COVID Odyssey
COVID19 in New Zealand
Findings
Alan Neil Grace
[Updates will be available soon only in an ebook;
Contact Alan Grace for details about the ebook]
Introduction
[Ro ~ 3, 4 or 6? See Update] Re ~ 5.8 and Ro ~ 6?
Please read Updates:
COVID Odyssey: [Pre]Summer Summary~ How many people may one person infect on average?
COVID Odyssey: Don’t underestimate Ro ~ How many people may one person infect on average?
COVID Odyssey Ro update: Occam’s Razor~ A close shave: In NZ is Ro ~ 3, 4, or 6? What are Ro values worldwide?
Originally: COVID19 is more than twice as virulent as most researchers believe. One person may on average infect 6 other people. The number of cases may double every two days. We explore an Excel spreadsheet simulation, CSAW (“SeeSaw”): COVID19 Sampling Analysis Worksheets, generating random samples of the spread of COVID19 in New Zealand.
Update: Historically here we built an incubation period into the simulations. This is no longer considered necessary since incubation periods are already built into the data.
We investigate whether one infected person may on average infect 6 others in New Zealand up until Lockdown Level 4 on 25 March at 11.59. First we investigate whether one infected person may on average infect 5.3782 (more than 5 and 3/8) other people in New Zealand using a simple model.
We collate information from posts when this site was a blog and examine the spread of COVID19.
Starting 1 August (or earlier), updates will only be found in the companion ebook (currently under development). Preliminary results can be found in this companion site (the page you are reading). Historically the site has been a blog (over 100 COVID19 posts); now it has become essentially a website.
We divide this section into six Chapters:
 Chapter 1: Scenario A. Infectious period One cycle (5 days)
 Chapter 2: Scenario B. Infectious period Two cycles (10 days)
 Chapter 3: The CSAW Excel spreadsheet
 Chapter 4: Worst case scenarios
 Chapter 5: Results
 Chapter 6: Conclusions
We apologise for the presentation of this material. This material was originally part of a blog. Accordingly we do not follow academic writing protocols or referencing. Mathematical symbols including superscripts and subscripts were not available.
For a widerpage (old) version of the page you are reading, see:
https://alangraceblog.wordpress.com/2020/05/27/covid19nzisre53782/
(note that the above page is an older version, particularly for conclusions, but may be more readable)
In the first wave of COVID19 we found that cases increased by a factor of 1.4 each day from Day 18 (16 March) up until Lockdown Level 4. This suggested that Re = 5.37824 for a five day cycle.
The factor a=1.4 clearly fits the actual (cumulative) number of cases well:
Note:
Ro is used when there is no quarantine. i.e. when the virus is unrestrained.
Re is used when there is quarantine (Re is the effective reproduction rate).
We may use R in general discussion.
For an explanation about the basic reproduction number Ro
(pronounced RNought or RZero) see:
https://aaamazingphoenix.wordpress.com/2020/05/26/covid19nzisrolikelytobe29/
In our discussions, the reader will need to decide where it may be appropriate to replace Ro by Re and vice versa.
The factor 1.4 means that the number of cases almost doubles every two days since 1.4^2 = 1.96.
If Ro = SQRT(2) then this would mean doubling every two days. The cases would follow the pattern 5, 7, 10, 14, 20, 28, … where every two days the number has doubled.
This would suggest that Ro is
SQRT(2)^5 ~ 5.65685
if infectivity is the same for each day once a case is symptomatic.
In Chapter 3 we look at the CSAW simulation spreadsheets where infectivity decreases (decays) by the same factor each day.
We find that Ro may be over 6 or even 6.3 or more.
New Zealand was very lucky during the first wave since New Zealand was almost too late going into Lockdown Level 4.
Scenario A and Scenario B both give the same estimated values for the total number of cases. We assume an infectious period of one cycle (Scenario A) or two cycles (Scenario B).
We adopt a cycle length of 5 days, and an incubation period one cycle long.
In Chapter 2 and Chapter 3, we use Excel to simulate an outbreak. We generate over 16,100 cases for each simulation (sample), and 40 simulations for each dataset, calculating the mean number of cases infected by each person in each simulation and dataset.
Where no source is indicated below or elsewhere on this site, data can be found at: https://www.worldometers.info/coronavirus/
We have found that
 One infected person may on average infect 5.3782 other people
(more than 5 and 3/8 and possibly 6) in New Zealand in March (even 5.3782 is twice as many people as most experts expect).  In reality a 5day cycle proves useful
 The total number of cases can be expected to almost double every two days (1.4^2 = 1.96). Note: ‘^‘ means ‘to the power of’
We collate information from previous posts.
We generate a curve fitting the total (cumulative) daily case numbers.
See the table of data and graph below.
Since our estimates for Ro are based on the actual New Zealand situation where there has always been at least some form of selfisolation, our estimates for Ro may only be high estimates for Re since the virus has not been unrestrained.
For an explanation about the basic reproduction number Ro
(RNought or RZero) see:
https://aaamazingphoenix.wordpress.com/2020/05/26/covid19nzisrolikelytobe29/
We assume the daily total number of cases increases by a factor of 1.4 (for the increase in the number of cases) since this fits the actual data well at least for most of March.
Chapter 1: Scenario A.
Infectious period One cycle (5 days)
In summary:
 The total number of cases increases daily by close to 40%
(by a factor of 1.4).
We start with an estimated 10 cases (total) on 16 March 2020
(actual number of cases = 8) and find a graph plotting an increase by a factor of 1.4 daily up to 26 March (the day after New Zealand went into Lockdown Level 4 at 11.59 pm on 25 March) fits the data well.  Therefore the daily number of new cases follows the same increase (factor 1.4).
 The spread follows 5day cycles (5day incubation period followed by one 5day infectious period).
 The 5day cycles have a factor of 1.4^5 = 5.3782.
 One infected person can therefore infect on average close to 5.4 other people (Re ~ 5.4 and Ro is therefore at least 5.4). This is at least double what most experts expect.
 We conclude that Ro = 5.65685 is also possible.
 We also see that Ro may be in the range 6 to 6.5.
 We looked at a twocycle infectious period (with Re near half the size) but have concluded the a onecycle period appears to work better.
 Re =5.3782 suggests cumulative case numbers almost double every two days (1.4^2 = 1.96).
A curve using a factor of 1.4 for total case numbers fits well for most of the second half of March (see table at the end of this post). i.e.
C[D] = 1.4 * C[D1] for D > 18
where C[D] is the estimated (total) number of case numbers on Day D and C[18] = 10.
We also see that
C[D] = SQRT(2) * C[D1] for D > 16
is possible where C[D] is the estimated (total) number of case numbers on Day D and C[16] = 5.
We found that a 5 day cycle worked well for the cycles up to 26 March and generated the following table:
The column before the black vertical line is generated first, starting with the value 10 for the estimated number of cases on 16 March.
The rest of the numbers in the last column before the black vertical line are generated by multiplying by 5.3782 (the daily factor is 1.4; a 5day cycle therefore has a factor of
1.4 ^ 5 = 5.3782).
The penultimate column before the black vertical line contains the differences of the last column.
This (penultimate) column is therefore the estimated daily number of new cases.
We see this (penultimate) column also has a factor of 5.3782.
It should not be surprising that the daily number of new cases also has a factor of a = 1.4.
The total number of cases starting on Day D on successive days are
C[D], a * C[D], and (a^2) * C[D].
This means that the daily number of new cases on the latest two days are
a * C[D] – C[D] = (a1) * C[D]
and
(a^2) * C[D] – a * C[D] = a * (a1) * C[D]
This is a factor of a times the number of new cases on the previous day.
When the total number of cases increases by a factor a then the daily number of new cases also increases by a.
i.e. The number of new cases also increases daily by a = 1.4.
This means that each new case infects on average 5.3782 other people.
This factor (in this column) means that Re ~ 5.3782 (= 1.4 ^5) from 16 March (8 actual cases) to 26 March (283 actual cases).
Rounding this number provides an estimate of at least 5.4 for Ro in New Zealand and hence worldwide for about the first four weeks after the first case appears.
We conclude that Ro is at least 5.4.
Initially a value half this amount was expected.
However after searching last night:
‘The COVID19 Coronavirus Disease May Be Twice As Contagious As We Thought‘
‘A single person with COVID19 may be more likely to infect up to 5 or 6 other people, rather than 2 or 3, suggests a new study of Chinese data from the CDC. It’s not clear if this higher number applies only to the cases in China or if it will be similar in other countries.’ See:
https://www.forbes.com/sites/tarahaelle/2020/04/07/thecovid19coronavirusdiseasemaybetwiceascontagiousaswethought/#4eecbbc729a6
Excerpts from the article are at the bottom of this post.
We see below how the curve with a daily factor of a = 1.4 fits the actual data:
Below is the same graph up to Day 35 (2 April):
Note that New Zealand went into Lockdown Level 4 on 25 March (Day 27) at 11.59 pm.
New cases on the next day(s) are likely to have been infected 5 days prior to being identified as a case.
Below is the table of data for March.
Note that New Zealand went into Lockdown Level 4 at 11.59 pm on 25 March.
Perhaps New Zealand’s curve flattened more than expected.
Perhaps New Zealand may have had by the end of March over 900 more cases (1,555 instead of 647; 2.4 times as many) if we had not gone into Lockdown L4.
This could have resulted in a total of over 3100 cases (2 x 1,555 cases, double the current 1504 cases) assuming a midpoint on 31 March (Day 33).
A midpoint two days later on Day 35 (2 April) may have meant a total of over 6,000 cases (2 x 3,049 cases) if the red curve had remained accurate.
Note that the curve is flipped (values up to the midpoint are rotated by 180 degrees) at the midpoint to create an S Shaped curve (“SCurve) to estimate the remaining values.
We obtain the graph below (estimated values are approximate after midpoint):
Note that the values are the same for the top two graphs up to Day 33 (31 March).
Note also that instead of doubling single values, historically two days’ values have been added together.
The graph below shows the result of “flipping” the actual cases at Day 34.5 (12 April) with the case midpoints giving a total case value 1505 = 708+797
Ideally we want our estimated curves to be higher than the actual data in most places so that some flattening on the actual curve may be indicated.
Using Day 33.5 (instead of 34.5) we may need to add on an extra 10% to account for the expected very long tail.
Day 33.5 gives a total case value of 647 + 708 = 1355.
Adding on 10% gives 1355 x 1.1 = 1490.
From the bar chart below Day 33.5 appears to be a likely midpoint (31 March – 1 April):
The bar chart is bimodal. It has two peaks and cannot be represented very closely after the first peak (even when the curve is flipped) by a curve which is increasing.
