solutions assignment 1 - pesticide decomposition...solutions assignment 1 –pesticide decomposition...

Solutions Assignment 1 –Pesticide Decomposition Systems Ecology: Principles and Modelling (Fischlin & Lischke) - FS 2015

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Solutions Assignment 1 - Pesticide Decomposition Remember, the model used in this case study has been deliberately chosen as simple as possible. This allows for focus on the essence of this case study, i.e. a methodology to analyze systems that is generally applicable.

Questions and answers

1) Problem A small story around the decomposition of the pesticide parathion: Let us assume you are a farmer who grows fruits and berries. In the last two years you had great difficulties, since your red currants were attacked by aphids. … an agent recommends to apply pesticide formulation E605 (contains 50% parathion), an unspecific insecticide (20 kg/ha per application) to control those pests … you are not certain: "Should I use the pesticide or not?" or "Is the use of the pesticide acceptable?"

See the general epistemological basic situation (handouts Fig. I-5) and please put yourself in the position of a systems analyst. This implies that your current horticulture represents the real system, and the goal for you is to find solutions for above problem. This is best done with the help of a model. According to the basic epistemological situation (handouts Fig. I-5), you will need a verbal and a mathematical model.

2) Reflect on the issue and try to answers as much as possible right here and now (in the lecture hall)

Try first to make good use of existing knowledge. From the real system you have available several observations and measurements, mostly from the small pilot study (see assignment text and below). Furthermore, make also good use of your theoretical knowledge that may be relevant and useful for finding solutions (see below). Perhaps this knowledge may be all you need to develop a verbal model. A solid qualitative model design will facilitate the construction of the mathematical model.

A useful mathematical model may help us to find solutions by using only the model, rather than working with the real system itself (see handouts Fig. I-5: Construction, Simulation vs. Experiment/Manipulation, Measurements, Observation). Here we assume that we will be able to find sufficient analogies between the real system and the model, which allows us to proceed with this approach.

3) That is likely not to suffice. Therefore, please write down briefly different steps that allow you to solve the problem (Note: Consider using a simple mathematical model to estimate the temporal pattern of the pesticide decomposition).

Follow the generally applicable recipe of a systems analysis (hand-outs p. I-7 and below). Note, while this recipe and the sequence of steps is recommended to follow through as closely as possible, deviations are possible, in particular if we need to overcome unforeseen difficulties or obstacles. In other cases individual steps make little sense and can then be omitted. Moreover, iterative sequences in which you go back by several steps to repeat them, or glimpsing at the subsequent steps are often beneficial. In the recipe the numbers at the right of a particular step indicate the most frequent steps to which you may go back from the current step in such iterative repetitions (see also below).

4) Now, please perform these steps according to your plan. Is there any crucial information missing that would allow you to solve the problem? Please ask now, maybe I can be of help.

The plan we follow adheres fully to the recipe for a “systems analysis” (hand-outs p. I-7).

Solutions Assignment 1 –Pesticide Decomposition Systems Ecology: Principles and Modelling (Fischlin & Lischke)

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Write down the problem question: The first, crucial step is to formulate the problem question. This should always be done in writing. This step is often underestimated in its importance; therefore, give him sufficient attention, please!

In our example, the question in its initial form “"Should I use the pesticide or not?" remains somewhat vague and open. It is therefore difficult to answer directly. We need first to reformulate it and a precise problem question will result. E.g.: "Will the tolerance limit1 r*S of 0.05 mg/kg of Parathion residues in red currants be achieved within 20 days?" This question can be simplified further by a slight amendment, for example:

Is the time t*S in which the parathion residue r in the red currants reaches the value r*S less than 20 days?

Prepare, collect, and sift facts, experimental data, and connections: Parathion is a xenobiotic insecticide (Fig. M1-1: a). This means that in a agro-ecosystem parathion is not a metabolic product and is therefore always of anthropogenic origin. Therefore, once parathion parathion has been applied, it can only decompose, but can’t be formed anew. Our model has thus only to model the process of a pesticide decomposition or pesticide degradation.

