Ecological modelling KarlineSoetaertPeter Herman

Netherlands Institute of Ecology(NIOO-CEME)Yerseke, the Netherlands

•Theoretical part: - course (85 pp)=>Text in boxes: not mandatory=> Examples: understand how they work, make equations.

•Practical part:=>Modelling in EXCEL

Ecological modelling

•Why modelling ?•What is a model ?•How do we make a model ?•Elementary principles•Examples

What is a model•A simplified representation of a complex phenomenon

•focus only on the object of interest•ignoring the (irrelevant) details•select temporal and spatial scales of interest

NH4

MEIO

MACRO

DETRITUS+BACTERIA

• Express quantitative relationships -> mathematical formulation => Predictions, tested to data=> Computers

Why do we use modelsBasic research: understand in a quantitative sense how a system works; test hypotheses.

•Experiments are more efficient if a model tells us what to expect•Some things cannot be directly measured (or too expensive)

•If model cannot reproduce even the qualitative aspects of an observation => it is wrong => change our conceptual understanding

Real world Conceptual world

Phenomena

Observations

Models (analysing)

Prediction

Why do we use modelsInterpolation, budgetting:

•measurements may not be accurate enough

•Black-box interpolation methods do not tell us anything about the functioning of the system.

Input

Output

understanding ??r 2 >0.9 r 2 <= 0.9

BLACK-BOX MODEL

Output

Input

Output

Why do we use modelsManagement tool:

•Model predictions may be used to examine the consequences of our actions in advance.

•What is the effect of REDUCING the input of organic matter to an estuary on the export of nitrogen to the sea ?

•MODEL ANSWER: it INCREASES the net export. O2 improves => denitrification lower => removal of N in estuary decreases

Why do we use models

Quantification: Fitting a model to data allows quantification of processes that are difficult to measure.

C

1.2 1.8 2.4

0

5

10

15

20

%

cm0

1

2

3

4O2

0 80 160

cm

NH30 4 8

0

5

10

15

20

cm

10

15

20NO3

0 20 40

0

5

cm

%

cm

1.2 1.8 2.4

0

5

10

15

20C

-1

0 80 160

0

1

2

3

4

cm

O2

0 20 40

0

5

10

15

20

cm

NO3

µmol liter-1 µmol liter-1

0 200 400 600 8000

20

40

60

80

day

O2 flux (model result)100

µm

ol

cm

-2

yr-1

C flux (sediment trap)

Highly reactive OM (>7 /yr)

Example: Westerschelde zooplanktonQuestion:

Is there net growth of marine zooplankton species in the estuary or do they deteriorate.What is the net import/export of marine zooplankton to the estuary

Fact: Difficult to measure directly (flow in/out estuary ?)Seasonal time scales, scale of km.

Tool: Simplified physics, simplified biology

Ex: Westerschelde zooplankton

1. Unknown: G

2. Data: monthly transect of zooplankton biomass along a transect from the sea to the river.Run the model assuming G=0 => Negative/positive growthEstimate G (calibration)

3. Calculating budgets

productionnet ort net transp t

ZOO

ZOOGx

ZOOAxK

xAZOOQ

xAt

ZOO

)()(

Ex: Westerschelde zooplankton

RESULT1. Marine zooplankton dies in the estuary; on average 5% per day; typical coastal species have lower loss rates than oceanic species

2. Over a year, some 1500 tonnes of zooplankton dry weight is imported into the estuary each year (~4000 dutch cows).

Environmental models• Environmental models deal with the exchange of energy, mass or momentum between entities

heat transfer from the air to the wateruptake of dissolved inorganic nitrogen by phytoplankton organisms transfer of movement from the air to the sea by the action of the wind on the sea surface

Elements of a model

•Differential equationTime dependent:problem can be expressed by means of sources and sinks

Source

Sink

NdN

dt

storage rate

source sink

ZOOksALGALG

ALGGrazingcALGrPARf

ksDIN

DINµ

dt

dA

)(

mortalitynRespiratioesisPhotosynth

SinkSourcelg

Modelling steps problem

Conceptual model main componentsand relationships

Mathematical model General theory

Parameterisation Literature, measurements

Mathematical solution

Calibration, sensitivityVerification,validation

Field data, laboratory measurements

Prediction / analysis

? good enough ?

Iterative process•=>Improve if wrong•=>Data

Conceptual model

COMPONENTS:

•State variable (biomass, density, concentration)•Flows or interaction•Forcing functions (light intensity, Wind, flow rates)•Ordinary variable (Grazing rates, Chlorophyll)

•Parameters (ks, pFaeces)•Universal constants (e.g. atomic weights)

TEMPORAL AND SPATIAL SCALE

MODEL CURRENCY (N, C, DWT, individuals,..)

Conceptual model

MODEL CURRENCY: N, -> mmol N m-3

Chlorophyll = PHYTO * [Chlorophyll/Nitrogen ratio]

dPHYTO/dt = 1-2-8-9dZOO/dt = 2-3-4-5dDetritus/dt = 3+8+6+12-7-10dFish/dt = 5-6-12dBottomDet/dt = 7+9-11dNH3/dt = 11+10+4-1

PHYTO

ZOO

FISH

DETRITUS

BOTTOMDETRITUS

NH3

Solar radiation

2

34

5

6

7

89

10

11

1

Chlorophyll

12

Conceptual model equations:

Ecological interactions•deal with the exchange of energy

•INTERACTION = MaximalINTERACTION * Rate limiting_Term(s)Compartment that performs the work controls maximal strength Rate limiting term:

a function of resource (Functional response)a function of consumer (Carrying capacity)

Prey

Predator

PREDATION = MaximalRate * Predator * f(Prey)

Nutrient

Algae

NUTRIENTUPTAKE = MaximalRate * Algae * f(Nutrient)

PREDATION (mmolC/m3/d) MaximalRate ( /d)Predator (mmolC/m3)f(Prey) (-)

Ecological interactions•Biochemical transformation: Bacteria perform work

•=> first-order to bacteria•=> rate limiting term = function of source compartment

Hydrolysis = MaximalHydrolysisRate * Bacteria * f(semilabile DOC)

Hydrolysis (mmolC/m3/d) MaximalHydrolysisRate ( /d)Bacteria (mmolC/m3)f(SemilabileDOC) (-)

semi-labile DOC

Bacteria

labile DOC

Ecological interactionsRate limiting term: functional response

how a consumption rate is affected by the concentration of resource

Functional response type Ic=0.01

0

0.5

1

1.5

2

0 50 100 150 200

Resource concentration

rate

(/ti

me)