Here is another graph:
Finally we take Day 34 as the midpoint and double it to get 2 x 708 = 1416
Here is the graph:
It is difficult to work out the midpoint even when the outbreak appears over.
D  Actual  a  
Date  Day#  Cases  1.4  
07/03/20  9  5  C=  
08/03/20  10  5  10a^(D18)  Diff 
09/03/20  11  5  
10/03/20  12  5  
11/03/20  13  5  
12/03/20  14  5  
13/03/20  15  5  
14/03/20  16  6  
15/03/20  17  8  
16/03/20  18  8  10.00  
17/03/20  19  12  14.00  4.00 
18/03/20  20  20  19.60  5.60 
19/03/20  21  28  27.44  7.84 
20/03/20  22  39  38.42  10.98 
21/03/20  23  52  53.78  15.37 
22/03/20  24  66  75.30  21.51 
23/03/20  25  102  105.41  30.12 
24/03/20  26  155  147.58  42.17 
25/03/20  27  205  206.61  59.03 
26/03/20  28  283  289.25  82.64 
27/03/20  29  368  404.96  115.70 
28/03/20  30  451  566.94  161.98 
29/03/20  31  514  793.71  226.78 
30/03/20  32  589  1,111.20  317.49 
31/03/20  33  647  1,555.68  444.48 
01/04/20  34  708  2,177.95  622.27 
02/04/20  35  797  3,049.13  871.18 
03/04/20  36  868  4,268.79  1,219.65 
Any difference in the estimated/actual totals up to 26 March is caught up on the next day(s).
The spread in New Zealand in March was worse than in the animated gif below which shows five cycles with R = 2.6.
New Zealand reached the same number of cases (368 as shown in the gif below) on 27 March one day after two cycles! Note that instead of 5 cases we started with 10 cases as our estimate in our model which provided a good fit for the actual number of cases in New Zealand.
How can a Coronavirus outspread from 5 to 368 people in 5 Cycles (Credit: The New York Times)?
If 5 people with new coronavirus can impact 2.6 others each, then 5 people could be sick after 1 Cycle, 18 people after 2 Cycles, 52 people after 3 Cycles and so on. See:
https://towardsdatascience.com/howbadwillthecoronavirusoNtbreakgetpredictingtheoutbreakfiguresf0b8e8b61991
We now look at abstract of the article referred to above:
Sanche S, Lin YT, Xu C, RomeroSeverson E, Hengartner N, Ke R. High contagiousness and rapid spread of severe acute respiratory syndrome coronavirus 2. Emerg Infect Dis. 2020 Jul [date cited]. https://doi.org/10.3201/eid2607.200282
DOI: 10.3201/eid2607.200282
Original Publication Date: April 07, 2020
https://wwwnc.cdc.gov/eid/article/26/7/200282_article?deliveryName=USCDC_333DM25287
Volume 26, Number 7—July 2020
Research
High Contagiousness and Rapid Spread of Severe Acute Respiratory Syndrome Coronavirus 2
Steven Sanche1, Yen Ting Lin1, Chonggang Xu, Ethan RomeroSeverson, Nick Hengartner, and Ruian KeComments to Author
Author affiliations: Los Alamos National Laboratory, Los Alamos, New Mexico, USA
Abstract
Severe acute respiratory syndrome coronavirus 2 is the causative agent of the 2019 novel coronavirus disease pandemic. Initial estimates of the early dynamics of the outbreak in Wuhan, China, suggested a doubling time of the number of infected persons of 6–7 days and a basic reproductive number (R0) of 2.2–2.7. We collected extensive individual case reports across China and estimated key epidemiologic parameters, including the incubation period. We then designed 2 mathematical modeling approaches to infer the outbreak dynamics in Wuhan by using highresolution domestic travel and infection data. Results show that the doubling time early in the epidemic in Wuhan was 2.3–3.3 days. Assuming a serial interval of 6–9 days, we calculated a median R0 value of 5.7 (95% CI 3.8–8.9). We further show that active surveillance, contact tracing, quarantine, and early strong social distancing efforts are needed to stop transmission of the virus.
In New Zealand Re = 5.3782 indicates that the number of cases almost doubles every two days (1.4^2 = 1.96).
If you look at the twoday ratio from 16 to 24 March (days 18, 19, … ,26) of
(total) number of cases two days ahead
current (total) number of cases
i.e.
C[D+2]/C[D] for D=18, 19, … , 26
The average is 2.053157507.
To get the daily increase we take the square root.
The SQRT of 2.053157507 is 1.432884331
This gives a 5day increase of
1.432884331^5 = 6.040260494
We note that a twoday increase using the factor 1.4 is
1.4^2 = 1.96
We conclude that a factor of SQRT(2) is also possible giving a 5day increase of
SQRT(2)^5 = 5.65685
We adopt Ro = 5.65685.
This means that it is possible for the total number of cases to double every two days.
We see that
C[D] = SQRT(2) * C[D1] for D > 16
is possible where C[D] is the estimated (total) number of case numbers on Day D and C[16] = 5.
This also gives C[18] = 10 as we had with the factor 1.4.
We also consider a 5.2 day cycle.
‘The mean incubation period was 5.2 days (95% confidence interval [CI], 4.1 to 7.0), with the 95th percentile of the distribution at 12.5 days.’ See:
https://www.worldometers.info/coronavirus/coronavirusincubationperiod/
A 5.2 day cycle has an increase of
1.432884331^5.2 = 6.490795035
This suggests that Ro in the range 6 to 6.5 is possible in New Zealand.
This is within the range in this graph:
The above graph was obtained from this article: taaa021 (Click to view PDF):
The reproductive number of COVID19 is higher compared to SARS coronavirus
published 13 February 2020, obtained from here:
https://academic.oup.com/jtm/article/27/2/taaa021/5735319
(Journal of Travel Medicine, Volume 27, Issue 2, March 2020)
Chapter 2: Scenario B.
Infectious period Two cycles (10 days)
In a previous post we saw that
C[D] = 1.14333333 * ( C[D2] + C[D3] ) for D > 20
(note C[D1] is missed out)
with starting values C[18] = 10, C[19] = 14, and C[20] = 19.6
i.e. a modified Fibonacci sequence.
For i>2 using C[2] = 8 and C[1] = 5, for given constant values of LAMBDA and Re, we can calculate
C[i] = Re* ( LAMBDA * C[i1] + (1 – LAMBDA) * C[i2] )
Where C[i] is the estimated number of new cases in cycle i.
See Chapter 6: Conclusions for calculations.
For Re = 5.8 we obtain:
In the above model each case infects people over two cycles (10 days).
Hence each cycle contains new cases infected by cases infected themselves during the two previous cycles.
Chapter 3:
The CSAW Excel Spreadsheet
We use Excel to simulate a (10day) scenario with infectivity decreasing (decaying) each day.
We randomly generated sets of 40 samples and their means in Excel using a factor of 1.4 each day for the increase in the number of cases.
Below are two sets of samples (the Mean is actually the mean of the means for 40 samples): The above tables used an infection rate P=0.50489 which halved on each successive day. The value of P was calculated so that in 10 days the sum was 1.
The values above suggest that Ro may be over 3.5 (see dataset for P=0.5) in this scenario.
However we understand that infection is highest on the first day and diminishes on successive days.
A scenario where the infection rate was the same each day (P=0.1), produced means over 13.7.
Below are the calculations for P= 0.75.
v=  3  
w=  4  
v/w=  0.75  
w/v=  1.333333  Factor 
1/Sum=  0.019891  
10  1  0.019891 
9  1.333333  0.026522 
8  1.777778  0.035362 
7  2.37037  0.04715 
6  3.160494  0.062866 
5  4.213992  0.083822 
4  5.618656  0.111762 
3  7.491541  0.149017 
2  9.988721  0.198689 
1  13.31829  0.264918 
Sum=  50.27318  1 
We work backwards, assigning a value of 1 on day 10, then multiplying each day by the inverse, in this case by 1/0.75 = 1.333333. We scale the values by the inverse of the Sum (see bottom row), so that the daily factors sum to 1.
On the spreadsheet each case was assigned a number, and each new case was randomly assigned an infection day using a table like the above (the day number adjusted up to 10 days prior), and a case from this day was randomly selected as the case causing infection. The number of people infected by each case was then counted, and the mean calculated for each new case infected on Day 12 (so that there were 10 prior days’ cases causing infections on each day for future development).
The means were calculated for the 40 samples, and the overall mean and standard deviation calculated.
Below we look at sample runs for P = 1/2 (0.5), 7/10 (0.7), 5/7 (0.714286), and 3/4 (0.75).
Mean=  3.207675  Mean=  5.50636 
Std Dev=  0.175297  Std Dev=  0.25242 
Max=  3.605263  Max=  6.236842 
Min=  2.921053  Min=  5.008772 
P=  0.5  P=  0.7 
Sample#  Mean  Sample#  Mean 
1  3.605263  1  5.807018 
2  3.131579  2  5.307018 
3  3.464912  3  5.570175 
4  3.061404  4  5.535088 
5  3.307018  5  5.622807 
6  3.447368  6  5.438596 
7  3.491228  7  5.438596 
8  3.149123  8  5.403509 
9  3.096491  9  5.912281 
10  3.087719  10  5.368421 
11  3.22807  11  5.868421 
12  3.298246  12  5.666667 
13  3.315789  13  5.675439 
14  2.921053  14  5.508772 
15  3.105263  15  5.482456 
16  2.921053  16  5.482456 
17  3.175439  17  6.236842 
18  3.377193  18  5.684211 
19  3.061404  19  5.429825 
20  3.175439  20  5.526316 
21  3.192982  21  5.526316 
22  3.394737  22  5.184211 
23  3.052632  23  5.236842 
24  3.245614  24  5.491228 
25  3.192982  25  5.5 
26  2.947368  26  5.324561 
27  2.929825  27  5.561404 
28  3.254386  28  5.350877 
29  3.438596  29  5.008772 
30  3.166667  30  5.175439 
31  3.333333  31  5.140351 
32  2.938596  32  5.421053 
33  3.061404  33  5.280702 
34  3.131579  34  5.807018 
35  3.035088  35  5.622807 
36  3.254386  36  5.675439 
37  3.263158  37  5.061404 
38  3.45614  38  5.27193 
39  3.175439  39  5.833333 
40  3.421053  40  5.815789 
Mean=  5.8  Mean=  6.555263 
Std Dev=  0.223263  Std Dev=  0.260832 
Max=  6.359649  Max=  7.04386 
Min=  5.394737  Min=  5.72807 
P=  0.714286  P=  0.75 
Sample#  Mean  Sample#  Mean 
1  6.22807  1  6.535088 
2  6.359649  2  6.403509 
3  5.815789  3  6.929825 
4  5.780702  4  6.473684 
5  5.561404  5  6.526316 
6  5.95614  6  6.798246 
7  5.780702  7  6.412281 
8  5.508772  8  6.614035 
9  5.842105  9  6.649123 
10  5.447368  10  6.315789 
11  5.807018  11  6.649123 
12  6.008772  12  6.307018 
13  6.192982  13  6.333333 
14  5.982456  14  7.04386 
15  5.649123  15  6.447368 
16  5.973684  16  6.622807 
17  5.868421  17  6.412281 
18  5.982456  18  6.54386 
19  6.008772  19  6.552632 
20  5.45614  20  6.964912 
21  5.710526  21  6.885965 
22  5.72807  22  6.605263 
23  5.666667  23  6.394737 
24  5.807018  24  5.72807 
25  5.675439  25  6.263158 
26  5.798246  26  6.473684 
27  5.842105  27  6.736842 
28  5.394737  28  6.54386 
29  5.447368  29  6.184211 
30  6.008772  30  6.973684 
31  5.552632  31  6.394737 
32  5.631579  32  6.631579 
33  5.912281  33  6.587719 
34  5.807018  34  6.947368 
35  5.833333  35  6.859649 
36  5.570175  36  6.596491 
37  5.885965  37  6.649123 
38  6.122807  38  6.666667 
39  5.719298  39  6.438596 
40  5.675439  40  6.114035 
We have seen scale factors of P = 1/2 (0.5), 7/10 (0.7), 5/7 (0.714286), and 3/4 (0.75).
