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Simple kuphindelelwa indlela ukuxazulula izinhlelo of zibalo lwento (Slough)
Simple kuphindelelwa indlela, ebizwa nangokuthi i-indlela isilinganiselo ngokulandelana - algorithm zezibalo lokuthola amagugu lenani engaziwa ngokusebenzisa kancane acacise. Ingqikithi le ndlela wukuthi, njengoba leli gama lisikisela, ngokuvamile baphuma kancane kancane ezwakalisa isilinganiselo kokuqala kwabangcwele okwalandela, beba Imiphumela sangempela. Le ndlela isetshenziswa ukuthola inani le-okuguquguqukayo umsebenzi esiwunikiwe, futhi ukuxazulula izinhlelo of zibalo, kokubili komugqa futhi non-eqondile.
Ake sibone ukuthi le ndlela ukusebenza ikhambi izinhlelo lwento. okungaguquki-iphuzu kuphindelelwa algorithm simiswe ngalendlela lelandzelako:
1. ukuqinisekiswa izimo Kuhlangana kuleli matrix kokuqala. A Kuhlangana theorem: uma sokuqala uhlelo matrix ngokuphambeneyo evelele (ie, irowu izakhi diagonal main ngamunye kumelwe ibe nkulu e ezidlula kuka isamba izakhi ohlangothini diagonals e inani eliphelele), indlela iterations elula - convergent.
2. matrix uhlelo yasekuqaleni akuyena ngaso sonke isikhathi predominance idayagonali. Ezimweni ezinjalo, uhlelo kungenziwa siguqulwe. I zibalo ukuthi ukwanelisa isimo Kuhlangana ishiywa beqinile, nge nezinganelisi futhi wenze inhlanganisela yomugqa, isb nande, kususa, ezothando eligoqiwe ndawonye nemiphumela ebesiyilindele.
Uma uhlelo awuthola idayagonali main izici esingesihle ke zombili izinhlangothi lokhu kwesibalo kunezelwa nge ngokuya ifomu i * x i, okumele kuhambisane izimpawu izimpawu izakhi idayagonali.
3. Ukuguqula uhlelo okuholela ekubukeni okujwayelekile:
x - = β - + α * x -
Lokhu kungenziwa ngezindlela eziningi, isib, kanje: ezothando wokuqala owabonisa x 1 nakwezinye angaziwa kusuka vtorogo- x 2, x 3 tretego- njll Ngakho sisebenzisa ifomula:
α IJ = - (a IJ / a ii)
i = b i / a ii
Qiniseka futhi ukuthi uhlelo umphumela uhlobo evamile oluhambisana isimo Kuhlangana:
Σ (j = 1) | α IJ | ≤ 1, futhi i = 1,2, ... n
4. Qala esetshenziswa, empeleni, indlela ziwukulinganisela ngokulandelana.
x (0) - isilinganiselo kokuqala, sisuke sibonisa ukuthi therethrough x (1), kulandele x (1) x express (2). Ifomula jikelele ifomu-matrix kanje:
x (N) = β - + α * x (n- 1)
Thina ukubala, kuze kufike ibanga ukunemba oyifunayo:
max | x i (k) -X i (k + 1) ≤ ε
Ngakho, ake sibheke practice, indlela kuphindelelwa elula. Ngokwesibonelo:
Xazulula izinhlelo lwento:
4,5x1-1.7x2 + 3.5x3 = 2
3.1x1 + 2.3x2-1.1x3 = 1
1.8x1 + 2.5x2 + 4.7x3 = 4 ngokunemba ε = 10 -3
Bheka busa uma izakhi idayagonali module.
Siyabona ukuthi isimo Kuhlangana wenelisekile ezothando lwesithathu. I lokuqala nelesibili uguqule, ezothando lokuqala sifaka ezimbili:
7,6x1 + 0.6x2 + 2.4x3 = 3
Khipha kusuka elesithathu:
-2,7x1 + 4.2x2 + 1.2x3 = 2
Esiliguqulile uhlelo yoqobo okulingana:
7,6x1 + 0.6x2 + 2.4x3 = 3
-2,7x1 + 4.2x2 + 1.2x3 = 2
1.8x1 + 2.5x2 + 4.7x3 = 4
Manje thina banciphise uhlelo ekubukeni okujwayelekile:
x1 = 0.3947-0.0789x2-0.3158x3
X2 = 0.4762 + 0.6429x1-0.2857x3
X3 = 0.8511-0.383x1-0.5319x2
Sibheka tekuhlangana inqubo iterative:
0,0789 + 0,3158 = 0,3947 ≤ 1
0,6429 + 0,2857 = 0,9286 ≤ 1
0.383+ 0.5319 = 0.9149 ≤ 1, isb isimo uhlangana.
.3947
Kokuqala isilinganiselo x (0) = 0.4762
.8511
Bamelele lezi zimiso phakathi kwesibalo yohlobo evamile, sithola amanani alandelayo:
0,08835
x (1) = 0.486793
0.446639
Obambele ukwamukela izindinganiso ezintsha, sithola:
0.215243
x (2) = 0.405396
0.558336
Siyaqhubeka ukubala kuze uze uthole eduze amagugu ukuthi ukuhlangabezana nemibandela eshiwo.
0,18813
x (7) = 0.441091
0.544319
0.188002
x (8) = 0.44164
0.544428
Hlola ngokunemba imiphumela:
4,5 * 0,1880 -1,7 * 0,441 + 3,5 * 0,544 = 2,0003
3,1 * 0,1880 + 2,3 * 0,441-1.1x * 0,544 = 0,9987
1,8 * 2,5 * 0,1880 + 0,441 + 4,7 * 0,544 = 3,9977
Ezenye etholwe abambele amagugu etholwe phakathi kwesibalo yokuqala, ukwanelisa ngokugcwele equation.
Njengoba singabona, elula kuphindelelwa indlela inika Imiphumela anembile, kodwa ukuxazulula le equation, saba ukuchitha isikhathi esiningi futhi wenze izibalo nzima.
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