Max Projection

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Here we try to calculate the "Max Projection" using the following assumotions

1. The peak day is day 56

2. The gradient (of daily growth) at the peak is -0.0031

3. The value of daily growth at peak is exactly 1.000000000000000000000000

4. The gradient is fixed at peak and remains constant for the next 24 days

5. After 24 days, the gradient is halved

6. The number of cases on peak day is 511

const peakday = 56.0

const gradient = -0.0031

const dailygrowth_atpeak = 1.0

function growth_rate(d)

    if d <= peakday + 24.0

        return dailygrowth_atpeak + (d - peakday) * gradient

    else

        return growth_rate(peakday + 24.0) + (d - (peakday + 24)) * (gradient/2.0)

    end

end

The Max Projection on peakday is 511

The Max Projection on day d is the growth rate on that day times the Max Projection of the day before

function max_projection(d)

    if d <= peakday

        return 511.0

    end

    if d > peakday

        return growth_rate(d) * max_projection(d-1.0)

    end

end

Create a floating point array starting at 56 and ending at 105

days = Float64.(collect(56:105))

Applying the function max_projection to a vector of floating point values then rounding it

maxpro = round.(  max_projection.(days)  )

Finally we obtain our array of "day,max projection value"

[ (days[k],maxpro[k]) for k=1:105-56+1 ]