The scale factors have themselves been scaled so that the sum for each (column) is 1.
Scales:  1/2  7/10  5/7  3/4 
1  0.500489  0.308721  0.295946  0.264918 
2  0.250244  0.216104  0.21139  0.198689 
3  0.125122  0.151273  0.150993  0.149017 
4  0.062561  0.105891  0.107852  0.111762 
5  0.031281  0.074124  0.077037  0.083822 
6  0.01564  0.051887  0.055026  0.062866 
7  0.00782  0.036321  0.039305  0.04715 
8  0.00391  0.025424  0.028075  0.035362 
9  0.001955  0.017797  0.020053  0.026522 
10  0.000978  0.012458  0.014324  0.019891 
This effect of this scaling can be seen in the graph below:
We use the sample below on the next page:
Mean=  5.818421 
Std Dev= 
0.244262 
Max=  6.307018 
Min=  5.342105 
and P=  0.75 
We normalise the values (subtract the mean from each and divide by the standard deviation) then, after rounding, create a frequency table.
As expected the graph roughly looks like a normal curve. The sampling distribution of the means is normal even when the underlying distribution is not normally distributed.
Sample#  Mean  N. Mean  Round  Dev  Frequency 
1  6.192982  1.834647  2  4  0 
2  5.95614  0.541293  1  3  0 
3  5.692982  0.89577  1  2  3 
4  5.824561  0.17724  0  1  8 
5  6  0.780803  1  0  17 
6  5.675439  0.99157  1  1  8 
7  5.491228  1.99751  2  2  4 
8  6.026316  0.924509  1  3  0 
9  6  0.780803  1  4  0 
10  5.745614  0.60836  1  Checksum  40 
11  6.017544  0.876607  1  
12  5.72807  0.70416  1  
13  5.719298  0.75206  1  
14  5.596491  1.42269  1  
15  6.166667  1.690941  2  
16  5.912281  0.301783  0  
17  5.921053  0.349685  0  
18  5.807018  0.27304  0  
19  5.578947  1.51849  2  
20  5.859649  0.014371  0  
21  5.780702  0.41675  0  
22  5.657895  1.08738  1  
23  6.04386  1.020313  1  
24  5.780702  0.41675  0  
25  5.824561  0.17724  0  
26  5.701754  0.84787  1  
27  5.780702  0.41675  0  
28  5.868421  0.062273  0  
29  5.833333  0.12934  0  
30  6.219298  1.978353  2  
31  5.903509  0.253881  0  
32  5.464912  2.14122  2  
33  5.789474  0.36885  0  
34  5.859649  0.014371  0  
35  5.789474  0.36885  0  
36  5.842105  0.08143  0  
37  6.140351  1.547235  2  
38  5.938596  0.445489  0  
39  6.087719  1.259823  1  
40  6.061404  1.116117  1 
Here are graphs from the data above:
It should now make sense if we talk in more depth about the spreadsheet.
We start Day 1 on 16 March when there were 8 cases in New Zealand. Day 1 is therefore 17 days after New Zealand identified its first case of COVID19 on 28 February 2020.
On the spreadsheet each case was assigned a number, and each new case was randomly assigned an infection day using a table like the above (the day number adjusted up to 10 days prior), and a case from this day was randomly selected as the case causing infection.
The number of people infected by each case was then counted, and the mean calculated for each new case infected on Day 12 (so that there were 10 prior days’ cases causing infections on each day). The spreadsheet also includes calculations for Day 10 and 11.
We assume spread by a factor of 1.4 on each day. i.e. the total number of cases increases by 1.4 each day.
The means were calculated for 40 samples, and the overall mean and standard deviation calculated.
For each sample 16,108 cases were generated randomly in Excel.
This was later increased to over 20,000 cases.
For the 115 cases infected on Day 12, the number of new cases infected by each case was counted and the mean calculated.
Samples were randomly generated 40 times, providing a set of 40 sample means.
The main spreadsheet contained over 1.44 million nonempty cells.
This was later increased to over 1.8 million nonempty cells.
Once a case became infectious the infection rate diminished over the next 10 days at the same factor (P) each day. Once a value of P (a fraction) was chosen, daily factors were calculated so that over 10 days the sum of the daily factors was 1 and each day the factor decreased by the factor P.
The infection rate was not uniform each day since we understand that infection is highest on the first day and diminishes on successive days.
A scenario where the infection rate was the same each day (P=1), produced means over 13.7.
We build in a two day incubation period (48 hours; any more made the numbers too big).
We explain the process using Case #290 (the first case generated on Day 12).
First we find (generate) who infected Case #290 (so that we can include results for Days 10 and 11). To do this we need to randomly choose a day number from 1 to 10 when Case #290 was infected. This is like throwing dice, we need to use a random number for this, generated so that it is equally likely to be any decimal number between 1 and 10.
We use this random number to choose a day using the scale factor as we have already discussed.
This infection day, numbered from 1 to 10, needs to be adjusted by adding on the difference between the current day number and 10. Case #290 was infected in Day 12 so we add 2 days to the number we have obtained already. The random day number from one to ten becomes relative to Day 12.
We use another random number to choose anyone infected on the chosen (adjusted) day to infect Case #290, using tables we have already created. This table was generated (once) starting with 10 cases on the first day (Day 1 has Case #1 to Case #10, and each day generating more to get 1.4 times as many on the next day, and repeating the process for each successive day. Day 2 would have 4 new Cases Case #11 to Case #14, Day 3 would have new Case #15 to Case#19 and so on.
Day 12 has new Case #290 to Case #404 (115 new cases).
The first case on Day 12 is Case #290. We record the case number of the person infecting #290 with the details for Case #290.
We then do the same for Case #291 and all the other cases for day 12, then for the other days.
At the end we have done this for over 16,100 (16,108) cases, Case #290 to Case #16398.
Once we have done this for all these cases (over 1,400 cases that may be infected by Case #290 over 10 days), we can count up the number infected by each case, starting with Case #290.
For Case #290, we can count how many other cases were infected by Case #290. During the process above anyone infected by case #290 had Case #290 recorded alongside their details. If Case #700, #987, and #11392 were infected by Case #290, they have case #290 (290) recorded alongside each of their case numbers so we only need to search for who has the number 290 next to their case number.
We can then do the same, for Case #291 and each of the other cases infected on Day 12.
We have now counted the number of people infected by Case #290 up to Case #404.
We then find the mean of all these numbers to get the average number of people infected in this sample.
This gives us a mean for Sample #1. We do exactly the same for Sample #3 up to Sample #40.
All the samples generate different results each time for Case#290 up to Case #404.
Each of these samples generates different results and hence we have different means for each sample.
Finally we find the overall mean and standard deviation for all of our (40) sample means.
This gives us the results for one execution (run) of the spreadsheet.
We can repeat the process over and over again using the same scale factor (fraction) as before or trying out different fractions. We prefer to use 5/7 since 7/5 = 1.4.
Chapter 4: Worst case scenarios
New Zealand currently has 22 deaths from COVID19 (and 1504 cases) on June 12.
In a final version of a report dated 24 March commissioned by the NZ Ministry of Health, a worst case scenario estimated 27,600 deaths in New Zealand by early July. This report uses Ro = 3.5 in the scenario. See bottom of p4 in the (link to the) report in this post:
https://aaamazingphoenix.wordpress.com/2020/05/16/covid19nzcouldtheresultsinthismodeloccur/
You may be able to get it directly below:
https://www.health.govt.nz/system/files/documents/publications/report_for_chief_science_advisor__health__24_march_final.pdf
The report includes this graph:
The peak appears to be on Day 104 = 13 June (104 Days from 1 March).
Using 5 July as an “early July” peak, the peak should instead be on Day 126.
Let’s look at the CFR (Case Fatality Risk/Ratio).
The report estimates that 0.83% of sick people die. i.e. CFR = 0.083.
The actual New Zealand CFR figure is 22/1504 ~ 1.46% (22 Deaths from 1504 cases).
In the report, the estimated number of deaths in a worst case scenario is 27,600 deaths in early July.
However to adjust the length (number of days) to the New Zealand midpoint, we need to divide by 2 twice to correct the number of days and hence take the square root twice (of 27,600) and double the answer. This calculation is OK since the estimate of 12.9 deaths before doubling is greater than the actual number (1 death).
This gives us a midpoint on Day 31.5 and an estimated total of 25.8 deaths.
This is a good estimate for the actual number of deaths (22).
However the estimated number of deaths on Day 31.5 is 12.9, a factor of 10 from the real number (1).
Using the report CFR (0.0083), the number of cases on Day 31.5 would be 12.9/0.0083 ~ 1,550 cases, about double the actual number of cases (after the adjustment to day numbers below), at least close to the right ballpark.
