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Kamis, 15 Februari 2018
MATA KULIAH:
PEMODELAN MATEMATIKA
Pengajar:
Heni Widayani, M.Si
GENERAL COMPARTMENT MODEL
Assumption :
 The population are sufficiently large
 Random fluctuation between individual are ignored and treat each individual
as being identical.
 Each individu has an equal chance of giving birth (per capita birth rate 𝛽 per
unit time)
 Each individual has an equal chance of dying within a given time interval.
(per capita death rate 𝛼 per unit time)
 Births and deaths are continuous in time
 Per capita birth and death rate are constant in time.
 We ignore immigration and emigration.
WORLD
Births deaths
town, organization, ocean
𝑟𝑎𝑡𝑒 𝑜𝑓 𝑝𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑠𝑖𝑧𝑒
𝑐ℎ𝑎𝑛𝑔𝑒
=
𝑟𝑎𝑡𝑒 𝑜𝑓
𝑏𝑖𝑟𝑡ℎ𝑠
−
𝑟𝑎𝑡𝑒 𝑜𝑓
𝑑𝑒𝑎𝑡ℎ𝑠
Place 𝜷
(per year per individu)
𝜶
(per year per individu)
World (1990) 0.027 0.010
EXPONENTIAL MODEL
 𝑥0 is initial value of population
 𝛽 is constant per capita birth rate
 𝛼 is constant per capita death rate
 𝑟 is the growth rate or the reproduction rate
(When 𝑟 > 0 a model describing exponential growth.
When 𝑟 < 0 a model describing exponential decay)
𝑑𝑋
𝑑𝑡
= 𝛽𝑋 − 𝛼𝑋 = 𝑟𝑋, 𝑋 0 = 𝑥0
An exponential distribution is the natural distribution for compartmen model.
Change in population at any time is proportional to the size of the population
at that time
𝑋 𝑡 = 𝑥0 𝑒 𝑟𝑡
Find the time for the population to double the size ?
INTERPRETATION OF PARAMETERS
𝑛𝑢𝑚𝑏𝑒𝑟
𝑜𝑓 𝑑𝑒𝑎𝑡ℎ𝑠
𝑖𝑛 𝑡𝑖𝑚𝑒 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 ∆𝑡
≈ 𝛼𝑋(𝑡)∆𝑡
Let us now suppose that 𝑥0 people will die in time 𝑡1
(𝑡1 𝑖𝑠 𝑡ℎ𝑒 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑙𝑖𝑓𝑒 𝑒𝑥𝑝𝑒𝑐𝑡𝑎𝑛𝑐𝑦). Let 𝑋 𝑡 = 𝑥0 and ∆𝑡 = 𝑡1, then we have
𝑥0 ≈ 𝛼𝑥0 𝑡1
𝛼 ≈
1
𝑡1
Per capita death rate in developing country 𝛼 ≈ 0.007(years-1 or 7 deaths
per 1000 persons per year). The average life expectancy of
1/𝛼=1/0.007=140 years. Is to high for human!
The real age distribution does not approximately follow an exponential
distribution (the population fall off rapidly at older ages)
DENSITY DEPENDENT GROWTH
Population are observe over long periods, they often appear to reach a
limit or to stabilise.
