Post on 31-Dec-2015
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 !