A life-saving party trick!

Exponential growth does not come naturally to most people but in the extreme, it can make us very rich, or it can kill us.

The former comes via the power of compound interest - over time, it is the interest on the interest on interest, that is so powerful, whereas the latter is being demonstrated by the spread of a virus like COVID-19, or was at the heart of Malthusian predictions of running out of food, as populations grew faster than farm output was seen to (before improvements in technique, fertiliser, food distribution, and slowing birth rates offset that).

Many years ago I had what seemed an impressive party trick, which was to answer almost instantaneously questions like:

a.       “If a company grew its sales 10X over 15 years, what was the annual growth rate?”

b.      “If a company grows sales at 10% for 20 years, how many times bigger will it be?”

c.      “How long does it take a company growing at 10% to see sales increase 10-fold?”

If we reframe these questions from revenues and profits to viruses and hospital beds, there is a practical application to this demonstration.

By sharing the secrets of my trick, I remove my ability to impress you, on cue, but I hopefully provide a helpful lesson to contextualise some of the COVID-19 information being presented to us.

To answer the questions above in one’s head, requires two pieces of knowledge, and an aptitude for mental arithmetic.

The first is the sequence you obtain from repeated doubling : 1,2,4,8,16,32,64,128,256,512,1024….

The second is the ‘Rule of 72” which states that if we have a growth rate of x%, it takes 72/x periods for the object growing, to double in size. The maths behind this requires an understanding of logarithms[1], but if one accepts that it works, then these two snippets are all we need.

Armed with this we can set about solving these three questions

a.      There are three steps to solving this question approximately:

                                I.           First solve for how many times sales doubled using the series 2,4,8,16…. In this case to grow 10X it doubles about 3 and a quarter times (2,4,8, … [10]….16) = 3.25

                              II.           Next work out how long it takes to double... In this case it doubles 3.25 times in 15 years, so it doubles in 15 / 3.25 or approx. 4.5 times

                            III.           Finally, use the Rule of 72 to work out the growth rate, or 72/ 4.5 = 16% pa (answer 16.6%pa)

The maths behind this is that part I is to find log2(X)[2] where X is the growth multiple. Part 2, then finds the doubling time as Y/ log2(X), where Y is the number of periods, and Part 3, uses the rule of 72 to find the growth rate Z = 72 / (Y/ log2(X))). With a bit of practice, these three steps become second nature.

b.      This time we do the three steps in reverse:

                                I.           First we work out how long it takes to double at the chosen growth rate, using the Rule of 72. In the example this is 72/10 or approx. 7

                              II.           Next we work out how many times it doubles in the period, Here this is 20/7 or almost 3.

                            III.           Finally we apply this to the series 2,4,8,16 and get somewhere much closer to 8 then 4 when we double almost 3 times, suggesting about 7. (answer 6.73X)

The maths behind this is that in the first part, we calculate the time to double at growth rate Z, or (72/Z), which we use in the second part to work out how many times we double in period of Y, or Y / (72/Z) which is (Y*Z)/72 and finally we apply this to the doubling series to get the growth multiple X= 2^(YZ/72)

To answer question c. we need to determine how many times does it need to double, and how long does it take to double

                                I.           We know it takes (72/Z) units of time to double or in this case where Z=10%, it takes just over 7 years

                              II.           To grow 10-fold, needs us to double 3.25 times (see a. i. above)

                            III.           Multiplying 7.2 * 3.25 gives us 23.4 years (answer = 24.16)

The maths this time is to work out how long it takes to double or (72/Z) and multiple it by the number of times it needs to double, which is log2(X), or the mapping of the number X to the series 2,4,8,16….

So here our answer Y = (72/Z) * log2(X) - I think this equation is critical today

Now to the applications for the virus.

While question a is helpful to work out how fast a virus might have grown historically, we are now familiar with the idea of the first wave and are trying to avoid a second wave, which makes it incumbent on authorities to track spread rates, and take preventative action if and when health care systems are under threat.

In the case of COVID-19, we are constantly asking questions of these types and dealing with virus growth rates on a daily basis.

For example, over a month (30-days) we will see growth in daily cases of 2 ^ (0.4*Z)[3] – or equivalently, cases will double if Z=2.5%, quadruple if Z=5%, go to 16X at Z=10%, etc…

Mostly importantly, in my focus on the value in an Early Warning System. For a given jurisdiction, we should be able to determine the X or growth multiple which puts the system under undue strain.

If we think in terms of what we can know:

-         We can determine a growth multiple X for the number of cases we can cope with on a daily basis in our hospital systems

-         We can also monitor the daily growth rates Z in the virus as it spreads using case data noting testing inconsistencies and a lag.

-         From that we can determine how long it is likely to take to hit this capacity at that current rate of growth and determine if we need to take action to reduce Z. ie increase restrictions again.

To show how this might look in practice. Today the Australian government suggests there are 7,000 ICU beds[4] - and currently <5% of active COVID cases are in ICU[5] . If we assumed we could allocate 5,000 beds to ICU for COVID-19 (assuming there are other causes requiring treatment) and each patient needed this for 10 days, we could accept 500 ICU admissions a day. At 5% of cases, this would allow for 10,000 new cases of COVID a day, or 500X the current level.  Please note this is hypothetical – just to give a worked example.

log2(500) is approx. 9  (2,4,8,16,32,64,128,256,512…..)[6]

So Y = 9 * (72/Z) = 650/Z.  So, if we track Z as the growth in new cases per day, we have 650/Z days before the system overloads.  Over the last month in Australia [7]new cases are not showing a clear trend of growth, so this is not a problem. (ie Z~0, or Y is infinite for now!)

However, if cases pick up not only does Z increase but X also decreases (ie the spare capacity is reduced)

So by the time we get to 100 cases per day, X becomes 100, and log X falls from 9 to 7, so we end up with 500/Z days to overload, and Z by that time will be positive not zero…

Understanding the maths in the section above is key to understanding the risks we face, and our leaders need to be all over understanding X, Z and keeping an eye on Y to manage the risks we face.

[1] If (1+x)n=2 then n ln(1+x) = ln 2. For small numbers ln (1+x) is approximately x, so n. x = ln 2 or n = ln 2/ x ln 2 = 0.693. To allow for the fact ln(1+x) < x and to simplify calculations 0.72/x is used, or 72/X where X is a growth rate, or 100x eg. if x=0.08, X=8 and (1.08)^9 = 1.999 This uses natural logarithms, beyond scope of article.

[2] log2(X) is the logarithm of X in base 2. If X=2, log2(X)=1, for X=4, log2(X)=2, for X=8, log2(X)=3, and so on….

[3] Here we substitute Y=30 and 30/72 is close enough to 0.4 for the purpose of a crude calculation

[4] Source: health.gov.au, Impact of COVID-19, published 6 April 2020

[5] Source: health.gov.au, Daily COVID-19 update, 16 May 2020

[6] As one performs mental gymnastics, approximations become important so for 512 read 500, then 1024 -> 1000, 2048 -> 2000 etc….. knowing that 2^10 = 1024 is approx. 1,000 is very important as 2^ 20 is then one million, 2 ^ 30 – 1 billion etc.

[7] Sources : covid19data.com.au or worldometers.info