gs.ppt - Solution of Linear and Non-linear System of...

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Solution of Linear and Non-linear System of Algebraic Equations (Simultaneous Equations)

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Methods for Solving Simultaneous Equations Solving Larger Numbers of Equations Some of the most widely used methods to solve a larger system of linear equations (n >3) are: – Gaussian Elimination Method – Gauss-Jordan Elimination Method – LU Decomposition / Factorization Method – TDMA (Tri-Diagonal Matrix Algorithm) Method – Gauss-Seidel Iteration Method
3 Naive Gauss Elimination Extension of method of elimination to large sets of equations by developing a systematic scheme or algorithm to eliminate unknowns and to back substitute. As in the case of the solution of two equations, the technique for n equations consists of two phases: – Forward elimination of unknowns – Back substitution

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4 Fig. 9.3
5 Naive Gauss Elimination Procedure Phase-1: Forward Elimination: 1. Set an index, p=1. 2. Pivot row = r p , pivot element = a pp [Pivot row and all preceding rows will remain unchanged] 3. Calculate factors, l ip = a ip /a pp , for i = (p+1) to n 4. Calculate, r i new = r i old - l ip * r p for i = (p+1) to n 5. Increment p by 1. 6. If p <n , go to step 2.

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6 Naive Gauss Elimination Procedure Phase-2: Back Substitution: 1. Calculate, x n = c n / a nn 2. Set an index, i = n – 1 3. Calculate, x i = ( c i – sumof (a i,j x j ) ) /a ii for j = n to i+1 by decrement, and j>i 4. Decrement i by 1 . 5. If i >=1 , go to step 2.
Gaussian Elimination – An Example 2x 1 + 4x 2 + 8x 3 + 6x 4 + 3x 5 = 73 9x 1 + 5x 2 + 3x 3 + 5x 4 + 4x 5 = 68 6x 1 + 7x 2 + 4x 3 + 8x 4 + 9x 5 =109 8x 1 + 5x 2 + 6x 3 + 4x 4 + 2x 5 = 62 4x 1 + 9x 2 + 2x 3 + 3x 4 + 6x 5 = 70

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Gaussian Elimination – An Example Problem x 1 x 2 x 3 x 4 x 5 c 2 4 8 6 3 73 9 5 3 5 4 68 6 7 4 8 9 109 8 5 6 4 2 62 4 9 2 3 6 70 1st Step: p = 1 i = (p+1) to n i = 2 to 5 Pivot row r p = r 1 Pivot element, a pp = a 11 = 2 l ip x 1 x 2 x 3 x 4 x 5 c -- 2 4 8 6 3 73 unchanged l 21 = (9/2) 0 -13 -33 -22 -9.5 -260.5 r 2 new =r 2 old - l 21 *r 1 l 31 = (6/2) 0 -5 -20 -10 0 -110 r 3 new =r 3 old - l 31 *r 1 l 41 = (8/2) 0 -11 -26 -20 -10 -230 r 4 new =r 4 old - l 41 *r 1 l 51 = (4/2) 0 1 -14 -9 0 -76 r 5 new =r 5 old - l 51 *r 1 1 st Step:
Gaussian Elimination – An Example x 1 x 2 x 3 x 4 x 5 c 2 4 8 6 3 73 0 -13 -33 -22 -9.5 - 260.5 0 -5 -20 -10 0 -110 0 -11 -26 -20 -10 -230 0 1 -14 -9 0 -76 2nd Step: p = 2 i = (p+1) to n i = 3 to 5 Pivot row r p = r 2 Pivot element, a pp = a 22 = -13 l ip x 1 x 2 x 3 x 4 x 5 c -- 2 4 8 6 3 73 unchanged -- 0 -13 -33 -22 -9.5 -260.5 unchanged l 32 = (-5/-13) 0 0 -7.3077 -1.5385 3.6538 -9.8077 r 3 new =r 3 old - l 32 *r 2 l 42 = (-11/-13) 0 0 1.9231 -1.3846 -1.9615 -9.5769 r 4 new =r 4 old - l 42 *r 2 l 52 = (1/-13) 0 0 -16.5385 -10.6923-0.7308 -96.0385 r 5 new =r 5 old - l 52 *r 2 2 nd Step:

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Gaussian Elimination – An Example x 1 x 2 x 3 x 4 x 5 c 2 4 8 6 3 73 0 -13 -33 -22 -9.5 -260.5 0 0 -7.3077 -1.5385 3.6538 -9.8077 0 0 1.9231 -1.3846 -1.9615 -9.5769 0 0 -16.5385 -10.6923 -0.7308 -96.0385 3rd Step: p = 3 i = (p+1) to n i = 4 to 5 Pivot row r p = r 3 Pivot element, a pp = a 33 = -7.3077 l ip x 1 x 2 x 3 x 4 x 5 c -- 2 4 8 6 3 73 unchanged -- 0 -13 -33 -22 -9.5 -260.5 unchanged -- 0 0 -7.3077 -1.5385 3.6538 -9.8077 unchanged l 43 = (1.9/-7.3) 0 0 0 -1.7895 -1 -12.1579 r 4 new =r 4 old - l 43 *r 3 l 53 = (-16.5/ -7.3) 0 0 0 -7.2105 -9 -73.8421 r 5 new =r 5 old - l 53 *r 3 3rd Step:
Gaussian Elimination – An Example x 1 x 2 x 3 x 4 x 5 c 2 4 8 6 3 73 0 -13 -33 -22 -9.5 -260.5 0 0 -7.3077 -1.5385 3.6538 -9.8077 0 0 0 -1.7895 -1 -12.1579 0 0 0 -7.2105 -9 -73.8421 4th Step: p = 4 i = (p+1) to n i = 5 Pivot row r p = r 4 Pivot element, a pp = a 44 = -1.7895 l ip x 1 x 2 x 3 x 4 x 5 c -- 2 4 8 6 3 73 unchanged -- 0 -13 -33 -22 -9.5 -260.5 unchanged -- 0 0 -7.3077 -1.5385 3.6538 -9.8077 unchanged -- 0 0 0 -1.7895 -1 -12.1579 unchanged l 54 = (-7.21/ -1.79) 0 0 0 0 -4.9706 -24.8529 r 5 new =r 5 old - l 54 *r 4

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• jane smith

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