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Logistic Growth

Logistic Growth. Nick, Ben and Vinny. The Logistic growth model is the per capita rate of increase approaches zero as the carrying capacity is reached. Formula -- ( dN / dt )= rmaxN ((K-N)/k)

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Logistic Growth

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  1. Logistic Growth Nick, Ben and Vinny

  2. The Logistic growth model is the per capita rate of increase approaches zero as the carrying capacity is reached. • Formula -- (dN/dt)=rmaxN((K-N)/k) • To construct the logistic model start with the exponential growth model which is (dN/dt)=rmaxN and add an expression that reduces the per capita rate of increase as N increases.

  3. Variables • N = Population Size • T= Time • ∆ N=change in population size • ∆T= Change in time • ∆T=dT • ∆N=dN • Rmax= maximum per capita growth rate of population • K= Carrying capacity – maximum population size that a particular environment can sustain • Determined by limiting factors: energy, shelter, refuge from predators, nutrient availability, water, and suitable nesting sites

  4. Logistic Growth of a hypothetical population

  5. Logistic growth model contunied • Maximum sustainable population size is K, so then k-N is the number if additional individuals and environment can support which means K-N/K is the fraction of K that is available for population growth by multiplying the exponential rate of increase RmaxN by K-N/K we modify the change in population size as N incrases

  6. A fisheries biologist is maximizing her fishing yield by maintaining a population of lake trout at exactly 500 individuals. Predict the initial instantaneous population growth rate if the population is stocked with an additional 600 fish. Assume that r for the trout is 0.005 individuals/(individual*day). For a populations growing according to the logistic equation, we know that the maximum population growth rate occurs at K/2, so K must be 1000 fish for this population. If the population is stocked with an additional 600 fish, the total size will be 1100. From the logistic equation, the initial instantaneous growth rate will be: DN/dt = rN [1- (N/K)] = 0.005(1100)[1-(1100/1000)] = -0.55 fish / day

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