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LECTURE 11:MATEMATICAL MODEL

The essence of mathematics is not to make simple things complicated,but to make complicated things simple. ~S. Gudder

LECTURE OUTCOMESAfter mastering the lecture materials at the end ofthe lecture, students should be able1. to develop mathematical, mechanistic models based on

assumptions on the biological process of biomassproduction of plants

2. to explain possible, biological processes in the biomassproduction of plants

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LECTURE OUTLINE1. INTRODUCTION Conceptual Model Approach Mechanistic Approach

2. BIOMASS PRODUCTION MODELS Assumption 1: The rate of biomass production is constant Assumption 2: The rate of biomass production is

dependent upon the growth machinery Assumption 3: The rate of biomass production is

dependent upon the substrate Assumption 4: The rate of biomass production is

dependent upon the growth machinery and the substrate

INTRODUCTION

1. A model is a simplification of a system2. A system is very complex and not easy to study3. The application of model to study a complex

system such as plant growth is one ofapproaches

4. Therefore, it is very important to develop agood model of a system, a model that couldsimulate the behavior of the system underconsideration

WHY DO WE NEED MODELS?

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x yA B A/B α

123456789

10

A/B

α

What is the bestmodel for thisrelationship, whatis your assumption

CH2OG

Temperature Light Photosyntheticefficiency

Rate

DevelopmentPhase

Rm

PHS

P

LeafBiomass

SeedBiomass

StemBiomass

RootBiomass

LA

e

How things happen: What are The Factors determining TheBiomass Production ?,. Make it simple, not too complicated

• Conceptual Model Approach

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BIOMASS PRODUCTION=

Productivity per unit timex

Growing period=

W/ t x T

W/t = Productivity per unit timeT = Growing period

• Mechanistic Approach

The main key for the analysisis

W/tthe amount of water (δw)falls per unit time (δt)

What is W/t dependent on ?

1. Is W/t constant with time, notdependent on anything (any factor) ?

2. Is W/t dependent upon something ?What ?

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3. A student analyzed his/her datafollowing his/her friends method

4. Unfortunately, the results of analysiswere not clearly understood by thestudents (what the results did mean)

5. Empirical Approach = the data collectedand analyzed directly to get the insightof plant system reflected by the resultsof analysis

6. Mechanistic Approach = the behavior ofplant system in a particular environmentis analyzed first, then data werecollected to test the results of analysis(hypothesis)

A simple example of mechanistic models: Waterflow from a base apertureA Outlet velocity (m.s-1) can be expressed as

v = Cv (2 g H )1/2

whereCv = velocity coefficient (water 0.97)g = acceleration of gravity (9.81 m/s2)H = height (m)

H

A

Volume flow (m3.s-1) can be expressed asV = Cd A (2 g H)1/2 (1b)Cd = dischare coefficient (Cd = Cc Cv)where

Cc = contraction coefficient (sharp edge apperture 0.62, wellrounded apperture 0.97)

A = area apperture (m2)

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BIOMASS PRODUCTIONMODELS

1. The system of plants in the production ofbiomass (dry weight) can be simplified by twoprocesses;i. Photosynthetic process that converts CO2 into

sugars (simple carbohydrates) using the energy ofsolar radiation in natural condition

ii. Metabolic processes that convert sugars intovarious molecules (plant biomass)

2. Growth can be defined as the biochemicalconversion of reserve substances into structuraldry matter (Penning de Vries et al., 1989)

3. Structural dry matter consists of the organiccomponents that remain at the end of the plant'slife, that is, are not normally broken down.

4. In contrast, 'reserves' are components that onlyexist temporarily and are used for maintenanceor growth within hours or days (availablereserves) or after a few weeks (shieldedreserves).

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http://biologytb.net23.net/text/chapter8/concept8.1.html http://manet.illinois.edu/pathways.php

Biomass, W

5. The production of biomass can besimplified to be

6. We may assume that a plant is a factoryusing machines to process substrates toproduce particular product (W)

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ASSUMPTION 1 If it is assumed that the production of biomass

is not limited by the capacity of machine andthe quantity of substrate, then

W/t = constantW/t = kwhere k = a constant value

t

0

W

W

tkW0

ktWW 0

W-W0 = k(t-0)

Apply the model to data

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Is this right ?

y = 0.219x - 3.643R² = 0.824

0

5

10

15

20

0 20 40 60 80

Tota

l Bio

mas

s (g

/pla

nt)

Days after sowing

Discussion Is the model good enough to describe the

growth of plant under consideration?

