I am in an odd position at the point of writing this. It is 15 years since I stopped teaching in Britain and at least 6 since I last taught anything at all. The lack of daily practice with maths creates brain rust and I'm afraid this rubs off only with repeated activity. My brains still work, I hope, but the maths bit is what has rusted.
Context at the time of writing is much fuss about coronavirus, so-called because the image of the virus looks a little like a corona In Mandarin it is a crown-virus for the same reason (so says the boss, who spends many hours reading about this, since Wuhan is what she thinks of as 'local'. Wuhan is, for those who haven't looked it up, halfway between these pairings: Shanghai and Beijing, Hainan and Tianjin, Chongqing and Shanghai. That places it pretty much in the middle of the heavily populated eastern side of the country and it is an excellent place for an epidemic to begin. Given that this is a flu-like virus it is easy to mis-identify it (and hence any urgency in what you might do about it individually) and it spreads easily since it is a lung infection. Worst, it hit the Chinese New Year holiday (CNY, which some western press decided to call the lunar new year), which is a time when the Chinese demonstrate that the idea of travelling to see family at Christmas is only a pale imitation of the Chinese concept of following the same idea. I think if the initial case had been just a few days earlier we could have seen a different order of magnitude in early infection.
Q1 The sensible media (by which I mean radio, actually) has chosen to describe transmittability as a number of people that each case will infect, a number around 3, plus or minus maybe 0.5. Time for some maths: what sort of model would you start with?
If we assume that each person infects exactly 3 others and that the population (people available to be infected) is large enough, then the number of people infected after n cycles of the infection period is 1+3+3²+3³+.....3ⁿ
This will continue to grow in the same way unless some changes in behaviour occur. One factor to consider immediately is just how quickly a nation state can react in comparison to the incubation cycle. With a new infection one of the significant difficulties is discovering how the infection propagates and in which period of the illness one might be infectious. For example, if one were transmitting the disease while unaware that one was yet ill, then many people will be acting in the belief that they are 'well' and will either go help the ill or will do very little (stay at home, continue as if there is no infection at all) or scarper rapidly to elsewhere. This last case causes a dramatic spread of disease and makes it very difficult for any sort of authority to reduce infection rates, Indeed, in this situation the spread of disease will be very rapid indeed.
Mathematically, this is a geometric series, so it would be correct to describe it as geometric growth.
Q2 I found it very easy to discover press contribution that talked about the growth being exponential (example). Decide, or better still prove, whether geometric growth is bigger, smaller or the same as exponential growth.
Suppose that the growth was exponential; then the rate of infection is directly related to the number infected, dx/dt = kx => x= Aeᵏᵗ.
At this point you might attempt a read of (1). My own thought was that this assumes an awful lot of the sort of content which would usually be explained in academic papers. So I looked instead to see if I could find the numbers of cases reported. A win for wikipedia here.
Q3 Put these figures into some sort of predictive graph and identify a suitable model.
Compare this with the R₀ figure from [1] of between 3.3 and 5.5 and comment.
Jan 2020 date 16 17 18 19 20 21 22 23 24 25 26
NHC cases 45 62 121 198 291 440 571 830 1287 1975 2744
You should find that there is a curve of the form y = Aeᵏˣ. Here, x is days and Y is cases declared by the National Health Commission. Try [3] for the current situation. I'd prefer to use x=Aeᵏᵗ; changing the initial day only changes the value of A (you should demonstrate that for yourself). A is what [1] is referring to as R₀, the initial condition, i.e what happens when t=0. It is less obvious what the value of 0.4094 indicates. However, given that the medics talk about each person infecting another 2.5 or so, that looks like the reciprocal of 0.4094, or 2.44.
Q4 Comment on that suggestion.
Q5 Imagine you are someone involved in recognising what is occurring. Imagine you are included on 18th January and that the figures for the cases you have are only the 45, 62, 121. What is your prediction of infection rate (people infected, 6k in Aeᵏᵗ) for the 20th and 21st? How does this change when you are told of 198 cases for the 19th?
Q6 Continuing the idea in Q4, put yourself very much closer to the problem and imagine you were aware that something might be happening as early as the 13th January, with the available case numbers being 2,3,5,9. Being experienced (as you are, now), you know to model this with an exponential. Find an approximate solution and predict the numbers for the 16th, being the moment when figures reached the notice of the international press. Unfortunately there is disagreement whether cases count towards the figures, so another equally valid count goes 2,3,4,6. Use these to produce a second set of figures and compare.
One of the significant problems you can now appreciate is that it takes time to ascertain that there is a new disease, to test for it and therefore there is a significant delay to actually discovering the early cases. You might imagine that a sharp medic might send material for testing and that this might build in a delay of, say, three days cycle time between suspicion that this is a different disease and having any confirmation that this is so. Thus, if in Q5 there are two cases from the 10th confirmed on the 13th, when there are perhaps 6-10 cases (4-8 new cases), by the time these are confirmed on the 16th there are 45 cases but you think there might be 25 to 50. By now you would hope that either testing can be speeded up or that some controls can be put in place. It is round about now that deaths begin to occur and that will raise the degree with which this is taken seriously, probably causing additional resources to become available.
The news tonight, the 27th, put the 1/k figure at between 2.5 and 6, so k is between 0.40 and 0.17 (rounding to the nearest .1), which suggests that some of the modellers think there is, as yet, an awfully large number of undiscovered cases.
Q7 Suppose there continues to be a 3-day delay in confirming new cases. Assume that the 45 cases on the 16th is now very much agreed by all concerned. On the 26th the 830 from the 23rd is confirmed, but there is doubt that this is a complete count. What sort of numbers of cases are suggested by k=0.16 to k=0.18 ? How does that change the apparent 830 figure from the 23rd? That is how many cases are perhaps to be discovered?
This morning the number of declared cases is 4545. This continues to be a very close fit to the original model, x=0.0713e^(0.4094t). That says that we have done very little to affect the spread of the outbreak. At this point we would begin to expect the virus to become modified by the population, as it changes in each subsequent transmission (so I understand); often one would expect a strain to weaken. News today says that the incubation period is 3-7 days with a maximum of 14 days. So quarantine periods will be set at 14 days and already there are schools in Australia asking travellers from infected parts of China to give themselves voluntary quarantine for 14 days. Followed by 'a medical certificate', which is asking quite a lot of the medical profession.
DJS 20200128
top pic from https://www.imperial.ac.uk/news/194815/coronavirus-outbreak-china-have-infected-thousands/
1 https://www.biorxiv.org/content/10.1101/2020.01.23.916395v1
3 https://gisanddata.maps.arcgis.com/apps/opsdashboard/index.html#/bda7594740fd40299423467b48e9ecf6
4 https://gisanddata.maps.arcgis.com/apps/opsdashboard/index.html#/bda7594740fd40299423467b48e9ecf6
Q1 I see this as of the form (1-xⁿ) / (1-x) = xⁿ - 1, so say at eight cycles we have 3⁸-1 = 6560 cases.
Q2 eˣ = 1+x+x²/2! + x³/3! +.... > 1+x+x²+x³+... while |x|>1. E.g. e³≈20. So geometric is bigger than exponential.
Q3 I shoved the data into Excel, added an exponential trend line and asked for the formula to be displayed. The R² value says this is a very good fit. Quite how we should interpret the values 0.07 and 0.409 is less clear. It is remarkable that x is the date for such a good fit.
Projecting these figures for a further four days gives 4500, 6800, 10,200, 15,300 cases respectively.
Changing the count for the first day so that there are 45 initial cases moves A to 33. Moving the initial cases to a single one makes day 1 the 7th of January and A=0.8312. In all cases the exponent stays at 0.4094.
You might play with your data to see how the prediction varies with far fewer days.
Q4 It cannot be the reciprocal, as that would suggest that infecting more people per carrier then reduces the number of people infected. There must be a multiplier instead, something like 6.
Q5 Still using Excel, then assuming an exponential and using the date for x, I got x = 0.0155e^(0.4096t) which predicts 306 and 503 for the 20th and 21st. Adding in the fourth data point moves the predictions to 323 and 539, actually worse. By this point you would hope to have found the numbers for. So 6k is around 2.5 people per case.
Q6 I got x = 0.0126 e ^(0.5023x) which predicts 2,3,5,9,14,24,39,64 - a remarkably good fit. On the other hand, if the figures began 2,3,4,6 they would continue 8 12 17 25, far less helpful.
Q7 I stuck new values of A and k into x=Aeᵏᵗ, using A=0.000005 and 6k=6 =>k=1, so as to have 45 cases on the 16th. This suggests that there are 49000 cases on the 23rd. That indicates that 6k is NOT representative of the number of people infected by any individual carrier.