201501 UEMX 3613 Topic3 Population

7/24/2019 201501 UEMX 3613 Topic3 Population

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Population growth&

Consumption of resources

1. Exponential growth model2. Logistic growth model

Chapter 3: Masters & Ela


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Out of 7 billions

Poverty 1/5 in bad health


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Cannot StandAlready!!!

Beh Tahan !!!


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Exponential growth model

Exponential growth is the growth of a system inwhich the amount being added to the system isproportional to the amount already present: thebigger the system is, the greater the increase.

Exponential function is a very useful mathematicaltool used in environmental studies, e.g. populationgrowth, resource consumption, pollutionaccumulation and radioactive decay.

It is a first-order rate process and overall theexponential growth can be given as N = N 0e rt


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Exponential growth model

N0 =initial number ofspecies,Nt = number ofspecies after time t ,

and r = growth rate,

General forms:

Nt = N0(1 + r)t

N = N0e rt


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Annual compounding (discrete, year-by-year increases)

Suppose something grows by a fixed percentage per year.

Imagine our saving at a bank earning 5% interest each year,compounded once a year, then the amount of increase in savings overany given year is 5% of the amount available at the beginning of thatyear .

If we start now with RM1000, then at the end of one year we wouldhave RM1050. At the end of two years, we would have RM1102.50(5% x 1050 + 1050); and so on.

N1= N 0(1+r); N 2 = N 1(1+r) = N 0(1+r) 2

General form: N t = N 0(1+r) t

N0 = initial amountNt = amount after t yearsr = growth rate (fraction per year)


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Continuous compounding (continuous growth)

For most events of interest in the environment, it is usually assumed

that the growth curve is a smooth, continuous function. This isreferred to as continuous compounding.

The rate of change of the quantity N is proportional to N. Theproportionality constant r is called the rate of growth and has units of

time-1

.N = N0e rt

integration

Example : Starting with the 2005 electricity consumption of 4.0 x10 12 kWhr/yr, what would consumption be in 2050 if the growthrate remains constant at 1.8 percent?

N = N 0e rt

N = 4.0 x 10 12 x e 0.018 x 45

N = 8.99 x 1012

kWhr/yr


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Doubling time

A quantity that is growingexponentially requires afixed amount of time todouble in size, regardless

of the starting point. It takes the same amount

of time to grow from N 0 to 2N 0 and from 2N 0 to4N0.

2N0 = N0e rt d , where t d isdoubling time (t d = ln2/r)


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Exponential DecayWhen the rateof decrease ofa quantity isproportionalto the amountpresent,exponential

growthbecomesexponentialdecay.

N = N 0 e-kt

k= reaction rateconstant (time -1)

Half life, t 1/2 = ln2 / k


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Half-life

Half-life, t ,is the time required for half of a substance to decayinto other elements. Example : if half life is 1 year and initial mass

of the substance is 100 grams, then after 1 year 50 grams will

remain; 25 g after another year and so on.This concept is especially useful for radioactive isotopes (Table 2.6,

Masters & Ela)

Exponential decay rate can be described using a reaction ratecoefficient ( k, time -1 ) or a half-life (t ).


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Disaggregated Growth Rates; a product of anumber of individual factors:

Affluence isenergydemand perperson

Technology isthe carbonemissions perunit of energy

If each factor grows exponentially, P i=p ie r it , then the total rate of growthis the sum of the individual rates (r=r 1 + r 2 + r 3+ r n).Final estimating growth: P=P

oe rt

GDP = Gross domestic product


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Resource consumption

Then time required to produceamount of resource ,Q, can beestimated: According to this model, if

resource production continues togrow exponentially for a longperiod of time, the numberbecomes unrealistically large.

When a mineral is extracted from the

Earth, we will say the resource isbeing produced.

Q = Total resource produced from

time 0 to time tP o= initial production rater = exponential growth rate inproduction


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Example 3.6 World Coal Production

World coal production in 2005 was estimated to be 6.1 billion (short)

tons per year, and the estimated total recoverable reserves of coalwere estimated at 1.1 trillion tons. Growth in world coal production inthe previous decade averaged 1.9 percent per year. How long wouldit take to use up those reserves at current production rates, and howlong would it take if production continues to grow at 1.9 percent?

At those rates (constant production rate), coal reserves would last

Reserves/production = (1.1 x 10 12 tons)/ (6.1 x 10 19 tons/yr) = 180 years

If production grows exponentially, we need to use

t = 78 years

Even though the growth rate of 1.9 percent might seen modest, comparedto a constant production rate, it cuts the length of time to deplete thosereserves by more than half.


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A radioactive chemical has an activity of 10,000Bq. What is theactivity of this chemical after 2 half-lives have passed?

After 2 half-lives, the activity is 2,500Bq.

