Lecture 5. Linear Systems. Gauss Elimination. Lecture in Numerical Methods from 24. March 2015 UVT. Lecture 5. Linear Systems. show Practical Problem.

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1 Linear Systems. Gauss Elimination. Lecture in Numerical Methods from 24. March 2015 FIGURE A S UVT

2 Agenda of today s lecture 1 Linear Systems. 2 Gauss Elimination FIGURE A S 1.2

3 We ve got a problem 210 LINEAR ALGEBRAIC EQUATIONS AND MATRICES 3 bungee jumpers connected by bungee cords. After they are released, ity takes hold we have (b). Compute the displacement of each e jumpers. cord linear spring x 1 0 x 2 0 x 3 0 (a) Unstretched (b) Stretched FIGURE 8.1 Three individuals connected by bungee cords. FIGURE k 1 x 1 A S m 1 g k 2 (x 2 his o FIGURE 8.2 Free-body dia 1.3 Using Newton s second law, force balances can

4 dt 2 We ve d 2 x 2 m 2 got a problem 0 dt 2 = m 2 g + k 3 (x 3 x 2 ) + k 2 (x 1 x 2 ) (8.1) d 2 x 3 m 3 = m dt 2 3 g + k 3 (x 2 x 3 ) 0 re Use m i Newton s = the of second jumper law i (kg), + t Hooke s = time (s), law: k j = the spring constant for cord j (N/m), x i = the displacement of jumper i measured k 1 x 1 downward k 2 (x 2 from x 1 ) quilibrium k 3 (x 3 x 2 ) position (m), g = itational acceleration (9.81 m/s 2 ). Because we are interested in the steady-state 0 solution, the second derivatives can be set to zero. Collecting terms gives (k 1 + k 2 )x 1 k 2 x 2 = m 1 g k 2 x 1 + (k 2 + k 3 )x 2 k 3 x 3 = m 2 g k 3 x 2 + k 3 x 3 = m 3 g Thus, the problem reduces to solving a m ree simultaneous equations for the three unknown displacements. Because we have used a linear law for the cords, these (a) Unstretched (b) Stretched m equations are linear algebraic equations. Chapters 8 1 g k through 2 (x 2 x 1 ) m 12 2 g k introduce 3 (x 3 x 2 ) m you to 3 g how MATLAB is used to solve such ms of equations. URE 8.1 FIGURE 8.2 e individuals connected by bungee cords. Free-body diagrams. Using Newton s second law, force balances can be written for each jumper: m 1 d 2 x 1 dt 2 = m 1 g + k 2 (x 2 x 1 ) k 1 x 1 (8.2) m 2 d 2 x 2 dt 2 = m 2 g + k 3 (x 3 x 2 ) + k 2 (x 1 x 2 ) (8.1) FIGURE A S m 3 d 2 x 3 = m 3 g + k 3 (x 2 x 3 ) 1.4

5 dt 2 We ve d 2 x 2 m 2 got a problem 0 dt 2 = m 2 g + k 3 (x 3 x 2 ) + k 2 (x 1 x 2 ) (8.1) d 2 x 3 m 3 = m dt 2 3 g + k 3 (x 2 x 3 ) 0 re Use m i Newton s = the of second jumper law i (kg), + t Hooke s = time (s), law: k j = the spring constant for cord j (N/m), x i = the displacement of jumper i measured k 1 x 1 downward k 2 (x 2 from x 1 ) quilibrium k 3 (x 3 x 2 ) position (m), g = itational acceleration (9.81 m/s 2 ). Because we are interested in the steady-state 0 solution, the second derivatives can be set to zero. Collecting terms gives (k 1 + k 2 )x 1 k 2 x 2 = m 1 g k 2 x 1 + (k 2 + k 3 )x 2 k 3 x 3 = m 2 g k 3 x 2 + k 3 x 3 = m 3 g Thus, the problem reduces to solving a m ree simultaneous equations for the three unknown displacements. Because we have used a linear law for the cords, these (a) Unstretched (b) Stretched m equations are linear algebraic equations. Chapters 8 1 g k through 2 (x 2 x 1 ) m 12 2 g k introduce 3 (x 3 x 2 ) m you to 3 g how MATLAB We can is used write to solve thissuch as ms of equations. URE 8.1 FIGURE 8.2 Ax = b. e individuals connected by bungee cords. Free-body diagrams. Using Newton s second law, force balances can be written for each jumper: m 1 d 2 x 1 dt 2 = m 1 g + k 2 (x 2 x 1 ) k 1 x 1 (8.2) m 2 d 2 x 2 dt 2 = m 2 g + k 3 (x 3 x 2 ) + k 2 (x 1 x 2 ) (8.1) FIGURE A S m 3 d 2 x 3 = m 3 g + k 3 (x 2 x 3 ) 1.4