𝑟𝑎𝑡𝑒 𝑜𝑓 𝑝𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑠𝑖𝑧𝑒
𝑐ℎ𝑎𝑛𝑔𝑒
=
𝑟𝑎𝑡𝑒 𝑜𝑓
𝑏𝑖𝑟𝑡ℎ𝑠
−
𝑛𝑜𝑟𝑚𝑎𝑙 𝑟𝑎𝑡𝑒 𝑜𝑓
𝑑𝑒𝑎𝑡ℎ𝑠
−
𝑟𝑎𝑡𝑒 𝑜𝑓 𝑑𝑒𝑎𝑡ℎ𝑠
𝑏𝑦 𝑐𝑟𝑜𝑤𝑑𝑖𝑛𝑔
Assumption :
• Individuals compete for the limited resources available (𝐾 as carrying
capacity for the population means as the limited number individuals
that environtment can support)
• Excluding external factors such as harvesting or interaction with
another population
𝑑𝑋
𝑑𝑡
= 𝛽𝑋 − 𝛼 + 𝛾𝑋 𝑋 = 𝑟𝑋 − 𝛾𝑋2
, 𝑋 𝑡 = 𝑥0
• 𝛽 is constant per capita birth rate
• 𝛼 is the per capita death rate due to natural attrition
• 𝛾 is the per capita dependence of deaths on the population size
• 𝑟 is the reproduction rate
𝑟𝑎𝑡𝑒 𝑜𝑓 𝑝𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑠𝑖𝑧𝑒
𝑐ℎ𝑎𝑛𝑔𝑒
=
𝑟𝑎𝑡𝑒 𝑜𝑓
𝑏𝑖𝑟𝑡ℎ𝑠
−
𝑟𝑎𝑡𝑒 𝑜𝑓
𝑑𝑒𝑎𝑡ℎ𝑠
𝑑𝑋
𝑑𝑡
= 𝛽𝑋 − 𝛼𝑋 − 𝛾𝑋2 = 𝑟𝑋 − 𝛾𝑋2, 𝑋 𝑡 = 𝑥0
INTERPRETATION
 𝐾 =
𝑟
𝛾
is the carrying capacity
 The equation is called the logistic equation / limited growth model /
the density dependent model
 𝑅(𝑋) represent a population dependent per-capita growth rate.
𝑅 𝑋 = 𝑟 1 −
𝑋
𝐾
 𝑅(𝑋) is a liner function of 𝑋
lim
𝑋→𝐾
𝑅 𝑋 = 0 and lim
𝑋→0
𝑅 𝑋 = 𝑟
 The equilibrium of (1) is 𝑋 𝑒 = 0 and 𝑋 𝑒 = 𝐾
 Find analytic solution and consider the second derivative !
𝑑𝑋
𝑑𝑡
= 𝑟𝑋 − 𝛾𝑋2 = 𝑟 1 −
𝑋
𝐾
𝑋, 𝑋 𝑡 = 𝑥0 (1)
𝑅 𝑋
Unstable stable
𝑿 𝒕 =
𝑲
𝟏 + 𝒎𝒆−𝒓𝒕 , 𝒘𝒉𝒆𝒓𝒆 𝒎 =
𝑲
𝒙 𝟎
− 𝟏
LIMITED GROWTH WITH HARVESTING
𝑟𝑎𝑡𝑒 𝑜𝑓 𝑝𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑠𝑖𝑧𝑒
𝑐ℎ𝑎𝑛𝑔𝑒
=
𝑟𝑎𝑡𝑒 𝑜𝑓
𝑏𝑖𝑟𝑡ℎ𝑠
−
𝑛𝑜𝑟𝑚𝑎𝑙 𝑟𝑎𝑡𝑒 𝑜𝑓
𝑑𝑒𝑎𝑡ℎ𝑠
−
𝑟𝑎𝑡𝑒 𝑜𝑓 𝑑𝑒𝑎𝑡ℎ𝑠
𝑏𝑦 𝑐𝑟𝑜𝑤𝑑𝑖𝑛𝑔
−
𝑟𝑎𝑡𝑒 𝑜𝑓 𝑑𝑒𝑎𝑡ℎ𝑠
𝑏𝑦 ℎ𝑎𝑟𝑣𝑒𝑠𝑡𝑖𝑛𝑔
𝑑𝑋
𝑑𝑡
= 𝑟𝑋 1 −
𝑋
𝐾
− ℎ, 𝑋 0 = 𝑥0
• ℎ is the constant rate of harvesting (total number caught per
unit time /deaths due to harvesting per unit time)
𝑑𝑋
𝑑𝑡
= −
𝑟
𝐾
𝑋2 − 𝐾𝑋 +
𝐾ℎ
𝑟
, 𝑋 0 = 𝑥0
Let 𝑟 = 1, 𝐾 = 10, ℎ = 9/10 and 𝑋 0 = 𝑥0. Find the solution and the
equilibrium point!
• Harvesting rate (can be some disaster) causes the population
stabilise to a value less than carrying capacity.
• In the case of harvesting, or some other disaster, causing
population to decrease below a critical level (the population will
become extinct).
DISCRETE POPULATION GROWTH AND CHAOS
Assumption :
 The population does not change except at dicrete intervals,
corresponding to breeding season.
 The population change occur from births or death alone.