Fig. M1-1: Degradation pathways of parathion in the body of a cow (after Korte, 1987)

When parathion is degraded or decomposing, the following chemical reactions play a key role (Fig. M1-1): In air, parathion is degraded rapidly (half-life ≈ 0.0035 d ≈ 5 min) by oxidation

1 According to the federal regulation for foreign substances and ingredients FIV (1997 update) the limit value and tolerance value fall together for Parathion (see Glossary).


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(photolysis of the aeorosol). In aqueous solution, such as in a plant tissue, it will degrade by hydrolysis, which is much slower. Data from the literature: Laboratory investigation: half-life ≈ 105-301 d (Jørgensen et al., 1979), Field investigations: Half-life ≈ 1-10 d (Howard, 1991). Transportation (downwash and run-off) and adsorption, particularly on soil particles, are other processes by which the content of parathion in the air and in organisms may decrease. Indicated half-life’s ≈ 16-240 d; e.g. after 16 a amounts in soils were found to have decreased to 0.1% of original concentrations (Jørgensen et al., 1979, Howard, 1991).

Finally, we have the following results from our pilot study available:

t

day (d)

measured parathion residue (mg/kg pulp)

0 1.2 4 0.9 7 .45

Note, these results represent accurately the given, possibly special circumstances in our real system.

Sketch a verbal/qualitative model by drawing a relational graph: The first step to develop a new model consists of the formulation of a so-called verbal model, which corresponds to a qualitative representation of a system. As for models, the following is generally valid:

A model represents a human-made construct or artifact, that was crafted after a real (or perhaps imagined) prototype (real system or example or model, as this in English sometimes confusingly called), which is built for simplification, illustration or idealization purposes.

There are three important model categories (handouts Fig. I-5): verbal models, mathematical models and simulation models. The role model of each model is the real system, i.e. a part of reality. The verbal or qualitative models are most common and in use by almost everybody. You have for sure already dealt with many of those. The concept is the following:

A verbal model is a model that defines a system solely with verbal descriptions of the structure, the components (system elements), and the qualitative properties of individual structural elements.

Note, the system concept is central. However, in this context we need, a somewhat more general and more precise understanding of the term system. So far we have regarded a system has been merely any object that has inputs and outputs.

A system is a set of objects that are linked to each other by a relation.

Anything can be a system component or system element, be its nature abstract, consist of matter or be it anything else. The only requirement is that these elements may be interrelated. Here we introduce the concept of a relation. A relation is understood as a predicate with two variables or also called a 2nd order predicate P(x,y), or short Pxy. In general a predicate is a statement, that is either true or false, depending on what you use for its variables . Most commonly there are first and second order predicates (for details see Czayka, 1974). An ecological example of a second order predicate P(x,y) may be one that defines a food web and could be defined as follows: P(x,y) ::= “Species x eats species y”


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Depending on what we insert for x and y the statement becomes true or false. Ex.: Statement “Lion eats zebra” is true, whereas “Zebra eats lion” is false.

Fig. M1-2: Relational graph, representing a qualitative model, to illustrate modeling. Given the fundamental set G of elements (all dots) a system border delineates the system core consisting of the set X of system components or system elements (dots within system border). A 2nd order predicate P(x,y) ::= “x influences y” (see text) defines dependencies among the system elements within the system border. An arrow or edge connects an ordered pair (x,y) of system elements. The set R of ordered pairs, i.e. the edges connecting system elements is also called the system structure. Depending on how system elements are connected, individual system components can then be understood as input, output or state variables. In the example shown there are only state variables, since all system components are knots of a closed path and no element from the environment outside the system border influences the system, therefore the shown system is called an autonomous system S characterized by the two sets X and R (S=(X,R)). Note, the choice of the system border may depend not only on the nature of the real system, but also on the precise problem question.

