Solutions Assignment 2 - Growth and Easy ModelWorks

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1 - FS 2015 Solutions Assignment 2 - Growth and Easy ModelWorks This assignment is used to learn using the general purpose modeling and simulation tool Easy ModelWorks. To this end we study the classical model logistic growth. Questions and answers 1) Unless preinstalled, please download and install Easy ModelWorks according to the instructions and always carry out all instructions immediately ( I hear and I forget. I see and I remember. I do and I understand. Confucius ). 2) Activate from the sample models the model logistic growth (choose menu command Modelling -> Sample MDPs 1 > Logistic Growth (cont.) ). It has a single state variable x1 2 that represents the population size x and the following conditions hold: x 0, a,b > 0. a) Examine the model behaviour for small values of the state variable x1 (Note: use Menu Command Modelling -> Edit model... to alter the initial condition of x1). Investigate the underlying differential equation: Why does the observed behaviour occur? What is the differential equation, which approximately describes such a growth? What do you call this kind of growth (Consult the Courseware «Drosophila» p. 2-5)? Which symbol is commonly used for the model parameter in this model? What is its unit and its ecological meaning? Continuous time logistic growth is an ordinary differential equation, i.e. a DESS (Differential Equation System Specification), of following form: x (t) = dx(t) / dt = a x(t) b x(t) 2 = x = a x b x 2 (1) The fact that the state variable x(t) is a function of time t, is often assumed implicitly: Instead of x(t) we write then just x and instead of x (t) just x. The shape of the curve of the solution is symmetrical-sigmoidal. The important model parameter a, often also called r, is the relative growth rate. Its unit is "per unit of time", in our case per day (d -1 ). It determines whether the population is growing (a > 0) or decaying (a < 0). However, normally a is considered to be positive only. In (1), b is a second model parameter (see also part 2c) To investigate small values of the state variables x, the initial values x 0 have to be reduced to a small value, say Use menu command Modelling -> Edit model... and enter the desired new value for x 0 (Figure 1). Figure 1: Dialogue to change the initial value x 0. of the population size, after having used the menu command Modelling -> Edit model... (population size in this simulation model: x10 = 0.001). The input field is located underneath x.(0), to the right of the differential equation of the state variable x (in this simulation model called x1). 1 MDP stands for Model Definition Program 2 In contrast to its larger sister ModelWorks, Easy ModelWorks has limited functionality and requires a specific naming of state variables: You may only use one of these 9 names x1, x2, x9. 1

2 Solutions Assignment 2 Growth and Easy ModelWorks The value x0 defines the initial state of a population. The smaller the size, the more time the population needs to reach the equilibrium. The reason being that the initial value does not change the actual shape of the curve. If you start the simulation with a smaller population, you have to allow for some time to elapse until the population will reach the value matching the former, larger initial value. Reducing the initial value results thus only in a temporal delay before stabilization (Figure 2 left). Figure 2: Simulation experiments with the model Logistic Growth (cont.) using different starting values x0 of the population size, i.e. x0 = 0.001, 0.5 and 1.0 (left) and simplified equations system, i.e. b = 0 (right, Menu command Set time... and tend = 10, b = 0 and 0.001). For small population sizes x << 1, the quadratic term in (1) 0, i.e. the logistic differential equation (1) reduces to an equation of exponential growth: x a x. This is true for the left parts of all curves in Figure 2 (left), i.e. for t < ~5. You can verify this by changing the equation and delete the self-inhibition term or by setting b = 0 (menu command Edit model... ) and shortening the simulation time to 10 d (menu command "Set time...", Figure 2, right). b) What is the role of the parameter a? How does the model behaviour change, if you increase the value of a from 0.7 to 1.2? Please sketch first the model behavior you expect before you make a simulation. Then make this parameter change by editing the model (choose menu command Modelling -> Edit model... ) and then solve the altered model again. Was your expectation correct? If not, don t worry, this is an excellent learning opportunity: try to understand what is going on. or? Figure 3: Two possible sketches for the prediction of the behaviour of the model by increasing the model parameter a from 0.7 to 1.2 (sub-question 2b, compare Figure 4). The parameter a is a relative growth rate and is to be interpreted as the net growth rate. It arises from the difference between the processes which make the population grow, i.e. natality and immigration, as well as the processes which cause the population to decrease, i.e. mortality and emigration (s. supplement to the assignment, i.e. Figure 1 in chapter 2 «Theory»). 2