Functional response type IIks=20

0

0.2

0.4

0.6

0.8

1

0 50 100 150

Resource concentration

rate

lim

itin

g

term

(-)

Functional response type IIIks=20; p=5

0

0.2

0.4

0.6

0.8

1

1.2

0 20 40 60 80 100

Resource concentration

rate

lim

itin

g

term

(-)

Rate=c . Resource conc

Blundering idiotrandom encounter

Monod/Michaelis-Menten

kssource

sourceRateLim

Re

Re

Low resource: ~linearHigh resource: handling time

pp

p

kssource

sourceRateLim

Re

Re

Low resource: ~exponential (learning,switch behavior)High resource: handling time

Ecological interactions•More than one limiting resource:

•Liebig law of the minimum: determined by substance least in supply•Multiplicative effect•preference factor for multiple food sources

kssource

sourceRateLim

Re

Re

ksFoodpFoodp

FoodpFoodpTermngRateLimiti

kssource

source

kssource

sourceTermngRateLimiti

kssource

source

kssource

sourceMINTermngRateLimiti

2*21*1

2*21*1_

22Re

2Re*

11Re

1Re_

22Re

2Re,

11Re

1Re_

Ecological interactions•CLOSURE TERMS

•Models are simplicifications, not everything is explicitly modeled => some processes are Parameterised.

kssource

sourceRateLim

Re

Re

•Closure on mesozooplankton•=> do NOT model their predators (fishes, gelatinous..)•=> take into account the mortality imposed by those predators

2.2

.1

nktonmesoZooplacMortality

nktonmesoZooplacMortality

Mesozooplankton: in mmolC/m3c1: /dayc2: /day/(mmolC/m3)Mortality : mmolC/m3/day

Ecological interactions•Carrying capacity model

Rate limiting term is a function of CONSUMER

NK

NGrowth

dt

dN

NngTermrateLimitiMaxGrowthdt

dN

)1(max-0.2

-0.15

-0.1

-0.05

0

0.05

0.1

0.15

0 5 10 15 20

N

Gro

wth

rat

e (/

day

)

02468

10121416

0 20 40 60 80 100

Time

N

Carrying capacity is a proxy for:•Resource limitation•Predation•Space limitation

Ecological interactions•Relationships between flows

One flow = function of another flow

onAssimilatiGrowthCostspirationActivity

nDefaecatioIngestiononAssimilati

IngestionpFaecesnDefaecatio

edeyksey

eyIngestionIngestion

Re

Pr)PrPr

Pr(max

Ingestion

Faeces production

Growth respiration

Assimilation

CO2

Chemical reactionsA

B

C

BARatedt

dB

dt

dA

dt

dC

CBAMaxRate

max

FEIDEKK

K

31

2

IKdt

dF

IKIKDEKdt

dE

IKIKDEKdt

dI

IKDEKdt

dD

3

231

321

21

Enzymatic reaction

0

20

40

60

80

100

120

0 200 400 600 800 1000

Time (Hours)

Co

ncn

en

tratio

n (m

ol/m

3)

Inhibition

•1-Monod denitrification inhibited by O2

kinhO

Kinh

KinhO

OInhibition

22

21

•Exponential NO3-uptake of algae inhibited by ammonium

Inhibition denitrificationkin=1

0

0.2

0.4

0.6

0.8

1

1.2

0 5 10 15 20 25

O2 concentration

Inh

ibit

ion

ter

m

(-)

Inhibition NO3-uptake

0

0.2

0.4

0.6

0.8

1

1.2

0 1 2 3 4 5

NH3-concentration

Inh

ibit

ion

ter

m (

-)

3exp NHCtInhibitionInhibition

Coupled reactions

..lg

..

akeNitrateUptdt

aedA

akeNitrateUptdt

dNitrate

INTERACTION = MaximalRate* WORK * Rate limiting_Term*Inhibition_Term

3exp3

3 NHCtInhibition

ksNitNO

NOAlgaermaxakeNitrateUpt

Nitrate

Algae

Ammoniumx

Coupled reactions

( 1 - )

r

P r e y

P r e d a t o rD e t r i t u s

DetrituskksPrey

PreyPredatorg

dt

dDetritus

PreyksPrey

PreyPredatorg

dt

dPrey

PredatorrksPrey

PreyPredatorg

dt

dPredator

)1(

Coupled reactions•Coupling via Source-sink (previous examples)•Stoichiometry: cycles of N, C, Si, P are coupled

(CH2O)106(NH3)16(H3PO4) +106 O2 ->106 CO2 +16 NH3 +H3PO4 + 106H2O

(CH2O)106(NH3)16(H3PO4) = C106H263O110N16P

C:H:O:N:P ratio of 106:263:110:16:1.

..]:[Re4

..]:[Re3

..]:[Re2

ratioCPspirationdt

dPO

ratioCNspirationdt

dNH

ratioCOspirationdt

dO

..Re spirationdt

dCMolar ratios:

O:C ratio = 1

N:C ratio = 16/106

P:C ratio = 1/106

Impact of physical conditions•Currents / turbulence

•pelagic constituents•benthic animals: supply of food / removal of wastes

Hydrodynamical models: coupled differential equations

z

TKEvk

zdl

TKEdz

v

z

umk

zzkg

t

TKE

2/322

0

2

2

2

2

2

21

0 z

uk

y

uk

x

uk

x

pvf

z

uw

y

uv

x

uu

t

u

Iz

I

z

Tzk

zz

I

cpt

T

where0

1

2

2

2

2

2

21

0 z

vk

y

vk

x

vk

y

puf

z

vw

y

vv

x

vu

t

v

Impact of physical conditions•Temperature

•Rates (Physical, chemical, physiological,..)•Solubility of substances -> exchange across air-sea

Forcing function / hydrodynamical models

Inidividual species response

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

0 10 20 30

Temperature (dgC)

Rat

e (/

tim

e)

Ecosystem response

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

0 10 20 30

Temperature (dgC)

Rat

e (/

tim

e)

Impact of physical conditions•Light

•Heats up water and sediment•PAR: photosynthesis

Forcing function (data/algorithm)

Model: prod=pmax*2*(1+bet)*(light/iopt)/((light/iopt)**2+2*be

y=(10.6727)*2*(1+(.411139))*(x/(231.897))/((x/(231.897))**2+2*(.411139)*x/(231.897)+1)

-100 0 100 200 300 400 500 600 700 800

LIGHT (mol.m -2.s-1)