Using the actual New Zealand CFR (.0146), the estimated number of cases for 12.9 deaths is 12.9/0.0146 ~ 880 cases, closer to the actual case figure and only a few days out (after the adjustment to day numbers below). The actual figure is 858 cases on 3 April (although New Zealand still only had one death on this date).
Since we are considering a worst case scenario we expect actual figures to lag behind the estimates.
Deaths usually lag behind cases by perhaps at least one cycle since obviously most deaths occur several days after the corresponding case is identified.
New Zealand had a total of 13 deaths on 21 April.
The model in the report uses Ro = 3.5 (i.e. no isolation).
However the low CFR in the report (0.0083 vs 0.0146), may (or may not) mean Re ~ 2 after adjusting for the actual CFR.
In the report model the CFR is assigned 0.83% but since the model assumes only 67% of infections lead to sickness (are the rest assumed asymptomatic?), we may need to add 50% and change the CFR to 1.25%. This may depend on whether a “case” in the CFR definition means an infected case (which may be asymptomatic) or a symptomatic case.
Day 31.5 assumes a starting date of 1 March.
The first death in New Zealand occurred on 28 February, two days before 1 March.
Therefore after adjusting Day 31. 5 to Day 33.5 to account for the actual start date, this is close to the actual midpoint.
The worst case scenario is close to the actual scenario when the length of time to the midpoint is adjusted to be close to the actual midpoint.
Sadly the actual model in the report estimates a total of 27,600 deaths, over 5,500 deaths per one million of population which is extremely unrealistic.
Below are the countries with the largest number of deaths per one million of population on 10 June:
Country,  Total  Total  Tot Cases/  Deaths/  Population 
Other  Cases  Deaths  1M pop  1M pop  
San Marino  688  42  20,279  1,238  33,927 
Belgium  59,437  9,619  5,130  830  11,586,764 
Andorra  852  51  11,028  660  77,258 
UK  289,140  40,883  4,260  602  67,865,632 
Spain  289,046  27,136  6,182  580  46,753,788 
Italy  235,561  34,043  3,896  563  60,466,601 
Sweden  45,924  4,717  4,549  467  10,095,634 
At the time of writing:
The USA has 345 deaths per one million of population.
China has 3 deaths per one million of population.
New Zealand has about 4.4 deaths per one million of population.
When we were estimating the number of deaths in New Zealand, we used Norway to estimate an upper limit. See:
https://aaamazingphoenix.wordpress.com/2020/04/06/covid19nzhowmanycasesanddeathsmayweexpect/
Below are today’s (10 June) statistics for Norway:
Country,  Total  Total  Tot Cases/  Deaths/  Population 
Other  Cases  Deaths  1M pop  1M pop  
Norway  8,576  239  1,583  44  5,418,762 
Norway currently has ten times as many deaths per one million of population as New Zealand (New Zealand has about 4.4 deaths per one million of population) and over five times as many cases per one million of population (1,583 vs 301 cases) and less than 10% more population (5,418,762 vs 5,000,000).
We consider Norway’s figures represent a realistic worst case scenario for New Zealand.
The deaths in Norway per one million of population (44) are less than one percent (0.8%) of the deaths estimated in New Zealand in the report (5,520 = 27,600/5).
Perhaps a cumulative normal distribution is also useful in predictions and case estimates. It is easy to read estimates for the total number of cases from cumulative graphs.
The mean (midpoint) and standard deviation apply to the number of Days and the cumulative normal distribution is scaled to fit actual cumulative case numbers in March.
We can roughly fit a scaled normal curve (cumulative normal distribution) to the New Zealand Case Numbers (total = 1504):
Below is the data table.
Date  Day  Normal 34.5,7.2  Normal 36.5,6.5  Actual Cases 
28Feb  1  0.00  0.00  1 
29Feb  2  0.00  0.00  1 
1Mar  3  0.01  0.00  1 
2Mar  4  0.02  0.00  1 
3Mar  5  0.03  0.00  1 
4Mar  6  0.06  0.00  3 
5Mar  7  0.10  0.01  3 
6Mar  8  0.18  0.02  4 
7Mar  9  0.30  0.03  5 
8Mar  10  0.50  0.07  5 
9Mar  11  0.83  0.13  5 
10Mar  12  1.34  0.24  5 
11Mar  13  2.12  0.44  5 
12Mar  14  3.32  0.79  5 
13Mar  15  5.09  1.39  5 
14Mar  16  7.66  2.37  6 
15Mar  17  11.34  3.98  8 
16Mar  18  16.49  6.52  8 
17Mar  19  23.56  10.46  12 
18Mar  20  33.10  16.41  20 
19Mar  21  45.72  25.20  28 
20Mar  22  62.07  37.87  39 
21Mar  23  82.88  55.73  52 
22Mar  24  108.85  80.28  66 
23Mar  25  140.64  113.28  102 
24Mar  26  178.81  156.57  155 
25Mar  27  223.77  212.05  205 
26Mar  28  275.72  281.49  283 
27Mar  29  334.59  366.36  368 
28Mar  30  400.04  467.69  451 
29Mar  31  471.42  585.83  514 
30Mar  32  547.78  720.37  589 
31Mar  33  627.90  869.99  647 
1Apr  34  710.37  1032.51  708 
2Apr  35  793.63  1204.92  797 
3Apr  36  876.10  1383.55  868 
4Apr  37  956.22  1564.29  950 
5Apr  38  1032.58  1742.92  1039 
6Apr  39  1103.96  1915.33  1106 
7Apr  40  1169.41  2077.85  1160 
8Apr  41  1228.28  2227.47  1210 
9Apr  42  1280.23  2362.01  1239 
10Apr  43  1325.19  2480.15  1283 
11Apr  44  1363.36  2581.48  1312 
12Apr  45  1395.15  2666.35  1330 
13Apr  46  1421.12  2735.79  1349 
14Apr  47  1441.93  2791.27  1366 
15Apr  48  1458.28  2834.56  1386 
16Apr  49  1470.90  2867.56  1401 
17Apr  50  1480.44  2892.11  1409 
18Apr  51  1487.51  2909.97  1422 
19Apr  52  1492.66  2922.64  1431 
20Apr  53  1496.34  2931.43  1440 
21Apr  54  1498.91  2937.38  1445 
22Apr  55  1500.68  2941.32  1448 
23Apr  56  1501.88  2943.86  1451 
24Apr  57  1502.66  2945.47  1456 
25Apr  58  1503.17  2946.45  1461 
26Apr  59  1503.50  2947.05  1469 
27Apr  60  1503.70  2947.40  1469 
28Apr  61  1503.82  2947.60  1472 
29Apr  62  1503.90  2947.71  1474 
30Apr  63  1503.94  2947.77  1479 
1May  64  1503.97  2947.81  1485 
2May  65  1503.98  2947.82  1487 
3May  66  1503.99  2947.83  1487 
4May  67  1504.00  2947.84  1486 
5May  68  1504.00  2947.84  1487 
6May  69  1504.00  2947.84  1486 
7May  70  1504.00  2947.84  1488 
8May  71  1504.00  2947.84  1489 
9May  72  1504.00  2947.84  1490 
10May  73  1504.00  2947.84  1492 
11May  74  1504.00  2947.84  1494 
12May  75  1504.00  2947.84  1497 
13May  76  1504.00  2947.84  1497 
14May  77  1504.00  2947.84  1497 
15May  78  1504.00  2947.84  1497 
16May  79  1504.00  2947.84  1498 
17May  80  1504.00  2947.84  1498 
18May  81  1504.00  2947.84  1499 
19May  82  1504.00  2947.84  1499 
20May  83  1504.00  2947.84  1503 
21May  84  1504.00  2947.84  1503 
22May  85  1504.00  2947.84  1503 
23May  86  1504.00  2947.84  1504 
24May  87  1504.00  2947.84  1504 
25May  88  1504.00  2947.84  1504 
26May  89  1504.00  2947.84  1504 
27May  90  1504.00  2947.84  1504 
The standard deviations are chosen to fit the actual data for as much of March/April as possible.
Let’s suppose we would like to investigate the effect of a scale factor of 1.4 per day.
The last column in the table above shows a curve with a mean two days later (36.5).
The new scale factor is 1504 x 1.4^2 ~ 2948.
We chose this value simply to try out scaling of the cumulative normal distribution to estimate cumulative case values.
We saw above in Chapter 1 that we could have had 1,555 case numbers by Day 33 giving a total of over 3100 cases.
In the table below using a factor of 1.4 we see that we could have had the same number of cases by Day 33.
Below is the graph (red curve) with a standard deviation chosen to roughly fit early actual case number values.
We also look at values to give a total close to 3008 (=1504 x 2) cases.
We also look at values to give a total close to 3008 (=1504 x 2) cases.
Total Cases  1504  2947.84  3008  1504  
Date  Day  Normal 34.5,7.2  Normal 36.5,6.5  Normal 36.5,6.5  Actual Cases 
28Feb  1  0.00  0.00  0.00  1 
29Feb  2  0.00  0.00  0.00  1 
1Mar  3  0.01  0.00  0.00  1 
2Mar  4  0.02  0.00  0.00  1 
3Mar  5  0.03  0.00  0.00  1 
4Mar  6  0.06  0.00  0.00  3 
5Mar  7  0.10  0.01  0.01  3 
6Mar  8  0.18  0.02  0.02  4 
7Mar  9  0.30  0.03  0.04  5 
8Mar  10  0.50  0.07  0.07  5 
9Mar  11  0.83  0.13  0.13  5 
10Mar  12  1.34  0.24  0.25  5 
11Mar  13  2.12  0.44  0.45  5 
12Mar  14  3.32  0.79  0.81  5 
13Mar  15  5.09  1.39  1.41  5 
14Mar  16  7.66  2.37  2.42  6 
15Mar  17  11.34  3.98  4.06  8 
16Mar  18  16.49  6.52  6.66  8 
17Mar  19  23.56  10.46  10.67  12 
18Mar  20  33.10  16.41  16.75  20 
19Mar  21  45.72  25.20  25.71  28 
20Mar  22  62.07  37.87  38.65  39 
21Mar  23  82.88  55.73  56.86  52 
22Mar  24  108.85  80.28  81.92  66 
23Mar  25  140.64  113.28  115.59  102 
24Mar  26  178.81  156.57  159.77  155 
25Mar  27  223.77  212.05  216.38  205 
26Mar  28  275.72  281.49  287.23  283 
27Mar  29  334.59  366.36  373.84  368 
28Mar  30  400.04  467.69  477.24  451 
29Mar  31  471.42  585.83  597.79  514 
30Mar  32  547.78  720.37  735.07  589 
31Mar  33  627.90  869.99  887.75  647 
1Apr  34  710.37  1032.51  1053.59  708 
2Apr  35  793.63  1204.92  1229.51  797 
3Apr  36  876.10  1383.55  1411.78  868 
4Apr  37  956.22  1564.29  1596.22  950 
For comparison for Days 26 to 28 we reproduce part of the table below which uses the daily factor 1.4 starting with 10 cases on Day 18 (actual values are in the middle column):
24/03/20  26  155  147.5789  
25/03/20  27  205  206.6105  
26/03/20  28  283  289.2547 
In the table below, we also look at what happens when we change the standard deviation to 6.55 (for 2947.84 column only) and also when we increase the total cases to 3110.
Total Cases  1504  2947.84  3110  1504  
Date  Day  Normal 34.5,7.2  Normal 36.5,6.55  Normal 36.5,6.5  Actual Cases 
28Feb  1  0.00  0.00  0.00  1 
29Feb  2  0.00  0.00  0.00  1 
1Mar  3  0.01  0.00  0.00  1 
2Mar  4  0.02  0.00  0.00  1 
3Mar  5  0.03  0.00  0.00  1 
4Mar  6  0.06  0.00  0.00  3 
5Mar  7  0.10  0.01  0.01  3 
6Mar  8  0.18  0.02  0.02  4 
7Mar  9  0.30  0.04  0.04  5 
8Mar  10  0.50  0.08  0.07  5 
9Mar  11  0.83  0.15  0.14  5 
10Mar  12  1.34  0.27  0.25  5 
11Mar  13  2.12  0.49  0.47  5 
12Mar  14  3.32  0.87  0.84  5 
13Mar  15  5.09  1.52  1.46  5 
14Mar  16  7.66  2.58  2.51  6 
15Mar  17  11.34  4.29  4.20  8 
16Mar  18  16.49  6.98  6.88  8 
17Mar  19  23.56  11.12  11.03  12 
18Mar  20  33.10  17.34  17.31  20 
19Mar  21  45.72  26.47  26.59  28 
20Mar  22  62.07  39.57  39.96  39 
21Mar  23  82.88  57.92  58.79  52 
22Mar  24  108.85  83.04  84.70  66 
23Mar  25  140.64  116.64  119.51  102 
24Mar  26  178.81  160.54  165.18  155 
25Mar  27  223.77  216.60  223.71  205 
26Mar  28  275.72  286.51  296.97  283 
27Mar  29  334.59  371.71  386.52  368 
28Mar  30  400.04  473.16  493.42  451 
29Mar  31  471.42  591.16  618.06  514 
30Mar  32  547.78  725.27  760.00  589 
31Mar  33  627.90  874.18  917.85  647 
1Apr  34  710.37  1035.72  1089.31  708 
2Apr  35  793.63  1206.94  1271.20  797 
3Apr  36  876.10  1384.23  1459.65  868 
4Apr  37  956.22  1563.61  1650.35  950 
We also used Goal Seek in Excel (Data tab, WhatIfAnalysis).
Using Goal Seek and least squares up to Day 28 we found the best fit was obtained using a standard deviation of 6.68 days.
We also obtained this graph using the same mean and standard deviation:
Note that this is not the actual mean and standard deviation.
Since the curve (for the New Zealand actual case data) is not close to being a normal curve, the actual mean and standard deviation are not useful for estimating case numbers. The normal curve does not represent the high values well.
The real mean and standard deviation are shown in the graph below:
Here are both graphs showing the normal curves sidebyside:
Later we also considered a standard deviation of 6.5.
The actual daily cases up to Day 37 are in the table below. A more complete table up to 27 May has already been listed above earlier in this section.
We look further at what may have happened if the original daily factor of 1.4 (starting with 10 cases on Day 18) had continued.
In the table below all columns except the factor 1.4 column have 1504 cases total (in the complete table).
Date  Day  Normal 34.5,6.68  Normal 34.5,7.2  Factor 1.4  Actual Cases 
15Mar  17  6.62  11.34  8  
16Mar  18  10.17  16.49  10.00  8 
17Mar  19  15.29  23.56  14.00  12 
18Mar  20  22.54  33.10  19.60  20 
19Mar  21  32.57  45.72  27.44  28 
20Mar  22  46.13  62.07  38.42  39 
21Mar  23  64.06  82.88  53.78  52 
22Mar  24  87.25  108.85  75.30  66 
23Mar  25  116.58  140.64  105.41  102 
24Mar  26  152.85  178.81  147.58  155 
25Mar  27  196.72  223.77  206.61  205 
26Mar  28  248.60  275.72  289.25  283 
27Mar  29  308.59  334.59  404.96  368 
28Mar  30  376.44  400.04  566.94  451 
29Mar  31  451.47  471.42  793.71  514 
30Mar  32  532.60  547.78  1,111.20  589 
31Mar  33  618.41  627.90  1,555.68  647 
1Apr  34  707.14  710.37  1,053.59  708 
2Apr  35  796.86  793.63  1,229.51  797 
3Apr  36  885.59  876.10  1,411.78  868 
4Apr  37  971.40  956.22  1,596.22  950 
For simplicity, in the table below we double the daily estimate (see last column) to get an estimated total number of cases. i.e. We assume the midpoint may be at each date.
Remember that the last two columns almost double (1.4^2 = 1.96) every two days.
If isolation (including selfisolation) did not already have an effect on slowing the rate (factor 1.4; see actual new cases after 26 March), we could have expected 2222 cases (30 March estimate total) since new cases on 30 March would are likely to have been infected before Lockdown Level 4 started at 11.59pm on 25 March.
It is likely there could have been a total of 3,000 to 6,000 cases even with Lockdown Level 4 starting at 11.59pm on 25 March.
New cases could easily have been infected at least five to eight days earlier (or more with a cycle length greater than five Days or an infection period two cycles long).
If Level 4 Lockdown had been delayed, we can look at the figures at least five to eight days later to see how this may have affected the total number of cases.
In the table below we double the daily estimate (see last column) to get an estimated total number of cases. i.e. We assume the midpoint may be at each date.
The table below also has the data for our worst case scenarios.
Note: SQRT(2) ~ 1.414213562
Date  Day# (D)  Actual Cases  a = SQRT(2)  a = 1.4  Estimated Total 
14Mar  16  6  5.00  
15Mar  17  8  7.07  
16Mar  18  8  10.00  10.00  
17Mar  19  12  14.14  14.00  
18Mar  20  20  20.00  19.60  
19Mar  21  28  28.28  27.44  
20Mar  22  39  40.00  38.42  
21Mar  23  52  56.57  53.78  
22Mar  24  66  80.00  75.30  
23Mar  25  102  113.14  105.41  
24Mar  26  155  160.00  147.58  
25Mar  27  205  226.27  206.61  
26Mar  28  283  320.00  289.25  
27Mar  29  368  452.55  404.96  
28Mar  30  451  640.00  566.94  
29Mar  31  514  905.10  793.71  1587.429547 
30Mar  32  589  1280.00  1111.20  2222.401365 
31Mar  33  647  1810.19  1555.68  3111.361911 
1Apr  34  708  2560.00  2177.95  4355.906676 
2Apr  35  797  3620.39  3049.13  6098.269346 
3Apr  36  868  5120.00  4268.79  8537.577084 
4Apr  37  950  7240.77  5976.30  11952.60792 
5Apr  38  1039  10240.00  8366.83  16733.65109 
6Apr  39  1106  14481.55  11713.56  23427.11152 
7Apr  40  1160  20480.00  16398.98  32797.95613 
8Apr  41  1210  28963.09  22958.57  45917.13858 
9Apr  42  1239  40960.00  32142.00  64283.99401 
Now we look at a normal curve which has a total of 27,600 cases (not deaths!).
The curve stays above the actual number of cases which means the cumulative number of cases in the normal curve is also above the actual cumulative number of cases.
At the peak in the above graph there are over 600 (629) cases per day.
Clearly even the number of cases in the normal curve is huge compared to the actual case numbers and extremely unlikely; let alone the same number of deaths.
If 27,600 cases did occur in New Zealand before early in July, the estimated number of resulting deaths using the report CFR of 0.083 would be
27,600 x 0.083 = 2,290.80 deaths.
This would mean almost 460 deaths per one million people in New Zealand.
Only eight countries in the world have 450 or more deaths per one million of population.
Therefore even 2,290.80 deaths (450 per million) is very unlikely in New Zealand even in a worst case scenario.
However since at least 450 deaths per million have occurred in eight countries we need to accept that this number is possible in a worst case scenario.
The report estimated 27,600 deaths in New Zealand, which is 5,520 deaths per one million of population. We saw above that the highest number of deaths per one million of population in the world is still less than 1,300 (1,238) deaths, less than one quarter of 5,520 deaths.
New Zealand’s total number of cases (1,504 cases) is less than the 2,290.80 death estimate with no active cases and no new cases for the last 20 days on 12 June.
Even the USA with over 2,000,000 cases and over 115,000 deaths has 353 deaths per one million of population, much less than 460.
The peak in the graph above is at the end of April.
The report mentioned a peak early in July.
Here is the normal curve in the report again:
We place the two curves side by side:
Most of the March data below was available before the final report was submitted.
Date  Day  Normal Curve  Actual Daily Cases 
28Feb  1  1.07  1.00 
29Feb  2  1.31  0.00 
1Mar  3  1.60  0.00 
2Mar  4  1.94  0.00 
3Mar  5  2.36  0.00 
4Mar  6  2.85  2.00 
5Mar  7  3.43  0.00 
6Mar  8  4.12  1.00 
7Mar  9  4.93  1.00 
8Mar  10  5.88  0.00 
9Mar  11  6.99  0.00 
10Mar  12  8.28  0.00 
11Mar  13  9.78  0.00 
12Mar  14  11.52  0.00 
13Mar  15  13.52  0.00 
14Mar  16  15.81  1.00 
15Mar  17  18.43  2.00 
16Mar  18  21.42  0.00 
17Mar  19  24.81  4.00 
18Mar  20  28.65  8.00 
19Mar  21  32.97  8.00 
20Mar  22  37.81  11.00 
21Mar  23  43.23  13.00 
22Mar  24  49.26  14.00 
23Mar  25  55.95  36.00 
24Mar  26  63.34  53.00 
25Mar  27  71.47  50.00 
26Mar  28  80.39  78.00 
27Mar  29  90.12  85.00 
28Mar  30  100.70  83.00 
29Mar  31  112.16  63.00 
30Mar  32  124.52  75.00 
31Mar  33  137.78  58.00 
1Apr  34  151.96  61.00 
2Apr  35  167.05  89.00 
3Apr  36  183.05  71.00 
4Apr  37  199.92  82.00 
5Apr  38  217.63  89.00 
Below is the cumulative normal distribution graph:
We are now ready to produce our own worst case scenario.