6 dt 2 We ve d 2 x 2 m 2 got a problem 0 dt 2 = m 2 g + k 3 (x 3 x 2 ) + k 2 (x 1 x 2 ) (8.1) d 2 x 3 m 3 = m dt 2 3 g + k 3 (x 2 x 3 ) 0 re Use m i Newton s = the of second jumper law i (kg), + t Hooke s = time (s), law: k j = the spring constant for cord j (N/m), x i = the displacement of jumper i measured k 1 x 1 downward k 2 (x 2 from x 1 ) quilibrium k 3 (x 3 x 2 ) position (m), g = itational acceleration (9.81 m/s 2 ). Because we are interested in the steady-state 0 solution, the second derivatives can be set to zero. Collecting terms gives (k 1 + k 2 )x 1 k 2 x 2 = m 1 g k 2 x 1 + (k 2 + k 3 )x 2 k 3 x 3 = m 2 g k 3 x 2 + k 3 x 3 = m 3 g Thus, the problem reduces to solving a m ree simultaneous equations for the three unknown displacements. Because we have used a linear law for the cords, these (a) Unstretched (b) Stretched m equations are linear algebraic equations. Chapters 8 1 g k through 2 (x 2 x 1 ) m 12 2 g k introduce 3 (x 3 x 2 ) m you to 3 g how MATLAB We can is used write to solve thissuch as ms of equations. URE 8.1 FIGURE 8.2 Ax = b. e individuals connected by bungee cords. Free-body diagrams. How do we solve such problems? Using Newton s second law, force balances can be written for each jumper: m 1 d 2 x 1 dt 2 = m 1 g + k 2 (x 2 x 1 ) k 1 x 1 (8.2) m 2 d 2 x 2 dt 2 = m 2 g + k 3 (x 3 x 2 ) + k 2 (x 1 x 2 ) (8.1) FIGURE A S m 3 d 2 x 3 = m 3 g + k 3 (x 2 x 3 ) 1.4

7 For few equations FIGURE A S 1.5

8 For few equations substitution FIGURE A S 1.5

9 For few equations substitution elimination of variables FIGURE A S 1.5

10 For few equations substitution elimination of variables Cramer s rule FIGURE A S 1.5

11 Summing up equations manipulated so as to result in one equation with one unknown FIGURE A S 1.6

12 Summing up equations manipulated so as to result in one equation with one unknown consequently: equation solved directly & result substituted back into original equations to solve for the rining FIGURE A S 1.6

13 Summing up equations manipulated so as to result in one equation with one unknown consequently: equation solved directly & result substituted back into original equations to solve for the rining This is the basic for an algorithm that works for large ms: FIGURE A S 1.6

14 Summing up equations manipulated so as to result in one equation with one unknown consequently: equation solved directly & result substituted back into original equations to solve for the rining This is the basic for an algorithm that works for large ms: eliminate unknowns FIGURE A S 1.6

15 Summing up equations manipulated so as to result in one equation with one unknown consequently: equation solved directly & result substituted back into original equations to solve for the rining This is the basic for an algorithm that works for large ms: eliminate unknowns substitute back FIGURE A S 1.6