10,000 5000 2500

228 Ac has a half life of 6.13 hours. How much of a 5.0 mgsample would remain after one day?

t1/2 = ln2 / k

Reaction rate coefficient, k = ln2 / t 1/2 = ln2 / 6.13 = 0.113 /hour

N=N oe -kt = 5mg x e -0.113 x 24 = 0.33 mg


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World population in 1850 has been estimated at about0.5 billion. World population reached 4 billion in 1975. If

we assume exponential growth at a constant rate overthat period of time, what would that growth rate be?


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Suppose world carbon emission are expressed as the followingproduct:

Carbon emission = (energy/person) x (carbon/energy) x (population)

If per capita energy demand increases at 1.5 percent per year,fossil fuel emissions of carbon per unit of energy increase at 1percent per year, and world population grows at 1.5 percent peryear. Total carbon emission in 1990 was 5 x 10 9 tonnes C/yr, whatwould the carbon emission rate in 2020.

The overall growth rate, r = 1.5 + 1.0 + 1.5 = 4.0 percent

Carbon emission rate in 2020 = 5 x 10 9 x e 0.04x30= 1.66 X 10 10 tonnes C/yr


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Logistic growth

Such growth indicates that initially the rate is exponentialfollowed by slower rates as the population reaches its carryingcapacity (K). It is a common successful method in biologicaland microbiological studies to reflect growth of living

organisms.


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The logistic curve is derivedfrom the following differentialequation

where N =the population size,K =the carrying capacity, r=the

exponential growth rate

(1-N/K) = environmentalresistance

t N (

For projections of population growth,

a logistic or S-shaped (sigmoidal)

growth curve is normally used.


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The factor (1 N/K) = the environmental resistance, asthe population grows, the resistance to furtherpopulation growth continuously increases.

To calculate the population at time t

To find r, a factor of instantaneous rate constant, R 0, is introduced:

If carrying capacity is known then it is possible toestimate time required for certain population increase.


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Example 3.8 Logistic Human Population Growth

Suppose the human population follows a logistic curve until itstabilizes at 15.0 billion. In 2006, the worlds population was 6.6billion, and its growth rate was 1.2 percent. When would thepopulation reach 7.5 billion one half of its assumed carryingcapacity?

Find r using

The time required to reach a population of 7.5 billion

The Earth would reach one-half its carrying capacity in 2017.


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For logistic growth, the maximum sustainable yield isobtained when the population is half the carryingcapacity, N* = K/2.The maximum sustainable yield is the maximum ratethat individuals can be harvested (removed) withoutreducing the population size.

Eq3.29

Eq3.30

Express in term of the current growth rate, R o and current size, N o:


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The maximum number of individuals that can be harvested right nowthat will allow the population to return to its carrying capacity as quicklyas possible, making it ready for another productive harvest.

When a population is exactly halfway to its carrying capacity, it isgrowing at its fastest rate. This means that if we stop harvesting at halfits carrying capacity, the population can quickly recover and allow for asuccessful harvest time after time.


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Example:

A population of 50 sheep doubled in population size after 1 year andreached stability at 3000 sheep after a few years. Considering that suchgrowth is logistic, we can estimate maximum sustainable yield:

1) Find R 0 from the doubling time equationR 0 = ln2/t d = ln2 / 1 year = 0.693/yr;

2) carrying capacity, K = 3000 ,since the size remains steady for sometime andN0 = 50

3) As the initial population, N 0 is small relative to the carrying capacity, K(N


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Human population growth, definitions:

Crude birth rate, b , which is the number of live births

per 1000 population in a given year. In thedeveloping countries this rate reaches 30-40, and indeveloped countries it is about 10.

Crude Death Rate, d , which is the number of deathsper 1000 population per year.

Infant mortality rate, the number of deaths to infants(under one year old) per 1000 live births in a givenyear. One of the best indicators of poverty in acountry.


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Example: In 2006, over 80 % of the worlds population, some 5.3billion people, lived in the less developed countries of the world.In those countries, the average crude birth rate was 23, crude

death rate was 8, and the infant mortality rate was 53. Whatfraction of the total deaths is due to infant mortality?

Total live birth = Population x crude birth rate

= 5.3 x 109

x (23/1000)= 121.9 x 10 6 per year

Infant death = Total birth x Infant mortality rate= 121.9 x 10 6 x (53/1000)= 6.5 x 10 6 per year

Total death = Population x Crude death rate= 5.3 x 10 9 x (8/1000)

= 42.4 x 10 6 per year

Fraction infants = 6.5 / 42.4 = 0.15 = 15 percent


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Rate of natural increase, r is the difference between

crude birth rate and crude death rate.r = b d

Net migration rate, m , is the difference betweenimmigration and emigration

r= b d + m

Total Fertility Rate (TFR) is the average number ofchildren that would be born alive to a woman, assumingthat current age-specific birth rates remain constantthrough the womans reproductive years (how manychildren each woman is likely to have in her lifetime).



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Replacement level fertility is the number of childrenthat a woman must have, on the average, to replaceherself with daughter in the next generation. Itaccounts for differences in the ratio of male tofemale births as well as child mortality rates

By definition, replacement is only considered to haveoccurred when the offspring reach 15 years of age.



