MA3264: Mathematical Modeling

Weizhu Bao

Department of Mathematics & Center for Computational Science and Engineering

National University of SingaporeEmail: bao@math.nus.edu.sg

URL: http://www.math.nus.edu.sg/~bao

Chapter 1 Introduction

Mathematical modeling– Aims:

• Convert real-world problems into mathematical equations through proper assumptions and physical laws

• Apply mathematics to solve real-life problems• Provide new problems for mathematicians

– History: • Started by the Egyptians and other ancient civilizations • Fairly recent named as mathematical modeling & a branch of applied and

computational mathematics – modeling, analysis & simulation • Rapid development in 20th centuries, especially after the computer • Mathematical modeling contest (MCM) – undergraduates & high school

Chapter 1 Introduction

– Wide applications in applied sciences• In physics --- Newton’s laws of motion, quantum physics, particle physics, nuclear

physics, plasma physics, ……..• In chemistry --- chemical reaction, mixing problems, first principle calculation, …….• In engineering --- mechanical engineering (fluid flow, aircraft, Boeing 777, …),

electrical engineering (semiconductor, power transport, …), civil engineering (building safety, dam analysis), ……

• In materials sciences – fluid-structure interaction, new materials, quantum dots, …….

• In biology --- cell motion, cell population, plant population, …… • In social sciences --- population model, traffic flow, president election poll, casino,

gambling, ……• ……..

Dynamics of soliton in quantum physics

Wave interaction in plasma physics

Wave interaction in particle physics

Vortex-pair dynamics in superfluidity

Vortex-dipole dynamics in superfluidity

Vortex lattice dynamics in superfluidity

Vortex lattice dynamics in BEC

A simple model—A saving certificate

The problem: Suppose you deposit S$10,000 into DBS bank as a fixed deposit. If the interest is accumulated monthly at 1% and paid at the end of each month, how much money is in the account after 10 years?

Solution: – Let S(n) be the amount in the account after nth month– Mathematical relation

– The result

( 1) ( ) 0.01 ( ) 1.01 ( ), 0,1,2,S n S n S n S n n

2

120

( ) 1.01 ( 1) 1.01 ( 2) 1.01 (0), 0,1,2,

(120) 1.01 (0) 3.3004*10,000 33,004

nS n S n S n S n

S S

A simple model

Related question: If the interest is accumulated yearly at 12% and paid at the end of each year, how much money is in the account after 10 years?

The solution: – Let S(n) be the amount in the account after nth year– Mathematical relation

– The result

Exercise question: If the year interest rate is at 12% and the interest is accumulated daily or instantly, how much money is in the account after 10 years, respectively???? 33,195 ( 33,201 )

( 1) ( ) 0.12 ( ) 1.12 ( ), 0,1,2,S n S n S n S n n

2

10

( ) 1.12 ( 1) 1.12 ( 2) 1.12 (0), 0,1,2,

(10) 1.12 (0) 3.1058*10,000 31,058 33,004

nS n S n S n S n

S S

Another example – Mortgaging a home

The problem: Suppose you want to buy a condo at $800,000 and you can pay a down payment at $160,000. You find a mortgage with a monthly interest rate at 0.3%. If you want to pay in 30 years, what is your monthly payment? If you can pay $4,000 a month, how long do you need to pay?

The solution: – Let S(n) be the amount due in the mortgage after nth month– Mathematic relation for the first part::

( 1) ( ) 0.003 ( ) 1.003 ( ) , 0,1,2,

(0) 640,000, (360) 0 ???

( ) ( )

( 1) 1.003 ( ) ( 1) 1.003 ( ) 0.003 1000 / 3

S n S n S n x S n x n

S S x

U n S n x

U n U n S n S n x

Another example – Mortgaging a home

– The result for the first part:

– Payment information• Total payment = 2909.7*360=1,047,500

360

360

( ) 1.003 ( 1) 1.003 (0), 0,1,2,...

(360) 0 1000 / 3 1.003 (640,000 1000 / 3)

1000 1640,000 [1 ] 2909.7($)

3 1.003

nU n U n U n

U x x

x x

Months 0 1 2 3 4

Amount owned 640,000 639010.3 638020 637020 636020

Premium paid 0 989.7 1982.4 2978 3976.7

Interest paid 0 1920 3837 5751.1 7662.1

5 12 60 120 180 240 300 360

635020 627930 575040 497300 404250 292870 159570 0

4978.3 12074

n

64956 142700 235750 347120 480430 639980

9570.2 22842 109630 206460 287990 351200 392480 407510

Another example – Mortgaging a home

– Mathematical relation for the second part:

– The result for the second part:

( 1) ( ) 0.003 ( ) 4000 1.003 ( ) 4000, 0,1,2,

(0) 640,000, ( ) 0 ??? ( ) ( ) 1000 4000 / 3

S n S n S n S n n

S S n n U n S n

( ) 1.003 ( 1) 1.003 (0), 0,1,2,...