16 is the most basic ese schemes. This section includes the matic techniques for forward elimination b stitution that comprise Gauss elimination. Although these techniques are ideally su implementation on computers, some modifications be required to obtain a algorithm. In particular, the computer program must avoid division by zero. The ing method is called naive Gauss elimination because it does not avoid this p Section 9.3 deal with the additional features required for an effective c program. Naive because it does not avoid division by zero! The approach is designed to solve a gen set of n equations: a 11 x 1 + a 12 x 2 + a 13 x 3 + +a 1n x n = b 1 a 21 x 1 + a 22 x 2 + a 23 x 3 + +a 2n x n = b 2.. a n1 x 1 + a n2 x 2 + a n3 x 3 + +a nn x n = b n FIGURE A S 1.7

17 a n3 x 3 + +a nn x n = b n re the double prime indicates that lements have been modified twice. The procedure can be continued using the rining pivot equations. The f Elimination manipulation of Unknowns: in the sequence First reduce is to use the ms (n 1)th matrix equation to eliminate the x n 1 t an upper triangular from the nth one. equation. At this point, the m have been transformed to an u triangular m: a 11 x 1 + a 12 x 2 + a 13 x 3 + +a 1n x n = b 1 (9. FIGURE A S a 22 x 2 + a 23 x 3 + +a 2n x n = b 2 (9. a 33 x 3 + +a 3n x n = b 3 (9.... a nn (n 1) x n = b n (n 1) (9. SubstituteBack back: Substitution. Equation (9.11d) can now be solved for x n : x n = b(n 1) n j=i+1 (9 x n = b(n 1) n a (n 1) b (i 1) nn i n a (i 1) ij x j j=i+1, x This result can be back-substituted into the (n 1)th equation to solve for x n 1. The a (n 1) i =, i = 1, n 1. nn a (i 1) cedure, which is repeated ii to evaluate the rining x s, can be represented by following formula: n b (i 1) i a (i 1) ij x j 1.8

18 Example Use Gauss elimination to solve: 3x 1 0.1x 2 0.2x 3 = x 1 + 7x 2 0.3x 3 = x 1 0.2x x 3 = 71.4 FIGURE A S 1.9

19 Example Use Gauss elimination to solve: 3x 1 0.1x 2 0.2x 3 = x 1 + 7x 2 0.3x 3 = x 1 0.2x x 3 = 71.4 Exact solution: x 1 = 3, x 2 = 2.5, x 3 = 7. FIGURE A S 1.9

20 Counting the operations Execution time depends on floating-point operations (or flops). Performing addition/subtraction multiplication/division is about the same. Summing up the ber of flops important in understing FIGURE A S 0

21 Counting the operations Execution time depends on floating-point operations (or flops). Performing addition/subtraction multiplication/division is about the same. Summing up the ber of flops important in understing which are the most time-consuming parts FIGURE A S 0

22 Counting the operations Execution time depends on floating-point operations (or flops). Performing addition/subtraction multiplication/division is about the same. Summing up the ber of flops important in understing which are the most time-consuming parts how computation time increases as ms get larger FIGURE A S 0

23 [135]:= Quit ; In[1]:= ut[2]= Gauss elimination algorithm SquareMatrixQ MatrixQ && Equal Dimensions &; SquareMatrixQ x True [16]:= Gauss A_?SquareMatrixQ, b_? ArrayQ : Module nb, Aug, n, factor, aux, x, If Length A Length b, Print "Error: dimensions of A b do not correspond" ; Abort ; n Length A ; nb n 1; Aug MapThread Append, A, b ; or Join A,Transpose b,2 For k 1, k n 1, k, For i k 1, i n, i, factor Aug i, k Aug k, k ; Aug i, k ;; nb Aug i, k ;; nb factor Aug k, k ;; nb ; Aug n, nb x Aug n, n ; For i n 1, i 1, i, Aug i, nb Aug i, i 1 ;; n.x aux ; x Prepend x, aux ; ; Aug i, i x [10]:= A 3, 0.1, 0.2, 0.1, 7, 0.3, 0.3, -0.2, 10 ut[10]= 3, 0.1, 0.2, 0.1, 7, 0.3, 0.3, 0.2, 10 [14]:= b 19.3, 71.4 ; FIGURE A S 1

24 Naive Gauss: counting operations Outer loop: n 1 iterations FIGURE A S 2

25 Naive Gauss: counting operations Outer loop: n 1 iterations Inner loop: one division; FIGURE A S 2

26 Naive Gauss: counting operations Outer loop: n 1 iterations Inner loop: one division; for every column element (from 2 to nb) - one subtraction & one multiplication FIGURE A S 2

