1 The Euler Forward scheme (schéma d'euler explicite)

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1 TP : Finite dierence metod for a European option M2 Modélisation aléatoire - Université Denis-Diderot Cours EDP en Finance et Métodes Numériques. December Te Euler Forward sceme (scéma d'euler explicite) We look for a numerical approximation of te European put function v = v(t, s), t [0, T ], s [0, S max ]. It satises in rst approximation te Black and Scoles backward PDE on te truncated domain Ω = [S min, S max ]: s2 2 s 2 v rs s v + rv = 0, t (0, T ), s (S min, S max ) t v σ2 2 v(t, S min ) = v l (t) Ke rt S min, t (0, T ) v(t, S max ) = v r (t) 0, t (0, T ) v(0, s) = ϕ(s) := (K s) +, s (S min, S max ). We will consider te following parameters: K = 100, S min = 0, S max = 200, T = 1, σ = 0.2, r = 0.1 In particular, we aim at computing v(t, s) at nal time t = T. We rst introduce a discrete mes as follows. Let := Smax S min I+1 be (spatial) mes step, and t := T N be te time step. Ten s := S min +, = 0,..., I + 1 are te mes points, and t n = n t, n = 0,..., N te time mes. We are looking for U n, an approximation of v(t n, s ). For any function v C 2 (or v C 3 for (4)), we recall te following approximations, as 0, v (s ) = v(s ) v(s 1 ) v (s ) = v(s +1) v(s ) v (s ) = v(s +1) v(s 1 ) 2 (1) + O() (2) + O() (3) + O( 2 ). (4) We terefore obtain several possible approximations by nite dierences for te rst order derivative: S v(t n, s ) U n U n 1 S v(t n, s ) U n +1 U n S v(t n, s ) U n +1 U n 1 2 (backward dierence approximation) (5) (forward dierence approximation) (6) (centered dierence) (7) Te rst two approximations are said to be consistent of order 1 (in space), wile te second one is consistent of order 2. 1

2 We also recall te approximation 2 SSv(t n, s ) U n 1 + 2U n U n +1 2, (8) wic is consitent of order 2 in space. Hence we obtain te so-called "Euler Forward sceme" (or Explicit Euler sceme), abreviated "EE" using te centered approximation, as follows: U n+1 U n t + σ2 2 s2 U n 1 + 2U n U n +1 2 rs U n +1 U n ru n = 0 n = 0,..., N 1, = 1,..., I U n 0 = v l (t n ) Ke rtn S min, n = 0,..., N (9) U n I+1 = v r (t n ) 0, U 0 = ϕ(s ) (K s ) +, n = 0,..., N = 1,..., I Let us remark tat we ave taken = 1 and = I as extremal indices in. For = 1, te sceme utilizes te known value U0 n := v l(t n ) (left boundary value). For = I, te sceme utilizes te known value UI+1 n := v r(t n ) (rigt boundary value). 2 Programming Euler Forward 2.1 Preliminaries. We coose to work wit te unkown te vector corresponding to (v(t n, s )) =1,...,I : U n = U n 1. U n I. We would like to write (9) under te vector form as follows: U n+1 U n + AU n + q(t n ) = 0, (10) t were A is a square matrix of dimension I and q(t) is a column vector of size I. Let us denote α := σ2 s 2 2 2, β := r s 2. We look for A and q(t) suc tat ( α i + β i )U n i 1 + (2α i + r)u n i + ( α i β i )U n i+1 (AU + q(t n )) i. By identication we see tat A is a tridiagonal matrix 2α 1 + r α 1 β 1 0 α 2 + β 2 2α 2 + r α 2 β A := α i + β i 2α i + r α i β i α I + β I 2α I + r 2

3 and q(t) will contain te known boundary values U 0 and U I+1 : ( α 1 + β 1 )v l (t) q(t) := ( α I β I )v r (t) 2.2 Getting into te program. a) Download te working program 1 (tp1.m if using Matlab, or tp1.sci if using Scilab). Tis program as some lines to be completed in order to work properly. In te case of te matlab program, you will also ave to download oter function les BS.m and ploot.m (you can also take te solution le tp1sol.m) b) Ceck te program for te payo function u0, ul (for v l ) and ur (for v r ) and program tem correctly. (notice tat te instruction y=max(k-s,0) works for s vector, K scalar, and return a vector of same size as s.) c) In te section MESH, complete te value of te mes step and program te vector s containing te (s ) 1 I values. Typically it sould look like: dt=t/n; =(Smax-Smin)/(I+1); s=smin+ (1:I)'*; // column vector (s 1,..., s I ) T d) Program te matrix A and test te program. Under te matlab command window, type tp1. (Under Scilab, type exec tp1.sci or exec('tp1.sci',-1)) e) Program te function q(t). For instance in matlab: q [(-alpa(1) + bet(1))* ul(t); zeros(i-2,1); (-alpa(end) - bet(end))* ur(t)]; In Scilab tis may look like function y=ul(t); y=k*exp(-r*t); endfunction function y=ur(t); y=0; endfunction; function y=q(t) y=zeros(s); // vector of zeros wit same size as te s vector y(1)= (-a(1)+b(1))*ul(t); u($)= (-a($)-b($))*ur(t); endfunction 2.2 Euler Forward sceme (or "Euler Explicit" sceme) a) Program te explicit form of te vector U n+1, in terms of U n, in te main loop, using te matrix A and te function q. In te end, it sould look like (in matlab) case 'EE' P = (Id - dt*a)*p - dt*q(t); 1 See ttps://lll.mat.upmc.fr/bokanowski/enseignement/ens2016.tml 3

