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Exam_1_-_repaired_-_flattened (1) - ‘ EML 3034 Modeling...

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Unformatted text preview: ‘ EML 3034: Modeling Methods in MMAE Name: lg: EXAM 1 - Spring 2011 Open books and open notes and put your answers on bubble sheet Problem #1: (60 points) Answer all 15 short answer questions: bubble in 11L choices that are correct (there may be more than one correct choice). l. The main sources of error in all numerical computations are Q round -off (b) segmentation Q truncation (d) phase error (6) 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. (Q triangulate the coefficient matrix into a lower triangular matrix and solving by backsubstitution. fl 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 190th iterations, 56"“ = 9(22”), we are guaranteed that the iteration will converge, if? (a) lg’(m“)| > lfor all :2” (b) l9 (36")I < 1 for all ac" (0) lg (16”)! > 1 for all 2:" ‘ la’(:c”)l < 1 for all :10” (e) none of the above 4. Machine precision is defined as (a) the smallest number you can add to one on the computer and still recognize the addition. a the 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 methods are ”multiplying a row by a non-zero constant 0 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: 1. Gauss elimination (b) Gauss—Seidel method 0 Gauss-Jordan elimination (d) SOR Q LU decomposition ‘7. For the function f (3:) = 3m —— Zea” 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) 0.0125 & 1.818 (c) 0.149 8. If A1: = 0.1, f(:co) = 10 , f(a:o + Am ) = 12, and flare — Ax ) = 9, the lst order backward difference for f’ (330) is: (a) —20 .M, 80350 (b) 20 i0“- AK 9. Iterative methods for the solution of an N x N system of simultaneous equations (a) solve the equations in K x ON) floating point operations, where K=number of iterations to convergence. (b) solve the equations in K x 0(N3) floating point operations, where K=number of iterations to convergence. (c) solve the problem exactly if there was no round off error. ((1) are characterized as schemes for which the dominant source of numerical error is truncation error. (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 (a) solve the equations in K x 0(N2) floating point operations, where K=number of iterations to convergence. (b) solve the equations in K X 0(N3) floating point operations, where K=number of iterations to convergence. (c) 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. (e) 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 (cc) = O, (a) requires only one initial guess. (b) requires only function evaluations. (c) requires both function and first derivative evaluations. 0 requires two initial guesses. (e) requires both function and second derivative evaluations. 12. In solving for the root(s) of a non—linear scalar equation, f (ac) = 0, which method converges fastest to the solution (assuming all methods converge to the root from the given initial guess), (a) Crout's method. (b) secant method. (c) Hilbert's method. (d) bisection method. '0 Newton-Raphson method. 13. Due to the finite accurac of the com uter, zero is ex ressed in single precision as (pick closest value) y P P O 10—38 (b) 10—108 .) 10—12 (d) 10—308 (e) none of the above, zero is written exactly on the computer. )7 ,9/14- The real number :1: = 0.465 is expressed to six decimal places in binary as (a) 1.010111 Q 0.011101 (0) —0.001 101 (d)101.001101 (e) 1.010101 15. The number of bits used to store a real number in single precision is (a) 16 O 32 1 (c) 64 ‘ (d) 128 (e) 256 Problem #2: (20 points) Find the roots of the function f (as) = 3:36” — 235. You are to solve for the roots of this equation by the Newton Rapshon Method using the initial guess :00 = 0.3 and carry out only the first two steps of the iteration to find the estimates of the root. Evaluate iterative convergence and residual at each of the two steps. Answer the questions below (there is only one correct anwer) ' It is helpful to report results in the following table n-———__ --——__— -—_—__ __-—_— 16. The Newton Raphson method is a fixed point iteration of the form :1; = g N R (x) for the root, and from problem 2, the 9NR($) is Q (a) gNR(m)=3xe”‘ —— 2m 911200: X — 4%ng) (b) 9NR($)=$ — 2:1: - 32:6” (c) gNR(:r)=x — 3276—m / (31136—3: — 36—” —~ 2) (d) g]\11.3(m)=3em — 2 + 3:53—35 UgNR($)=:c — (3336—3 — 2:1:)/(3e‘z — 3:66” —— 2) 17. With the intial guess $0 = 0.3, first and second iteration estimates for the root are two 5 decimal places, (3) 0.32028 and 0.4757 respectively. (b) — 1.65028 and — 5.30757respectively. w 0.45021 and 0.40867 respectively. ((1) — 0.45021 and — 0.40857 respectively. ’(e) none of the above. 18. The iterative convergence criteria after two iterations is (a) 2.616 x 10’3 Q 4.154x 10_2 (c) 3.186 x 10—3 (d) 4.132 x 10‘5 (e) none of the above 19. The residual after two iterations is Q 2.616 x 10-3 (b) 4.154X 10—2 (c) 3.186 x10~3 (d) 4.132 x10—5 (e) none of the above. 0. After 2 iterations strating from the intial guess, x0 = 0.3, we expect the iteration to diverge because IgNR (562)] = 0.405 < 1 and indeed the iteration is converging. (b) the iteration to diverge because I g’NR (1:2)} = 0.012 < 1 and indeed the iteration is converging. (c) the iteration to converge because lgNR(:cz)] = 1.45 > 1 and indeed the iteration is diverging. (d) the iteration to converge because lg'NR (3:2)l = 1.45 > 1 and indeed the iteration is converging. (e) none of the above. Problem #3: (20 points) Given the matn'x—vector system 41 g = {Q as: 1 3 2 2:1 1 -5 1 1 x2 = 2 3 2 1 $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 g; through backward substitution. Answere the following questions pertaining to this process(there is only one correct anwer): 21. Upon completion of the triangulation procedure, the element of the triangulated matrix a2,3 is (a) 2.45 (b) — 0.188 (c) 16 V 22. Upon completion of the triangulation procedure, the element of the right hand side vector by, is (a) 2.788 ' (b) — 7.3 72 (c) 3 .531 O) 3.063 (e) 12.000 23. The determinant is (a) — 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. fl zero and the system of equations can be solved. (e) car-mot compute the determinant as it is imaginary. 24. The solution vector is ‘ (a) ($1,332,133) 2 (2.00, 11.00, - 0.189) (b) (m1,x2,x3) 2: ( —— 1.33, — 11.67, — 16.33) Q (901, $2,953) = ( — 1.33,11.67, — 16.33) ((1) (:01, 932,303) = (1.00,2.00,3.00) (e) ($1,332, 2:3) = (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. (b) yes, because the matrix is skew Hermetian. (c) yes, because the matrix is not diagonally dominant. ((1) no, because pivoting is too complicated and takes up computer time. (e) no, because the matrix is diagonally dominant. ...
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