27 Naive Gauss: counting operations Outer loop: n 1 iterations Inner loop: one division; for every column element (from 2 to nb) - one subtraction & one multiplication total n subtractions & n + 1 multiplication/division FIGURE A S 2

28 Therefore, the limits on the inner loop are from i = 2 to n. According to Eq. (9.14d), this means that the ber of iterations e inner loop be Naive n Gauss: counting operations 1 = n = n 1 (9.15) i=2 For Outer every one loop: ese n iterations, 1 iterations there is one division to calculate the factor. The next line then Inner performs loop: a multiplication a subtraction for each column element from 2 to nb. Because nb = n + 1, going from 2 to nb results in n multiplications n subtractions. Together with one the division; single division, this amounts to n + 1 multiplications/divisions n addition/subtractions for every column for every element iteration of (from the inner 2 to loop. nb) The - one total for subtraction the first pass through the outer loop is therefore (n 1)(n + 1) multiplication/divisions (n 1)(n) addition/subtractions. & one multiplication total Similar nreasoning subtractions can be used & n to + estimate 1 multiplication/division the flops for the subsequent iterations e outer loop. These can be summarized as Outer Loop Inner Loop Addition/Subtraction Multiplication/Division k i Flops Flops 1 2, n (n 1)(n) (n 1)(n + 1) 2 3, n (n 2)(n 1) (n 2)(n).. k k + 1, n (n k)(n + 1 k) (n k)(n + 2 k).. n 1 n, n (1)(2) (1)(3) Therefore, the total addition/subtraction flops for elimination can be computed as n 1 n 1 (n k)(n + 1 k) = [n(n + 1) k(2n + 1) + k 2 ] (9.16) k=1 k=1 FIGURE A S 2

29 addition/subtractions. Naive Similar Gauss: reasoning can counting be used to operations estimate the flops for the subsequent iterations e outer loop. These can be summarized as Outer Loop Inner Loop Addition/Subtraction Multiplication/Division k i Flops Flops 1 2, n (n 1)(n) (n 1)(n + 1) 2 3, n (n 2)(n 1) (n 2)(n).. k k + 1, n (n k)(n + 1 k) (n k)(n + 2 k).. n 1 n, n (1)(2) (1)(3) Therefore, the total addition/subtraction flops for elimination can be computed as n 1 n 1 Total addition/subtraction flops for elimination: (n k)(n + 1 k) = [n(n + 1) k(2n + 1) + k 2 ] (9.16) k=1 k=1 n 1 (n k)(n + 1 k) = n3 3 + O(n). k=1 For multiplication/division flops: n O(n2 ). Rrk: As n gets large, the lower terms become negligible! FIGURE A S 3

30 Thus, the total ber of flops is equal to 2n /3 plus an additional comp portional Naive Gauss: to terms counting of order operations n 2 lower. The result is written in this way becaus large, the O(n 2 ) lower terms become negligible. We are therefore justified in ing that for large n, ffort involved in forward elimination converges on 2n 3 For Because elimination: only a single loop is used, back substitution is much simpler to eva ber of addition/subtraction 2n flops is equal to n(n 1)/2. Because of xtr O(n2 ). prior to the loop, the ber of multiplication/division flops is n(n + 1)/2. Th added For substituting to arrive at back, a total total of ber of flops: n 2 + O(n) n 2 + O(n). Thus, Total ffort total ineffort naive in Gauss naive Gauss elimination: elimination can be represented as 2n O(n2 ) }{{} Forward elimination + n 2 + O(n) } {{ } Back substitution as n increases 2n3 3 + O(n2 ) Two useful gen conclusions can be drawn from this analysis: FIGURE A S 1. As the m gets larger, the computation time increases greatly. As in Tab amount of flops increases nearly three orders of magnitude for every order 4

31 Naive Gauss: counting operations Conclusions: as the m gets larger, the computation time increases greatly the amount of flops increases nearly three orders of magnitude for every order of magnitude increase in the ber of equations; FIGURE A S 5