-18

-16

-14

-12

-10

-8

-6

-4

-2

0

2

4

6

8

10

12

PR

OD

UC

TIO

N

linear

saturation inhibition

Light in water

0

50

100

150

200

250

300

350

0 50 100 150 200

µmol / m2 / s

Wat

er d

epth

(m

)

0.0

50.0

100.0

150.0

200.0

250.0

300.0

350.0

400.0

450.0

0 200 400 600 800

Day

RA

DIA

TIO

N (

W/m

2)

Impact of physical conditions•Wind

•Turbulence in water•Exchange of gasses at air-sea interface

Forcing function

Model formulation-summary•Ecological interactions:

•first-order to work compartment•rate limiting terms (functional responses, carrying capacity terms)•inhibition terms•closure terms - proxy for processes not modeled

•Chemical reactions•inhibition terms

•Coupled models•source-sink compartments•stoichiometry

•Physical conditions•currents•temperature•light•wind

NPZD model•4 state variables - mmol N/m3 - rates per day•1 ordinary variable: chlorophyll (calculated based on PHYTO)•1 Forcing function: Light

PHYTO

ZOO

DETRITUS

DIN

Solar radiation

2

34

56

1

Chlorophyll dPHYTO/dt = 1-2dZOO/dt = 2-3-4-5dDetritus/dt = 3+5-6dDIN/dt = 4+6-1

NPZD model

PHYTO

ZOO

DETRITUS

DIN

Solar radiation

2

34

56

1

Chlorophyll

PHYTOksDINDIN

DIN

ksPARPAR

PARMINMaxUptakeUptakeN

),(_

ZOOksGrazingPHYTO

PHYTOMaxGrazingGrazing

PHYTONratioChllChlorophyl _

ZOOteExretionRaExcretion

2ZOOateMortalityRMortality

pFaecesGrazingoductionFaeces Pr

NPZD modelNPZD model

0.00.51.01.52.02.53.03.54.04.55.0

0 200 400 600 800

Day

Ch

lo

ro

ph

yll, m

g/m

3

•Too simple: no temperature dependence•No sedimentation of algae / detritus

2 fundamental principles

More robust model applications•Dimensional homogeneity and consistency of units•Conservation of energy and mass

All quantities have a unit attachedS.I. units: * m (length) * kg (mass) * s (time) * K (temperature) * mol (amount of substance)Derived units: * C = K - 273.15 (C-1=K-1) * N = kg m s-2 (force) * J = kg m2 s-2 = N m (energy) * W = kg m2 s-3 = J s-1 (power)

An equation is dimensionally homogeneous and has consistent units if the units and quantities at two sides of an equation balance

1. Consistency of units

To a certain extent units can be manipulated like numbers•J kg-1 = (kg m2 s-2)/kg = m2 s-2

(units mass-specific energy)

•Relative density of females in a population(number of females m-2) / (total individuals m-2) = (-)

it is not allowed to add mass to length, length to area, ..

It is not allowed to add grams to kilogramsBefore calculating with the numbers, the units must be written to base S.I. Units

Consistency of units

•Units on both sides of the ‘=‘ sign must match=> can be used to check the consistency of a model

ex: the rate of change of detrital nitrogen in a water column

mmol N m-3 d-1 = ( d-1) * (mmol C m-3) NOT CONSISTENT !

Consistency of units

.. PHYCrPHYmortdt

dNDET

.. PHYCrPHYmortdt

dNDET

rPHYmort = phytoplankton mortality rate (d-1)PHYC = phytoplankton concentration (mmol C m-3)NDET= detrital Nitrogen (mmol N m-3)

dNDET

dtrPHYmort NCrPHY PHYC + . . . .

rPHYmort = phytoplankton mortality rate (d-1)PHYC = phytoplankton concentration (mmol C m-3)

NCrPHY = Nitrogen/carbon ratio of phytoplankton (mol N (mol C)-1)

dNDET

dtrPHYmort NCrPHY

PHYC + . . . .

Units = d-1 mmol C

m 3mol N

mol C+. . .

= d-1 mmol C

m 3mmol N

mmol C =

mmol N

m 3 d-1

Consistency of units

mmol N m-3 d-1 = mmol N m-3 3 d-1 CONSISTENT !

2. Conservation of mass and energy

neither total mass nor energy can be created or destroyedSum of all rate of changes and external sources / sinks constant

3 state variables: FOOD, DAPHNIA, EGGS (mmolC/m3)

2 external sinks

dDaphnia/dt+dEggs/dt+dFood/dt = Basalresp+GrowthResp+FaecesProd

Ingestion

Growth respiration,Faeces production

Assimilation

basal respiration

Reproduction

Somaticgrowth

FOOD

DAPHNIA

EGGS

2. Conservation of mass and energy

If no external sources/sinks:total load must be constant.

PHYTO

ZOO

FISH

DETRITUS

BOTTOMDETRITUS

NH3

Solar radiation

2

34

5

6

7

89

10

11

1

Chlorophyll

12

Phyto+Zoo+Fish+ Detritus+NH3+Bottom detritus = Ct

AQUAPHY•Physiological model of unbalanced algal growth:Algae have variable stoichiometry due to uncoupling of

photosynthesis (C-assimilation)protein synthesis (N-assimilation)

N/C (mol/mol)

0.02 0.03 0.05 0.1 0.2 0.3

IrradianceNutrient avail

1: photosynthesis2: exudation3: storage4: catabolism5: protein synthesis6: respiration7: lysis

LMWCarbohydratesReserve

carbohydrates

Biosynthetic and photosyntheticapparatus

CC DIN

C, N,Chl

1

2

43

57

6

AQUAPHY

1: photosynthesis2: exudation3: storage4: catabolism5: protein synthesis6: respiration7: lysis

LMWCarbohydratesReserve

carbohydrates

Biosynthetic and photosyntheticapparatus

CC DIN

C, N,Chl

1

2

43

57

6

P DIN

INPUT

Dilution

d LMW / dt = Photosynthesis -exudation +Catabolism -Storage -Respiration -LMWLysis -LMWdilutiondReserve/dt = Storage - Catabolism - ReserveLysis - ReservedilutiondSynth/dt = ProteinSynthesis - Synth_Lysis - SynthDilution

dDIN/dt = -ProteinSynthesis * NCratio_Synth -DINdilution + Input

Conceptual model equations:

•4 state variables: Reserve, LMW, proteins (mmol C/m3), DIN (mmolN/m3)

Homogeneously mixed models: simple but not always realistic

•Estuary: spread of contaminants, invasion of species: at least 1-D

•Biogeochemical cycles of upper oceaneuphotic: autotrophic

•Simple predator-prey models:different in spatial environment

PAR

0

20 60 100 140

0

10

20

30

40

µE/m2/s

Depth (m)

Spatial components

Landscape and patch models•Dynamics described in large number of cells

Each cell: properties from GIS-> data requirements large•Real-world phenomena•Large animals•Usually statistical modelling approach

Taxonomy of spatial models I

Taxonomy of spatial models II

Cellular automaton models

• Large number of cells• Each cell: occupied or not -> data

requirements small• Interaction between neighbouring cells

• Explore ecological dynamics in spatial context

Microscopic rules

Emerging patterns

Taxonomy of spatial models IIIContinuous spatial models

• Macroscopic approach: do not resolve individual molecules, individuals, etc.. but average over appropriate space/time and describe dynamics of the average.