Is the assumption used sound biologically? The production of biomass per unit time (day) is

constant for the entire life of plants?

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Kesimpulan1. Aplikasi dari pers dengan Asumsi 1

menghasilkan suatu koefisien korelasi yangcukup tinggi, sehingga model dari segi statistikcukup baik untuk menggambarkan hubunganantara biomassa tanaman dengan umur.

2. Dari segi biologis dan kenyataan di lapangan,model tersebut tidak cukup baik untukmenggambarkan keadaan yang sesungguhnya.

1. A growing plant from molecular to plant levelperforms the characteristic of autocatalyticsystem (self-reproduction) which uses itsproducts for its own formation. Molecular level. The photosynthetic carbon reduction

(PCR) or Benson-Calvin cycle is autocatalytic withregeneration of the CO2 acceptor (Ribulose 1,5bisphosphate).

Cellular level. New cells are produced from theexisting cell by cell devition

Organ level. An increase in leaf are, determining thequantity of light interception and carbohydrateproduction, is dependent on the supply ofcarbohydrate for leaf formation.

ASSUMPTION 2

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2. The autocatalytic system of plants leads tothe notion that the machinery of plantsproducing biomass is the plants themselves.

3. If total dry weight (W) is proportional toleaves, roots and metabolic capacity, then itmay be assumed that the quantity of growth machinery is proportional

to dry weight (W), the growth machinery works at a maximal rate so

long as there is any substrate available at all growth is irreversible and stops once the

substrate is exhausted

4. With the assumptions, the growth of plants(dry weight increase) is dependent uponthe initial magnitude of W (growthmachinery), so that The conversion of substrate to biomass is

controlled by the present total biomass (W)

dW/dtdW/dt

rWt

W

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The form of equation to describe thephenomenon is

where W is an increase in plant biomass, t is anincrease in time (day or week), w = plant biomass, and r= specific growth rate or relative growth rate (RGR) as

The integration of the differential equationgives

This an exponential equation

rWt

W

t

0

w

wtr

WWt

0

rtWW

ln0

t

rt0t eWW

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Is this right ?y = 0.119e0.071x

R² = 0.844

0

5

10

15

20

0 20 40 60 80

Tota

l Bio

mas

s (g

/pla

nt)

Days after sowing

Kesimpulan Aplikasi dari pers dengan Asumsi 2

menghasilkan suatu koeffisienkorelasi yang cukup tinggi .

Model ini cukup baikmenggambarkan keadaan awalpertumbuhan tanaman hinggaumur tertentu (mis. 40 hst), tetapitidak untuk masa pertumbuhanberikutnya.

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It may be assumed that1. the quantity of growth machinery is constant and

independent of dry weight (W),2. the growth machinery works at a rate proportional to

the substrate level (S)3. growth is irreversible

With the assumptions, the growth of plants (dryweight increase) is dependent upon substratelevel (S), so that

Assumption 3

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where k is constant The assumptions defines

plants as systems thatconvert substrate (S) to plantbiomass (W) as illustrated inthe figure.

kSt

W

tW

tW

tW

With the assumption that there is no gain or lossof material from the system, therefore

or

W+S = a constant = W0+S0 = Wf + Sf = C W0 and S0 are the initial values of W and S at

time t = 0; Wf and Sf are the final values of Wand S approached t (assuming that a steadystate is eventually reached); C is a constant.

tS

tW

0tS

tW

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With the assumption of W+S = W0+S0 = Wf + Sf,then at the final stage

Sf = 0 W+S = Wf + 0so that

S = Wf – W.

