Resource Systems

Engineering Sciences 22 Lumped Analysis of Resource Systems Summer, 03 p.1 _____________________________________________________________________________

GENERAL INTRODUCTION. Analyzing the dynamic behavior of resource systems is important for understanding and managing populations in contexts such as fisheries, forests, wildlife populations and agriculture. In addition, resource systems analysis is important when applied to human populations and their resource demands. With respect to the latter, there is a widely (although not universally) held belief that providing for our resource needs on a sustainable basis is one of the most significant challenges facing humanity in the 21st century. As we shall see, resource systems can be analyzed as lumped systems using approaches similar to those applied to other systems. There are however several distinctive features of resource systems that are useful to recognize at the outset. In particular: a) Component equations are often descriptive rather than mechanistically-based. b) The values of constants appearing in component equations are often hard to come by in practice. c) System behavior featuring growth (over a finite time period) as well as decay is common. d) Most resource systems, even simple ones, are non-linear. Because of features a) and b), resource system modeling tends to be somewhat more concerned with gaining insight into characteristic system behaviors as opposed to quantitative prediction. Because of features c) and d), the behavior of even simple resource models is often difficult to intuit in the absence of a model and its solution. VARIABLES. Many resource systems may be usefully thought of as consisting of a one or more resources, one or more consumers, and a specified set of functional relationships between these. We will begin with examples involving a single resource and one consumer. These examples will illustrate some of the characteristic (and distinctive) behavior of resource systems, and will also make clear how one would go about analyzing systems with more components. The following nomenclature will be used: R = number or amount of resource per system unit C = number of amount of consumers per system unit Examples of resources for which R might be defined include the mass of plankton per m3 ocean water (available as a resource for fish to feed on), the tons of grass per hectare (available for terrestrial herbivores to feed on), the number of hares per hectare (available for carnivores), and the mass of soil organic matter per acre (which enhances soil fertility and increases the growth rate of plants). Examples of consumers for which C might be defined include one or more species of fish that feed on plankton, one or more herbivores that feed on plankton, one or more carnivores that feed on hares, and humans that harvest organic matter.


Sometimes it is convenient to take the system unit to be the system itself; then the units become simply mass of plankton (in the system), tons of plant matter, number of hares, or mass of organic matter rather than these quantities per some system unit. Also, a given entity may be both a resource and a consumer. For example, hares eat grass and are eaten by lynx, and fish may eat plankton (or little fish) and be harvested and eaten by humans or bigger fish. Components and component equations. We will call the functions that specify the rate of consumption of a particular resource rate laws as we did for reacting systems. In so doing it should be recognized that rate laws for resource systems may or may not be describing an actual reaction, and that the form of the rate law may or may not stem from mechanistic considerations1. Accounting equation. The form of the accounting equation often used for resource calculations is essentially the same as the material balance:

=dtdC

dtdR or rate of regeneration or addition rate of consumption or removal [1]

Several examples will be given below. It may be noted that when the units of R or C are concentration (material/volume) then equation [1] is truly a material balance. However, when the units of R or C are numbers of individuals (e.g. of hares or fish), then equation [1] becomes a population balance. In this case, we are actually modeling a discrete system as a continuous system. This is a good approximation if the numbers are sufficiently large. Non-Renewable Resources. When a resource is not regenerated or added to the system on an ongoing basis, the resource may be said to be non-renewable. A fixed initial quantity of the resource is consumed according to the rate laws involving that resource, and eventually is exhausted. This corresponds to the situation for batch reacting systems as previously considered. Example 1. Utilization of a non-renewable resource. Consider a non-renewable resource R and a consumer C for which the rate of resource consumption and consumer production are given by: rR RC= - kRC [1]

rC RCkY = Y- = RCC/RRC/R r [2] where kRC is a rate constant (consumer units-1time-1)

YC/R is the yield of consumers produced per unit of resource consumed (consumer units/resource units)

1 Empirically-determined rate laws that do not reflect mechanistic considerations are often used for reacting systems as well as resource systems. The growth rate of microorganisms is a common example. We also saw something similar in thermal systems, where thermal resistance for convection is typically derived from experimental measurements.


Note that the form of the rate expression for consumption of R in equation [1] is consistent with the rate being proportional to the probability of the resource and consumer being in the same location at the same time as discussed previously for reacting systems. Other forms may also be applicable for a particular case (e.g. when the rate per consumer approaches a constant value at high R). These two equations describe the common situation for which the yield of C with respect to resource consumption is other than (an typically less than) unity. We can use material balance to write state-space equations for this system:

RCkdtdR

RCkYdtdC

RC

RCRC

=

= /

We can solve this system using ode45 in MATLAB, by writing a function that calculates the derivative: function y = dXdt(t,x) %L. Lynd 1/00, rev. C. Sullivan 6/00 R= x(1); C= x(2); Kr = 0.04; %1/(consumer units)/(time) Ycr = 0.2; %(consumer units)/(resource units) y(1,1) = -Kr*R*C; y(2,1) = Ycr*Kr*R*C;

Results with initial conditions R(0) = 10, C(0) = 0.1 are plotted in Figure 1 below. The shape of the resource curve in Figure 1 is characteristic of non-renewable resource utilization and is called the S curve.

R

C

Consumer or Resource Units


Renewable Resources. Renewable resources are either regenerated or continuously added to the system. For some renewable resources, such as solar energy, the resource exists as a flow rather than a stock. There is no need for a state variable and accounting to keep track of it. Putting a solar panel in the sun today does not decrease the amount of sunlight that will be available tomorrow. Thus, there will rarely be interesting system dynamics involved in describing such resources. (The systems that use these resources, such as electrical power converters for photovoltaic systems, or buildings that use solar heating, involve some very interesting system dynamics, some of which we will see later in this course.) Renewable resources involving a flow may also have storage of a stock. For example, a hydroelectric system such as Wilder Dam involves a large quantity of potential energy stored in water behind the dam, and a flow replenishing that stored energy. Perhaps the most interesting system dynamics in resources systems are those of biological renewable resources. With a dam, the rate of replenishment is independent of the rate at which the resource is used. In a biological system, however, the rate of regrowth of a resource depends on the level of that resource. For a wolf, a rabbit is a resource, and the growth rate of rabbit population depends, of course, on how many rabbits are around producing offspring. For a first approximation, we can assume, then, that the regeneration rate is proportional to R (in this case, the population of rabbits). rg = kgR when rg is limited only by R [3a] where

kg = a rate constant Equation [3a] can be valid for a time when a population is growing under resource-plentiful conditions and is limited by its rate of reproduction. Alternatively, rg may be limited by some resource other than R, or may be self-limiting. In the self-limiting case, rg may be positive at low values of R but have a limiting value at high values of R. An expression describing this behavior is

rg = sat

g,

R + kRr max [3b]

where

rg,max is the maximum rate of formation of R (occurring at high R. The units of rR,max are the units of R per time.

ksat is a saturation constant, which has the same units as R and is equal to the value of R

when rg/ rg,max = 1/2. A common example of a situation for which the rate of resource-generation is self-limiting is plant growth, where more plant matter (e.g. per hectare) allows increasing rates of plant production up to a point until the available sunlight is fully utilized and addition of more plant matter would shade plants (or parts of plants) that are already there.


Example 2. Utilization of a self-limiting renewable resource by a single consumer. Consider the following system:

RRC - k - k

R + kRr

=

=dtdR

dRCsat

R,max

decay rate -n consumptio rate -on regenerati rate [4]

and

( )( )( )( )mRCC/R

mRCC/R

C/R

C/R

R - kkC = YCRC - kk = Y

= Y

= YdtdC

e"maintenanc"for requiredn consumptio R rate - consumed is R rate gross

growthfor available is R ratenet

[5]

where

kd is a rate constant for resource decay (time-1)

km is a rate constant reflecting the maintenance rate of resource consumption (resource unitsconsumer units-1time-1)