• Example: diffusion of molecules Microscopic, stochastic model

Macroscopic, continuous model

Macroscopic description by integration

Realisation of stochastic stationary process

Probability of occurrence under this process

Count no. passing in 2 directionsDifference = net fluxRepeat many timesAverage=expected flux

Express expected flux as function of concentrations (=expected occurrence prob.)

Flux = - K dC/dx (Fick’s law)

Taxonomy of spatial models

1-D advection-diffusion equation

M

o

d

e

l

unit surface

x

Storage

InFlux

OutFlux

Storage = InFlux - OutFlux

x

Consider slice with unit surface and thickness x

Flux = mass passing a surface per unit time+ in direction x-axis- in opposite direction

Fluxes and concentration changesC = Mass / Volume

When the length of the box becomes very small, this expression becomes:

boxlength

OutFluxInFlux

t

C

volume

tlostMasstreceivedMass

t

C

11 ..

x

Flux

t

C

Advective / diffusive fluxes

• Advective fluxes: – caused by directional movement. – Movement independent of concentration.

• Diffusive fluxes: – caused by random movement. – Net movement of mass dependent on

concentration gradient

Advective fluxes

downstream

upstream

CuOutFlux

CuInFlux

.

.

x

Cu

x

Flux

t

C

x

Cu

x

Flux

x

Cu

.

u

Cup Cdown

Examples:Net current through systemSinking velocity of particles

Sediment accretion

Diffusive fluxes

x

CDFlux

Fick’s first law:

-> mass balance:

x

CD

xx

CD

xx

Flux

t

C

)(

Examples:Tidal mixing in estuary

Molecular diffusion in sediment

Advective-diffusive transportreaction

x

CD

xx

Cu

t

C

actionQsxAx

sEA

xAt

sRe)(

1)(

1

With changing cross-sectional surface A(x) :x

A

Spatial boundary conditions

Mo

d

el

Boundary

x

Boundary

Boundary

Consider first (or last) compartment/slice-> what is influx / outflux?-> dependent on concentrations outside model

domain !

-> boundary conditions specify what happens in the outside world

E.g.: fixed concentration in freshwaterimposed flux through air-sea interfaceimposed deposition flux on sedimentetc..

Boundary conditions 1D org. C model

x

Sed

imen

t

Non-reactive layer

Water column

Depositional fluxD

ecay

(1-

orde

r)

No-flux

Model elements:• x=0 at sed.-water interface

•Sediment accretion -> advective flux

•Bioturbation: diffusional flux (D constant)

•1-order decay of org. C

Ckx

CD

x

Cv

t

C

2

2

Boundary conditions 1D org. C modelUpper boundary: depositional flux of organic carbonThis flux enters the model domain through advection and bioturbation

000

x

CDvCFlux x

Lower boundary: all organic C consumed -> concentration = 0

0xC

These conditions fully specify the model

Example: competition in a 1-D grid

2 species in 1D grid

Rules: individuals stay forever in their cellReproduction to xi cells apart, if this is emptyReproduction with certain probability i

Interference mortality (probabilities 12 and 21) when different species in neighbouring cellsMortality with probability i

Example

Competition in 1-D grid: Implementation

Generation of a random event: •Generate random number x [0,1]•If x <= p : event happens; else does not happen

Test of model implementation:•Reshuffle spatial arrangement at each step•Compare results with non-spatial model

Competition in 1-D grid: results

Homogeneous space:Species 1 wins

In space: species 2 survives ! Species 2 disperses better: wins!

Model parameterisation

•Problem:• Values for constants (parameters) in

equations?

• Both models the same dN / dt = a * N• Both models start at same value• but: a1 = 2 * a2

Sources for parameters

• Direct observation• Literature• Calibration

Direct observationModel: prod=pmax*2*(1+bet)*(light/iopt)/((light/iopt)**2+2*be

y=(10.6727)*2*(1+(.411139))*(x/(231.897))/((x/(231.897))**2+2*(.411139)*x/(231.897)+1)

-100 0 100 200 300 400 500 600 700 800

LIGHT (mol.m -2.s-1)

-18

-16

-14

-12

-10

-8

-6

-4

-2

0

2

4

6

8

10

12

PR

OD

UC

TIO

N

Example: P-I relationFitting of Eilers-Peeters eq.-> pmax, Iopt, beta

12)(

)1(2*max

2

IoptI

IoptI

IoptI

pprod

Estimate Standard error

t-value p-level Lo. Conf Up. Conf

pmax 10.6727 0.29478 36.20622 0.000000 10.0059 11.3395beta 0.4111 0.17141 2.39856 0.039992 0.0234 0.7989iopt 231.8965 10.02050 23.14220 0.000000 209.2286 254.5645

Direct observation: beware

Consider different sources of error !• experimental error (see previous)• spatial variability• temporal variability

-> base estimates on data base geared to time and space scalesof model !

Literature-derived parameters

• Direct: use published values for exact purpose• Indirect: through relationships with other (master)

variables or parameters

y = 1.44*3.398x-0.571

R2 = 0.6777N=528

0.0001

0.001

0.01

0.1

1

10

0.1 1 10 100 1000 10000

Water depth (m)

SC

OC

(m

mol

O2 c

m-2

yr-1

)

y = 0.2664x-0.2109

R2 = 0.63061.E-03

1.E-02

1.E-01

1.E+00

1.E+00 1.E+02 1.E+04 1.E+06 1.E+08

Zooplankton volume (µm3)

Max

imal

net

gro

wth

rat

e (/

hr)

Literature : beware !

0

100

200

300

400

500

600

0 100000 200000 300000 400000 500000 600000-3

-2

-1

0

1

2

3

4

5

6

7

-5 0 5 10 15

Log-log Linear scale

Parameters from log-log relationships are excellent when used in a similar context. Otherwise large uncertainty may arise !