WWkt

Wf

In other words, the quantity ofsubstrate available converted toplant biomass at any time isproportional to a differencebetween the final dry weight andthe dry weight that has beenreached. This substitutes S inthe above equation which gives

Max

volu

me

=W

f

water= W

Ava

ilabl

e sp

ace

=W

f-W

The integration of the above equation gives

If the initial dry weight W0 = 0, thenor

tW

W f

tkWW

W

00

ktWWWW

f

f

0ln

ktff eWWWW 0

ktff eWWWW 0

ktff eWWWW 0

ktff eWWW kt

f eWW 1

Wf = Wmax

kteWW 1max

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Kesimpulan Aplikasi dari pers (6b) pada data yang

sama dengan yang disajikan pada Gambar7 menghasilkan suatu koefisien korelasiyang rendah

Jadi secara statistik, model tersebut tidakcukup baik untuk menggambarkanperkembangan biomassa tanaman denganwaktu, sekalipun dapat meliput sebarandata pada bagian akhir pertumbuhantanaman.

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Dengan demikian, produksi biomassatanaman secara umum tidak berhubungandengan kapasitas aktual tanaman(perbedaan kapasitas maksimum dengankapasitas yang sudah dipakai) untukmenampung produk biomassa.

Sekalipun asumsi ini cukup logis secarafisik tetapi tidak secara dinamikapertumbuhan tanaman karena kapasitasyang belum terpakai pada awalpertumbuhan besar yang tidak akanmembuat produksi biomassa tanaman yangbesar.

Ini didasarkan atas kenyataan bahwakuantitas organ fotosintesis (daun), yangberkembang dengan waktu, sangat rendahpada awal pertumbuhan.

Pengaruh dari kapasitas tanaman yang belumterpakai untuk menampung produk biomassadapat nyata pada bagian akhir pertumbuhan

Kapasitas yang belum terpakai pada bagianakhir pertumbuhan pada sebagian tanamandidominasi oleh bagian generatif seperti biji

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Hasil analisis diatas mengisyaratkankombinasi asumsi 2 dan 3 mungkin dapatmenghasilkan model pertumbuhan yangdapat menjelaskan kinerja sistem tanamandalam produksi biomassa.

Jadi produksi biomassa tanamanditentukan oleh kapasitas mesin dankapasitas tanaman yang tersedia untukmenampung produksi biomassa tersebut.

Assumption 4.

Model dari sistem tanaman dalam bentukdiagaram alir sama dengan Gambarsebelumnya dan model dalam bentukpersamaan matematis adalah

)WmaxW(rWt

W

rWSt

W

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Integrasi persamaan diatas dapat dilakukanlebih mudah dengan cara parsial setelahpengaturan persamaan seperti berikut.

max)W/W1(Wt

W

t)W/W1(W

W

max

tW)W/)WW(W

1

maxmax

If r.Wmax =mthenmax

maxmax

...

WWWrW

WrWt

W

Integrasi dari persamaan diatas akanmenghasilkan suatu persamaan yangdikenal dengan persamaan logistik

tW)WW(W

W

max

max

tWW1

WW1

max

t

0

w

w maxtW

W1

WW1t

0

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atau

Pembagian pembilang dan penyebut dari ruaskanan pers (7) dengan W0 akan menghasilkan

dimana a = (Wmax - W0)/W0.

ut00max

utmax0

eWWW

eWWW

ut0max0

max0

e)WW(W

WWW

ut

0

0max

max

eW

WW1

WW

utmax

e.a1

WW

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1. Aplikasi dari pers (8) menghasilkan suatukoefisien korelasi yang tinggi

2. Jadi secara statistik dan biologis, modellogistik merupakan yang paling baik untukmenggambarkan kinerja dari sistemtanaman dalam pertumbuhan untukmenghasilkan biomassa.

Kesimpulan

3. Dengan demikian kinerja sistem tanaman dalambiomassa total tanaman dikendalikan olehkemampuan mesin pertumbuhan dalam menghasilkanbiomassa (fotosintesis dan metabolisme) dankapasitas total lubuk (sink) tanaman untukmenampung produksi biomassa.

4. Kapasitas total lubuk (sink) tanaman yang tersediauntuk menampung biomassa pada Gambar 12 adalah405 kali dari biomassa awal (W0) yangmenggambarkan potensi tanaman pada lingkungantertentu.

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PM

W

PM

W

PM

W

WmaxdW

PM

W

WmaxdW

Linear Model Exponential Model

Monomolecular Model Logistic Model

1. Linear Model

2. Exponential Model

3. Monomolecular Model

4. Logistic Model

5. Others?

utmax

e.a1

WW

)e1(WW rtmaxt

rt0t eWW

ktWW 0

SUMMARY

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