My Test 11111 - [ML 3034 Modeling Methods in MMAE Name...

Info icon This preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
Image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: [ML 3034: Modeling Methods in MMAE Name: Kt (' /\ (3 #gm EXAM 1 - Spring 2012 Open books and open notes and put your answers on bubble sheet mmm #1: (60 points) Answer all 15 short answer questions: bubble in g choices that are m (there may be more than one correct choice). 1. Direct methods for the solution ofan N x N system ofsimultaneous equations fsolve the equations in M x 0(N’) floating point operations, where M = number of iterations to convergence. % solve the equations in M x 0m") floating point operations, where M = number of iterations to convergence. c) 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. - solve the problem exactly if there was no round off error. 2. The solution ofa linear system ofequations by Gauss elimination consists of using ERO’s on the system to (a) triangulate the coeflicient matrix into aw matrix and solving by backsubstitution. (b) W the coefiicient matrix and solving by backsubstiurtion. (c) triangulate the coefiicient matrix into an upper triangular matrix and solving byw. (d) triangulate the coefficient matrix into a W and solving by backsubstitution. . triangulate the coefficient matrix into an upper triangular matrix and solving by backsubstitution. 3. For fixed point iterations, 3,,“ = g(:r:,.), for g(z) = fie: and with an initial guess so = 1.0, we have a sufficient condition for the iteration will converge? I. _ . yes, because |g’(zo)l = 0.544 and all subsequent derivatives are smaller than one. $&\\~\\R\ ._ ’ _ (b) W " .4‘ (c) W (d) no, becasue Lq’(£9)l > 1 for all 2:,I and all subsequent derivatives are greaterthan one. (e) none of the above ‘ 7’ 4. Machine precision is defined as the smallest number you can add to one on the computer and still recognize the addition. the smallest number you can substract fi'om one on the computer and still recognize the substraction. (c the range over which integers that can be expressed exactly on the computer. ((1) the range over which real numbers that can be expressed exactly on the computer. (c) none of the above. 5. The elementary row operations (EROs) that are used to solve simultaneous linear equations by direct methods are row switches. ) switching diagonal coefficients. . adding a multiple of a row to another. . multiplying a now by a non-zero constant. (e) none of the above. . 6. Direct methods for solution of simultaneous equations include: Gauss elimination. _Q~ LU decomposition. (e) none of the above. For the firnction f (z) = 5:1: — 2ez with an initial bracket [a,b] = [0, 1.5], the residual at the lst estimate of the root .18ng bisection is to 3 decimals places: (a) 0.378 (b) 1.633 (c) 0.013 0. H7, ll (d) 2.I85 @lnone of the above 8. In order to evaluate the first derivative of f (1:) at 1., using a backward first order finite difference, “30) = W). +0031.) on the computer (a) we can always increase the accuracy by taking A2: to be increasingly “MW!!- ‘flwe can always increase the accuracy by taking A1: to be increasingly small within the limitation posed by mtmdofi‘ error amplification. (c) we can always increase the accuracy by taking A2: to be increasingly small within the limitation posed by Mention. is. mrmd ofi‘andmr competition places an optimal limit on how small we can take A1: and beyond which decreasing A2: causes e error to stop decreasing and to grow. (e) none of fire above. 9. With an initial bracket of [2.5,3.75] for the root of the function f (z) = 1:3 - 192: + 30, using bisection, after 10 bisections the error is estimated to be: (a) 2.232 x 10‘5 a 1 .x |‘- . _ _ \ ' ‘ '_ 8.1.2233: 502...... 