Question

An epidemic reported by the British Communicable DiseaseSurveillance Center in the British Medical Journal (March 4, 1978, p. 587) took place in a boarding school with 763 residents. The statistics for the infected population are shown in the graph in Figure 5.25.

Assuming that the average duration of the infection is 2 days, use a numerical differential equation solver (see Appendix G) to try to reproduce the data. Take S(0) = 762, I(0) = 1, R (0) = 0 as initial conditions. Experiment with reasonable estimates for the average number of contacts per day by the infected students, who were confined to bed after the infection was detected. What value of this parameter seems to fit the curve best?

Short answer

As a result at\(a = 4\), we get a peek at\(t = 7\).

Step-by-step solution

Use given values.

Here S(0) =762,I(0)=1.

Since the total population is 763 residents. Then,

\(\begin{aligned}{c}{\bf{s(0) = }}\frac{{{\bf{S(0)}}}}{{{\bf{763}}}}\\{\bf{ = }}\frac{{{\bf{762}}}}{{{\bf{763}}}}\\{\bf{ = 0}}{\bf{.0086}}\\{\bf{i(0) = }}\frac{{{\bf{I(0)}}}}{{{\bf{763}}}}\\{\bf{ = }}\frac{{\bf{1}}}{{{\bf{763}}}}\\{\bf{ = 1}}{\bf{.310 \times 1}}{{\bf{0}}^{{\bf{ - 3}}}}\end{aligned}\)

We have the system.

\(\begin{aligned}{c}\frac{{{\bf{ds}}}}{{{\bf{dt}}}}{\bf{ = - asi}}\\\frac{{{\bf{di}}}}{{{\bf{dt}}}}{\bf{ = a}}\left( {{\bf{s - }}\frac{{\bf{k}}}{{\bf{a}}}} \right){\bf{i}}\end{aligned}\)

Find the values of k and a.

Since the average duration of the infection is 2 days then\({\bf{k = }}\frac{{\bf{7}}}{{\bf{2}}}{\bf{ = 3}}{\bf{.5}}\).

Apply the same procedure of values of up to 1 to 10 shows the peak at t=7. And plotting on the graph.

Plotting graph.

The peak at time t=7 at a=4.

Therefore, the result at\(a = 4\), we get a peak at\(t = 7\).