 Deaths are due to normal causes and also to crowding effects
𝑋 𝑛+1 − 𝑋 𝑛 = 𝛽𝑋 𝑛 − 𝛼𝑋 𝑛 − 𝛾𝑋 𝑛
2
, 𝑛 = 0,1,2,3, …
• Set 𝑟 = 𝛽 − 𝛼 and 𝛾 =
𝑟
𝐾
, we get
𝑋 𝑛+1 = 𝑋 𝑛 + 𝑟𝑋 𝑛 1 −
𝑋 𝑛
𝐾
, 𝑛 = 0,1,2,3, …
• This equation called discrete logistic equation.
LOGISTIC EQUATION WITH TIME LAG
If we suppose that the logistic equation applied at the earlier time 𝑡 − 𝜏,
where 𝜏 represents the time delay between increased deaths and the
resulting decrease in population reproduction, we obtain
𝑑𝑋
𝑑𝑡
= 𝑟𝑋(𝑡) 1 −
𝑋(𝑡 − 𝜏)
𝐾
We called the equation as a differential-delay equation.
The effect of the time delay has been to induce some oscillations in the
population.
TUGAS
1. Pada sebuah tambak ikan, ikan dipanen dengan laju konstan sebesar 2100
ikan per minggu. Laju kematian ikan per kapita sebesar 0.2 ikan per hari per
ikan dan laju bertelurnya ikan sebesar 0.7 ikan per hari per ikan.
(a) Tuliskan model matematika dari laju perubahan populasi ikan ini dalam
persamaan diferensial? (Note:tuliskan definisi dari simbol /parameter yang
Anda gunakan)
(b) Jika pada awalnya terdapat 240.000 ekor ikan, berapakah jumlah ikan yang
ada dalam satu minggu ke depan ?
(c) Tentukan titik kesetimbangan dari populasi ikan tersebut (jika ada)!
2. Model dari penyebaran teknologi sama dengan model logistik dari pertumbuhan
populasi. Misalkan N(t) adalah jumlah dari individu masyarakat yang telah
menerima kursus/pelatihan pembuatan website di UIN. Laju N(t) akan mengikuti
persamaaan
𝑑𝑁
𝑑𝑡
= 𝑎𝑁 1 −
𝑁
𝑀
dengan M adalah total populasi dari masyarakat. Diasumsikan bahwa laju
penyebaran informasi tersebut sebanding dengan banyak individu yang telah
menerima pelatihan dan sebanding dengan banyak interaksi individu yang telah
dllatih dengan individu yang belum dilatih
(a) Bagian mana dari model tersebut yang menggambarkan banyaknya individu
yang belum menerima pelatihan?
(b) Misalkan M=17015, 𝑎 = 0.490, dan 𝑁0 = 141. Tentukan berapa lama waktu yang
dibutuhkan untuk melatih /memberikan informasi tentang website ke 80% dari
populasi ?

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Single Population Model

  • 1. Kamis, 15 Februari 2018 MATA KULIAH: PEMODELAN MATEMATIKA Pengajar: Heni Widayani, M.Si
  • 2. GENERAL COMPARTMENT MODEL Assumption :  The population are sufficiently large  Random fluctuation between individual are ignored and treat each individual as being identical.  Each individu has an equal chance of giving birth (per capita birth rate 𝛽 per unit time)  Each individual has an equal chance of dying within a given time interval. (per capita death rate 𝛼 per unit time)  Births and deaths are continuous in time  Per capita birth and death rate are constant in time.  We ignore immigration and emigration. WORLD Births deaths town, organization, ocean 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑝𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑠𝑖𝑧𝑒 𝑐ℎ𝑎𝑛𝑔𝑒 = 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑏𝑖𝑟𝑡ℎ𝑠 − 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑑𝑒𝑎𝑡ℎ𝑠 Place 𝜷 (per year per individu) 𝜶 (per year per individu) World (1990) 0.027 0.010
  • 3. EXPONENTIAL MODEL  𝑥0 is initial value of population  𝛽 is constant per capita birth rate  𝛼 is constant per capita death rate  𝑟 is the growth rate or the reproduction rate (When 𝑟 > 0 a model describing exponential growth. When 𝑟 < 0 a model describing exponential decay) 𝑑𝑋 𝑑𝑡 = 𝛽𝑋 − 𝛼𝑋 = 𝑟𝑋, 𝑋 0 = 𝑥0 An exponential distribution is the natural distribution for compartmen model. Change in population at any time is proportional to the size of the population at that time 𝑋 𝑡 = 𝑥0 𝑒 𝑟𝑡 Find the time for the population to double the size ?