First, possible system elements are listed. What results is the fundamental set G of elements (Fig. M1-2 und M1-3a). The set X to be considered consists then of the following elements:

X = {e, s, n, p, o, l, R, r, d, I, g, m, T, w} (for meaning see M1-3a)

Then we draw the system structure in form of a relational graph, a directed graph, similar to the one shown in Fig. M1-2. You may first do this intuitively, later it is recommended to use a 2nd order predicate P(x,y). The 2nd order predicate P(x,y) :: = "Changes in y are caused by x" is a very useful and widely applicable predicate. Systematically applying P(x,y) for every possible ordered pair (x,y) where x and y are contained in set G yields the system structure, i.e. we add the ordered pair (x,y) to the structure set R whenever P(x,y) becomes true (see also Fig. M1-2).

In our case study the application of the insecticide e affects the quantity of parathion p in the agro-ecosystem (quantity of parathion outside of fruits), i.e. for the ordered pair (e,p) above predicate P(x,y) becomes true (L), i.e. P(e,p) = L (see also Fig. M1-3). Other factors that may influence p are: l - Wind transport; R - rainfall (eluviations); s - antagonists such as lady beetles; p - the concentration of parathion itself (reflexivity resulting from 1st order degradation kinetics); and T - the temperature T. This means it holds: P(l,p) = L, P(R,p) = L, P(s,p) = L, P(p,p) = L, P(T,p) = L. On the other hand we assumed that once parathion has entered a fruit, it does not exit that fruit, which means P(x,y) for (r,p) becomes false (0), i.e. P(r,p) = 0. This all defines a first part of the system structure R', a set which consists of the following six ordered pairs: R’ = {(e,p), (l,p), (R,p), (s,p), (p,p), (T,p)} (see


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Fig. M1-3a and b). We could add other edges to the graph by considering the fact that the rate of formation of the end products that result from the degradation of the reactant parathion, i.e. system elements o, d, i, and g, depend on the reactant’s concentration (see Fig. M1-1 and Fig. M1-3b). If one continues such considerations, a complete relational graph of the system results. The set of all edges contained in the relational graph form the final system structure, the set R of all ordered pairs given by the relational graph (see Fig. M1-3b).

In a next step we can classify the nodes (dots, system elements) in our system universe X. Let us first determine the state variables (Zustandsvariablen): To find them use the general rule that on any closed path in the relational graph at least one node (dot) has to serve as a state variable. Any nodes on a closed path that do not serve as state variables are called auxiliary variables (Hilfsvariablen). There are also open (non-closed) paths: Nodes in the feeding part of an open path are input variables (Eingangsgrössen), the elements at the end parts of open paths leading away from the system core are the output variables (Ausgangsgrössen).

In our example the elements e, l, R, T are input variables. The element m is an output variable. All other elements, i.e. p, o, i, g, d, s, n, r, and w (Fig. M1-3b) are state variables (because of reflexivity P(x,y) = L for all those elements, there can be no auxiliary variables).

In a next step we can determine the inner system boundary. It surrounds all state and all auxiliary variables, i.e. all system elements on some closed path. Note, it may well be that the relational graph contains some parts that are only loosely coupled to each other. If we subdivide a relational graph into several partial graphs, this might permit a separate modelling of each partial graph or sub-graph. All rules apply then for each sub-graph with the exception that some state or auxiliary variables serve as output and input variables for the sub-graphs they are connected with. We can also get a chain of sub-graphs that are not mutually interdependent. Working with sub-graphs or sub-systems may substantially simplify an investigation.

