Using the realistic life history parameters for sea turtles given below. Calculate R0 for the following scenarios:
- Increase survivorship of eggs/hatchlings to 1.0.
- Absolute increase of 0.1 in juvenile survivorship.
- Absolute increase of 0.1 in survivorship of breeders.
| Class | Ages (x) | Age-Specific Survivorship (gx) | Fecundity (mx) |
|---|---|---|---|
| Eggs/Hatchlings | <1 | 0.6747 | 0 |
| Juveniles | 1-15 | 0.7308 | 0 |
| Sub-Adults | 16-21 | 0.7425 | 0 |
| Novice Breeders | 22 | 0.8091 | 127 |
| Mature Breeders | 23-54 | 0.8091 | 80 |
Although this assignment won't be graded, I encourage you to come to class Wednesday with your answers. Below is the R code for the spotted owl example which you can use as a template.
## Age-specific survivorship
## Note total age classes = 20, i.e., from 0-19
g0 <- .26
g1 <- .94
gx <- c(g0, rep(g1,17), 0) # One less gx value than age classes!
## Function to convert gx to lx
gx2lx <- function(gx) {
lx <- vector()
lx[1] <- 1
for(i in 1:length(gx)) lx[i+1] <- lx[i]*gx[i]
lx
}
## Survivorship schedule
lx <- gx2lx(gx)
## Fecundity
m1 <- .07
m2 <- .21
m3 <- .68
mx <- c(0, m1, m2, rep(m3,17))
## Visualize lifetable
data.frame(lx,mx)
## Calculate R0
R0 <- sum(lx*mx)

