Assignment 1: Optimization and mathematical modeling

Transcription

1 Math 360 Winter 2017 Section 101 Assignment 1: Optimization and mathematical modeling 0.1 (Due Thursday Sept 28, 2017) Let P = (, y) be any point on the straight line y = 4 3. (a) Show that the distance between P and the origin is epressed as d() = 2 + (4 3) 2. (b) Find the point on the line that is closest to the origin. (c) Actually, the distance squared is given by Solution: s() = 2 + (4 3) 2. Find the absolute minimum of this function and show that the result is identical to that found in (b). Write one sentence eplaining why you think the two results are identical. (a) d = 2 + y 2 But for each point on the straight line, y = 4 3. Substitute into the equation above d() = 2 + (4 3) 2. (b) (c) d () = 1 2d() (2 6(4 3)) = 0 = 6 5 and y = 4 3 = 2 5. Since d() is defined on the open interval (, ) and that d as ±. So, the only etremum at = 6/5 must be both the local and absolute minimum of the function. s () = 2 6(4 3) = 0 = 6 5. When d() is at min, s() = d 2 () must also be at min. It was a little bit simpler just to minimize s() in this case. 1

2 0.2 Optimal age of reproduction. (Roff, D.A The Evolution of Life Histories ) Semelparous organisms breed once during their lifetime. Eample of this type of organisms are pacific salmon and bamboo. The per capita rate of increase, r, is considered as a measure of their fitness. Larger r represents more offsprings an average individual reproduces. r is typically a function of the age,, at which breeding happens, i.e. r = r(). Models of age-structured population predict that this function can be epressed as r() = ln[l()m()], where l() is the probability of surviving to age and m() is the average number of female offsprings produced by each individuals at age. (a) Find the optimal age,, at which the fitness measure r() is maimized provided that where a, b, c are positive constants. l() = e a, and m() = b c (b) For a = 0.1, b = 4, c = 0.9, calculate the optimal age. (c) Using the same values of a, b, c given in (b), plot the graph of r() using a graphic calculator or an online graph plotter (e.g. a pretty good one can be found here at: graphs/) to plot the graph of r() and compare it to the results you obtained. Make a one sentence comment on the comparison. Solution: (a) Substitue l() = e a and m() = b c into r(), Now, r() = ln[e a b c ] = a + ln b + c ln. r () = ( a + c ) ( a + ln b + c ln ) 2 = c ln b c ln 2 = 0 1 ln b1/c = e. It is easy to check that r () > 0 for < and r () < 0 for >. Thus, this is indeed a local ma. r() is defined on the interval (0, ). As 0, r ; as, r a < 0. Thus, the unique CP is both the local and absolute ma. It is independent of a value. (b) For a = 0.1, b = 4, c = 0.9, = e 1 ln 41/

3 (c) It shows that there is one single CP which is both a local and absolute ma and it happens at a value of bit larger than 0.5 consistent with the result found in (b). 0.3 Optimal age at first reproduction (Lloyd, D.C.1978 Selection of offspring size at independence and other size-versus-number-strategies, American Naturalist, 129:800.) Iteroparous organisms breed more than once during their lifetime. We consider a model where the intrinsic rate of increase (or fitness), r, depends on the age of the first reproduction, denoted by, and satisfies e (r()+l) (1 e k ) 3 c 1 e (r()+l) = 1 where k, L, c are positive constants (parameters) describing the life history of the organism. Our goal is to find the optimal age of first reproduction,, that maimizes the fitness, r(). Since r() is implicitly defined as a function of, implicit differentiation is needed to find the critical points (CPs). (a) Calculate r () = dr. (Hint: Taking logarithms on both sides of the equation before differentiation can facilitate the d calculation.) (b) Show that r () = 0 leads to the following equation that determines the CPs r() = 3ke k L. 1 e k 3

4 (c) Substitute the result in (b) back into the original equation and show that the optimal age is independent of L. Also for c = k = 1, show that the equation that determine the optimal value of is determine by the equation (1 e ) 3 = 1 3 (e 1) ln This is a transcendental equation that we cannot solve in closed forms. However, one can always plot the graphs of the functions on the two sides and try to locate the nonzero point of intersection between the two. Do the graph plotting and estimate the optimal value of based on the graph (with an accuracy up to 1 digit after decimal point.) (d) (Not required! It is a challenge only for bonus marks!) Think if you can find any method to solve the equation obtained in (c) numerically using a programmable calculator to determine the optimal value of with an accuracy up to 6 digits after the decimal point. If you can, you get 10% bonus mark over the whole assignment mark. Solution:. (a) [ ] r 1 e (r()+l) 3ke k () = (r() + L). + (1 )e (r()+l) 1 e k (b) Let r () = 0, since 1 e (r()+l) cannot be zero because both r() and L are positive, one finds 3ke k 3ke k (r() + L) = 0 r() = L. 1 e k 1 e k (c) Substitute the epression for r() in (b) back into the original equation and notice that parameter L only occurs in as r() + L which always cancels the L term, L will never show up in the final equation for optimal value of. Making k = c = 1, one obtains Take logarithms on both sides: e 3e (1 e ) 3 =. 3e 1 e + ln[(1 e ) 3 ] = ln[1 e 3e ] Put on the left-hand-side and the other terms on the other side, one yields = 1 (1 e ) 3 3 (e 1) ln. Below is the graph of the two functions. 4

5 Based on the graph, optimal value of is approimately 2.2. (d) When we try to solve the equation found in (c) using iteration directly, the nonzero fied point turns out to be unstable and cannot be solved. One can use Newton s method learned in calculus to solve the equation numerically, but the derivative of a complicated function must be calculated (a good chance to test your calculus skills). Let (1 e ) 3 F () = 1 3 (e 1) ln be the equation that we want to solve using Newton s method. [ F () = 1 1 (1 e ) 3 3 e ln ] 13 (e 1) 3(1 e )e [(1 e 3e )(1 e ) + e 3e ] (1 e ) 3 2 = 0 = 1 1 (1 e ) 3 3 e ln (1 e )( ) + e 3e (1 e )( = Now, we obtain Newton s iteration e 3e (1 e )( ) n+1 = n F ( n) F ( n ) 5 ) 1 (1 e ) 3 3 e ln

6 This map is either too complicated for the online iteration equation solver to handle or when one of the solvers does solve it but only allows an accuracy of up to 5 digits after the decimal point. If one wants to avoid using computer programs or softwares such as MatLab, Ecel spreadsheet is one widely-owned tool that one can use to handle this. Here are the results: The result above is accurate up to 9 digits after the decimal point. required accuracy, the optimal value should be = So, if 6 digits is the 6