Test 111111 - W 3034 Modeling Methods in MMAE EXAM 1 Spring...

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Unformatted text preview: W 3034: Modeling Methods in MMAE _,- EXAM 1 - Spring 2011 Open books and open notes and put your answers on bubble sheet MM ay be Problem #1: (60 points) Answer all 15 short answer questions: bubble in fl choices that are correct (there m more than one correct choice). ' l. The main sources of error in all numerical computations are © round -off " . segmentation (G truncation \ (d) phase error (e) none of the above 2. The solution of a linear system of equations by Gauss elimination consists of using ERO's on the system to (a) diagonalize the coefficient matrix and solving by backsubstitution. (b) triangulate the coefficient-matrix into an upper triangular matrix and solving by forward substitution. c triangulate the coefficient matrix into a lower triangular matrix and solving by backsubstitution. triangulate the coefficient matrix into an upper triangular matrix and solving by backsubstitution. (e) triangulate the coefficient matrix into a tridiagonal matrix and solving by backsubstitution. 3. For fixed point iterations, x"+1 = g(:r"), we are guaranteed that the iteration will converge, if? (a) lg’(a:")| > lfor all as” (b) '9 (33”)I < lfor all 2:" 9) l9 (113")1 > l'for all 1:" I9'(1‘")l < lfor all x” (e) none of the above 4. Machine precision is defined as the smallest number you can add to one on the computer and still recognize the addition. e smallest number you can substract from one on the computer and still recognize the substraction. (c) the range over which integers that can be expressed exactly on the computer (d) the range over which real numbers that can be expressed exactly on the computer (e) none of the above. 5. The elementary row operations (EROS) that are used to solve simultaneous linear equations by direct I ethods are multiplying a row by a non-zero constant 3 row switches (c) switching diagonal coefficients adding a multiple of a row to another e) none of the above 6 Direct methods for solution of simultaneous equations include: Gauss elimination O Gauss-Seidel method (Q Gauss-Jordan elimination (d) SOR LU decomposition I v Iil-n-II-III'_-_l_- ‘ .._..._.___-.._._‘-...m.._...w .. .1 t . - 7. For the fimction f(x) = 3a: — Qe“J with an initial bracket [a,b] = [0,05], given that the lst estimate of the root p1=0.25, the residual at the 1st iteration for bisection is: (a) 0.378 (b) -0.378 0.0125 1.818 (e) 0.149 8. If A1: = 0.1, f(za) = 10 , f(:1:o + A1: ) = 12, and f(a:o — Ax ) = 9, the lst order backward difference for fl($o) is; O .. (q) (a) — 20 ‘ 5 (b) 20 T (c) — 10 10 ~ e 15 9. Iterative methods for the solution of an N x N system of simultaneous equations (a solve the equations in K X 0(N2) floating point operations, where K==number of iterations to convergence. olve the equations in K x 0(N3) floating point operations, where K=number of iterations to convergence. . solve the problem exactly if there was no round off error. ‘7? d are characterized as schemes for which the dominant source of numerical error is truncation error. comm. (e) are characterized as schemes for which the dominant source of numerical error is round-off error. 10. Direct methods for the solution of an N x N system of simultaneous equations @ solve the equations in K x W) floating point operations, where K=number of iterations to convergence. vb) solve the equations in K x 0(N3) floating point operations, where K=number of iterations to convergence. [email protected] solve the problem exactly if there was no round off error. ‘ ' d are characterized as schemes for which the dominant source of numerical error is truncation error. @ are characterized as schemes for which the dominant source of numerical error is round-off error, 11. The secant method to find the root(s) of a non-linear scalar equation, f (9:) = 0, (a requires only one initial guess. @equires only function evaluations. c requires both function and first derivative evaluations. requires two initial guesses. (6) requires both function and second derivative evaluations. 