The rate expression for resource consumption in equation [4] follows directly from equation [1]. Many resources are subject to some sort of decay. Note that without this term R would increase without bound in the absence of C, which is not realistic. In equation [5], the quantity kRCR corresponds to rate of resource consumption per consumer. If kRCR > km, then dC/dt is positive and most resource consumption is used to produce more C rather than for maintenance. In the limit if kRCR >> km, then essentially all resources are used for growth of C and km may be neglected (this was implicitly assumed for example 1). If kRCR < km, then dC/dt is negative which for many systems corresponds to starvation. In the limit if there is no R present, then C declines with decay constant = YC/Rkm. In order to simulation this system numerically, we can use ode45 with the following derivative function.


function y = dXdt(t,x) %Lynd/Sullivan % modified 7/00 R= x(1); C= x(2); rgmax = 2; %resource units/system units/time Ksat = 1; %1/2 saturation constant (resource units/system units) Krc = 0.03; %resource consumption constant

%(system units/consumer units/time) Krd = 0.2; %resource decay constant (1/time) Ycr = 0.5; %resource -> consumer conversion yield;

%consumer units/resource units Kmc = 0.5; %Maintenance constant

%(resource units/consumer units/time) rg = rgmax*R/(R + Ksat); y(1,1) = rg - Krc*R*C - Krd*R; y(2,1) = Ycr*C*(Krc*R - Kmc);

Simulations will be performed using the above file with the values of the constants varied to demonstrate different situations of interest. In these simulations, R(0) = 5 and ODE45 is used to obtain the solution. First, let us consider the systems behavior in the absence of consumers, which can be simulated if we let Ycr = 0 and C(0) = 0. Results are shown in Figure 2a below. We find that in the absence of consumers, R rises to a steady-state value of 9.

0 50 100 1501

2

3

4

5

6

7

8

9

10

time

Res

ourc

e un

its

Fig. 2a. Resource growth with no consumers


If we next put the consumers back into the simulation by letting Ycr = 0.5 and C(0) = 0.1 we obtain the results shown in Figure 2b. We find that the steady state values of R and C are Rss = 2.5, and Css = 1.9. Observe that while the resource level is high, the consumers overshoot their final population, depleting the resource slightly below its final value, but then with the lower resource level, the population of consumers decreases again, allowing the resource level to come up to its final value, although not up to a level nearly as high as without consumers.

0 10 20 30 40 500

1

2

3

4

5

6

7

time

Res

ourc

e or

con

sum

er u

nits

RC

. Fig. 2b Results with krc = 0.2.

If we next increase the value of krc from 0.2 to 1, representing a higher rate of resource consumption, we obtain the results shown in Figure 2c., where Rss = 0.5 and Css = 1.1. The same overshoot phenomenon occurs, but much more severely. When the consumer population is high, the resource gets almost completely depleted, and then the consumer population decays rapidly. One would not want to be a consumer during this time period, but it is not as bad as the situation with a non-renewable resource, in that with the reduced population of consumers, the resource recovers, and neither the consumers nor the resource are wiped out completely. However, the resource recovers to a lower level than before, and the population of consumers supported in steady state is just over half of what it was with the lower value of krc.


0 10 20 30 40 500

1

2

3

4

5

6

time

Res

ourc

e or

con

sum

er u

nits

RC

Fig. 2b Results with krc = 1.

It is interesting that changing the value of the parameter kRC affects the steady state value of C, even though we have kept the resource requirement to sustain a given population, km, constant. The strategy with which we use a resource determines the population that resource can sustain, independent of how efficiently we use the resource. The steady-state value of C, Css is referred to in ecology as the carrying capacity of the system. We could continue a series of simulations to study the carrying capacity, but doing repeated simulations gets tedious, and, in fact, we can easily calculate the carrying capacity by finding a steady-state solution of our system equations.