The data for the graph below (up to Day 39 = 6 April) is two tables above the table with these words above it:
Note: SQRT(2) ~ 1.414213562
We consider two daily factors a = 1.4 and a = SQRT(2).
These factors correspond to five day (cycle) factors of
Re = 5.37824 and
Ro = 5.656854249
We obtain the graph below:
Note:
Day 1 is 28 February (first NZ case).
Day 27 is 25 March (Lockdown Level 4).
Day 35 is 2 April.
Day 65 is 2 May.
Below is the Daily Graph showing the estimated daily number of new cases.
Note that the mean (midpoint) differs by one halfday.
Could New Zealand have coped with 3350 to 4250 new cases daily around Day 39 (6 April)?
Even our previous normal curve had over 600 new cases for several days near the peak.
The actual number of cases was 89 the previous day.
Compare this with the actual number of cases on the graph (maximum 89 cases).
The worst case scenarios estimated 23,427.11 to 24,721.55 cases with over 23,000 cases in two worst case scenarios by 16 April (Day 49). Recall that one report commissioned by the NZ Ministry of Health estimated more deaths than this (27,600 deaths) by early July.
The graph below shows the number of daily cases for the above worst case scenarios.
Note: Peak for blue curve is not shown (the halfday value is not plotted at the midpoint). We use the values plotted (3347). The actual value for the blue curve at the midpoint (39.5 days) is 3960.
We will use the midpoint new case values (3,350 – 4,250 cases) as an estimate for the total number of deaths. See table below.
The midpoints are at Day 39 and Day 39.5 (near Day 40) in the graph above.
Recall that for actual case values we calculated the mean (midpoint) at 35 days and a standard deviation of 9.155 days in Excel.
So long as the the Case Fatality Risk/ Ratio (CFR), the number of cases per million (CpM), and deaths per million (DpM) are realistic, the total number of deaths can be up to at least three times the maximum daily number of new cases in a worst case scenario. See the USA graph below.
We consider if our daily maximum number of new cases may be acceptable to be an estimate for the number of deaths.
Look at the table below comparing NZ worst case scenarios with statistics for Belgium.
We will assume the curves are symmetrical about the midpoints and adopt the original totals of 23,400 and 24,700.
Our own worst case scenarios estimated 23,427.11 to 24,721.55 cases with over 23,000 cases in two worst case scenarios by 16 April (Day 49). Recall that one report in the last Chapter estimated more deaths than this (27,600 deaths) by early July.
We compare our worst case scenarios with the actual numbers for Belgium
(14 June: Belgium population 11.6 million; NZ population 5 million):
NZ  CpM  4685.42  4944.31  
Belgium  CpM  5181.00  5181.00  
NZ  Daily max Cases/ Estimated Deaths  3346.73  4241.55  CFR 
NZ  DpM  669.35  848.31  0.171572875 
Belgium  Dpm  833.00  833.00  0.160779772 
We consider the CpM and DpM acceptable in our scenario.
The worst case NZ CFR (last column) is only a little higher than the Belgium CFR (see last column) so we deem it acceptable.
We note that the NZ C FR estimate is over 11.7 times the actual CFR (0.01462766).
We accept a worst case scenario of 24,700 cases and 4250 (4241.55) as an estimate for the total number of deaths.
Below is a graph of daily case number for Belgium:
Source:
https://www.worldometers.info/coronavirus/country/belgium/
The USA graph below demonstrates that the number of cases can take many times the number of days it took to reach the peak to go down to zero.
We now consider a slow drop after the midpoint. i.e. the graph is not symmetrical.
This will increase the number of cases.
The limit to the number of cases depends on what may be acceptable for the number of cases per million (CpM).
We will look at how large the increase may be later.
For now, read the discussion below but consider at this stage that the limit for any increase may be small. i.e. We may already be close to the limit already.
The values in the above table in the symmetrical case may be very close to the limit for a realistic number of cases (the CpM in the table may not allow for much of an increase).
The discussion below may only have theoretical value as far as case numbers in New Zealand are concerned and little if any practical application for case numbers in New Zealand.
We leave the discussion here since it may have a practical application outside New Zealand even if not for New Zealand.
This method may be useful for USA cases.
In round figure let’s suppose the peak is at Day 40 (actually Day 39 and Day 39.5) and it takes twice as long (80 days) for the number of cases to go down to zero.
Let’s consider the downward slope a triangle (on average).
On Day 40 the lower value has 3350 cases, with a total of about 15,000 (15,060) cases up to Day 40.
We first look at a rectangle 3350 cases high an 80 (2 x 40) days long.
For the rectangle 3350 x 80 = 268,000 cases.
The triangle has half this number, 134,000 cases.
Adding on the 15,000 cases in the first half (up to Day 40) we get a total of 149,000 cases in this scenario.
This shows that 150,000 cases (instead of 23,400 if the curve is symmetrical) is quite possible 120 (=40+80) Days from 28 February (27 June) since we have only considered the scenario with the lower number of cases.
In reality the curve may take far more than double the time to come down.
In a worst case scenario, a figure of 150,000 cases may be conservative.
Instead of 23,400 to 24,700 cases, 150,000 cases or more may be likely.
It is difficult to work out the midpoint even when the outbreak appears over.
New Zealand’s CFR is
22/1504 = 0.01462766.
Using the 24,721 figure above, we may have expected
24,721 x 0.01462766 = 362 deaths, 340 more deaths or over 16 times our present number (22).
With 150,000 cases, this becomes
150,000 x 0.01462766 = 2,194 deaths, over 6 times as many.
Even 362 deaths is 72 deaths per million of population.
In a worst case scenario, New Zealand could easily have over 2,200 deaths.
Our daily case estimate range at the midpoint (3350 to 4250) in a worst case scenario could translate to a (total) death estimate (3350 to 4250) if the curves are not symmetrical. i.e. As with the USA, the curve takes much more than twice as long to come down to zero.
In the USA graph below, the maximum number of daily cases is 39,072 on April 24, yet the total number of deaths is three times this figure on 14 June (117,527).
For lower values for NZ consider again our discussion above using triangular shape for after the midpoint.
Note that how far we can take this discussion depends on what is considered acceptable for the CpM.
If it takes five times longer to come down, the death estimate in New Zealand is almost 4,150 using 3,350 cases and clearly more if we use 4,250.
Fortunately we only have a total of 22 deaths (4.4 per million), 1504 cases (none active), and no new cases for over 20 days.
In our worst case scenarios, New Zealand could have had 3350 to 4250 new cases daily around the midpoint (Day 39 = 6 April). See the graphs above.
Even the lower figure (3350) near the daily peak is more than double our total number of cases (1504).
Would it have been easy for New Zealand to cope with near this many new cases on each day for a number of days?
The actual number of cases in New Zealand was 89 the previous day.
Fortunately the graphs rise and fall sharply but a sharp fall may not happen in reality.
Consider the slow decline in cases in the USA:
Source: https://www.worldometers.info/coronavirus/country/us/
For using linear regression to create a straight line for the top of a triangle, see:
https://aaamazingphoenix.wordpress.com/2020/05/09/covid19usacouldthetotalnumberofconfirmedcasesexceed3million/
Once the line is created, all we need to do is change the constant (c) so that it matches the actual value (add on the difference to c in the equation y=mx+c).
We can then find the xaxis intercept to estimate the number of days it takes to get down to zero cases.
Finally we can use the rectangle/triangle method above to estimate the total number of cases.
You may like to try this out using USA figures.
Chapter 5: Results
Both Scenario A and Scenario B generate the same estimates for the daily total number of cases.
We are confident in our models since the estimates (exactly the same in both scenarios) closely fit the actual data.
Our approach has been mathematical.
Scenario A and Scenario B both give the same estimated values for the total number of cases.
In Chapter 2 and Chapter 3, we used Excel to simulate an outbreak. We generate over 16,100 cases for each simulation, and 40 simulations for each dataset, calculating the mean number of cases infected by each person in each simulation and dataset. We prefer to use P=5/7 [~0.7143] since 7/5 = 1.4. We will also consider P=1/SQRT(2) [~0.7071].
We found that Re>3.2 (usually a mean over 3.2 was achieved with P=0.5; and mean over 4 with P=0.6). Re>5.5 was usually achieved with P=0.7 or less, where each day for 10 days the infectivity for each case reduces by the same (scaled) factor close to P. Recall that P is scaled so that the sum of the daily factors sum to 1. Recall also that where a mean is mentioned, it is an average of 40 sample means.
We note that with P=0.5, on the sixth day the probability of infection is only 0.01564, halving each successive day, so essentially a twoweek cycle is in reality just practically one week.
In our discussions, the reader will need to decide where it may be appropriate to replace Ro by Re and vice versa.
We started with estimates for Re of 2.9 in Scenario B and 5.3782 in Scenario A. Even Re = 2.9 is greater than most experts expect.
The spread of COVID19 in New Zealand is worse than in the above animation (which uses Re = 2.6).
Both estimates for Re and Ro are within the values in the above graph in the article by Sanche et al.
By contacttracing we hope to reduce the spread to a onecycle situation.
i.e. With an infectious period two cycles long, we anticipate/ hope that contact tracing will have identified and isolated potentially infected people before the second cycle starts.
We conclude that Re = 5.3782 is possible.
In the abstract for the article by Sanche et al. above we see that “Assuming a serial interval of 6–9 days, we calculated a median R0 value of 5.7 (95% CI 3.8–8.9).”
We conclude that Ro = 5.65685 is also possible in Scenario A. So is a range of values for Ro from 5.67 to 6.27. in Scenario B.
We note that an infected person may become infectious two days before symptoms appear.
With a daily factor of 1.4, the values almost double every two days (1.4^2 = 1.96). This can significantly change estimates for the total number of cases even if the midpoint date estimate is only one or two days later.
Our daily case estimate range at the midpoint (3350 to 4250) in a worst case scenario could translate to a (total) death estimate (3350 to 4250) if the curves are symmetrical or not. i.e. As with the USA, the curve takes more than four times as long to come down to zero.
The number of cases at the midpoint could be less than the total number of deaths in a worst case scenario.
In the previous section we saw that in a worst case scenario 24,700 cases were possible in New Zealand with 4,250 deaths.
It is very easy to get it wrong when looking at live updates on each day as the outbreak progresses.
However near the midpoint, we can add two consecutive actual days’ case numbers together or double the case number for single days to get an estimate of the total number of cases.
We may need to add on 10% to allow for a very long tail.
We may have ended up with two or four times the actual total number of cases.
It is likely there could have been a total of 3,000 to 6,000 cases even with Lockdown Level 4 starting at 11.59pm on 25 March. New cases could easily have been infected at least five to eight days earlier (or more with a cycle length greater than five Days or an infection period two cycles long).
New Zealand should consider itself very lucky.
On June 11, New Zealand had not had any new cases for 20 days and had no active cases.
Sadly “Six rest homes had Covid19 cases, two of which — Rosewood Rest Home in Christchurch and CHT St Margaret’s in Auckland’s Te Atatu — account for 16 of New Zealand’s 22 Covidlinked deaths.”
Source:
New Zealand Herald 12 June pA2:
“Delays allowed Covid19 to spread in rest homes, review finds”
New Zealand had 16 significant clusters including the six rest homes above.
It is easy to see that from these clusters the number of deaths in New Zealand could have been double the actual number (22) of deaths from COVID19 without any extra cases.
New Zealand has 4 deaths (4.4) per one million of population and is ranked #132 (from the top for deaths per one million of population) of the 215 countries etc listed on this website: https://www.worldometers.info/coronavirus/
On 12 June, about 36% have 3 deaths or less per one million of population, 60% have 7 or more, 40% have 15 or more, and 30% have 28 or more.
We leave it to epidemiologists to consider which model best suits the spread of COVID19.
Chapter 3 shows how even expert predictions may be extremely unlikely (impossible?) in reality. Could New Zealand ever have had 27,600 deaths by early July?
It is easy to have 20/20 hindsight vision.
The spreadsheet has demonstrated that Re>5.3 is also possible in a two cycle scenario. Consider also the maximum and minimum mean values below. For Ro we may need to add three standard deviations to the mean.
For P=1/SQRT(2) we have, for example:
Cases:  59  83  115  257 
Day  Day  Day  Day  
10  11  12  1012  
Mean=  5.67161  5.659639  5.732609  5.695039 
Std Dev=  0.280813  0.219948  0.202852  0.110989 
Max=  6.186441  6.084337  6.104348  5.996109 
Min=  4.915254  5.120482  5.33913  5.470817 
P=  0.707107  0.707107  0.707107  0.707107 
Each column uses 40 sample means. Each of the sample means are the mean of the number of new cases (from the number of cases at the top of each column).
For P=1/1.4:
Cases:  59  83  115  257 
Day  Day  Day  Day 1012  
10  11  12  99  
Mean=  5.722034  5.800602  5.820652  5.791537 
Std Dev=  0.340042  0.260213  0.21057  0.123064 
Max=  6.457627  6.445783  6.304348  6.062257 
Min=  5.016949  5.204819  5.417391  5.509728 
P=  0.714286  0.714286  0.714286  0.714286 
We note that for P=1/2 we have an infectivity “halflife” of one day and for P=1/SQRT(2) the halflife is two days.
Below is the rest of the table above (using P=1/1.4):
Sample#  Mean  Mean  Mean  Mean 
1  5.694915  6.228916  5.913043  5.964981 
2  5.355932  5.349398  5.704348  5.509728 
3  5.745763  5.939759  5.878261  5.867704 
4  5.864407  5.891566  5.730435  5.81323 
5  6.389831  5.759036  5.495652  5.785992 
6  5.830508  5.614458  5.852174  5.770428 
7  5.694915  5.86747  5.791304  5.793774 
8  5.491525  5.759036  6.304348  5.941634 
9  5.389831  5.578313  5.713043  5.595331 
10  5.440678  6  6.104348  5.918288 
11  5.542373  5.879518  5.913043  5.817121 
12  6.084746  5.566265  5.913043  5.840467 
13  5.745763  5.891566  5.930435  5.875486 
14  6.135593  5.783133  5.826087  5.883268 
15  5.677966  6.445783  5.982609  6.062257 
16  5.525424  5.373494  6.104348  5.735409 
17  6.016949  5.879518  5.730435  5.844358 
18  5.847458  5.204819  5.721739  5.583658 
19  6.457627  5.86747  5.556522  5.863813 
20  5.813559  5.855422  5.747826  5.797665 
21  5.101695  6.228916  5.53913  5.661479 
22  5.864407  5.819277  5.86087  5.848249 
23  5.915254  5.506024  6  5.821012 
24  5.254237  6.060241  5.721739  5.723735 
25  6.016949  5.927711  5.713043  5.85214 
26  5.644068  6  6.113043  5.968872 
27  5.813559  5.60241  5.426087  5.571984 
28  5.559322  5.759036  5.93913  5.793774 
29  5.050847  5.951807  5.869565  5.708171 
30  5.474576  5.710843  5.713043  5.657588 
31  6.186441  5.73494  5.695652  5.821012 
32  5.711864  5.578313  6.017391  5.805447 
33  6.101695  5.686747  5.417391  5.661479 
34  6.033898  5.457831  5.426087  5.575875 
35  5.254237  6.204819  5.817391  5.81323 
36  5.542373  6.048193  5.913043  5.871595 
37  5.864407  5.638554  5.556522  5.653696 
38  5.016949  5.975904  6.017391  5.774319 
39  6.033898  5.46988  6.13913  5.898833 
40  5.694915  5.927711  6.017391  5.914397 
Below are the scale factors for Scenario B (10day; two cycle infection) for various values of P:
Scales:  1/SQRT(2)  7/10  5/7  3/4 
1  0.302341  0.308721  0.295946  0.264918 
2  0.213788  0.216104  0.21139  0.198689 
3  0.151171  0.151273  0.150993  0.149017 
4  0.106894  0.105891  0.107852  0.111762 
5  0.075585  0.074124  0.077037  0.083822 
6  0.053447  0.051887  0.055026  0.062866 
7  0.037793  0.036321  0.039305  0.04715 
8  0.026723  0.025424  0.028075  0.035362 
9  0.018896  0.017797  0.020053  0.026522 
10  0.013362  0.012458  0.014324  0.019891 
Sum  1  1  1  1 
Note that the sum of the weights is equal to 1.
Below is the graph of the scale factors.
If the Scale factor (P) is chosen so that it is a multiple of the inverse of the Daily Factor (F), then the scale factor for each day is constant (using a daily Factor of F=1.4, we can choose P = 1/1.4 and then scale it so that as above the weights sum to 1).
In the table below, the Scaled column is the value for Day 1 in the table above the graph.
The other 9 days end up with the same value.
We multiply this by 10, then as in Scenario B (Chapter 2), we multiply the result by the square of the Factor to get an estimate of the value of Ro (see last row in the table below).
Factor  1.428571  1.4  1.414214  
Factor  10/7  7/5  SQRT(2)  
Factor  Squared  100/49  49/25  2 
Factor  Inverse  7/10  5/7  SQRT(2)/2 
Factor  Inverse  0.7  0.714286  0.707107 
Factor  Scaled  0.308721  0.295946  0.302341 
Scaled x 10 x  Factor Squared  6.30042  5.800534  6.046828 
Let x = Factor
e.g. x = SQRT(2)
Then
FactorScaled = x^9 (x – 1)/(x^10 – 1)
and
Ro = 10x^2 * x^9 (x – 1)/(x^10 – 1) or Ro = 10x^2 * [FactorScaled] or
Ro = 10x^11 * (x – 1)/(x^10 – 1)
For the second row (1.4), we see that 5.8005 is close the mean values in the previous table:
Cases:  59  83  115  257 
Day  Day  Day  Day  
10  11  12  1012  
Mean=  5.722034  5.800602  5.820652  5.791537 
For SQRT(2), P = 1/SQRT(2), a run produced:
Cases:  66  94  132  292 
Day  Day  Day  Day 1012 

10  11  12  99  
Mean=  6.083333  6.000798  6.032765  6.033904 
Std Dev=  0.306238  0.296  0.182605  0.095343 
Max=  6.757576  6.606383  6.333333  6.246575 
Min=  5.606061  5.457447  5.598485  5.859589 
P=  0.707107  0.707107  0.707107  0.707107 
Since we have generated distributions rather than sampling from a distribution it is doubtful whether the Central Limit Theorem applies.
Regardless all we wished to do was show that Re around 5.4 or greater was possible in a worst case two cycle infection timeframe. The values in the tables above achieve this.
We conclude that Re = 5.3782 or Re (or Ro) = 5.65685 is possible in a one or two cycle infection timeframe.
We also see that values for Ro over 6 are possible and are in fact likely since the factor 1.4 reflects the actual increase in New Zealand where cases are isolated once they have been identified.
We find it interesting in the bottom row, that the Daily Factor = SQRT(2), with an estimate for Ro of 6.0468, means that the number of infected people doubles every two days.
We have used an incubation period of only two days (we miss out a day) for most of our modelling. This created a good fit for the actual data.
In reality we assume the incubation period is one cycle long. This means the value of Ro is likely to be greater than we have estimated.
We have seen that R0=6.3 or Ro=6.4 is also possible.
Our values for Re and Ro are also twice as big as most experts consider for COVID19.
This would appear to make Sweden’s attempt at “herd immunity” doomed to failure. A
new study, published in the Emerging Infectious Diseases journal, shifts the R0 for COVID19 from about 2.2 to about 5.7. With the lower number, only 55% of a population needs to be immune from COVID19 to stop its spread through herd immunity. Herd immunity refers to enough of a population being immune to a disease that the disease cannot travel through it. …
But if more people get infected from a single person with COVID19, then more people need to be protected from the disease to stop it from continuing to spread. With an R0 of 5.7, approximately 82% of the population needs to be immune to reach herd immunity and stop the disease from spreading easily through the population, the researchers concluded.
See:
https://www.forbes.com/sites/tarahaelle/2020/04/07/thecovid19coronavirusdiseasemaybetwiceascontagiousaswethought/#103db97329a6
In January this year the World Health Organization (WHO) estimated the R number of the COVID19 virus to be between 1.4 and 2.5, whereas researchers in China, Germany and Sweden estimated it to be at least as transmissible as SARS, and calculated it to be 3.28. In China, other researchers have found the R number to be much higher, putting it at 4.7 – 6.6.
See:
https://www.gavi.org/vaccineswork/whatcovid19srnumberandwhydoesitmatter
At some point early in the outbreak, some cases generated humantohuman transmission chains that seeded the subsequent community outbreak prior to the implementation of the comprehensive control measures that were rolled out in Wuhan. The dynamics likely approximated mass action and radiated from Wuhan to other parts of Hubei province and China, which explains a relatively high R0 of 22.5.