This is precisely the case in our case study, since we find a chain of sub-systems (Fig. M1-3b) formed from the following sets of elements: {p,s,n}, {d}, {o}, {i}, {g}, {w}, {r}. For example d is state variable of a subsystem Sd = ({d}, {(d, d)}), whose input variables are p and r. The rest of the system is then merely the environment of Sd, and elements p and r are output variables of Sd. Moreover, since Sd does not influence the remaining system, Sd can be examined independently of the remaining system. The influence of the latter can be treated in a greatly simplified manner, i.e. through p and r, for which we may merely need scenarios of their temporal behaviour. If we would be interested only in Sd, all would become very simple, we could neglect all the complex dynamics taking place in the remainder of the system. However, our problem question tells us clearly that this is now not the case.

Let us investigate the relational graph some more and perhaps we find a simple sub-system that still allows us to answer the problem question (parsimony). Similar arguments to what we found for Sd can be made for the subsystem Sr = ({r},{(r,r)}). Element r affects the output variable m, which is of greatest interest to us with respect to the problem question. What results finally is a pleasingly simple model (Fig. M1-3c).


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a)

e E605 (insecticide with parathion as active reactant) s pest insects n antagonists (lady beetles) p parathion in the agro-ecosystem (outside fruits, cf. r) o paraoxon (degradation product of parathion ) l wind transport R Rain r parathion residues in the fruits (red currants) d diethylthiophosphoricacid (degradation product of parathion ) i p-Nitrophenol (degradation product of parathion ) g glucuronid (degradation product of parathion ) m measured residue in the fruits (parathion) T air temperature w host plants (red currant bushes)

b)

e

l

R

o

p

s

n

T

i

r

d

g

w

m


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c)

Fig. M1-3: a) Tabulation of potential system elements; b) Construction of the relational graph using the 2nd order predicate Pxy :: = "change in y is caused by x"; c) The final system boundaries are laid such a system as small as possible results (Occam's Razor, parsimony).

Building the mathematical model: While the previous step yielded in a qualitative model, there may still be several mathematical models fitting the same qualitative model. Fortunately, considerations we made under step showed also that we have in the field a dominant decomposition process, namely the hydrolysis in aqueous solution (Fig. M1-3c). This is a 1st order kinetic, thus

r· (t) = - γ r(t) Dynamic equation (1) m(t) = r(t) Output equation where r(t) State variable γ Model parameters (constant, i.e., time-independent) m(t) Output variable

Now that we have built a mathematical model2, but we can already answer our problem question? No, because g is unknown (see step ). The general solution

r(t) = Ce-γt

(C – integration constant) (2)

of the differential equation (1) also describes an infinite set of solutions (steps and ), not necessarily all of interest.

Calibration: For the state variable r(t) is no calibration required, because our model describes directly the wanted metric. I.e. the state variable r(t), the parathion concentration in the fruits, has the same unit as the measured pesticide residue m(t). The only difference between r(t) and m(t) being sampling or measurement errors. In the absence of any systematic errors, i.e. m(t) is a bias-free estimate of r(t) that may vary stochastically around the true r(t).

However, the model parameter γ is still unknown. We need an exact value for it that fits our field situation. We can use the measurements from our pilot experiment. Assumed our model approach (1) is correct, then the following must apply:

2 The model is autonomous, because it has no inputs (no input variables)


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The solution of (1) using r(t0) for the integration constant can be written as

r(t) = r(t0) e-γ (t - t0) (3)

and must apply for any time interval, e.g. t0 to t1, e.g. the time elapsed till the first measurement

r(t1) = r(t0) e-γ (t1 - t0) or 0.9 = 1.2 e-γ (4 - 0) (4)

ln 0.9 = ln 1.2 - 4γ => γ = ln 1.2 - ln 0.9

4 ≈ 0.072 (/d) (5)

We can linearize (4) by using logarithms (5). This gives us a first estimate of γ that we may have to revise later (see step ), since instead of using sampling times t0 and t1 in (4) we could equally well have used sampling times t1 and t2 or t0 and t2.