CalibrationPrinciple: find those parameters for which model fits data best

i i

ii

error

lueObservedvaModelvalueModelCost

2)(= minimise

Straightforward for linear models (Y=a+bx)

Non-linear models: beware of local (<> global) minima

‘Model cost landscape’

Local minimumGlobal minimum

Iterative methodsDirect search: go in direction of steepest descent, or some variations

Random methods: • simulated annealing: accept better fit, but continue to

accept worse fits also, as a basis for exploring other parts of parameter space

• genetic algorithms: make a breeding population of parameter sets, let them reproduce, with some mutation and cross-over, select the fittest.

Model solution

•Differential equationproblem expressed by means of sources and sinks (time-dependent)

dC

dt storage rate source sink

Source Sink

C

dPHYTO C

dtphotosynthesis respiration mortality

_

transport along a physical dimension- Advective-diffusive equation

Some elementary maths

0

5

10

15

20

25

0 10 20 30 40 50

time (d)

N (

ind

.m-2

)

-1-0.5

00.5

11.5

22.5

0 10 20 30 40 50

time (d)

dN

/dt

(in

d.m

-2.d

-1)

Differentiation

Integration

N

dNdt

Specify rate of change from process knowledge

Solve (integrate) model

Model solving •Differential equationsExpress a rate of change (in time or in space) of a quantity (e.g. concentration, density,..) Must be converted to the quantity itself at a certain time or position

C(z0) = Cz0

C(t0) = Ct0

dC

dzC z

dC

dtC t

( )

( )

from to Extra condition

This requires the introduction of a boundary (spatial) or initial condition (temporal model)

Importance of initial conditions

- ‘My salary has increased by 10 % per year over the past three years’

- ‘So, you are a rich man now ?’- ‘No, you should know where it started three years ago!’

Rate of increaseActual value?

Dependent on initial conditions!

0

50

100

150

200

250

300

350

400

0 20 40 60 80

time

N

Same model,Different initial conditions-> different time course!

0

50

100

150

200

250

300

350

400

0 20 40 60 80

time

N

Same model,Different initial conditions-> different time course!

Model solving •If the differential equation is simple an analytical solution may exist

-> State variable = f(variables, parameters,forcings..)

Model: dN/dt = .N

N(t0) = N0

Solution: N(t) = N0.e(.t-.t0)

•Complex models: approximated numerically.-> Discretised in time and space t+tt

Paper, pen, brains

Brains,Computertime

Analytical solution General approach: Integrate differential equations (or look up solution)

-> General solution (up to one or more constants)Use boundary (initial) conditions to derive constants

-> Particular solution

Example:

Exponential growth:

with initial condition: N=N0 at t=t0

Ndt

dN

General solution

Ndt

dN

dtN

dN Reorder terms

dtN

dN Integrate both sides

ln dtNd Remember d(lnN)=dN/N

'ln AtN Solve integrals')( At eAwhereeAtN Take exponents

Particular solutionWhat is value of constant A?Take initial condition: 0

0)( teANtN

0

- t0 Therefore: A = N0 e

- t0 tAnd the particular solution is: Nt = e N0 e

(t - t0 )

Nt = N0 e

t And when t0 = 0: Nt = N0 e

2nd order differential equationExample: carbon in sediment

kCx

Cw

x

CDb

xt

C

0

x

Sed

imen

t

Non-reactive layer

Water columnDepositional flux

Dec

ay (

1-or

der)

No-flux

000

x

CDvCFlux x

0xC

orgC: general solution

xbxa

xD

kDwwx

D

kDww

x

eBeA

BeAeC

2

4

2

4 22

Lookup method: from textbooks one finds that the general solution is:

orgC: particular solution

Db

kDbwwb

eBeAC bax

2

4

0

2

Db

kDbwwa

2

42

Apply boundary condition at x=

b>0, a<0--> B=0

Particular solution - cont’d

Now solve for A, from upper boundary condition:

AwAaDb eAwAeaDbwC

x

CDbFlux aa

xx

000

0

DbkDbwwa

242 withwaDb

FluxA

From which:

xaewaDb

FluxxC

)(And:

Analytic analysis of equilibrium

Consider e.g. the Lotka-Volterra predator-prey model:

predatormeyedatordt

edatord

eyedatorKey

eyrdt

eyd

PrPrPr

PrPr)Pr

1(PrPr

1

predatormeyedator PrPr0

At equilibrium, the rates of change in time are zero, we have:

eyedatorK

eyeyr PrPr)

Pr1(Pr0 1

Analytic analysis of equilibrium

Consider the predator equation first.predatormeyedator PrPr0

Fulfilled when:Predator=0

or

a

meyPr

Analytic analysis of equilibrium

Consider the prey equation then

Fulfilled when:Prey=0

or

eyedatorK

eyeyr PrPr)

Pr1(Pr0 1

eyK

rr

K

eyredator Pr)

Pr1(Pr 111

Analytic analysis of equilibriumWe can plot these conditions simultaneously in a phase plane

Intersections: neither prey nor predator change -> equilibrium

0

2

4

6

8

10

12

0 5 10

Prey

Pre

da

tor

prey isocline

prey isocline

predator isocline

predator isocline

equilibrium

0

2

4

6

8

10

12

time

De

nsi

ty

Prey

Predator

Stability of equilibrium

<- Depart from (0,10) equil. Unstable !

Depart from 0,0 equil. Unstable ! 0

1

2

3

4

5

6

7

8

time

De

nsi

ty

Prey

Predator

Stability and phase plane

Trajectories in phase plane show equilibrium stability: only stable for (2,4) equilibrium0

2

4

6

8

10

12

0 5 10

Prey

Pre

da

tor

prey isocline

prey isocline

predator isocline

predator isocline

equilibrium

Reeks6

Reeks7

Reeks8

Stability properties

Unstable

Stable

Neutrally stable

Stable limit cycle

Examples of stability of equilibrium

0.0

1.0

2.0

3.0

4.0

5.0

6.0

0 20 40 60 80 100

Time

0.0

5.0

10.0

15.0

20.0

25.0

30.0

0 20 40 60 80 100

Time

2.96

2.98

3.00

3.02

3.04

3.06

3.08

3.10

3.12

3.14

0 5 10 15 20 25 30

Time

Unstable oscillatory Stable non-periodic

Stable limit cycle Oscillatory stable

Model solving

ANALYTICAL SOLUTIONOnly for simple modelsCan be calculated at any point of interestCan be obtained by means of a calculator or spreadsheet Is an exact solution

N(t) = N0.e(.t-.t0)

t+ttNUMERICAL SOLUTIONModels can be very complexSolution only obtained at a set of pointsFrequently requires the use of a computerApproximate solution ->Numerical errors

Numerical solutionComplex models:

•non-linear terms•complex forcing function•..