To find the steady-state solution, we set the derivatives equal to zero in [4] and [5]. From [5], either C = 0, or

( )RCm

mRC

k kR R - kk = /

0=

[6]

Substituting this in [4] with the derivative set to zero,

Ckk

- k + kk

r

Ckkk

- kk + kk

kr

kk

C - kkk

- k + k

kk

kk

r

RC

d

RCsatm

R,

mRC

md

RCsatm

mR,

RC

md

RC

mRC

satRC

m

RC

mR,

=

=

=

max

max

max

0

[7]

We can now use a simple MATLAB script (no differential equation solutions involved) to plot the functions in [6] and [7] as a function of krc % Plot steady state solution of renewable resource example % Charlie Sullivan, 7/00 rm = 2; %max resource growth rate (resource units/system units/time) Ks = 1; %1/2 saturation constant for resource growth (resource units/system units) Krc = linspace(0.06,2,100); %resource consumption constant %(system units/consumer units/time) Krd = 0.2; %resource decay constant (1/time) Km = 0.5; %Maintenance constant %(resource units/consumer units/time) R = Km./Krc; C = rm./(Km + Ks*Krc) - Krd./Krc; plot(Krc,R,Krc,C,'--') legend('R','C') ylabel('Consumer or resource units') xlabel('k_{rc} [system units/consumer units/time]')


0 0.5 1 1.5 20

1

2

3

4

5

6

7

8

9

Con

sum

er o

r re

sour

ce u

nits

krc

[system units/consumer units/time]

RC

Fig. 2c. Steady state populations (carrying capacity) vs. krc

In this plot, we see that the carrying capacity has a maximum near krc = 0.2, and decreases above or below that. In essence, below that point, the resource does not support a large population because it is not being utilized to the consumers full advantage. Above that point, however, the resource is being over-used, and depleted to the point that it does not regenerate as rapidly as it would be able to if it were used with more restraint. These considerations are important in formulating and evaluating strategies to sustainably support a population of consumers. Examples of such consumer populations include wildlife, food sources such as fish, and human beings. Example 3. Predator-prey population cycles. Consider the same system model represented by equations [4] and [5] and used to simulate example 2 but with a different set of parameters: function y = dXdt(t,x) %x(1) = R %x(2) = C % L. Lynd 1/00 Krc = 1.5e-4;%resource consumption constant (system %units/consumer units/time) Kd = 0.1; %resource decay constant (1/time) Ycr = 0.6; %resource -> consumer conversion efficiency Km =1; %Consumer maintenance constant (resource % units/consumer units/time) Ksat = 5e5; %saturation constant (resource units) rgmax = 7.5e5; rg = rgmax*x(1)/(x(1) + Ksat); y(1,1) = rg - Krc*x(1)*x(2) - Kd*x(1); y(2,1) = Ycr*x(2)*(Krc*x(1) - Km);


Results when this m-file is used with ode45 and with initial conditions R(0) = 1000, C(0) = 500 are plotted in Figure 3a. We see that a slowly decaying oscillation occurs with a period of roughly 10 years. First, both R and C are low; next R grows (because there are not many C); then C grows (because R is plentiful); R declines (because C is plentiful), and finally C declines (because R is low), bringing us back to the original point. It is instructive at this point to examine further equation [3b]:

rg = sat

g,

R + kRr max [3b].

If ksat >R, then the limiting behavior of [3b] is described by

rg = Rkr

sat

g,max [6b]

0 10 20 30 40 50 60 70 80 90 1000

1

2

3

4

5

6x 104

time (years)

Figure 3a. Simulated preditor and prey populations, rgmax = 7.5e5, Ksat = 5e5

Population

Prey Predator


Equation [6b] is of the same form as equation [3a] rg = kgR [3a] Thus we see that the equation we have used to describe the rate or resource regeneration by self-limiting renewable resources [3b] approaches a constant when ksat > R. When the size of renewable resource R is limited by C, it may be permissible to use equation [3a] to describe the rate of regeneration of R. rg = kgR [3a] This corresponds to the case when the rate of resource regeneration is not self-limiting. Replacing equation [3a] with equation [3b] in the system represented by equations [4] and [5] gives:

RRC - kR - k = kdtdR

dRCg [4b]

and

( )mRCC/R R - kkC = YdtdC [5]

The following MATLAB derivative function can be used to solve an example of a system modeled by equations [4b] and [5] with constants and initial conditions as for the example presented in Figure 3a. (The only change is that we have changed the resource term so that it is not self-limiting). function y = dXdt(t,x) %x(1) = R %x(2) = C %L. Lynd 1/00 Kg = 1.5; %resource units/system units/time Krc = 1.5e-4;%resource consumption constant (system %units/consumer units/time) Kd = 0.1; %resource decay constant (1/time) Ycr = 0.6; %resource -> consumer conversion efficiency Km =1; %Consumer maintenance constant (resource %units/consumer units*time) y(1,1) = Kg*x(1) - Krc*x(1)*x(2) - Kd*x(1); y(2,1) = Ycr*x(2)*(Krc*x(1) - Km);

Sustained oscillation occurs with no decay in this case, as shown in Fig. 3c.


Figure 3b below presents historical data over a 90-year period for the number of lynx and snowshoe hare pelts received by the Hudson Bay Company. The actual data are of course less uniform than the simulation results due to factors not in the model such as variations in the weather, variations in the numbers of hunters etc. However, the same overall trendsstrongly oscillating populations with peaks every 10 years and the peaks in lynx population following the peaks in hare populationare exhibited by both the model and the data.

Figure 3b.

If you are a consumer, the down-slopes in those oscillations are undesirable, and good resource management policies often are intended to avoid such oscillations; to keep the system at a stable steady state. Although you now have the ability to try any system model and see from a simulation

0 10 20 30 40 50 60 70 80 90 1000

1

2

3

4

5

6x 104

time (years)

Figure 3c. Simulated populations with non self-limited resource regeneration

Prey Predator

Population


whether it oscillates, a direct calculation that shows what system parameter values will lead to a stable steady state, and what values will lead to oscillation, can be useful in design and policy making. The frequency-domain techniques that will be developed in the second half of this course can be used to determine whether or not a system will have a stable steady-state operating point, or will oscillate. Example 4. Considering more variables. Although the downslopes in an oscillation are undesirable for consumers, in the systems we have seen so far, the population eventually recovers. Unfortunately, that is not the case for all types of systems. In the homework, you will consider a system of plants (the resource) and consumers, similar to what we have modeled, analyzed, and simulated above, but in which there is the possibility of soil erosion as well. To model this, we can use three state variables: S, the amount of soil (it could also represent the amount of a particular nutrient in the soil), and R and C as before. The model for consumers is unchanged, but the growth rate of plants is depends on the soil. Once there is enough soil, more soil doesnt help, so there is a saturation effect for soil as well as for R:

rg = satsat

g,

S + SS

R + kRr max

Now we need a model for soil. Its erosion rate is retarded by the presence of plants. We could model it as simply inversely proportional to R, but we dont expect the erosion rate to be infinite in the absence of plants, so we can modify that idea as follows:

re = x

eR + k

Sk

where ke is an erosion constant and kx is the threshold level of plant matter (R) below which further decreases in R dont speed erosion much. If we also assume soil growth is proportional to the amount of plant matter (some of which dies, decays, and enriches and adds to the soil),

x

es R + k

SkRkdtdS

= .

You will explore the implication of this in the homework. Example 5. Resource Management. Fisheries are subject to oscillations due to the same underlying factors that give rise to the oscillations in the preceding examples. Indeed, same simulations can equally well be for the oscillations in the population of fish (R) in relation to the number of fishing boats (C), assuming that the number of the fishing boats increases when the fish are plentiful and decreases when the fish are not. It would be desirable to avoid such oscillations since they are accompanied by economic hardship for fisherman and can put the entire fish population at risk (e.g. if some random challenge to the fish population occurs when population levels are already low).