Build the simulation model (Implementation of the mathematical model in form of a simulation model): For our mathematical model (1) one may argue that we do not really need a simulation model, because this ordinary differential equation is analytically solvable (equation (2), see basic math courses, in particular «Analysis»). However, the purpose of this case study is to learn the full recipe of systems analysis and therefore we nevertheless perform this step. Note, most ecological models are very complex and therefore this step is in general quite elaborate.

This step consists of translating the mathematical model into an equivalent simulation model. Equivalent means here we require an isomorphic mapping between the mathematical model and the simulation model. I.e. we can relate both ways components of the mathematical model to components of the simulation model and vice versa.

We do this step here with the general modelling- and simulation tool Easy ModelWorks, but could have used any other suitable methodology. There are many simulation languages and simulation tools available, which are misleadingly often also called modelling tools.

The following identifiers to be used in the simulation model are introduced in addition to the mathematical symbols we used so far:

Math. symbol Identifier in the simulation model Value Range Unit

t t [t0.. t*] d

r· x1Dot3 ? mg/kg/d

r(t) x1 [0..1.2] mg/kg

r(t0) x10 1.2 mg/kg

γ gamma 0.0724 /d

3 Here we use the syntax of EMSL, i.e. the «Easy ModelWorks Simulation Language». This means firstly that we must use xi for all state variables, where i stands for a number between 1 and 9. E.g. x8 is the identifier or name for the eight state variable EMSL knows. Derivatives over time are named by adding the suffix Dot, i.e. xiDot. E.g. x8Dot is the identifier to be used for the derivative over time of the state variable x8. Similarly initial values of state variables xi(0) are named by adding the suffix 0, i.e. xi0. E.g. x80. 4 Preliminary estimates from the previous step


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Next we have to formulate the model equation of our ordinary differential equation system, i.e. a DESS5, according to the syntax of EMSL3 from «Easy Model Works":

x1Dot = –gamma*x1 (1')

Fig. M1-3: Dialog for entering the model equation (1') in Easy ModelWorks to specify the simulation model, which is isomorphic to the mathematical model (1).

Finding initial and/or boundary conditions, domains, and simulation parameters:

The initial value r(t0) can be assumed to be given by the first measurement: r(t0) = r(0) = m(t0) = m(0) = 1.2 (mg/kg).

No boundary conditions are to be met in our case study. Note, if in this case study the state variable r(tx) at a particular time tx would have to have a certain given concentration r(tx)=bc, we would call bc a boundary condition.

Finding simulation parameters means first of all specifying the domain or range for the simulation time t. For instance time t should remain in the finite interval [t0.. tend]. Actually, since our problem question needs to find the value of the unknown t* and tend = t*, we need to find here a rough estimate for tend, say 45 d. This is rather a maximum value for tend, because we are anyway no longer interested to solve our system beyond such a date.

Other simulation parameters are the size of the integration step ∆t6 and the integration method. We leave both as predefined by Easy ModelWorks.

Finally, we have to choose useful ranges for all involved variables. We are interested in the pesticide residue r(t) in the interval [0..1.5] (mg/kg), i.e. we assume the state variable x1 ≈ m ∈ [0,1.5].

Simulation (conducting experiments with the simulation model):

The model equation (1) describes infinitely many solutions varying with values of the integration constant. Thanks to step , we can focus on a single, specific solution, i.e. running a simulation means finding the solution of the initial value problem using the initial condition r(t0). When we

5 Acronym for Differential Equation System Specification 6 In EMSL deltat


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execute a simulation run within Easy ModelWorks using the simulation model with the initial conditions and simulation parameters as described above, we get the following simulation results:

Fig. M1-4: A first result obtained with the simulation model. In addition to the simulation results of main interest, i.e. the dynamics of the pesticide residue (state variable r(t) ~ x1), the graph shows also two horizontal lines, the threshold value for adults ( x1 = r*E) and infants ( x3 = r*S)7.

Again note, an analytical solution would have been possible as well as using a simulation thanks to equation (2). However, to follow as much as possible the recipe of a systems analysis, we executed this step and used our simulation model implemented by means of Easy ModelWorks. Find further details on how a simulation tool such as Easy ModelWorks functions in the documentation «System Theoretical techniques» that will be discussed later in this course.