Numerical errors:•Roundoff error (binary format of computer:).. 0.0625, 0.125, 0.25, 0.5 , 1, 2, 4, 8, 16, 32, .. Exactif randomly -> will cancel -> stableif systematic -> magnification of error -> unstable

•Truncation error (discretisation of continuous equations)numerical accurate method: solution near to exact solution

•An accurate method can be unstable

t+tt

Numerical solution

Problem:•Values of state variables at time t•Rate of change at time t

•Values at t+t ?

Start

Initial condition of

state variablesT=T0

Rate of change at T

Update state variables Update time T

Write resultsStop

T= Tend ?

t+tt

Update formula: TAYLOR expansion

Numerical solution t+tt

Taylor expansion:What is the value of f(x+h), when f(x), f’(x), f’’(x),.. known ?

~ What will be the weather tomorrow ?

•If weather does not change: the same as today.

•If it is getting colder during the day (f ’(x)):similar as today but colder

•Accounting for higher-order derivatives:

)()( xfhxf

)()()( ' xfhxfhxf

..)('''6

)(''2

)(')()(32

xfh

xfh

xfhxfhxf

..)('''6

)(''2

)(')()(32

xfh

xfh

xfhxfhxf

Numerical solution t+tt

Taylor expansion:

..)('''6

)(''2

)(')()(32

xfh

xfh

xfhxfhxf

For very small h: higher-order terms become insignificant -> ignore• n-th order accuracy: when truncated at hn+1

step h -> h/2 truncation error -> divided by 2n+1

..)('''6

)(''2

)(')()(32

xfh

xfh

xfhxfhxf

First-order:

Second-order:

..)('''6

)(''2

)(')()(32

xfh

xfh

xfhxfhxf

Numerical integration t+tt

First-order = EULER:

t0 t1=t0+t t2=t1+t t3 …. tn. Ct0 Ct1 Ct2 Ct3 …. Ctn.

)(

..2 2

22

tOt

CtCC

t

Ct

t

CtCC

tttt

ttttt

Assumption: rate of change constant in interval

t -> small t

Analytical/numerical solution

0

0.2

0.4

0.6

0.8

1

1.2

0 20 40 60 80 100

Time

Co

nce

ntr

atio

nTrue dt=8

Numerical integrationt+tt

To increase accuracy: Reduce t

•More complex integration routines•Runge-Kutta methods:

known point at 1:extra evaluations 2, 3, 4Extrapolate to new point

•Predictor-corrector:simple formula to obtain a ‘prediction’more complex formula to obtain a ‘correction’

•Implicit methods:Taylor expansion evaluated at next time step

1

2

3

4

T+tT

Approximating spatial derivatives

Numerical methods deal with space by calculating the values in a finite number of layers

-> advection diffusion reaction equation:

Diffusion2nd orderderivative

kCx

Cw

x

CDb

xt

C

Advection1st orderderivative

1-st order spatial derivatives

kCx

Cw

x

CDb

xt

C

Taylor expansion:

..)('''6

)(''2

)(')()(32

xfh

xfh

xfhxfhxf

)( xOx

CxCC xxx

x

CC

x

C xxx

Forward differencing formula:non-monotone (negative conc)unstable

=> NEVER USED FOR ADVECTION

1-st order spatial derivativeskC

x

Cw

x

CDb

xt

C

Taylor expansion formula 2:

..)('''6

)(''2

)(')()(32

xfh

xfh

xfhxfhxf

)( xOx

CxCC xxx

x

CC

x

C xxx

Backward differencing formula:Monotone (No negative conc)Stable

=>But: only 1st order accurate !

2nd order spatial derivatives

kCx

Cw

x

CDb

xt

C

Taylor expansion:

..)('''6

)(''2

)(')()(32

xfh

xfh

xfhxfhxf

Centered differencing formula

..)('''6

)(''2

)(')()(32

xfh

xfh

xfhxfhxf

(1)

(2)

(1)-(2): )()('2)()( 2hOxfhhxfhxf

h

hxfhxfxf

2

)()()('

2nd order spatial derivatives

x

CK

xt

C

211

11

2/12/1

)2(

][][][

x

CCKx

xC

Kx

CK

xx

Kx

K

xK

x

iii

iiii

ii

C

CC

CCC (Outer gradient)

(Inner gradient)

Centered differences

Truncation error:Numerical diffusion

• = artificial diffusion (even in absence of true diffusion: gradients are smoothed)•Cause: truncation error in approximation of first-order derivative

(backward differences, only 1-st order accurate)

x

CC

x

C xxx

1 10time

Numerical diffusion Population model:•T= 0 to 10 day: 10 newborns /day•growth 1 mm/day, no mortality

•T>10:

Size

IndteTransferRaIndteTransferRa

dt

dIndSize

IndteTransferRa

t

Ind

iii

1

Size (mm)

cohort:

100 indiv.

10 size classes1mm/day

Ind.

Numerical diffusion•True solution after 100 days: <-> Backward differencing:

square pulse Numerical smoothing

True solution

0

2

4

6

8

10

0 50 100 150 200Size

# in

div

idu

als

Backward differences

0

2

4

6

8

10

0 50 100 150 200Size

# in

div

idu

als

Solution ?Use of more difficult schemes

Pragmatic: •Biological variability ~ Diffusion•as long as ‘numerical diffusion’ is but a fraction of ‘true’ physical diffusion: OK

Testing and validating

• Dimensional consistency• Conservation laws: neither individuals,

mass, energy nor momentum can be created or lost except by exchange between the model and the outside world use principle to obtain unknown fluxes use principle to test model consistency by

constructing mass budget

Mass budgets to solve unknowns

Sea 12

River

outflow

A concentration of a non-conservative substance has been measured at the mouth of the bay and at the outflow.

Question: what is the production or decay rate of the substance ?

The following assumptions are valid:The system is at steady-state.Each section is well mixed.The evaporation and rain rate balance.