What if the number of boats is to be determined by the number of licenses issued rather than as a response to the widely fluctuating fish population? We will examine this situation by looking at the dynamics of the fish population, the steady-state harvest, and the harvest per boat per year with the number of boats fixed at various levels. Changing the model used to generate the output shown in Figure 3a so that B is at a fixed level we obtain:

function y = dXdt(t,x) %x(1) = R % Lee Lynd rgmax = 750e3; % fish/yr ksat = 500e3; % fish kd = 0.1; %1/yr krc = 0.15e-3; %fish/(boat*fish*year) B = 500; %Number of fishing licenses (boats) issued rg = rgmax*x(1)/(x(1) + Ksat); y(1,1) = rg - Krc*x(1)*B - Kd*x(1);

Results when this mfile is used with ode45 to simulate the system with initial conditions R(0) = 1000 are plotted in Figure 5a below.

The steady-state fish population with 500 boats is 3.8 million.

0 10 20 30 40 50 60 70 80 90 1000

0.5

1

1.5

2

2.5

3

3.5

4 x 10 6

time (years)

Figure 5a. Simulated fish populations, 500 boats

Fish Population


If we now change to 1000 boats, we have the situation shown in Figure 5b, with a steady-state fish population of 2.5 million.

This can be repeated for different numbers of boats, but running many simulations is not necessary in this case, because we can easily do a little bit of simple math, involve no solutions of differential equations, and find the steady-state values. All we have to do is set the derivative equal to zero:

RRB - k - kR + k

Rr =

dtdR

dRCsat

R,max0=

Dividing by R results in

dRCsat

R, B - k - kR + k

r = max0

which can be solved for the stead-state value of R, Rss, to obtain

satdRC

R,ss - k kB k

rR

+= max [5.1]

As consumers, we might care more about the total number of fish caught be year, rather than the number of fish in the population. That is simply

ssRC BRkcatchAnnual = [5.2]

0 10 20 30 40 50 60 70 80 90 1000

0.5

1

1.5

2

2.5 x 10 6

time (years)

Figure 5b. Simulated fish populations, 1000 boats

Fish Population


We can plot [5.1] and [5.2] with basic MATLAB commandsno differential equation solvers like ode45 are needed. The result is shown in Figure 5c.

0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

Number of boats

fish

[see

lege

nd]

Fish Population, tens of millionsAnnual catch, millions

Figure 5c. Fish population and annual catch as a function of number of boats,

computed analytically from steady-state solution. Code for this plot is on the following page. We see that the harvest goes through a maximum at about 2000 boats (1915, to be exact). Both the numbers of fish and the number of fish per boat decline with increasing number of boats. It is reasonable to expect there to be some minimum threshold of boats/fish that would determine economic viability. A more extensive model could add other factors such as the price of fish. If we use the same model, but allow the number of boats to fluctuate in response to the fish supply, oscillations in the number of boats and in the fish population result. Instead of a constant 1.4 million fish, (as with the optimum number of boats, above), we get a maximum fish population of 55,000 and much lower average and minimum values. Thus we see that by fixing the number of boats we accomplish: 1) elimination of fluctuations in fish population, boats, and harvest; and 2) an increase in the size of the fish population by over 20-fold. The dynamic behavior indicated by our model has in fact been observed in real life, as recounted in the article A Tale of Two Fisheries available as a handout separate from this one. This August 2000 New York Times article contrasts the decline of lobster and tuna fisheries in Point Judith Rhode Island with the thriving fisheries of Port Lincoln Australia.


%HW 8 p 2, engs22 01X % C. Sullivan % Plot steady state solution for fish population and number of fish rgmax = 750e3; % fish/yr ksat = 500e3; % fish kd = 0.1; %1/yr krc = 0.15e-3; %fish/(boat*fish*year) B = linspace(0,10000); R = rgmax./(kd+krc*B) -ksat; catchn = krc*R.*B; plot(B,R/10e6,B,catchn/1e6,'--') big(16) xlabel('Number of boats') ylabel('fish [see legend]') legend('Fish Population, tens of millions','Annual catch, millions') grid

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Resource Systems