3 Increasing the parameter a from 0.7 to actually resulted in the following curves (Figure 4): a b c Figure 4: The increase of the relative growth rate a from 0.7 to 1.2 d -1 results in a faster growth as well as an increased equilibrium position (cf. Figure 3). a: default axis scaling; b: exactly the same simulation results as in a, however, customized presentation by changing the axis scaling. It was made by use of the dialogue (c), increasing the maximum from 1000 to 1200 (input field below max). This dialogue is selected by the menu command Set monitoring... in the menu Simulation. Depending on the prediction you have made (Fig. 3, left or right), it may or may not match the simulation result shown in Fig. 4. Be assured, it doesn t matter if your prediction does not match the simulated results. This happens often and is quite normal. Having erred with such a prediction is even to be welcomed, since it is an excellent opportunity to learn. Try to understand why your prediction has been wrong and you can gain often many insights. The following will help you find an explanation for this perhaps quite surprising model behavior. c) Study empirically the model behaviour for values of the state variable near the value of the ratio a/b = 0.7/0.001 (Note: First discard all changes you have made by activating once more the same sample model and by answering No when asked to save changes). Explain again by analytical examination of the underlying differential equation why the observed behaviour occurs. What differential equation describes approximately this form of behaviour? Increasing the initial value x 0 of the state variable x(t) up to a value of 0.7/0.001 = 700 results in a horizontal straight line along the equilibrium position x(t ). The net growth of the population is obviously equal to 0, i.e. growth and decline compensate each other and the population size does not change (Figure 5). 3 Use the same dialog, you have used for the change of the initial values, menu command "Edit Model" (see Figure 1) 3

4 It holds: dx(t )/dt = a*x(t ) b*x(t ) 2 (1) = 0.7* *700 2 = = 0 This equilibrium condition can be interpreted to be a dynamic equilibrium such that for the population density x(t ) the growth term (a*x(t )) is equal to the density-dependent selfinhibition term (b*x(t ) 2 ). We can not say whether births and immigration (g) have decreased, whether death rate and emigration (m) have increased, or whether both changed in the aforementioned way with increasing population size, because the model describes merely the net effect of any or all of these changes. The differential equation, which describes the behaviour of this model, is: dx(t)/dt = g m = 0 where: g m with x 0 = x(t ) = 700 g: absolute rate of growth 4 m: absolute rate of mortality 5 If you create "artificially" exceptionally high population sizes, x(t) > x(t ), the self-inhibition term m will be higher than the growth term g. Such high values could for instance be the result of immigration into a population in the equilibrium position x(t ). In that case, the model will even describe a population decline. Please note that this will occur, although the growth rate a is positive (Figure 4, e.g. x 0 = 900)! Figure 5: If the initial value x 0 is set at the equilibrium value x(t ) = 700, you get a horizontal straight line. Other initial values x 0 are also shown here: x(t ) = 900, 800, 675, 300 and 1. 4 in contrast to relative growth rate r, a model parameter, which is a per capita growth rate 5 in contrast to relative mortality/emigration rate, which is a per capita mortality/emigration rate 4

5 d) Please derive from the original differential equation a new equivalent differential equation with only the following two model parameters: a and a/b. What is the name of the new parameter? = a/b? What is its unit and its ecological meaning? Please modify in Easy ModelWorks your MDP according to the obtained equation. Do you really get the same model behaviour? dx(t)/dt = a x b x2 (1) = a (1 b/a*x) x = a (1 1 a/b *x) x = a ( a/b a/b x a/b ) x = a ( a/b x a/b ) x dx(t)/dt = a ( K x K ) x = r (K x K ) x (2) = a*([k x(t)]/k)*x(t) = r*([k x(t)]/k)*x(t) (see «Theory») respectively implemented with a = r in Easy ModelWorks (Figure 6): Figure 6: The formula of the differential equation is to be entered in the field right to the text x1dot = (menu command Edit model... in the menu Modelling ). x1dot: derivative of the state variable x = x1 The new parameter K (= a/b) represents the carrying capacity of a population, resulting from dividing the growth coefficient a, respectively the growth rate r, through the self-inhibition coefficient b. It corresponds exactly to the equilibrium value x(t ). The greater the growth coefficient a or the smaller the self-inhibition coefficient b, the greater is K and the higher is the maximum population size (a/b = K = x(t )). The unit of K as carrying capacity must of course be the same as that of the population. Therefore, the unit equation is: a b = d 1 d 1 1 = Individuals * Individuals You can also see easily, why the equilibrium position x(t ) exists in the new form of the differential equation (2). If the population x = K, the numerator and thus the value of the whole equation will become zero (see sub-question 2c). e) Repeat the task b). What is now the role of the parameter a? Please sketch again the expected model 5

6 behavior before you run the model. What happens if you increase a from 0.7 to 1.2? Implement this change now and solve the new model. Was your expectation correct? Figure 7: Sketch of the prediction of the behaviour of the new model variant when increasing the model parameter a from 0.7 to 1.2 (sub-question 2e, see Figure 8). The sketch shows that it is assumed that because the carrying capacity K remains the same, the curve will run steeper, but that the equilibrium position does not change. Because K was formulated independently from a (= r) (see Figure 6). Increasing the model parameter a from 0.7 to 1.2 shows that a actually affects only the slope of the curve and not the equilibrium position anymore (Figure 8). The curve is steeper and the value of K is reached earlier. Please note that this is not just a termporal shift, but also the shape of the curve changes. Please note also that the maximum slope in the population size x(t) = K/2 is at a = 1.2 d -1 greater than at a = 0.7 d -1. Figure 8: By varying the growth rate, i.e. the model parameters a from 0.7 to 1.2 d -1, the slope of the population size becomes steeper. However, the carrying capacity K remains the same. (see Figure 7). 6