Given: system characteristics

Volume section 1 V1 4*105 m3

Volume section 2 V2 12*105 m3

Dispersion coefficient river-section1 E’01 0 m3 d-1

Dispersion coefficient section 1- 2 E’12 3*104 m3 d-1

Dispersion coefficient section 2- sea E’2S 3*104 m3 d-1

Total Outflow from section 1 Qout 5*104 m3 d-1

Total Inflow by the river Q0,1 104 m3 d-1

Concentration of substance in sea Ssea 1 mg/l

Concentration in outflow Sout=S1 0.5 mg/l

Concentration of substance in river S0 0.0 mg/l

Model

KsQsxAx

sAE

xAt

sx

)(1

)(1

0

)(')(' 1,11,11,1,1

SinksSourcessKVsQsQssEssEdt

dsV iiiiiiiiiiiiiiii

ii

1,1,1,'

iiii

ii x

AEE

1, ii

outOutin sQsKVsQsQssEssEdt

dsV 1101,012,1122,111,0

11 )(')('0

Advective-diffusion-reaction eq. with variable cross-section A

The equation is discretised as:

For compartment 1 we have:

1 Q 0,1

Q 0ut

Q 1,22 Q 1,2Q 2,Sea

1 1e4

5e4

-4e42 -4e4-4e4

Assuming sout = s1, this reduces to:

1101,012,1122,11

1 )()('0 sKVsQsQoutQssEdt

dsV

2212,122,2,2212,12

2 )(')('0 sKVsQsQssEssEdt

dsV SeaseaSea

Water balance:

Q0,1-Q1,2 - Qout=0

Q1,2= - 4 104 m3 d-1

For box 2:

Q2,Sea = Q1,2 = - 4 104 m3 d-1

Remains: two equations in two unknowns (K, S2) -> solve

Correctness of model solution

• Compare with published model results• Compare numerical with analytical solution

0

50

100

150

200

250

300

350

400

0 0.5 1 1.5 2

Concentration

De

pth

t=1

t=2

t=3

t=4

t=5

t=6

t=7

t=25

0

50

100

150

200

250

300

350

400

0 0.5 1 1.5 2

Concentration C (mmol.m-3)

De

pth

(m

)

analytical numerical

0

50

100

150

200

250

300

350

400

0 0.5 1 1.5 2

Concentration

De

pth

t=1

t=2

t=3

t=4

t=5

t=6

t=7

t=25

Check numerics

Max t=1 !

Numerical diffusion !!

Check numerics

-1

0

1

2

3

4

5

6

0 10 20 30 40 50

time (days)

um

olN

.dm

-3

Nutrients

Algae

-0.2

0

0.2

0.4

0.6

0.8

1

0 1 2 3 4 5

Model verification

0

2

4

6

8

10

12

14

16

18

20

0 5 10 15

time (d)

um

olN

.dm

-3 DIN observed

Algae observed

DIN modelled

Algae modelled

Mismatch model -data

• parameters wrong?

• Formulations wrong?

• Data too noisy?

• Initial conditions wrong?

Verification: data quality

• Estimate, where possible, data uncertainty• Compare model mismatch to uncertainty

-5

0

5

10

15

20

25

0 5 10 15

time (d)

um

olN

.dm

-3 DIN observed

Algae observed

DIN modelled

Algae modelled

0

2

4

6

8

10

12

14

16

18

20

0 5 10 15

time (d)

um

olN

.dm

-3 DIN observed

Algae observed

DIN modelled

Algae modelled

CalibrationPrinciple: find those parameters for which model fits data best

i i

ii

error

lueObservedvaModelvalueModelCost

2)(= minimise

Straightforward for linear models (Y=a+bx)

Non-linear models: beware of local (<> global) minima

‘Model cost landscape’

Local minimumGlobal minimum

Verification: parameters

Vary parameters and initial conditions

check if better solution can be obtained

0

5

10

15

20

25

0 5 10 15

time (d)

um

olN

.dm

-3 DIN observed

Algae observed

DIN modelled

Algae modelled

Sensitivity analysis

• Vary parameters in the model within reasonable range

• Look at change in model outcome as a consequence

• Large model change for small parameter change -> sensitive parameter !

• Concentrate efforts on sensitive parameters

Change model formulations?

0

2

4

6

8

10

12

14

16

0 5 10 15

time (d)

um

olN

.dm

-3 DIN observed

Algae observed

DIN modelled

Algae modelled

0

2

4

6

8

10

12

14

16

0 5 10 15

time (d)

um

olN

.dm

-3 DIN observed

Algae observed

DIN modelled

Algae modelled

• If parameter optimisation (calibration) cannot resolve differences: -> model may not be appropriate -> change model !

Model validity

• Good fit model - data : Not possible to proof that model is wrong

proof that model is right !!

• Test by applying to other system, other datasets etc..

Example

Grazing

Respiration+Mortality

Food

Bosmina Daphnia

Bosmina and Daphnia compete for food

Daphnia best competitor at high food concentration

Bosmina superior at low food concentration

batch cultures in the dark with regular food additions

-> model outcome

umFoodInMediFOOD

:timetransfer at

osminaIngestionB-aphniaIngestionDdt

dFOOD

nBosminaRespiratioBosminaAdt

dBosmina

nDaphniaRespiratioDaphniaAdt

dDaphnia

nssimilatio

nssimilatio

Conceptual model:

cyonEfficienAssimilatiaphniaIngestionDonDaphniaAssimilati

DAPHNIAnDaphniaRespiratio

DAPHNIAFOOD

FOODaphniaIngestionD

arespDaphniksDAPHNIA

niaMaxIngDaph

Mathematical model formulation:•Ingestion regulated by Monod kinetics

•Respiration first-order in biomass

•Fixed fraction of food converted into animal biomass

0 0.2 0.4 0.6 0.8 10

0.5

1

1.5

2

2.5

3

Food concentration (g C/m3)

Food assimilation (/d)

Parameterisation:

Parameters obtained from laboratory experiments

ex.: functional response of both species

Steadyrun1Food in medium=1 gC/m3

0.0

1.0

2.0

3.0

4.0

5.0

6.0

7.0

8.0

0 10 20 30 40

Time (hours)

gC

/m3

0.0

0.2

0.4

0.6

0.8

1.0

1.2

gC

/m3

Cladocera Food

Steadyrun 2Food in medium=0.1 gC/m3

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

0 10 20 30 40

Time (hours)

gC

/m3

0.0

0.0

0.0

0.1

0.1

0.1

0.1g

C/m

3

Cladocera Food

Solution: numericalVerification:

all state variables remain positive alwayssteady state behaves as expected:

Bosmina dominates at low food Daphnia at high food

TransferTime (hours)

FoodInMedium (gC/m3)

0 10 20 30 400.5

0.6

0.7

0.8

0.9

1

Sensitivity analysis:

Two important parameters:

Food added at transferTransfer time

Coexistence /dominance dependent on both

A taxonomy of ecological models

• Strategic models: highly simplified. Abstract. Minimalist. Explore consequence of certain type of dynamics.