Interpretation of simulation results

These first simulation results show: The threshold value for infants, i.e. r*S, is reached only many days after the harvest date. However, the threshold value for adults, i.e. r*E, is reached early enough, i.e. before 20 d. However, these results depend strongly on the choice of the model parameter γ. If we vary γ slightly, the results look quite different. For example, the value of γ differs for each successive pair of measurements (see table in step ). Similar to equation (5), but using m(t1), m(t2), t1, and t2, we get a conflicting estimate for γ: γ01 = 0.072/d (see step ),

7 The drawing of the threshold values was done with following trick: Two additional, uncoupled differential equations were added to the simulation model: x2Dot = 0.0 and x3Dot = 0.0, where the initial values of these additional state variables were set at the threshold values, i.e.. x2(t0) = x20 = r*E = 0.5 , and x3(t0) = x30 = r*S = 0.05 mg/kg, respectively.


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whereas γ12 = (ln 0.9 - ln 0.45)/3 /d = 0.231 /d. This considerably faster decomposition rate leads to a very different result (Fig. M1-5):

Fig. M1-5: Simulation results of the simulation model for two different estimates of the model parameter γ: γ01 = 0.072/d (upper black curve) and γ12 = 0.231/d (lower black curve), respectively, (further explanations s. Fig. M1-4).

Given our mathematical model, both estimates of the model parameter γ should be valid. However, the difference between the two estimates γ01 and γ12 of the model parameter γ is apparently sufficiently big, to make all the difference: The answer to our problem question is the opposite from the other, depending which estimate we use. Consequently, these previously used methods to determine the value of the model parameter γ are not good enough. The value of γ needs to be determined in a less arbitrary manner before we can draw any robust conclusions.

Model and parameter identification:

The calculation of γ should not be based on arbitrarily selected pairs of consecutive measurements. It should rather take all measurements into account. According to the model approach it must hold:

r(t1) = r(t0) e - γ (t1 - t0)

r(t2) = r(t0) e -γ (t2 - t0)

...

r(tn) = r(t0) e - γ (tn - t0)


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When using the present measurements we have 2 equations and one unknown, i.e.:

0.9 = 1.2 e - γ 4

0.45 = 1.2 e - γ 7

The problem is over determined and we need to resort to curve fitting (Ausgleichsrechnung). To this end it is advantageous to linearize equation (3) by using logarithms and to set t0 = 0 so that a linear regression becomes possible (see also eq. (5) and (6)). We consider m(tn) to be a random variable that may differ from r(tn) by an error term ε (the latter normally distributed with mean 0):

ln m(tn) = ln m(t0) - γ tn + ε (6)

The unknown coefficients µ = γ respectively α = ln( r(t0)) are then to be estimated from:

ln m(tn) = α + µ tn + ε (7)

It follows (see sheets «How to do a linear regression? »):

Sxx = (02 + 42 + 72) - 1/3((0+4+7)2) = 65 - 40.3_

= 24.6_

Syy = (ln(1.2)2 + ln(0.9) 2 + ln(0.45)2) - 1/3((ln(1.2)+ln(0.9)+ln(0.45))2) = 0.682 - 0.174 = 0.508

Sxy = (0·ln(1.2) + 4·ln(0.9) + 7·ln(0.45)) - 1/3(0+4+7)(ln(1.2)+ln(0.9)+ln(0.45)) = -6.011-2.646 = -3.365

x_= 1/3(0+4+7) = 3.6

_

y_= 1/3(ln(1.2)+ln(0.9)+ln(0.45)) = -0.241

From this follows µ = Sxy / Sxx = -0.136 and we can estimate the model parameter as γ = 0.136 / d. To compute the 90% confidence interval of the estimated model parameter γ we need its variance sµ

