Question

2. Let q(x) -Je^Ar + b'x + c, where A20 100 4 2 (a) Calculate the condition number of the matrix A and hence find an upper bound to the rate of convergence of the steepest decent method (b) Use the steepest descent method to find the minimiser, starting from ao = (2.2. 2. 2)T. Tabulate the values of æk that you find.

Answer 1

%%Matlab code for steepest descent clear all close all A and b Matrix for solving linear equation A-[1 0 1 1;0 2 0 0;1 0 3 0;1 0 0 4]; b-131:2 fprintf Here A matrix n disp(A) fprintf Here b vector is\n disp b) fprint f ( 'condition number of A is %f.\n",cond(A)) %Initial guess x0-2:2; 2; 2];%taking initial guess as 0 z-1;k-0; %counter and initialisation of while loop e-10A-4; &Desired accuracy %Loop for steepest descent method fprintf( Steepest descent iterations n') while z>e k-k+1; %iteration count do-(A*x0-b); dad-do' A d0 dd- (norm (d0)) 2; x1=x0+ ( dd/dad)*d0; %desired x %gradient value value z-norm (b-A*x0) ; %Checking error n',k,x0 (1), x0 (2), x0 (3), x0 (4)) end fprintf( "\nSolution matrix is x=\n') disp (x0) fprint f ("Total number of iteration is %d for accuracy %2.2e\n',k,e) Here A matrix = 2 Here b vector is Condition number of A is 12. 545486. Steepest descent iterations

1. x1=0 . 8 01 724 x2=1.760345 x3=0 . 3 2 2 4 1 4 x4=0 .082759. 2. x1-0.737351 x2=1.598339 x3=0 .083 16 1 x4=0 . 352589. 3. x1=0.672438 x2=1.524584 x3=0 .0 8 8 098 x4-0.297198 . 4. x1-0.649766 x2-1.505276 x3-0.112943 x4-0.351692. 5. x1=0 . 5 9 8 40 9 x2-1.500539 x3=0 . 118063 x4=0 . 326313. 6, x1=0 . 5 8 1 295 x2-1.500108 x3=0 . 137024 x4=0 . 364849. 7. x1=0 . 543935 x2-1.500011 x3=0 . 140453 x4=0 .346570. 8. x1=0 . 531552 x2-1.500002 x3-0.154336 x4-0.374484 . 9. x1=0 . 504433 x2-1.500000 x3=0 . 156 780 x4=0 . 361238. 10. x1=0.495452 x2-1.500000 x3-0. 166871 x4=0.381485. 11. x1=0.475773 x2=1.500000 x3=0.168638 x4-0.371876 . 12. x1=0.469258 x2=1.500000 x3=0.175963 x4=0.386566. 13. x1=0.454980 x2=1.500000 x3=0.177244 x4=0.379594. 14. x1=0.450252 x2=1.500000 x3-0.182559 x4=0.390252. 15. x1=0.439892 x2=1.500000 x3=0.183489 x4=0.385194. 16. x1=0.436462 x2=1.500000 x3=0.187346 x4=0.392927. 17. x1=0.428945 x2=1.500000 x3=0.188020 x4=0.3892 57. 18. x1=0.426456 x2=1.500000 x3=0.190818 x4-0.394868 . x4-0. 392205 19. x1=0.421002 x2=1.500000 x3-0.191308 . 20. xI-0.419196 x2-1. 500000 x3-0.193338 x4 0. 396276 21. x1-0.415238 x2-1. 500000 x3-0.193693 x4 0. 394344. 22. x1-0.413928 x2-1. 500000 x3-0. 195166 x4 0. 397298 23. x1-0.411057 x2-1. 500000 x3-0. 195424 x4-0. 395896. x4-0.398040 24. xī=0.410106 x2 1.500000 x3=0.196493 . 25. x1=0.408022 x2=1.500000 x3=0.196680 x4=0.397022. 26. x1=0.407333 x2 1.500000 x3=0.197455 x4=0.398578. 27. x1-0.405821 x2=1.500000 x3=0.197591 x4=0.397840.

28. x1=0.405320 x2=1.500000 x3=0.198153 x4=0.398968. 29. x1=0.404224 x2=1.500000 x3=0.198252 x4=0.398432. 30. x1=0.403860 x2=1.500000 x3=0.198660 x4=0.3992 51. 31. x1=0.403065 x2=1.500000 x3=0.198732 x4=0.398863. 32. x1-0.402801 x2=1.500000 x3=0.199028 x4=0.399457. 33. x1=0.402224 x2=1.500000 x3=0.199080 x4=0.399175. 34. x1-0.402032 x2=1.500000 x3=0.199295 x4=0.399606. 35. x1=0.401613 x2-1.500000 x3=0.199332 x4=0.399401. 36. x1=0.401475 x2-1.500000 x3-0.199488 x4=0.399714. 37. x1=0.401171 x2-1.500000 x3-0.1995 15 x4=0.399566. 38, x1=0.401070 x2-1.500000 x3-0.199629 x4=0.399792. 39. x1=0.400849 x2=1.500000 x3=0.199648 x4=0.399685. 40. x1=0.400776 x2=1.500000 x3=0.199731 x4=0.399849. 41. x1=0.400616 x2=1.500000 x3-0.199745 x4=0.399771. 42. x1=0.400563 x2=1.500000 x3=0.199805 x4=0.399891. 43. x1=0.400447 x2=1.500000 x3=0.199815 x4=0.399834. 44. x1=0.400409 x2=1.500000 x3=0.199858 x4=0.399921. 45. xl 0. 400324 x2-1.500000 x3-0. 199866 x4-0. 399880. 46. xī=0.400297 x2=1.500000 x3=0.199897 x4=0.399942. 47. xl 0. 400235 x2-1.500000 x3-0. 199903 x4-0. 399913. 48. xl 0. 400215 x2-1.500000 x3-0. 199925 x4-0. 399958. 49. xl-0. 400171 x2-1.500000 x3-0. 199929 x4 0. 399937 Solution matrix is x- 0.4002 1.5000 0.1999 0. 3999 Total number of iteration is 49 for accuracy 1.00e-04 Published with MATLABR R2018a