Numericl nlysis f ngineers Germn Jdnin University ITRTIV SOLUTION RFINMNT Numericl solution of systems of liner lgeric equtions using direct methods such s Mtri Inverse, Guss limintion, Guss-Jdn limintion, nd LU-Decomposition my include lrge round-off errs, especilly f ill-conditioned systems. To improve the ccurcy of the solution, n itertive solution refinement cn e used. Consider the following system of equtions: Let the pproimte solution otined y direct method is given s,, T Sustituting in the iginl system of equtions, new right-hnd-side vect is otined Sutrct equtions from equtions to get sites.google.com/site/ziydmsoud/numericl
Numericl nlysis f ngineers Germn Jdnin University sites.google.com/site/ziydmsoud/numericl 4 Let nd let Then, the system of equtions ecomes 4 Δ The refinement procedure is done s follows:. Otin n pproimte solution of the system of equtions using direct method.. Sustitute the pproimte solution in the sme system of equtions to otin new right-hnd-side vect, s in system.
Numericl nlysis f ngineers Germn Jdnin University. Clculte the vect nd sustitute in the system of equtions 4 nd solve f the crections vect Δ. 4. Otin new pproimte solution s Δ. 5. Clculte the pproimte err the reltive pproimte err nd compre to preset tolernce. 6. Repet steps 5 until the required tolernce is met. Since system 4 hs to e solved severl times f different right-hnd-side vects, the most suitle method f otining n pproimte solution is either the Mtri Inverse method the LU-Decomposition method. mple Using 4 significnt figures clcultions, determine the solution of the following system of liner lgeric equtions to the est possile 4 significnt figures..000 0. 0. 0.500 0. 0.500 0.000 The est 4 significnt figures mens tht the reltive err in the pproimte solution must ecome zero. The reltive errs vect is Δ which is the crections vect. This mens tht itertions hve to e perfmed until there ecome no crections to e mde. In this emple, we will use the direct Mtri Inverse method, tht is The inverse of the mtri cn e otined using severl methods including the Guss- Jdn limintion nd the LU-Decomposition methods. The inverse of the mtri is sites.google.com/site/ziydmsoud/numericl 5
Numericl nlysis f ngineers Germn Jdnin University 9.8 6.8 0.68 6.8 96. 84. The infinity nms of the mtri nd its inverse re m n in j m n ij in j The condition numer of the mtri is 0.68 84. 8.9.000 0..8 ij 6.8 96. 84. 47. cond.8 47. 764.5 The mtri condition numer cn e rounded to the nerest power of 0 numer s cond 0.8 47. 764.5 000 ssuming minimum reltive err in the nm of the mtri, which is 4 0 using 4 significnt figures clcultions, leds to minimum reltive err in the nm of the solution vect s cond 0 4 0 0 This mens tht the minimum mesurle reltive err in the solution vect is 0 which indictes tht the solution cn e trusted up to significnt figure only. Therefe itertive solution refinement is needed to rech the 4 significnt figures ccurcy. sites.google.com/site/ziydmsoud/numericl 6
Numericl nlysis f ngineers Germn Jdnin University Using the direct mtri inverse method, the first pproimtion of the solution cn e otined s 9.8 6.8 0.68 6.8 96. 84. 0.68 7.58 84. 96.7 8.9 4. Sustituting this pproimtion ck in the iginl system new right-hnd-side vect s.000 0. 0. 0.500 0. 7.58 0.6 0.500 96.7.78 0.000 4..85, we cn determine the Notice the ig difference etween the new right-hnd-side vect [0.6,.78,.85] T nd the ect right-hnd-side vect [,, ] T The errs vect is 0.6 0.8.78 0..85 0.5 Use the errs vect to determine the crections vect s Δ Δ sites.google.com/site/ziydmsoud/numericl 7
Numericl nlysis f ngineers Germn Jdnin University Δ 9.8 6.8 0.68 6.8 96. 84. The new crected pproimtion ecomes 0.68 0.8 0.008 84. 0..55 8.9 5.6 7.58 0.008 7.57 Δ 96.7.55 95. 4..6.9 The reltive err in this itertion is the Δ vect which is still not zero. Therefe me itertions re needed. Sustitute the new pproimtion ck in the iginl system right-hnd-side vect s.000 0. 0. 0.500 0. 7.57 0.9 0.500 95..94 0.000.9.97 to determine the new Notice tht the difference etween the new right-hnd-side vect [0.9,.94,.97] T nd the ect right-hnd-side vect [,, ] T is decresing. The errs vect is 0.9 0.07.94 0.06.97 0.0 Use the errs vect to determine the crections vect s sites.google.com/site/ziydmsoud/numericl 8
Numericl nlysis f ngineers Germn Jdnin University Δ 9.8 6.8 0.68 6.8 96. 84. The new crected pproimtion ecomes Δ 0.68 0.07 0.6456 84. 0.06.670 8.9 0.0.85 7.57 0.6456 6.9 Δ 95..670 9.5.9.85 09.5 The reltive err in this itertion is the Δ vect which is still not zero. Therefe me itertions re still needed. Crrying out me itertions, the reltive errs vect crections vect Δ ecomes zero in the th itertion. The solution vect of tht itertion is 7.5 94.6.4 Using 4 significnt figures clcultions, the ove solution is considered ect since it produces the ect right-hnd-side vect. However, using infinite numer of digits, the ect solution of the system is 7 9 0 sites.google.com/site/ziydmsoud/numericl 9