Optimal Virulence of Pathogen (homework)

This is related to the function we learned in Disease Ecology
!hw, p.10
!hw, p.11

Disease reproduction and virulence

$α$ -> Rate of death (virulence)
$v$ -> rate of recovery
rate of dissapearance = $α + v$ (ignoring natural death which is small)

average length of infection: $1 / (α + v)$ time units
$β$ -> number of other hosts infected per unit time
$R_{0}$ -> basic disease reproduction number

R_{0} (β) = \frac{β}{α + v}

Relation of virulence to transmission:

\alpha (\beta) = a \beta {#2} + b \beta + c

Where $a > 0$ and $b, c \geq 0$

Optimal rate of infection for reproduction

Combine the two functions:

R_0 (\beta) = {\beta \over (a \beta {#2} + b \beta + c) + v}

Take derivative (quotient rule):

R'_0(\beta) = {-a\beta^2+c+v \over a \beta {#2} + b \beta + c + v}

Optimum will be when derivative is zero. We can ignore the denominator and solve for $β$ to get

β_{o p t} = \pm \sqrt{\frac{c + v}{a}}

Virulence at optimal rate of infection:

α (β_{o p t}) = 2 c + v + b \sqrt{\frac{c + v}{a}}

Optimal virulence increases with treatment

Since increasing rate of recovery $v$ makes $β_{o p t}$ (and thus $α$ ) increase.

Probability of death with changing recovery rate

$p = α / (α + v)$ is the probability of death.
If $α$ assumes its optimal value with increasing recovery rate $v$ , the formula is: