Background

There is a tradeoff faced in managing abalone harvest. The infant population must be protected since it represents future harvests. On the other hand, the harvest should be designed to be efficient with a yield to justify the effort. The following will use VOLUME to form binary decision rules to guide harvesting. If VOLUME is below a "cutoff" (i.e. a specified volume), that individual will not be harvested. If above, it will be harvested. Different rules are possible. Management needs to make a decision to implement 1 rule that meets the business goal.

Next Steps

The next steps will require consideration of the proportions of infants and adults harvested at different cutoffs. For this, similar "for-loops" will be used to compute the harvest proportions. These loops must use the same values for the constants min.v and delta and use the same statement "for(k in 1:10000)."Otherwise, the resulting infant and adult proportions cannot be directly compared and plotted as requested. Note the example code supplied below.

A series of volumes covering the range from minimum to maximum abalone volume will be used in a "for loop" to determine how the harvest proportions change as the "cutoff" changes. Code for doing this is provided below.

```

idxi <- mydata$TYPE == "I"

idxa <- mydata$TYPE == "ADULT"

max.v <- max(mydata$VOLUME)

min.v <- min(mydata$VOLUME)

delta <- (max.v - min.v)/10000

prop.infants <- numeric(10000)

prop.adults <- numeric(10000)

volume.value <- numeric(10000)

total.infants <- sum(idxi)

total.adults <- sum(idxa)

for (k in 1:10000) {

value <- min.v + k*delta

volume.value[k] <- value

prop.infants[k] <- sum(mydata$VOLUME[idxi] <= value)/total.infants

prop.adults[k] <-sum(mydata$VOLUME[idxa] <= value)/total.adults

}

```

(1)

Our first "rule" will be protection of all infants. We want to find a volume cutoff that protects all infants, but gives us the largest possible harvest of adults. We can achieve this by using the volume of the largest infant as our cutoff. You are given code below to identify the largest infant VOLUME and to return the proportion of adults harvested by using this cutoff. You will need to modify this latter code to return the proportion of infants harvested using this cutoff. Remember that we will harvest any individual with VOLUME greater than our cutoff.

#Code for largest infant volume (provided below)

(max_inf_vol <- max(mydata$VOLUME[mydata$TYPE == "I"]))# [1] 526.6383

#Code for proportion of adults harvested (provided below)

sum(mydata$VOLUME[mydata$TYPE == "ADULT"] > max_inf_vol) /

total.adults# [1] 0.2476573

(1a) #Add code to calculate the proportion of infants harvested:

Note, if we use the largest infant volume, we harvest approximately 24.8% of adults and 0%, as expected, of infants.

(2)

Our next approaches will look at what happens when we use the median infant and adult harvest VOLUMEs. Using the median VOLUMEs as our cutoffs will give us (roughly) 50% harvests. We need to identify the median volumes and calculate the resulting infant and adult harvest proportions for both.

(2a) #Add code to determine the median infant volume:

(2b) #Add code to calculate the proportion of infants harvested

sum(mydata$VOLUME[mydata$TYPE == "I"] > max_inf_vol) /

total.infants

(2c) # Add code to calculate the proportion of adults harvested

sum(mydata$VOLUME[mydata$TYPE == "ADULT"] > max_inf_vol) /

total.adults# [1] 0.2476573

Note, if we use the median infant volume as our cutoff, we harvest almost 50% of our infants and a little more than 93% of our adults.

(2d) #Add code to determine the median adult volume:

(2e) #Add code to calculate the proportion of infants harvested:

(2f) #Add code to calculate the proportion of adults harvested:

Note, if we use the median adult volume as our cutoff, we harvest almost 50% of adults and approximately 2.4% of infants.