5‘an HM \. Dike; (d) 6.104 x 10’4 -—-‘_ . cannot be estimated. ‘ I ' ' / g) P ’ Pn \ é I M 10. The secant method to find the root(s) of a non-linear scalar equation, f (z) = 0, Q A (a) requires oneinitialguess. - |'~ 31¢ A (b) requires function evaluations. ’ m - (J 00 1:2, (c)requires both function and first derivative evaluations. \ r (d) requires both function and second derivative evaluations. . requires two initial guesses. 1]. 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), (a) The Thomas algorithm (b) secant method. (c) Hilbert's method. (d) bisection method. . Newton~Raphson method. l2. Due to the finite accuracy of the computer, zero is expressed in double precision as (pick closest value) (a) 10-38 61"“ (b) 10'108 (c) 10"2 Q 10‘308 (e) none of the above, zero is written exactly on the computer. 13. The main sources of error in all numerical computations are ’round 011'. (b) segmentation. . truncation. (1:!) phase error. (e) none of the above. (8) 0.010111 0 0.110010 - (c)-0.001101 (d) 10.001101 / . (e) 1.011101 (V10 13411 M 15. The number of bits used to store a real number in double precision is (a) 16 (b) 32 . 64 (d) 128 (e) 256 Prom #2: (20 points) Find the roots of the function f (2:) = 7360-53 — 3. You are to solve for the roots of this equationbytheMMMethod usingtheminal guecsx, = 0.10mdcanyouton1ythemmof ml 16. The Newton Raphson method is a fixed point iteration of the form :1: = gNR(z) for the root, and from problem 2 the 9013(3) is (a) 9NR($)=73€—°5‘ — 3-77 ‘gng(z)=a: — (he-0'5“ — 3) / (7e‘°'5’ - 3.5134“) (c) gNR(:1:)=z — 3/(7e‘°‘5‘ — 3.5:ce‘0'5‘) (d) 9NR(3)=3 -‘ 7wed/(76“ — 3e" ‘ 3) (e) 91.”;(:1:)=3ez - 2 + 336“ 17.Withthe intialgueesz, = 0.10,firstandsecond iterationeetimatesfortherootaretwo3decimal places, \. 0.469 and 0.564 respectively. (b) 1.350 and 3.7] 7respectively. (c) 4.514 and 0.467 respectively. (d) 0.7023 and 0.357 respectively. (e) none of the above. 8. The iterative convergence criteria after two iterations is (at)2.616x10--3 (b) 4.154x 10-2 (c) 3.186 x10‘3 d“ 9.50 x 10‘2 (e) none of the above 19. The residual alter two iterations is w“ 2.158 x 10‘2 (b) 4.154 x 10" (c) 3.186 x10'3 (d) 4.132 x 10‘5 (e) none of the above. 20.AfierZ iterationsstratingfi'omtheintinlguess,:r:o = 0.10,weexpect (a) the iteration to diverge because I gm; (2:1)I = 0.4052 < 1 and indeed the iteration is converging. -.the iteration to diverge because lg’NR(zz)| = 00068 < 1 and indeed the iteration is converging. (c) the iteration to converge because |9N1r(12)l = 1.4563 > 1 and indeed the iteration is diverging. (d) the iteration to converge because I g‘NR(zz)| = 1.4563 > 1 and indeed the iteration is converging. (e) none of the above. w: (20points)Giventhematrhr—vectorsystem4g =15 as: 10 3 5 2:1 1 S — 9 l 172 = 3 3 7 ‘15 1:3 8 21. Is it necessary to use pivoting to solve this set ofequations? (a) no, because the matrix is symmetric positive definite. *no, because the matrix is diagonally dominant. (c) yes, because the matrix is skew Hermetian. (d) yes, because the matrix is not diagonally dominant. (e) no, because pivoting is too complicated and takes up computer time. 23. Upon completion of the triangulation procedure, the element of the right hand side vector b3 is ‘9152 m—mn (c) 3.531 (d) 3.063 (e) none of the above la) - 3 and the system of equations cannot be solved because it is negative. 9:1, 824 and the system of equationsean be solved. (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. (e)cannot compute the dotenninant as it is imaginary. 25. The solution vector is 3(zl,z2,z3) = (0.412, —o.163, -o.527) (b (zl,z2,zs) = (2.00, 11.00, - 0.189) (c (31,22,23) = ( — 1.33, — 11.67, — 16.3) (d (31,32,233) = ( — 1.33, 11.67, — 16.33) (e (21,22,33) = (1.00, 2.00, 3.00) ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern