R0 and virus spread math in the film Contagion..

C

cjard

Registered Senior Member
There's a scene in this film where the annoying blogger says:

"This virus has an r-nought of 2, so if 2 people have it, then 4 people have it, then 16 people have it, then 256, then 65 thousand"

I don't get how he arrived at his math. Rx refers to the number of new cases (x) an infected individual will generate during the contagious period. R2 means one person will give the virus to 2 other people..

So one person has it.. They (in this film) infect 2 others then die
2 people each give to 2, realiising 4 cases, then they die.. So far his math checks out

But then, how do 4 infected people generate 16 infected people? They only infect 2 people each.. and I'm fairly sure 4x2 is 8, not 16..

He's squaring his math.. But why? Did i miss something about R-nought when i read up on it? Did the script writers miss something? I can't believe that they did it deliberately to make him look like a misinformed sensationalist moron (even though that's his character) because noone (like the CDC director on the talk show he was opposing) then refuted his math (which would have been a logical thing to do)

His math is true, if you perform the interim steps he left out (he does 2 4 16 256 65k, all those numbers are encountered in a sequence of powers of 2) but you'd get them for an R0 of 1.1 also.. I'd have just settled on the "he skipped some steps" explanation if it hadn't been a sequence of squaring the previous term for 5 examples in a row..
 
The exponential growth is the simplest and is typically true in the initial stages of an outbreak but other factors come into play which lead to more interesting phenomena.

For example, we might wish to consider healthy H, infected but not contagious, I, and then contagious, C people. Considering just short time scales we can neglect the fact everyone dies at some point and say only contagious people die, $$C \to_{\lambda} \empty$$, with rate $$\lambda$$. For a healthy person to become infected they must meet a C and they get the disease with some rate $$\mu$$, $$H+C \to_{\mu} I+C$$. An infect person turns into a contagious person with rate $$\nu$$, $$I \to_{\nu} C$$.

From this you can write down the differential equations the populations will satisfy : $$\dot{H} = -\mu HC$$ , $$\dot{I} = \mu HC - \nu I$$ and $$\dot{C} = + \nu I - \lambda C$$. We might start with initial conditions of there being $$H(0) = H_{0}$$, $$I(0) = 0$$ and $$C(0) = 1$$. This can then be solved using various methods, depending on whether you want to consider expected rates or do everything Monte Carlo style.

It's immediately clear that this isn't an exponential system. The number of people getting infected (ie the decrease in H) depends not just on the number of people already infected but also on the number of healthy people left. If a population is already entirely ill then you can't continue with exponential growth of the infected, there's no one left to infect! This is like exponential population growth in animals, it's limited by the 'food' supply. When a disease runs out of people to infect it stops increasing in number. In the above equations since there's no growth term in H everyone will eventually die* if I or C are non-zero, the population is always decreasing in such cases. That's only in the long time future though and in such cases you might include birth rates in the $$\dot{H}$$ term to get a balanced system arising, in certain circumstances.

* This is only true from the differential equation point of view. In a stochastic system you could have all the ill people drop dead before they infect all the healthy people. This is more likely to happen if $$\lambda$$ is very big (quick death) and $$\mu$$ is very small (not very contagious). The disease burns itself out. If the opposite is true and a 'slow burning' but highly contagious disease arises then it's big problems.

These sorts of equations arise a lot in dynamical systems. The simplest ones, which cannot describe disease spreading, are the Lotka Volterra equations, otherwise known as predator-prey equations. More elaborate systems with structure like the ones I just wrote down arise in the Lorenz attractor, one of the best known examples of chaotic dynamics.
 
The problem is that the quote isn't describing exponential growth, it's describing something faster... 2^(2^n) rather than 2^n.

I just realised (after reading the movie plot in Wikipedia) that this might not be a case of "Writer's can't do math", since the annoying blogger is not supposed to be a reliable character.

Also, it's worth noting that Ro=2 is high, but not that high.
The Wikipedia page on Ro lists Measles as the highest example, with Ro=12-18.
 
The math behind the infections numbers in Contagion

The blogger uses the sequence 2, 4, 16, 256, 65 thousand, etc. These numbers are not made up. He obtains these by assuming that with an R0 of 2, on day 1, 2 people have the virus. On day 2 those two people each got two more people infected bringing the total to 4. On day three each of the four infected 2 more each. This continues on. This comes from the formula #infected=n^(R0); with n being number infected previously and R0 being the R-nought value assigned to the particular outbreak. In this particular example this means the number infected equation =n^2. We must assume on day 1 that two people have the infection, because only one person being infected for any length of time is nearly impossible. Thus on day two, #inf=2^2, so 4. On day three, #inf=4^2, so 16. Day four, #inf=16^2, so 256. Then 256^2=65536, then after only 65536^2=nearly 4.3 billion.
In conclusion, the numbers from the movie are not made up or a writer not knowing how to do math but instead a series based on exponential growth. All you do is take the number of infected people from the day before and raise that number to the power that is equivalent to the outbreaks R-nought value.
To go a little more in depth, some simple calculus can show you how fast the rate of infected people is changing and a little more calculus can show you when the maximum infection value will occur.
 
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