We introduce the motivation and process to modelling some problem. Further, we also include some problem for simple mathematical modelling.

1. Selasa, 30 Januari 2018
Pengajar:
Heni Widayani, M.Si

2. Melalui pengamatan dan proses berpikir (reasoning), kita mampu
menjelaskan apa yang terjadi di alam secara sederhana
 Ahli pikir Yunani yang pertama kali menemukan bahwa
matematika tidak hanya dapat digunakan untuk menghitung,
namun juga untuk mempelajari alam semesta.
 Penemu Pertama sistem berpikir logis (logical reasoning) yang
merupakan akar dari pemodelan
 Salah satu dari The Seven Sages of Greece
 Beberapa hasil pemikiran:
1. Memprediksi gerhana matahari pada 25 Mei 585 SM dengan
tepat (dari The Historis of Herodotus).
2. Menggunakan geometri untuk menghitung tinggi piramida
dan jarak kapal dari pelabuhan.
PHYTAGORAS
THALES of MILETUS
(624-546 SM)

3. • Model matematika adalah jembatan antara dunia nyata (real world) dengan dunia
berpikir (thinking) untuk memecahkan suatu masalah.
• Pemodelan (Modelling) adalah proses menerima, memfomulasikan, memproses,
dan menampilkan kembali persepsi dunia nyata.
• Pemodelan dapat menghindari atau mengurangi biaya yang dibutuhkan, biaya yang
tidak perlu ataupun eksperimen yang tidak mungkin dilakukan di dunia nyata.
Neglected variables
Exogenous variables Endogenous variables

4. 1. Simplify reality
2. Created for a particular purposes
3. Attempt to mimic nature
4. Able to consider only certain effects, the object being to see which effects
account for given observations and which effect are immaterial.
5. Never ends process, since the mathematical model is continually revised
and improved
Modelling is essential to “explain” the world.

5. REAL-WORLD PROBLEM Well-defined
math problem
Solution to
model behavior
Predictions/
Explanations
simplification
inform
interpretation
observation
Governing
equations
DATA
MATH WORLD‘REAL’ WORLD
Make assumptions,
Start simple
Solve with whatever it
takes
(analytic/asymptotic/
numerical)
validate

6. PROPERTIES OF THE MODEL
FIDELITY
The preciseness of a
model’s representation of
reality
Real-world observations
Experiments
Simulations
Constructed Models
Selected Models
Real-world observations
Experiments
Simulations
Constructed Models
Selected Models
Real-world
observations
Experiments
Simulations
Constructed Models
Selected Models
COSTS
The total cost of the
modeling process
FLEXIBILITY
The ability to change and control
conditions affecting the model as
required data are gathered

7. Model Dinamik Stokastik
Model yang dihasilkan dari interaksi
antara skala waktu dan ketidakpastian
Tanpa mengandalkan data riil
Dari pengamatan empiris data riil
Empiris
Analitis
DATA RIIL
tidak mempertimbangkan aspek waktu
mempertimbangkan aspek waktu
Dinamik
Statis
SKALA WAKTU
tidak mempertimbangkan aspek ketidakpastian
mempertimbangkan aspek ketidakpastian
Stokastik
Determinisitik
KETIDAKPASTIAN
Replication of
behavior
Mathematical
Representation
Phenomenon
of interest
Model construction
Model Selection
Experimentation
Simulation

8. 1. A spring-mass system
2. Modeling change
a. A saving certificate
b. Sewage Treatment
c. Mortaging Home
d. Growth of a Yeast Culture
e. Growth of a Yeast Culture Revisited
f. Spread of a contagious disease
g. Decay of Digoxin in the bloodstream
h. Heating of a cooled object
3. Systems of difference equations
a. A car Rental company
b. The Battle of Trafalgar
c. Traveler’s Tendencies at a Regional Airport

9. Consider a spring-mass system, such as the one shown in figure below. We conduct an
experiment to measure the stretch of the spring as a function of the mass (measured
as weight) placed on the spring. Consider the data collected for this experiment,
displayed in Table.
Mass ∆𝒙
50 1.000
100 1.875
150 2.750
200 3.250
250 4.375
300 4.875
350 5.675
400 6.500
450 7.250
500 8.000
550 8.750
𝑆𝑙𝑜𝑝𝑒 =
4.875 − 3.25
300 − 200
= 0.01625
∆𝒙 = 𝟎. 𝟎𝟏𝟔𝟐𝟓 𝒎

10. Change = Future Value – Present value
Future Value = Present Value + Change
𝑎 𝑛+1 = 𝑎 𝑛 + ∆𝑎 𝑛
If the behavior is taking place over iscrete time period, the preceding construct
lead to a difference equations.
If the behavior is taking place continuously with respect to time, then the
conctruct leads to a differential equation

11. Consider the value of a savings certificate initially worth $1000 that accumulates interest
paid each month at 1% per month. No deposits or withdrawal occured in the account. The
following sequences of number represents the value of the certificate month by month.
𝐴 = (1000,1010,1020.10,1030.30, … )
𝒂 𝒏+𝟏 = 𝟏. 𝟎𝟏𝒂 𝒏, 𝒏 = 𝟎, 𝟏, 𝟐, …
𝒂 𝟎 = 𝟏𝟎𝟎𝟎

12. Six years ago, your parents purchased a home by financing $80.000 for 20 years,
paying monthly payments of $880.87 with a monthly interest of 1%. They have
made 72 payments and wish to know how much they owe on the mortgage,
which they are considering paying off with an inheritance they received.
Answer :
The change in the amount owed each period increases by the amount of interest
and decreases by the amount of the payment :
∆𝑏 𝑛 = 𝑏 𝑛+1 − 𝑏 𝑛 = 0.01𝑏 𝑛 − 880.87
with initial condition 𝑏0 = 80000.

13. The data in table below were collected from an experiment measuring the growth of a
yeast culture. The graph represents the assumption that the change in population is
proportional to the current size of the population. That is,
∆𝑝 𝑛 = 𝑝 𝑛+1 − 𝑝 𝑛 = 𝑘𝑝 𝑛,
where 𝑝 𝑛 represents the size of the population biomass after 𝑛 hours, and 𝑘 is a
positive constant. The value of 𝑘 depends on the time measurement.
Time in
hours
(n)
Observed yeast
Biomass
(pn)
Change in
Biomass
(pn+1-pn)
0 9.6
1 18.3 8.7
2 29.0 10.7
3 47.2 18.2
4 71.1 23.9
5 119.1 48.0
6 174.6 55.5
7 257.3 82.7

14. From the third column of the data, note that the change in population per hour
becomes smaller as the resources become more limited or constrained. From the
graph of population versus time, the population appears to be approaching a limiting
value, or carrying capacity. Based on our graph, we estimate the carrying capacity to
be 665. Nevertheless, as 𝑝 𝑛 approaches 665, the change dose slow considerably.
Because 665 − 𝑝 𝑛 gets smaller as 𝑝 𝑛 approaches 665, we propose the model
∆𝑝 𝑛 = 𝑝 𝑛+1 − 𝑝 𝑛 = 𝑘 665 − 𝑝 𝑛 𝑝 𝑛
Time in
biomass
n
Yeast
Biomass
pn
Change
per hour
pn+1-pn
0 9.6 8.7
1 18.3 10.7
2 29.0 18.2
3 47.2 23.9
4 71.1 48.0
5 119.1 55.5
6 174.6 82.7
pn+1-pn pn(665-pn)
8.7 6291.84
10.7 11834.61
18.2 18444.00
23.9 29160.16
48.0 42226.29
55.5 65016.69
82.7 85623.84

15. Suppose that there are 400 students in a college dormitory and that one or more
students has a severe of the flu. Let 𝑖 𝑛 represent the number of infected students after
𝑛 time periods, Assume that some interaction between those infecte and those not
infected is required to pass on the disease. If all are susceptible to the disease, then
400 − 𝑖 𝑛 represents those susceptible but not yet infected. If those infected remain
contagious, we can model the change of those infected as a proportionality to the
product of those infected by those susceptible but not yet infected, or
∆𝑖 𝑛 = 𝑖 𝑛+1 − 𝑖 𝑛 = 𝑘𝑖 𝑛 400 − 𝑖 𝑛