( ) 0 1000 4000 / 3 1.003 (640,000 1000 4000 / 3)

4000 / 31.003 218.3 (months)

4000 / 3 640

n

n

n

U n U n U n

U n

n

Another example--Optimization of Profit

Economics problems:– Marco-economics: economic policy – Micro-economics: profit of a company

An example: optimization of profit

Consider an idealized company: – Object of the management: to produce the best possible dividend for

the shareholders. – Assumption: The bigger the capital invested in the company, the bigger

will be the profit (the net income)

Optimization of profit

Two strategies to spend the profit:– Short term management: The total profit is paid out as a dividend to

the shareholders in each year. The company does not grow and shareholders get the same profit in each year.

– Long term management: The total profit is divided into two parts. One part is paid out as a dividend to the shareholders and the other part is to re-invest annually in the company so that the subsequent profits in future years will increase.

Question: What part of the profit must be paid out annually as a dividend so that the total yield for the shareholders over a given period of years is a maximum ???

Optimization of profit

Variables: t: time– u(t): the total capital invested in the company in time t– w(t): total dividend in the period [0,t] to the shareholders

Parameters:– k: constant fraction of the profit which will be re-invest ( )– a: profit rate ( profit per time per capital investment)

Assumptions– The capital and profit are continuous and the process of re-investment and

dividends is also continuous. (Normally the capital and profit will be calculated at the end of the financial year of the company!!!)

– The profit is directly proportional to the capital invested.

0 1k

Optimization of profit

Balance equation:

Consider the time interval – Profit: – Change of the investment: – Change rate:– Rate of change:

rate of change production rate loss rate

of quantity of quantity of quantity

[ , ]t t t

( ) ( ) ( )u t t u t k a u t t

( )a u t t

( ) ( )( )

u t t u tk a u t

t

0

( ) ( ) ( )lim ( )t

du t u t t u tk a u t

dt t

Optimization of profit

– Dividend paid to shareholders: – Change of the total dividend: – Change rate:– Rate of change:

Mathematical model:

( ) ( ) (1 ) ( )w t t w t k a u t t

(1 ) ( )k a u t t

( ) ( )(1 ) ( )

w t t w tk a u t

t

0

( ) ( ) ( )lim (1 ) ( )t

dw t w t t w tk a u t

dt t

( )( ), (0) ,

( )(1 ) ( ), (0) 0.

du tk a u t u

dtdw t

k a u t wdt

Optimization of profit

Solution

Interpretation– If k=0:all profit is paid to shareholders, total dividend increases linearly, total

investment doesn’t change & the company doesn’t grow!!!

– If k=1: all profit is re-invest, total dividend is zero, total investment increases

exponentially & the company grows in the fastest way. – If 0<k<1: both total investment & dividend increase

( ) , 0,

(1 )( 1) for 0 1,

( )for 0.

a k t

a k t

u t e t

ke k

w t ka t k

Optimization of profit

Central issue: Given a period of time [0,T], how must k be chosen so that the total dividend over the period [0,T] is a maximum?

Total dividend:

Question: Find k in [0,1] such that w(k;T) to be maximum?

For simplicity, introduce new variables:

Find x in [0,aT], such that y to be maximum

(1 )( ; ) ( 1) ( )( 1)a k T a k Tkw k T e e

k k

( ; ) &

w k Tx a T k y

( 1)xa T xy e

x

Optimization of profit

Find the derivative of y:

Different cases:– If a T=2: y is a decreasing function of x with the maximum of y at x=0

– If a T<2: y is a decreasing function of x with the maximum of y at x=0

– If a T>2: y increases and then decreases & attains its maximum at x* which is the root of

22

2

1( 1) [ ( 1 )]

1 1[ ...]

2 3! 4!

x x x x

x

dy aT aT xe e aTe x x e

dx x x aT

x xaTe

aT

2

1 xxx eaT

Optimization of profit

Interpretation:If a T<=2: then k=0 produces the largest total dividend over the period of T years, which means that all the profit is paid out as a dividend. It does not pay to re-invest money in the company because either a or T or both are too small. In this case, the maximum profit is

If a T >2: there exists a unique number k=x*/a T such that k u(t) must be re-invested. In this case, the maximum profit is

a T

2*(1 ) * ( *)

( 1) with , 1 *

a k T xk x xe k x e

k a T aT