ma3264: mathematical modeling weizhu bao department of mathematics & center for computational...
TRANSCRIPT
MA3264: Mathematical Modeling
Weizhu Bao
Department of Mathematics & Center for Computational Science and Engineering
National University of SingaporeEmail: [email protected]
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