2015. While everything I presented above was done with differential calculus, the original paper showed a more-complete, stochastic solution. I’ve noted before that difference calculus is better. Stochastic calculus shows that, if you start with only one or two zombies, there is still a chance to survive even if ß/k is high and there is no immunity. You’ve just got to kill all the zombies early on (gun ownership can help). Here’s my statistical way to look at this. James Sethna, lead author of the Cornell paper, was one of the brightest of my Princeton PhD chums.
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