4 Note tat te grapic function ploot also plots te exact Black and Scoles formula (see te given function BS in te le BS.m). b) Correct te lign errli=0 in te main loop, in order to compute correctly te maximum norm. between te sceme values and te Black and Scoles values: U V BS = max 1 i I U i V BS (s i ) 2.3 Solution. A solution le is given in tp1sol.m 3 First numerical tests a) Test te Euler forward sceme (EE). First x N = 10 and take I = 10, 20, 50,.... Ten take te following N = I values : 10, 20, 50, 100. Observe tat: - te sceme is not always numerically stable - it does not always give a positive solution (i.e. we do not always ave U n 0 n, ). b) Try to understand te origin of te oscillations wen tey occur. One sould look at te "amplication" matrix dened as B := I d ta (Te coecients of B are tey positive? Do tey ave a modulus smaller tan 1?) c) Fill in te CFL number dened ere as µ := t 2 /( 1 σ 2 S max 2 ) and print it. Ceck tat tere is no stability problem wen ν is suciently small. d) In te case wen σ = and for instance wit N = 10 and I = 50: observe tat te sceme is stable but tat te solution given by te sceme is not positive. e) Order of te sceme. We rst consider te following values of I and of N: I = 10, 20, 40, 80, 160, and N = I 2 /10 (tat is, N = 10, 40,...). Tis is in order to ave t 2. Fill in te following error table and compute te corresponding (spatial) order of te metod. I N e k order α k More precisely, if te error is e k for a given parameter I = I k, we will compute te order at step k by te formula α k := log(e k 1/e k ) log( k 1 / k ) log(e k 1/e k ). log(2) Te idea is to try to detect a beavior of te form e k = C α k, were C est a constant and were k is te spatial mes size corresponding to I k. 4

5 Te numerical (spatial) order sould be close to two. But ere setting N const I 2 is costly in terms of number of operations. More precisely, since t = const 2 one can see tat error also beaves as O( t), tat is, a rst order error beavior wit respect to te time discretisation. Tis motivates te use of implicit scemes in order to avoid te time-step condition (or "CFL" condition). Exercice: backward and forward dierence approximations a) Program in te same way te oter approximations (backward dierences / décentrage droit : 'DROIT' and te forward dierences - décentrage gauce 'GAUCHE'). Te parameter CENTRAGE is dened at te begining of te program (section NUMERICAL DATA) and determines te type of nite dierence approximation tat is used. b) Using te forward dierences, x temporarily σ = and N = 10, I = 50, and ceck tat now te solution keeps positive. c) Using te backward dierences, x temporarily σ = and N = 10, I = 50, and ceck tat now te numerical solution is unstable (understand te problem by looking at te coecients of te amplication matrix). 4 Implicit Euler sceme (scéma d'euler implicite) Te CFL constraint imposes some restriction on te time step t. Implicit scemes may allow us to get rid of tis restriction. For instance, te implicit euler sceme, ereafter abrieviated "EI" (for Frenc "Euler Implicit"), wit centered dierence approximation for te rst spatial derivatives reads U n+1 U n t + σ2 2 s2 U n+1 1 n+1 + 2U U+1 n+1 U+1 n+1 2 rs U 1 n ru n+1 = 0 n = 0,..., N 1, = 1,..., I U0 n+1 = v l (t n+1 ) Ke rt n+1 S min, n = 0,..., N 1 (11) U n+1 I+1 = v r(t n+1 ) 0, n = 0,..., N 1 U 0 = (K s ), = 1,..., I a) Write te sceme in vector form, in analogy wit (10). b) Program EI: set te parameter SCHEMA='EI' at te begining of te program, and complete accordingly, in te main loop, te part case 'EI'... Note tat in order to solve a linear system of te form Ax = b one may use te linear solver in matlab (resp. Scilab) as follows x=a\b; c) Ceck tat wit te EI sceme tere is no more stability problems (wit for instance N = 10 and I = 50). 5 Crank-Nicolson sceme a) Program te Crank-Nicolson sceme (CN) (tis is te θ-sceme wit θ = 1 2 ). 5

6 b) Numerically study te convergence of te centered CN sceme. To do so, one can compute te numerical error obtained wit te following values N = I + 1 {10, 20, 40, 80, 160}. Ten do te same study as for te Euler implicit sceme and compare by making an error table (and wit an order estimation) One sould observe a second order beavior for CN, wile it sould be rst order for EI. c) Ceck tat tese results are coerent wit te teory. 6 Exercices Exercice. 1 ("Call option") a) Using te put-call parity formula, do a function tat also computes te black and scoles formula for te call. b) Propose adequate left and rigt boundary conditions (tat is, at s = S min and s = S max ) for te call option, in te form (v(t, S min ) = v l (t) and v(t, S max ) = v r (t), were v l, v r are function to determine. 2 c) Propose an apropriate PDE for te call option, using tese boundary conditions. d) Write an Euler explicit sceme for te call option, and program it. Exercice. 2 (Improved eciency using sparse matrices) Te matrix A as only a few non zero elements (about 3I non zero elements). It is possible to code only te non-zero elements of A by using te matrix type sparse. Type elp sparse or doc sparse for documentation in matlab. use A=spzeros(I,I) to initialize a sparse 0-matrix of size I I. type A or full(a) to ceck te values, sparse(a) corresponds to te sparse matrix of a full matrix A. use speye(i,i) to get te sparse identity matrix. Program te EI or CN sceme by using only sparse matrices, and compare te speed of te new code wit respect to te old one for large I values. Execution time is in general reduced. It can be evaluated by using te commands tic; toc; (resp. time() and time() in Scilab) as follows: tic; % INSTRUCTIONS; %... t=toc; printf('cpu time : t=%5.2f, t); (ten te variable t contains te time elapsed between te calls of tic and toc.) 2 v l (t, s) 0, v r(t, s) s Ke rt. 6