2 = Smin/(n-2)/Sxx = (Syy - Sxy·γ)/(n-2)/Sxx = (0.508 - (-3.365 · -0.136))/(3-2)/24.6 = 0.049/24.6 = 0.002 and the standard deviation sµ = 0.002 = 0.045. Thus the 90% confidence limits for γ can be calculated by µ ±sµ·tα=0.1;υ=n-2=1 = 0.236 ±0.045·6.3138. We get the following estimate for the decomposition rate γ together with its upper and lower confidence limits: γ = 0.136± 0.282 /d

The initial condition r(t0) estimated by m(t0) contains measurement errors. Since α = ln( r(t0)) = y_

- µ x

_ = -0.241 - (-0.136·3.667) = 0.260 follows r(t0) = e0.260 = 1.297 mg/kg. An estimate of the

confidence interval of r(t0) is based on the variance of the estimate, i.e. s 2y|x' = Smin · (1/n + (x' - x_

)2/Sxx) = (Syy - Sxy·γ)·(1/n + (x' - x_

)2/Sxx). At the point x' = t0 = 0 we get s2y|0 = (0.508 - (-3.365 · -0.136))·(0.3

_ + 0.545) = 0.049·0.878 = 0.043. The 90% confidence interval for r(t0) is calculated

from G = s2y|x'·tα=0.1;ν=n-2=1 = 0.043·6.3138 = 1.314, i.e. the upper and lower confidence limits of r(t0) are eα±G = e0.26±1.314 or r(t0) = 1.297 with rmin(t0) = 0.349 and rmax(t0) = 4.824 mg/kg

If we use the estimated expected value in the simulation model (repeat step ), we get results similar to Fig. M1-6. Using the upper and lower limits of the 90% confidence interval shows that trajectories lie all in an astonishingly large area of possible results (Fig. M1-7), which is mainly due to the rather small sample size of the data we have been able to use for the parameter identification.


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Fig. M1 6: Standard run of the identified simulation model. γ = 0.136/d based on measurements during 7 days (for further explanations s. Fig. M1-4).

Fig. M1-7: Several runs (solid curves) of the identified simulation model: Best estimate is 2nd curve (from bottom) as in Fig. M1-6 (note altered scale [0..10 mg/kg]) plus 90% confidence interval va-lues for model parameters γ, i.e. γ, and the initial value r(t0), i.e. r(t0): Lowest curve: γ = 0.136 + 0.228 = 0.364 /d; r(t0) = 0.349; Top curve: γ = 0.136 - 0.228 = -0.092 /d; r(t0) = 4.824 (note, any trajectories above the solid horizontal line (γ = 0) represent only a theoretical result, since γ < 0 assumes increasing parathion (broken horizontal lines explained in Fig. M1-4).


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Model Validation:

Before we actually apply the model we need to test its validity. The validity of a model can only be satisfyingly answered with respect to the problem question (cf. step ). The correlation coefficient of the linear regression to estimate the model parameters is ρ = -0.95. This rather high value can be interpreted as a first confirmation of the suitability of the model. Another consideration is to test whether the model behavior matches other measured data than those that were used to develop the model. In our case study such data would be measurements from the field or laboratory that were not used to identify or otherwise estimate model parameters. If the simulated model behavior matches well such independent observed data (the meaning of “well” typically depends strongly on the problem question), we may consider the model as validated. If the deviations are too large, the model is rejected as unsuitable to help answering the problem question. Although such validations never conclusively “prove" the validity of a model, the more such validation tests a model "endured" the more we can consider it to be fit for the final application (see step ).

In our pilot study we have continued the measurements and obtained following results (cf. step ):

t days (d)

r(t) ~ m(t) residues parathion

(mg/kg) 14 0.215 21 0.09

Since these data were not used to estimate the model parameters (step ), they are perfect for testing the validity of the model. Comparing simulated model predictions as obtained in the repeated step using the identified simulation model (Fig. M1-6 u. M1-7) with the additional measurements tabulated above show an almost perfect fit between model behavior and measurements. We may conclude the identified model (step ) to be valid for being used in the envisaged model application (step ).