• Example: dynamics of interaction between diatoms, silt and sediment erosion on tidal flats

Model

SDSeIdtdS

),,()(

DDSlDDSgdtdD

),,(),(

Van de Koppel, Herman, Thoolen & Heip, Ecology, 2001

Generic systemcontaining

“silt”

and

“diatoms”

Generic properties of functions

0)(

I

0),,(DSe ; 0

),,(S

DSe; 0

),,(D

DSe;

e(0,S,D) = 0; e(,S,D) = ; e(,S,) = 0; e(,,D) = 0;

0),(S

DSg; 0

),(D

DSg; g(S,K) 0

0),,(DSl ; 0

),,(S

DSl; 0

),,(D

DSl;

l(0,S,D) = 0; l(,S,D)

Deposition

Erosion

Algal growth

Algal loss

Graphical analysis

Silt content

Dia

tom

de

nsit y

SB SD

S

D

Multiple stable states

Bottom shear stress Bottom shear stress

Equ

ilib

rium

silt

con

ten

t

Equ

ilib

rium

dia

tom

den

sity

Field evidenceMarchS

ilt content (%)

0

20

40

60

80June

0

20

40

60

80

Logarithm of max

A B

Log( max ) 0Log( max ) 0Frequency

Silt Silt

Logarithm of max

Tactic models

• Forecast quantitatively the state of a particular ecosystem under particular circumstances

• Example: coupled physical-biological-sediment model for the Goban Spur, E.Atlantic (Soetaert et al., 2002)

Model

• 1-dimensional vertical model• Hydrodynamics resolved: stratification, vertical

mixing, etc.• Primary production, zooplankton grazing, detritus

formation, detritus breakdown, nutrient dynamics• Sediment model: organic matter breakdown,

oxygen consumption, nutrient release, biological mixing by macrobenthos

400 500 600 700

D ay

0

50

100

150

200

De

pth

(m

)

Chlorophyll, µg/l

0.0

0 .1

0 .5

1 .0

3 .0

5 .0

10 .0

400 500 600 700

D ay

0

50

100

150

200

De

pth

(m

)

Ammonium, µmol/l

0.00

0.10

0.25

0.50

1.00

1.50

2.00

2.50

3.00

Euphotic zone primary production

200 300 400 500 600 700 800 900 1,0000

100

200

300

400

Day

mmol C/m /d2

f-ratio

200 300 400 500 600 700 800 900 1,0000

0.2

0.4

0.6

0.8

1

Day

-

In situ

Satellite

Wind speed

0

5

10

15

20

m s -1

J F M A M J J A S O N DJ A S O N D J F M A M J

400 500 600 700

D ay

-5

-4

-3

-2

-1

0

De

pth

(cm

)

Sedim ent oxygen, µm ol/l

0

40

80

120

160

200

240

280

400 500 600 700

Day

-10

-9

-8

-7

-6

-5

-4

-3

-2

-1

0

De

pth

(cm

)

Sedim ent am m onium , µm ol/l

0

5

10

15

20

25

30

35

40

45

50

Oxygen

0 100 200 300

0

0.5

1

1.5

2

µmol/l

cmNitrate

0 5 10 15 20 25 30

0

5

10

15

20

µmol/l

cmAmmonium

0 10 20 30 40 50 60

0

5

10

15

20

µmol/l

cm

-2 -1Sediment oxygen flux

200 300 400 500 600 700 800 900 1,0000

2

4

6

8

10

12

Day

mmol m d

Model use

• Interpolate scarce and expensive data• Estimate annual budgets• Test hypotheses about ecosystem

functioning at continental slope (e.g. effect of enhanced mixing in water column)

Remarks

• Dichotomy tactic-strategic is not absolute in terms of model use:– Application of strategic model to understand patterns in

the real world– Application of tactic model to test hypotheses on

potential structuring forces

• Dichotomy is not absolute in terms of model construction, e.g.:– Use of experimental data to constrain strategic model– Use of abstract closure terms in strategic model

Continuous vs. discrete time models

Continuous time: • differential equations

• Note: time may be discretized for numerical solution, but model is formulated in continuous time

Ndt

dN

Continuous vs. discrete time modelsDiscrete time models:

• time is split up in discrete steps• • model is formulated as transition steps from one

time step to the next: difference equation :

Ex. : Nt+t = R + Nt

• used in population dynamics, where a time step may be a season, a generation,..

Continuous vs. discrete time modelsContinuous time model:

Discrete time model: Nt+t = Nt

• value of parameter fundamentally depends on value of time step:

• finite rates ( ) different from instantaneous rates (λ).

Ndt

dN

Ex.: Leslie matrix models

01

1

iforNpN t

it

ii

1

110

i

ti

ti

NfN

survival

birth

tt

N

N

N

N

N

s

s

s

s

sfsfsf

N

N

N

N

N

5

4

3

2

1

5,4

4,3

3,2

2,1

1,01,01,0

1

5

4

3

2

1

0000

0000

0000

0000

54300

SolutionTime T

0

0.05

0.1

0.15

0.2

0.25

0.3

1 2 3 4 5 6 7 8 9

AGE CLASS

(-)

• Derive stable age distribution• Derive population consequences of changes in age-

specific survival or fecundity characteristics• Applied models mainly for large animals (e.g. hunting

models)

Advanced age-structured models

• Consider density a function of both age and time: N(a,t)• Then the fundamental equation is the McKendrick-von

Forster equation:

• Solution: usually through transformation into delay-differential equation

),(),( taNtama

N

t

N

dttaNtaftN ),(),(),0(

Stochastic vs. deterministic models• Stochastic: model individual behaviour at microscopic scale.

Chance elements important• Deterministic approach: do not resolve individual molecules,

individuals, etc.. but average over appropriate space/time and describe dynamics of the average.

Microscopic, stochastic model

Macroscopic, continuous model

Macroscopic description by integration

Level of abstraction

• physiological processes (e.g. biochemical processes determining the growth of an algal cell)

• Individual behaviour (e.g. migration decisions, predator behaviour)

• Population and community processes (e.g. predator-prey relationship with movement)

• Ecosystem processes (e.g. ecosystem model for N.E. Atlantic)

What level to choose?

• Basically depends on the question• Models tend to describe dynamics at one

level based on processes at a lower level• Models cannot cross all levels without the

penalty of being untractable• Upscaling from one level to the next is

always problematic, as process descriptions need to be parameterised

Concluding remark

• Modelling may make sense of difficult data sets

• But don’t get trapped – think clearly !