32 Naive Gauss: counting operations Conclusions: as the m gets larger, the computation time increases greatly the amount of flops increases nearly three orders of magnitude for every order of magnitude increase in the ber of equations; most of ffort is incurred in limination step efforts to make the method more efficient should probably focus on this step. FIGURE A S 5

33 How does work for 2x 2 + 3x 3 = 8 4x 1 + 6x 2 + 7x 3 = 3 2x 1 3x 2 + 6x 3 = 5? FIGURE A S 6

34 How does work for 2x 2 + 3x 3 = 8 4x 1 + 6x 2 + 7x 3 = 3 2x 1 3x 2 + 6x 3 = 5? The first step would involve a division by the pivot a 11 = 0. FIGURE A S 6

35 How does work for 2x 2 + 3x 3 = 8 4x 1 + 6x 2 + 7x 3 = 3 2x 1 3x 2 + 6x 3 = 5? The first step would involve a division by the pivot a 11 = 0. Rrk: problems arise also n the pivot element (the element in Gauss elimination that we re dividing by) is close to zero roundoff errors introduced! FIGURE A S 6

36 Partial complete pivoting Solution: before each row is worked, determine the coefficient with the largest absolute value in the column below the pivot element; switch rows such that the largest element is the pivot element. This method is called partial pivoting. FIGURE A S 7

37 Partial complete pivoting Solution: before each row is worked, determine the coefficient with the largest absolute value in the column below the pivot element; switch rows such that the largest element is the pivot element. This method is called partial pivoting. If both columns rows are searched for the largest element, then switched: complete pivoting FIGURE A S 7

38 Partial complete pivoting Solution: before each row is worked, determine the coefficient with the largest absolute value in the column below the pivot element; switch rows such that the largest element is the pivot element. This method is called partial pivoting. If both columns rows are searched for the largest element, then switched: complete pivoting rarely used because most significant improvement comes from the partial pivoting. FIGURE A S 7

39 An example Use Gauss elimination to solve x x 2 = x x 2 = Repeat the computation with partial pivoting. The exact solution is x 1 = 1/3, x 2 = 2/3. FIGURE A S 8

40 ue Anto example subtractive cancellation, the result is very sensitive to the ber of significant ote With how naive the solution Gauss for elimination x gures carried in the computation: 1 is highly dependent on the ber of significant figures. This because in Eq. (E9.4.1), we are subtracting two almost-equal bers. On the other h, if quations are solved in reverse Absolute order, Value the row of with the larger ivot Significant element is normalized. The equations are Percent Relative Figures x 2 x 1 Error for x x x 2 = x x 2 = limination 5 substitution again yields x 2 = 2/3. For different 10 bers of significant gures, x6 1 can be computed from the first equation, as in x 1 = 1 (2/3) 1 te With case how is partial the much solution pivoting less for sensitive x 1 is highly to the dependent ber of on significant the ber figures of significant in the computation: figures. This because in Eq. (E9.4.1), we are subtracting two almost-equal bers. On the other h, if quations are solved in reverse Absolute order, Value the row of with the larger ivot Significant element is normalized. The equations are Percent Relative Figures x x x 2 = 2 x Error for x x x = limination 5 substitution again yields x 2 = 2/3. For different 0.001bers of significant gures, x 1 6can be computed from the first equation, as in x 1 = 1 (2/3) FIGURE A S 9

41 with Gauss elimination For a triangular matrix (matrix having zeros either above or below the main diagonal) det A = a 11 a 22 a a nn. FIGURE A S 1.20

42 with Gauss elimination For a triangular matrix (matrix having zeros either above or below the main diagonal) For : det A = a 11 a 22 a a nn. det A = a 11 a (1) 22 a(2) 33 a(n 1) nn. FIGURE A S 1.20

43 with Gauss elimination For a triangular matrix (matrix having zeros either above or below the main diagonal) For : det A = a 11 a 22 a a nn. det A = a 11 a (1) 22 a(2) 33 a(n 1) nn. For Gauss elimination with partial pivoting: det A = ( 1) p a 11 a (1) 22 a(2) 33 a(n 1) nn, p - the ber of times that rows are pivoted. FIGURE A S 1.20