16.  Digoxin is used in the treatment of heart disease. Doctors must prescribe an amount
of medicine that keeps the concentration of digoxin in the bloodstream above an
effective level without exceeding a safe level (there is a variation among patients).
For an initial dosage of 0.5 mg in the bloodstream, Table below shows the amount of
digoxin 𝑎 𝑛 remaining in the blood stream of particular patient after 𝑛 days, together
with the change ∆𝑎 𝑛 each days.
n 0 1 2 3 4 5 6 7 8
𝑎 𝑛 0.5 0.345 0.238 0.164 0.113 0.078 0.054 0.037 0.026
∆𝑎 𝑛 -0.155 -0.107 -0.074 -0.051 -0.035 -0.024 -0.016 -0.011
𝑎 𝑛+1 = 0.69𝑎 𝑛, 𝑎0 = 0.5

17.  Suppose a cold can of soda is take from a refrigerator and placed in a warm classroom and
we measure the temperature periodically. The temperature of the soda is initially 400 F
and the room temperature is 720F. Temperature is a measure of energy per unit volume.
Because the volume of the soda is small relative to the volume of the room, we would
expect the room temperature to remain constant. Furthermore, we assume the entire can
of soda is the same temperature, neglecting variation within the can. We might expect the
change in temperature per time period to be greater when the difference in temperature
between the soda and the room is large and the change in temperature per unit time to
be less when the differences in temperatire is small.
Answer:
 Letting 𝑡 𝑛 represent the temperature of the soda after 𝑛 time periods, and letting 𝑘 be a
positive constant of proportionality we propose
∆𝑡 𝑛 = 𝑡 𝑛+1 − 𝑡 𝑛 = 𝑘 72 − 𝑡 𝑛 , 𝑡0 = 40

18. A sewage treatment plant processes raw sewage to produce usable fertilizer and clean
water by removing all other contaminants. The process is such that each hour 12% of
remaining contaminants in a processing tank are removed. What percentage of the sewage
would remain after 1 day? How long would it take to lower the amount of sewage by half?
How long until the level of sewage is down to 10% of the original level?
Answer :
Let the initial amount of sewage contaminants be 𝑎0 and let 𝑎 𝑛 denote the amount after 𝑛
hours. We then build the model as
𝑎 𝑛+1 = 𝑎 𝑛 − 0.12𝑎 𝑛 = 0.88𝑎 𝑛
Solusi : 𝑎 𝑘 = 0.88 𝑘
𝑎0
a. n = 1 days = 24 hours => 𝑎24 = 0.88 24
𝑎0 ≈ 45% 𝑎0
b. 𝑎 𝑛 = 50% 𝑎0 => 𝑛 = 5.4223 hours
c. 𝑎 𝑛 = 10% 𝑎0 => 𝑛 = 18.01 hours

19. A car rental company has distributorships is Orlando and Tampa. The company
specializes in catering to travel agents who want to arrange tourist activities in both cities.
Consequently, a traveler will rent a car in one city and drop the car off in the second city.
Travelers may begin their itinerary in either city. The company wants to determine how
much to charge for this drop-off convenience. Because cars are dropped off in both cities,
will a sufficient number of cars ends up in each city to satisfy the demand for cars in that
city? If not, how many cars must the company transport from Orlando to Tampa or from
Tampa to Orlando? The answers these questions will help the company figure out its
expected costs.
The historical records reveal that 60% of the cars rented in Orlando are returne to Orlando,
whereas 40% ended up in Tampa. Of the cars rented from Tampa office, 70% are returned to
Tampa, whereas 30% end up in Orlando.
Answer :
Let 𝑛 represent the number of business days. Define
𝑂 𝑛= the number of cars in Orlando at the end of day 𝑛
𝑇𝑛= the number of cars in Tampa at the end of day 𝑛
Thus the historical records reveal the system
 𝑂 𝑛+1 = 0.6𝑂𝑛 + 0.3𝑇𝑛
 𝑇𝑛+1 = 0.4𝑂 𝑛 + 0.7𝑇𝑛

20. In the battle of Trafalgar in 1805, a combined French and Spanish naval force under
Napoleon fought a British naval force under Admiral Nelson. Initially, the Frech-Spanish force
had 33 ships and British had 27 ships. During an encounter, each side suffers a loss equal to
10% of the number of ships of the opposing force. For full-force engagements, who will be
the winner of this battle ? (Fractional value are meaningful and indicate that one or more
ships are not at full capacity.)
Answer :
Let 𝑛 denote the encounter stage during the course of the battle and define
𝐵 𝑛 = the number of British ships at stage 𝑛
𝐹𝑛 = the number of French-Spanish ships at stage 𝑛
Then, after an encounter at stage 𝑛, the number of ships remaining on each side is
𝐵 𝑛+1 = 𝐵 𝑛 − 0.1𝐹𝑛, 𝐵0 = 27
𝐹𝑛+1 = 𝐹𝑛 − 0.1𝐵𝑛, 𝐹0 = 33

21. 3th day2nd day1st day
Napoleon’s force of 33 ships was arranged essentially along a line separated into three groups as
shown in figure below.
Lord Nelson’s strategy was to engage force A with 13 British ships (holding 14 in reserve). He
then planned to combine those ships that survived the skirmish against force A with the 14 ships
in reserve to engage force B. Finally, after the battle with force B, he planned to use all
remaining ships to engage force C. Assuming each side loses 15% of the number of ships of the
opposing force for each of the three battle. Who is the winner of this battle?
Force C =13Force B = 17Force A = 3
Force A = 13 Force B = 14
Force C =
remaining ships

22. Consider a regional airport that is supported by three major airlines, American
airlines, united airlines, and US airways, each flying out to respective hubs. We
survey the weekly local business travelers and find 75% of those who traveled on US
Airways traveled again on US Airways, 5% switched to fly United and 20%
switched to fly American. Of those who traveled of United, 60% traveled again on
United but 20% switched to US Airways and 20% switched to American. Of those
who traveled on American, only 40% remained with American, 40% switched to US
Airways and 20% switched to United. We assume these tendencies continue week to
week an that no additional local business travelers enter or leave the system. If the
system has 4000 weekly travelers, how are the long-term behaviour of traveler
number?
Answer :
Let 𝑛 represent the 𝑛th week of traveling and define
𝑆 𝑛= the number of US Airways travelers in week 𝑛
𝑈 𝑛= the number of United Airlines travelers in week 𝑛
𝐴 𝑛= the number of American Airlines travelers in week 𝑛
We have the following dynamical system as
𝑆 𝑛+1 = 0.75𝑆 𝑛 + 0.20𝑈 𝑛 + 0.40𝐴 𝑛
𝑈 𝑛+1 = 0.05𝑆 𝑛 + 0.60𝑈 𝑛 + 0.20𝐴 𝑛
𝐴 𝑛+1 = 0.20𝑆 𝑛 + 0.20𝑈 𝑛 + 0.40𝐴 𝑛

23. Suppose a species of spotted owls competes for survival in a habitat that also supports
hawks. Suppose also that in the absence of the other species, each individual species
exhibits unconstrained growth in which the change in the population during an interval of
time (such as 1 days) is proportional to the population size at the beginning of the interval.
The effect of the presence of the second species is to diminish the growth rate of the other
species, and vice versa. We will assume that this decrease is appriximately proportional to
the number of possible interactions between two species (although there are many ways to
model the mutually detrimental interaction of the two species). If 𝑂 𝑛 represents the size of
the spotted owl population at the end of day 𝑛 and 𝐻 𝑛 represents the competing hawk
population, then contruct mathematical model of Owl and hawk population !
Answer :
𝑂 𝑛=represents the size of the spotted owl population at the end of day 𝑛
𝐻 𝑛=represents the competing hawk population
∆𝑂 𝑛 = 𝑘1 𝑂𝑛 − 𝑘3 𝑂 𝑛 𝐻 𝑛
∆𝐻 𝑛 = 𝑘2 𝐻 𝑛 − 𝑘4 𝑂 𝑛 𝐻 𝑛
With
 𝑘1 and 𝑘2 are the constant positive growth rates.
 𝑘3 and 𝑘4 are the constant positive competition rates
.
Do the simulation with 𝑘1 = 0.2, 𝑘2 = 0.3, 𝑘3 = 0.001, 𝑘4 = 0.002 and variate
the Initial condition
𝑂 𝑛+1 = 1 + 𝑘1 𝑂𝑛 − 𝑘3 𝑂𝑛 𝐻 𝑛
𝐻 𝑛+1 = 1 + 𝑘2 𝐻 𝑛 − 𝑘4 𝑂𝑛 𝐻 𝑛