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Age structure

Age structure reflects a countrys population trends (apopulation pyramid showing year of the births andnumber of women and men born in those years).

A graphical presentation of the data indicating

numbers of people (or % of the population) in each agecategory is called age structure or population pyramid.

We can picture a populations age structure as a pile ofblocks, one for each age group, with the size of eachblock representing the number of people in that group.

Four general types: a pyramid, a column, an invertedpyramid, and a column with a bulge


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Slightly more boys in the younger age groups than girls;

however, the ratio tends to reverse in the upper age groups, asfemales tend to outnumber males. Many countries have afemale majority as a result of the longer life expectancy forfemales.

Notice that at about age 35, the majority changes.

Age and sex distribution


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A Pyramid

The pyramid age structure occurs in a population that has manyyoung people and a high death rate at each age, and therefore ahigh birth rate, characteristic of a rapidly growing population andalso of a population with a relatively short average lifetime.


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A Column

A column shape occurs where the birth rate and death rateare low and a high percentage of the population is elderly.


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Japan has the largest elderly population in the world. In one respect

that is good. But less good is not having babies, and a situation ofout-migration almost as large as in-migration into the country. Japanesemarriage customs make most young Japanese women disinterested inmarrying into a family where they might be dominated by their motherin law. As a result, Japans birth rate is far below a replacement level.By 2055 the age pyramid will likely be inverted.

An Inverted Pyramid

An inverted pyramid occurs when a population has more older thanyounger people.


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A column with a bulge

A bulge occurs if some event in the past caused a high birthrate or death rate for some age group but not others.China had an extreme youth bulge when it sharply curbedpartly as an effect of the one-child policy.


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Kenyas pyramid -shaped reveals a rapidly growing population heavily

weighted toward youth. About 34% of the populations are under 15 years ofage. Such an age structure indicates that the population will grow veryrapidly in the future, when the young reach marriage and reproductive ages.

For the U.S., the age structure is more like a column, showing a populationwith slow growth.

Italys top -heavy pyramid shows a nation with declining growth.


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Factors limiting growth of population:

Short-term factor is the disruption of fooddistribution in a country, commonly caused bydrought, energy shortage for food transportation.

Intermediate-term factors include desertification,dispersal of toxic pollutants; disruption of energysupplies;

Long-term factors include soil erosion, a decline ingroundwater supplies, and climate change.

Natural disasters of high magnitude (e.g. tsunami,earthquake) can result in significant suddendecrease of population.


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Actual situationon humanpopulationgrowth

Example: Human population in 1960was 3 billion with a growth rate of1.2%. Estimate population size in2010.N0 = 3x10 9; r=0.012; t=50 yrs

N = N 0xe rt = 5.5 x10 9

Use the same data and assume thathuman population growth follows alogistic model where K=150billions.


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The following statistics are for India in 1985: population, 762 million;crude birth rate, 34; crude death rate, 13; infant mortality rate, 118(rates are per thousand per year). Find(a) The fraction of the total deaths that are infants less than 1 year old

(b) The avoidable deaths, assuming that any infant mortality above 10could be avoided with better sanitation, food, and health care; and

(c) The annual increase in the number of people in India

Total birth = 762 x 10 6 x (34/1000) = 25.9 million/yrInfant death = Total birth x Infant mortality rate = 25.9 x 10 6 x (118/1000)

= 3.06 million/yrTotal death = 762 x 10 6 x (13/1000) = 9.91 million/yrFraction infants = 3.06/9.91 = 0.31 = 31 percent

Infant death = 25.9 x 10 6 x (10/1000) = 0.259 million/yr

Avoidable deaths = 3.06 0.259 = 2.801 million/yr

Annual increase = population x net of natural increase= 762 x 10 6 x ((34-13)/1000)= 16 million/yr


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Suppose some sewage drifting down a streamdecomposes with a reaction rate coefficient k equal to

0.2/day. What would be the half-life of this sewage? Howmany percent of waste would be left after 5 days?

t1/2 = ln2 / k = 3.466 days

N=N oe -ktN/N o = 0.368

About 37% sewage remains.


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Resource consumption

World reserves of chromium are about 800 million tons, and currentusage is about 2 million tons per year. If growth in demand forchromium increases exponentially at a constant rate of 2.6 percentper year, how long would it take to use up current reserves? Supposethe total resource is five times current reserves; if the use ratecontinues to grow at 2.6 percent, how long would it take to use up theresource?

Time = 93.6 years

If the total resource is five times current reserves,

t = 152.7 years

Multiply reserves by five times only increase the lifetime by a factorof 1.6.


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Suppose human population grows from 6.3 billion in 2000to an ultimate population of 10.3 billion following thelogistic curve. Assuming a growth rate of 1.5 percent in2000, when would the population reach 9 billion? Whatwould the population be in 2050?

r = 0.0386

t = 38.4 years

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201501 UEMX 3613 Topic3 Population