See:
https://www.who.int/docs/defaultsource/coronaviruse/whochinajointmissiononcovid19finalreport.pdf
Chapter 6: Conclusions
[We will be consulting with colleagues soon and these conclusions may be updated]
We have used an incubation period of only two days in most of our modelling. This created a good fit for the actual data.
The incubation period was created by missing out one day before having transmission occur. Since transmission may occur in reality two days before an infected person becomes symptomatic, this may simulate an incubation period of two extra days.
In reality we assume the incubation period is one cycle long. This means the value of Ro is likely to be greater than we have estimated.
Our Excel simulations use mean values. For better estimates for Re and Ro, perhaps we need to add two or three standard deviations to the means.
We decide to stay with the mean values since these reflect our calculated values.
We are not sure whether our Excel simulations may be called a stochastic process.
We hypothesised that Re = 5.3782 or Re (or Ro) = 5.65685 is possible in a one or two cycle infection timeframe.
We conclude that Re is at least this amount.
Let r be the daily infection rate.
A person becomes less infectious over successive days. A person is most infectious on the first day they can transmit COVID19.
For any value of r, we can use c/r (chose an appropriate value for the constant c) for the reduction in infectivity (is this the right word?) over successive days.
For r = 1.4, we can choose c/1.4 and for r = SQRT(2), choose c/SQRT(2). This means the daily rate and the reduction in infectivity create a constant value once the incubation period is over.
The reduction in infectivity is inversely proportional to the daily rate in our Excel simulations.
We saw that r = 1.4 fits New Zealand data well until after Lockdown level 4.
We also saw that r = 1.4 leads to Re just over 5.8 when the infectious period (the time when a person can transmit COVID19) is two cycles long. In reality the infectious period is at least this long.
We adopt Re = 5.8.
Since 1.4 x 1.4 = 1.96, we see that the number of cases almost doubles every two days with r=1.4.
Since SQRT(2) x SQRT(2) = 2, we consider r = SQRT(2) [ = 1.414214] a suitable value for r when estimating Ro. This means that the number of cases will double every two days.
With r = 1.414214 [=SQRT(2)], we obtain a value for Ro just over 6.
Clearly since in practice we have r = 1.4 (with isolation occurring once new cases are identified), for Ro we require r > 1.4.
Hence we consider r = 1.414214 [=SQRT(2)] a suitable value for r when estimating Ro, and we conclude that Ro is at least 6.
We adopt Ro = 6 in New Zealand and hence worldwide. Once we add on three standard deviations for all of Days 1012 for r=SQRT(2), we commonly get close to or over 6.4.
All the values for Re and Ro are more than twice as big (2.3 times for Ro = 6) as the value used in the animation (2.6).
Here is the animation again:
Source: New York Times.
How can a Coronavirus outspread from 5 to 368 people in 5 Cycles (Credit: The New York Times)?
If 5 people with new coronavirus can impact 2.6 others each, then 5 people could be sick after 1 Cycle, 18 people after 2 Cycles, 52 people after 3 Cycles and so on. See:
https://towardsdatascience.com/howbadwillthecoronavirusoutbreakgetpredictingtheoutbreakfiguresf0b8e8b61991
When we looked at calculations for this outspread, we used a 6day cycle. See:
https://aaamazingphoenix.wordpress.com/2020/05/18/covid19nzcanonepersoninfect26others/
In our New Zealand simulations, we start with 10 cases and use a 5day cycle length. With Ro = 6, in three cycles (15 days) we have 10, 60, and then 360 cases.
We conclude that in New Zealand up until Lockdown Level 4 (25 March; Day 27), one person may infect on average at least 6 other people if there is no isolation (Ro = 6) and at least 5.8 other people if cases are isolated (Re = 5.8) once they are identified.
Most researchers believe that values are less than half of the above figures. Consequently their models may need to be adjusted once our results are independently verified.
We note that normally once a Lockdown is introduced, we may not expect a fall in the Case number rate (Re) for at least once cycle since new cases could easily have been infected at least five to eight days earlier. In New Zealand there is a much quicker change.
Perhaps prior isolation was already having an effect?
If the daily rate of increase (r) for case numbers matches the rate of decay of infectivity, then for each day during the infectious period (two cycles = 10 days), the daily contribution towards R is the same.
To calculate R, we can simply multiply by ten the value for the first day of the infectious period.
We add in a factor of x^2 for incubation and realise that
1 + x + x^2 + … + x^10 = (x^10 – 1)/ (x – 1)
For Day i (i=1,2, …, 10) and x = r (the daily rate of increase), we have a daily factor of
r’ = x^(i+1) * x^(10i) * (x – 1)/(x^10 – 1)
= x^11 * (x – 1)/(x^10 – 1)
i.e. The same value for each of the ten days.
Let f(x) = 10x^11 * (x – 1)/(x^10 – 1) then
Ro = f(SQRT(2))
and for the ten days on and before 26 March (the first full day of Lockdown L4)
Re = f(1.4)
i.e. Ro ~ 6 and Re ~ 5.8.
Our values for Re and Ro are twice as big as most experts consider for COVID19.
We also looked at worstcase scenarios.
We looked at a normal curve which has a total of 27,600 cases (not deaths!).
The curve stays above the actual number of cases which means the cumulative number of cases in the normal curve is also above the actual cumulative number of cases.
We also considered another worst case scenario of 24,700 cases:
While the above number of cases may be possible in a worstcase scenario, 27,600 deaths (as estimated in a report commissioned by the NZ Ministry of Health) would not appear to be possible by July in New Zealand in a worstcase scenario.
We now use our simple model to verify our calculation Re = 5.8.
We modify the model so that we have a 2cycle infection period.
For i>2 using C[2] = 8 and C[1] = 5, for given constant values of LAMBDA and Re, we can calculate
C[i] = Re * ( LAMBDA * C[i1] + (1 – LAMBDA) * C[i2] )
Where C[i] is the estimated number of new cases in cycle i.
The number of new cases follows the sequence [5], 8, 44, 231.
We note that LAMBDA = 1 gives 231 = Re * 44, Re = 5.25, and r ~ 1.393259.
We need to find LAMBA so that
231 = 5.8 * (LAMBDA * 44 + (1 – LAMBDA) * 8)
i.e.
LAMBDA = ( 231/5.8 – 8 )/36.
However the value of LAMBDA obtained gave a value near 233 instead of 231, so we recalculated LAMBDA using 229.5 instead of 231.
We obtained LAMBDA = 0.87691571 and rounded this value to use
LAMBDA = 0.877 and (1 – LAMBDA) = 0.123 to generate the Table below:
This table and the table below suggest a total of 7800 to 7900 cases may have been possible by 5 April. New Zealand currently (29 July) has 22 deaths from 1557 cases with a CFR of about 1.4%. This suggests 112 deaths may have been possible by 5 April.
Each estimate for the number of new cases is obtained by multiplying the previous number of cases (previous cycle) by LAMBDA and the number of cases in the cycle before by (1LAMBDA).
The value 5 (highlighted in red) is only used to start the sequence. Hence we obtain
44.26 = 5.8 * (LAMBDA*8 + (1 – LAMBDA)*5) and
230.84 = 5.8 * (LAMBDA*44.26 + (1 – LAMBDA)*8).
The values obtained are very close to the actual case numbers.
We earlier used Goal Seek in Excel to get LAMBDA = 0.876. The above value gave better results.
We see the 5day value (0.877) above and the cumulative 5day value (0.8432) for the 2cycle CSAW Excel scenario are reasonably close (r~1.4) considering the first calculation has two levels and the second has ten levels with over half the infections below occurring in the first two days and over threequarters in the first four days:
1  0.2959 
2  0.5073 
3  0.6583 
4  0.7662 
5  0.8432 
6  0.8982 
7  0.9375 
8  0.9656 
9  0.9857 
10  1 
A better approach is to estimate LAMBDA to fit the values [5], 8, 44, 231.
We start with Re = 5.8 (the cell highlighted in gold below) and use Goal seek in Excel to set the number of new cases for 26 March to 231 by changing Re. We obtain Re ~ 5.7115:
The values match.
We also calculate Re in the CSAW model using the formula
Re = 10 * r^11 * (r1)/(r^10 – 1)
Using values of r in the table below:
For r = SQRT(2) [~1.4142] we obtain
Ro ~ 6.0468.
We therefore estimate Re up to 26 March to be in the range 5.7 to 5.9 and estimate Ro = 6.
By 31 March the above tables estimate totals of 1489 and 1497 cases.
This number of cases was not achieved until the beginning of May.
Let C[i] be the number of new cases in cycle i.
Let R[i] = C[i]/C[i1] and r[i] = R[i]^(1/5]
Then R[i] is the ratio of the number of new cases in cycle i divided by the number of new cases in the previous cycle, and r[i] is the daily rate.
In April for Re = 5.8, R[i] becomes 5.223183, and r[i] becomes 1.391833.
In April, for Re = 5.7115386, R[i] eventually is 5.254786 and r[i] is 1.393513.
The above are the limits.
We consider the above close to our estimated r = 1.4.
These results help validate our calculated Re = 5.8 for the CSAW model.
We conclude that in New Zealand up to 26 March, Re = 5.8 and Ro = 6.
Our values for Ro are likely to be conservative. The CSAW modelling uses mean values. Isolation of identified cases in New Zealand is likely to underestimate calculations for Ro.
Update:
For an updated value for Ro and worldwide values for Ro, see:
COVID Odyssey Ro update: Occam’s Razor~ A close shave: In NZ is Ro ~ 3, 4, or 6? What are Ro values worldwide?
COVIDWorldAvNewRanked
COVIDWorldAvNewAlpha
COVIDWorldAvNewRanked4
COVIDWorldAvNewAlpha4
COVID Odyssey: Vir[tu]al World Tour ~ How many people can one person infect in your country?
My other COVID19 posts can be found here:
https://aaamazingphoenix.wordpress.com/tag/coronavirus/
Data for my posts can be found at:
https://www.worldometers.info/coronavirus/
https://en.wikipedia.org/wiki/2020_coronavirus_pandemic_in_New_Zealand
https://www.ecdc.europa.eu/en/publicationsdata/downloadtodaysdatageographicdistributioncovid19casesworldwide
https://ourworldindata.org/coronavirussourcedata
https://www.webmd.com/lung/news/20200507/100daysintocovid10wheredowestand
https://projects.fivethirtyeight.com/covidforecasts/?
https://ourworldindata.org/coronavirussourcedata
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https://guestdailyposts.wordpress.com/guestpingbacks/
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