  • 4. INTERPRETATION OF PARAMETERS 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑑𝑒𝑎𝑡ℎ𝑠 𝑖𝑛 𝑡𝑖𝑚𝑒 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 ∆𝑡 ≈ 𝛼𝑋(𝑡)∆𝑡 Let us now suppose that 𝑥0 people will die in time 𝑡1 (𝑡1 𝑖𝑠 𝑡ℎ𝑒 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑙𝑖𝑓𝑒 𝑒𝑥𝑝𝑒𝑐𝑡𝑎𝑛𝑐𝑦). Let 𝑋 𝑡 = 𝑥0 and ∆𝑡 = 𝑡1, then we have 𝑥0 ≈ 𝛼𝑥0 𝑡1 𝛼 ≈ 1 𝑡1 Per capita death rate in developing country 𝛼 ≈ 0.007(years-1 or 7 deaths per 1000 persons per year). The average life expectancy of 1/𝛼=1/0.007=140 years. Is to high for human! The real age distribution does not approximately follow an exponential distribution (the population fall off rapidly at older ages)
  • 5. DENSITY DEPENDENT GROWTH Population are observe over long periods, they often appear to reach a limit or to stabilise. 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑝𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑠𝑖𝑧𝑒 𝑐ℎ𝑎𝑛𝑔𝑒 = 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑏𝑖𝑟𝑡ℎ𝑠 − 𝑛𝑜𝑟𝑚𝑎𝑙 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑑𝑒𝑎𝑡ℎ𝑠 − 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑑𝑒𝑎𝑡ℎ𝑠 𝑏𝑦 𝑐𝑟𝑜𝑤𝑑𝑖𝑛𝑔 Assumption : • Individuals compete for the limited resources available (𝐾 as carrying capacity for the population means as the limited number individuals that environtment can support) • Excluding external factors such as harvesting or interaction with another population 𝑑𝑋 𝑑𝑡 = 𝛽𝑋 − 𝛼 + 𝛾𝑋 𝑋 = 𝑟𝑋 − 𝛾𝑋2 , 𝑋 𝑡 = 𝑥0 • 𝛽 is constant per capita birth rate • 𝛼 is the per capita death rate due to natural attrition • 𝛾 is the per capita dependence of deaths on the population size • 𝑟 is the reproduction rate 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑝𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑠𝑖𝑧𝑒 𝑐ℎ𝑎𝑛𝑔𝑒 = 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑏𝑖𝑟𝑡ℎ𝑠 − 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑑𝑒𝑎𝑡ℎ𝑠 𝑑𝑋 𝑑𝑡 = 𝛽𝑋 − 𝛼𝑋 − 𝛾𝑋2 = 𝑟𝑋 − 𝛾𝑋2, 𝑋 𝑡 = 𝑥0
  • 6. INTERPRETATION  𝐾 = 𝑟 𝛾 is the carrying capacity  The equation is called the logistic equation / limited growth model / the density dependent model  𝑅(𝑋) represent a population dependent per-capita growth rate. 𝑅 𝑋 = 𝑟 1 − 𝑋 𝐾  𝑅(𝑋) is a liner function of 𝑋 lim 𝑋→𝐾 𝑅 𝑋 = 0 and lim 𝑋→0 𝑅 𝑋 = 𝑟  The equilibrium of (1) is 𝑋 𝑒 = 0 and 𝑋 𝑒 = 𝐾  Find analytic solution and consider the second derivative ! 𝑑𝑋 𝑑𝑡 = 𝑟𝑋 − 𝛾𝑋2 = 𝑟 1 − 𝑋 𝐾 𝑋, 𝑋 𝑡 = 𝑥0 (1) 𝑅 𝑋 Unstable stable 𝑿 𝒕 = 𝑲 𝟏 + 𝒎𝒆−𝒓𝒕 , 𝒘𝒉𝒆𝒓𝒆 𝒎 = 𝑲 𝒙 𝟎 − 𝟏
  • 7. LIMITED GROWTH WITH HARVESTING 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑝𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑠𝑖𝑧𝑒 𝑐ℎ𝑎𝑛𝑔𝑒 = 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑏𝑖𝑟𝑡ℎ𝑠 − 𝑛𝑜𝑟𝑚𝑎𝑙 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑑𝑒𝑎𝑡ℎ𝑠 − 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑑𝑒𝑎𝑡ℎ𝑠 𝑏𝑦 𝑐𝑟𝑜𝑤𝑑𝑖𝑛𝑔 − 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑑𝑒𝑎𝑡ℎ𝑠 𝑏𝑦 ℎ𝑎𝑟𝑣𝑒𝑠𝑡𝑖𝑛𝑔 𝑑𝑋 𝑑𝑡 = 𝑟𝑋 1 − 𝑋 𝐾 − ℎ, 𝑋 0 = 𝑥0 • ℎ is the constant rate of harvesting (total number caught per unit time /deaths due to harvesting per unit time) 𝑑𝑋 𝑑𝑡 = − 𝑟 𝐾 𝑋2 − 𝐾𝑋 + 𝐾ℎ 𝑟 , 𝑋 0 = 𝑥0 Let 𝑟 = 1, 𝐾 = 10, ℎ = 9/10 and 𝑋 0 = 𝑥0. Find the solution and the equilibrium point! • Harvesting rate (can be some disaster) causes the population stabilise to a value less than carrying capacity. • In the case of harvesting, or some other disaster, causing population to decrease below a critical level (the population will become extinct).
