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MA3264: Mathematical Modeling Weizhu Bao Department of Mathematics & Center for Computational Science and Engineering National University of Singapore Email: [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 S (n 1) S (n ) 0.01S (n ) 1.01S ( n), n 0,1, 2, – The result S (n) 1.01S (n 1) 1.012 S (n 2) 1.01n S (0), n 0,1,2, S (120) 1.01120 S (0) 3.3004 *10,000 33,004 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 S (n 1) S (n ) 0.12 S (n ) 1.12 S (n ), – The result n 0,1, 2, S (n) 1.12S (n 1) 1.122 S (n 2) 1.12n S (0), n 0,1,2, S (10) 1.1210 S (0) 3.1058*10,000 31,058 33,004 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 ) 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 $4,000 a month, how long do you need to pay? The solution: – Let S(n) be the amount due in the mortgage after nth month – Mathematic relation for the first part:: S (n 1) S (n ) 0.003S (n ) x 1.003S (n ) x, S (0) 640,000, n 0,1, 2, S (360) 0 x ??? U ( n) S ( n) x U (n 1) 1.003U (n ) S (n 1) 1.003S (n ) 0.003 x 1000 / 3 Another example – Mortgaging a home U ( n ) 1.003U ( n 1) 1.003n U (0), n 0,1, 2,... – The result for the first part: U (360) 0 1000 x / 3 1.003360 (640,000 1000 x / 3) – Payment information 640,000 1000 1 [1 ]x 360 3 1.003 x 2909.7($) • Total payment = 2909.7*360=1,047,500 Months n 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 60 120 180 240 300 5 12 635020 627930 575040 497300 404250 292870 159570 360 0 4978.3 12074 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: S (n 1) S (n ) 0.003S (n ) 4000 1.003S (n ) 4000, S (0) 640,000, S ( n ) 0 n ??? n 0,1, 2, U ( n) S ( n) 1000 4000 / 3 – The result for the second part: U ( n ) 1.003U ( n 1) 1.003n U (0), n 0,1, 2,... U ( n ) 0 1000 4000 / 3 1.003n (640,000 1000 4000 / 3) 4000 / 3 1.003 4000 / 3 640 n n 218.3 (months) 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 ( 0 k 1 ) – 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. Optimization of profit Balance equation: rate of change production rate loss rate of quantity of quantity of quantity Consider the time interval [t , t t ] – Profit: a u (t ) t u (t t ) u (t ) k a u (t ) t – Change of the investment: – Change rate: u(t t ) u (t ) k a u (t ) t – Rate of change: du (t ) lim u (t t ) u (t ) k a u (t ) dt t 0 t Optimization of profit – Dividend paid to shareholders: (1 k )a u (t ) t – Change of the total dividend: w(t t ) w(t ) (1 k ) a u(t ) t w(t t ) w(t ) – Change rate: (1 k ) a u (t ) t – Rate of change: dw(t ) lim w(t t ) w(t ) (1 k ) a u(t ) Mathematical model: dt t 0 du (t ) k a u (t ), u (0) , dt dw(t ) (1 k ) a u (t ), w(0) 0. dt t Optimization of profit Solution u (t ) e a k t , t 0, (1 k ) a k t (e 1) w(t ) k a t for 0 k 1, for k 0. 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 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: w(k ; T ) (1 k ) a k T (e 1) ( )(e a k T 1) k k Question: Find k in [0,1] such that w(k;T) to be maximum? For simplicity, introduce new variables: x aT k & y w(k ; T ) Find x in [0,aT], such that y to be maximum y aT x x (e 1) x Optimization of profit Find the derivative of y: dy aT aT x x 1 2 (e x 1) e aTe x [ x 2 (1 x e x )] dx x x aT 1 1 x x2 x aTe [ ...] aT 2 3! 4! 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 x2 1 x e x aT 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 a T 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 (1 k ) a k T x* ( x*) 2 (e 1) with k , 1 x * e x* k aT aT