12. In solving for the root(s) of a non-linear scalar equation, f (1:) = 0,which method converges fastest to the solution (assuming all methods converge to the root from the given initial guess), ’3 (a) Crout's method. (b) secant method. (c) Hilbert's method. (d bisection method. @ Newton-Raphson method. 13. Due to the finite accuracy of the computer, (a) 10‘38 (1)) 10—108 (c) 10‘12 ,(d) 10‘308 @ none of the above, zero is written exactly on the computer. zero is expressed in single precision as (pick closest value) . 14. The real number a: = 0.465 is expressed to six decimal places in binary as ( 1.010111 ,' ) 0.011101 ,5 O (c)-0.001101 I (d) 101.001101 1 (e) 1.010101 , 15. The number of bits used to store a ‘re‘ah number in sinOgle precision is (a) 16 e 32 c 64 (d) 128 (e) 256 (x) = 32c“ '— 22:. You are to solve for the roots of this ‘ Problem #2: (20 points) Find the roots of the function f carry out only the first two steps of equation by the Newton Rapghon Method using the initial guess 2:0 = 0.3 and the iteration to find the es ' ates of the root. Evaluate iterative convergence and residual at each of the two steps. Answer the questions below (there is only one correct anwer) - 3 It is helpful to report results in the follow‘ 0 T ' Esau-«term Root form. :1: = 91m (:5) for the root, and from problem 2, 16. The Newton Raphson method is a fixed point iteration of the the gNR (97) is (a) 9NR($)=3$e—$ ’ 2x , _,< .- (b)QNR($)=$-2$—3$€_3 5; ()0 : 35 + 8de X0 - Qn. 3 .5. (c) gNR(x)=x —— 3me"$/(3:1:e‘$ .. 36—3 _. 2) , : 38'X {.gxe-‘Y __ 3‘ 967‘”): X 2 3‘3 16" .72 ‘ (d) gNR(a:)=3e” — 2 + 327e’ ' , . @N3(z)=m — (3:06“z — 22:)/(3e‘“ ~.— 3313’“5 — 2) 31061 + 36,)(Xxe' X) - 5L 'Jxe'x: fig 32. ‘r‘ x: . 36.x '3><€_x'“’9.. i . -x 17. With the intial guess :co = 0.3, first and second iteration estimates for the root are two 5 decimal; p aces, (a) 0.32028 and 0.4757 respectively. ' (b) — 1.65028 and — 5.30757respectively. 3(_ ,7 _ ->< @ 0.45021 and 0.40867 respectively. ( 36 )4” (d) — 0.45021 and — 0.40857 respectively. (e) none of the above. 18. The iterative convergence criteria after two iterations is . 2.616 x10‘3 @ 4.1541110—2 c) 3.186x10’3 (d) 4.132 x 10’5 (e) none of the above 19. The residual after two iterations is 2.616 x 10"3 (b) 4.154x 10’2 (c) 3.186 x10“3 (d) 4.132 x 10‘5 (6) none of the above. 20. After 2 iterations strating from the intial guess, 1:, = 0.3, we expect (a) the iteration to diverge because | gNR(x2)| = 0.405 < 1 and indeed the iteration is converging. @the iteration to diverge because lg’NR(:c2 )l = 0.012 < 1 and indeed the iteration is converging. , i c) the iteration to converge because lg N R (2:2)l = 1.45 > 1 and indeed tlieiteration is diverging. (d) the iteration to converge because lg'NR (x2)| = 1.45 > 1 and indeed the iteration is converging. (e) none of the above. ' __________/—————————— Problem #3: (20 points) Given the matrix—vector system A a: = '1‘), as: =~ 1 3 2 .271 1 -5 1 1 $2 = 2 3 2 1 11:3 3 Find the corresponding upper triangular matrix [U] and modified right hand-side vector '13, produced by Gauss Elimination implemented without pivoting, from the resulting upper triangular matrix compute the determinant, and calculate the solution 55 through backward substitution. Answere the following questions pertaining to this process(tliere is only one correct anwer): 21. Upon completion of the triangulation procedure, the element of the triangulated matrix 0.23 is . The determinant is Q — 3 and the system of equations can be solved. (b) — 3 and the system of equations cannot be solved because it is negative. (c) 345 and the system of equations cannot be solved because it is too large. (d) zero and the system of equations can be solved. (6) cannot compute the determinant as it is imaginary. 24. The solution vector is (a) (x1,x2,z3) = (2.00,11.00, — 0.189) b ($1,x2,x3) = ( —- 1.33, —— 11.67, — 16.33) ($1,372, :53) = ( —- 1.33, 11.67, - 16.33) ((1) ($1,172,133) = (1...00,200,300) (e) (3:1,x2,a:3) = (1.00, 7.00, 5.06) 25. In practice, should you use pivoting to solve this set of equations? (a) no, because the matrix is symmetric positive definite. yes, because the matrix is skew Hermetian. @yes, because the matrix is not diagonally dominant. (d) no, because pivoting is too complicated and takes up computer time. (e) no, because the matrix is diagonally dominant. ...
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