  • 8. DISCRETE POPULATION GROWTH AND CHAOS Assumption :  The population does not change except at dicrete intervals, corresponding to breeding season.  The population change occur from births or death alone.  Deaths are due to normal causes and also to crowding effects 𝑋 𝑛+1 − 𝑋 𝑛 = 𝛽𝑋 𝑛 − 𝛼𝑋 𝑛 − 𝛾𝑋 𝑛 2 , 𝑛 = 0,1,2,3, … • Set 𝑟 = 𝛽 − 𝛼 and 𝛾 = 𝑟 𝐾 , we get 𝑋 𝑛+1 = 𝑋 𝑛 + 𝑟𝑋 𝑛 1 − 𝑋 𝑛 𝐾 , 𝑛 = 0,1,2,3, … • This equation called discrete logistic equation.
  • 9. LOGISTIC EQUATION WITH TIME LAG If we suppose that the logistic equation applied at the earlier time 𝑡 − 𝜏, where 𝜏 represents the time delay between increased deaths and the resulting decrease in population reproduction, we obtain 𝑑𝑋 𝑑𝑡 = 𝑟𝑋(𝑡) 1 − 𝑋(𝑡 − 𝜏) 𝐾 We called the equation as a differential-delay equation. The effect of the time delay has been to induce some oscillations in the population.
  • 10. TUGAS 1. Pada sebuah tambak ikan, ikan dipanen dengan laju konstan sebesar 2100 ikan per minggu. Laju kematian ikan per kapita sebesar 0.2 ikan per hari per ikan dan laju bertelurnya ikan sebesar 0.7 ikan per hari per ikan. (a) Tuliskan model matematika dari laju perubahan populasi ikan ini dalam persamaan diferensial? (Note:tuliskan definisi dari simbol /parameter yang Anda gunakan) (b) Jika pada awalnya terdapat 240.000 ekor ikan, berapakah jumlah ikan yang ada dalam satu minggu ke depan ? (c) Tentukan titik kesetimbangan dari populasi ikan tersebut (jika ada)! 2. Model dari penyebaran teknologi sama dengan model logistik dari pertumbuhan populasi. Misalkan N(t) adalah jumlah dari individu masyarakat yang telah menerima kursus/pelatihan pembuatan website di UIN. Laju N(t) akan mengikuti persamaaan 𝑑𝑁 𝑑𝑡 = 𝑎𝑁 1 − 𝑁 𝑀 dengan M adalah total populasi dari masyarakat. Diasumsikan bahwa laju penyebaran informasi tersebut sebanding dengan banyak individu yang telah menerima pelatihan dan sebanding dengan banyak interaksi individu yang telah dllatih dengan individu yang belum dilatih (a) Bagian mana dari model tersebut yang menggambarkan banyaknya individu yang belum menerima pelatihan? (b) Misalkan M=17015, 𝑎 = 0.490, dan 𝑁0 = 141. Tentukan berapa lama waktu yang dibutuhkan untuk melatih /memberikan informasi tentang website ke 80% dari populasi ?