Remark: The new measurements came of course too late for the model application in the current year. Consequently above validation comes also too late, and we could have relied only on the argument of a high correlation coefficient. However, should we wish to use the same model in coming years, our confidence in the model has increased. In addition we could reestimate γ and r(t0) using those new data. Similarly reestimating the 90% confidence interval would yield smaller limits, meaning a somewhat reduced model uncertainty.

5) What is your solution? Please try to answer above question precisely, accept uncertainties, yet find a clear answer by a useful approximation (home work). Assuming all your assumptions were true in reality, justify or reject the use of the pesticide given your plan to sell the harvest for baby food?

Model application:

We come back to the original problem: "Should I use the pesticide or not?" or "Is the use of the pesticide acceptable?". Notably the problem question is now most relevant: Is the time t*S in which the parathion residue r in the red currants reaches the value r*S less than 20 days? The standard run (Fig. M1-6) shows that the pesticide residue (state variable x1) reaches the threshold value for adults ( x2 = r*E) in time, i.e. around day 7. Whereas, the threshold value for infants ( x3 = r*S) will be reached only after the harvest time, around t*S = 23 d. As Figure M1-7 shows the uncertainties are considerably big. Based on the result of our simulation model und the remaining considerably uncertainties, we can tentatively answer the initial problem question as follows:


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The use of the pesticide for growing red currants for the production of baby food is not acceptable (t*S > 20 d). In contrast, for adult consumption of the fruits the use of the pesticide would be in accordance with the regulation FIV (t*E << 20 d).

Literature Cited Czayka, L., 1974. Systemwissenschaft: eine kritische Darstellung mit Illustrationsbeispielen aus den

Wirtschaftswissenschaften. ETH-BWI 9758, 185: 110. Howard, P.H. (ed.) 1991. Handbook of environmental fate and exposure data for organic chemicals. Volume III

pesticides., Lewis Publishers, Chelsea, Michigan. Jørgensen, S.E., Jørgensen, L.A. & Hendriksen, J. (eds.), 1979. Handbook of environmental data and ecological

parameters., ISEM, Copenhagen, 1162 pp.

Glossary

MAXIMUM CONCENTRATION

Concentration of a substance and its toxicologically significant derivatives, which may be present in or on a particular food at the time of delivery to consumers. The maximum concentration of a substance is specified as a tolerance or threshold value. It should be noted: In the EU there there are no tolerance values, but only threshold values!

TOLERANCE VALUE

The tolerance value is the maximum concentration, above which the food is considered contaminated or otherwise reduced in value. Tolerance values are the maximum tolerated concentrations of substances (additives, ingredients and impurities), or microorganisms defined by the Schweizerische Bundesrat, whose value was not determined due to immediate or compelling health hazards, but because of a safety margin whose implementation is technically feasible in production and storage (Article 10 Lebensmittelgesetz). Oversteps are objected by the enforcing authority (cantonal laboratories). Goods may also be withdrawn from the market, if e.g. frequent oversteps occur and responsible actors make insufficient attempts to remedy the problem.

THRESHOLD VALUE

The threshold value is the maximum concentration, above which the food is considered inappropriate for human consumption. Limits are maximum concentrations of substances set by the Schweizerische Bundesrat due to toxicological or epidemiological assessment. By oversteps the enforcing authority imposes measures necessary for the protection of health. Thus, goods could be immediately withdrawn from the market (see e.g. tins; for pesticide residues, this is usually impossible, since the analysis of pesticide residues requires typically too much time to prevent selling and distribution of the products, notably when they need to be consumed fresh as is the case for produce. Only supplies still stored could be affected)

solutions assignment 1 - pesticide decomposition...solutions assignment 1 –pesticide decomposition...

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