Reproductive number of polio.

Ro = 6

Generate sequence of numbers for fraction of population vaccinated from 0 to 1 with 0.1 interval.

fraction_vaccinated = seq (0, 1, 0.1)

Print fraction of population vaccinated.

cat ("Fraction of population vaccinated: ", fraction_vaccinated)
## Fraction of population vaccinated:  0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Compute effective reproductive number.

Re = Ro * (1 - fraction_vaccinated)

Print effective reproductive number.

cat ("Effective reproductive number:", Re)
## Effective reproductive number: 6 5.4 4.8 4.2 3.6 3 2.4 1.8 1.2 0.6 0

Compute herd immunity threshold.

herd_immunity_threshold = 1 - (1/Ro)

Print herd immunity threshold.

cat ("herd immunity threshold = ", herd_immunity_threshold)
## herd immunity threshold =  0.8333333

Plot fraction of population vaccinated (versus) effective reproductive number.

subtitle = paste ("Ro = ", Ro, ", herd immunity threshold = ", round (herd_immunity_threshold, digits = 4), "; Polio", sep="")
plot (fraction_vaccinated, Re, main = "Polio" , sub = subtitle, xlab = "Fraction of Population Vaccinated \n", ylab = "Effective Reproductive Number (Re)")

Results and Discussion

Reproductive number of polio is 6. The graph illustrates that as fraction of population vaccinated increases, effective reproductive number decreases. Herd immunity threshold is 83.33%; that is, at this level of vaccination coverage, effective reproductive number is 1 (Re = 1). When vaccination coverage is above 83.33%, effective reproductive number is less than 1 (Re < 1); thereby, polio will be eliminated at these higher levels of vaccination coverage.

Public Health Implications

Recommend polio vaccination at coverage levels of above 83.33%.