Question 1 20 points,You are given a matrix u of size 4 1. in Matlab u 2 1 1 1 and a second matrix v of size 4 1. in Matlab v 0 5 1 0 0 Here the size of a matrix m n refers to the number of rows m. and number of columns n in the matrix In order to avoid any ambiguity we shall refer solely to. matrices in this question although of course some of the entities are vectors. We now perform the matrix matrix product u times v T. where recall that T denotes transpose,i 3 points The size of the matrix w is. ii 3 points The operation of equation 1 is implemented in Matlab as. iii 4 points The value of w 1 1 is, We now perform the matrix matrix product uT times v. where again T denotes transpose,iv 3 points The size of the matrix z is. v 3 points The operation of equation 2 is implemented in Matlab as. vi 4 points The value of z 1 1 is,Question 2 20 points. We run the script,begin script,A zeros 1000 1000,A 400 404 1 0. A 678 2 8 0,A 678 999 4 0,A 705 678 2 0,A 705 999 10 0. w ones 1000 1,end script, Note that A is a 1 000 1 000 matrix with only ve non zero entries and that w is a 1 000 1. vector of all ones,i 5 points The value of v 678 is. Hint Consider the row interpretation of the matrix vector product. ii 5 points The value of v 999 is, Hint Consider the row interpretation of the matrix vector product. We now clear all variables and de ne a 1 000 5 matrix B in Matlab B and a 5 1 vector. w in Matlab w as, the j th column of B consists of all j s We then evaluate. and nd the maximum of the vector v as,M max vi 3, If you wish to see the corresponding Matlab script refer to Appendix A at the end of the sampler. iii 10 points The value of M from equation 3 is, Hint Consider the column interpretation of the matrix vector product. Question 3 20 points, We consider linear regression for a particular simple case a single independent variable p 1. which we denote x as always a single dependent variable y a model with only n 2 terms. Ymodel x 0 1 x 4, hence h0 x 1 h1 x x and m 3 noisy measurements given by x1 0 Y1 0. x2 1 Y2 0 x3 2 Y3 2 We denote by Y the 3 1 column vector 0 0 2 T and by. the 2 1 column vector 0 1 T,i 3 points The regression matrix X is given by. Recall that X i the ith component of the vector X is equal to Ymodel xi for the measurement. points xi i 1 2 3,ii 3 points The equation, has for our particular noisy measurements and hence Y vector 0 0 2 T indicated above. a no solution for,b a unique solution for,c many solutions for. Hint Consider the relationship of equation 5 to the problem of putting a straight line. through three points plot the experimental data xi Yi i 1 2 3. iii 5 points The matrix X T X is given by,iv 4 points The matrix X T Y is given by. v 5 points The least squares solution to our problem is. a 0 0 3333 1 1 0,b 0 0 1667 1 1 5,c 0 0 1 1 0,d 0 0 5 1 1 2. Recall that the least squares solution is the particular which minimizes the norm of the. residual squared i e IY X I2, Hint Recall the simple formula for the inverse of a 2 2 matrix. Question 4 20 points, A naturalist studying a pair of geese devises a camou aged scale to provide weight data for each. of the geese henceforth denoted Goose 1 and Goose 2 Sometimes a single goose will walk onto. the scale but on other occasions both geese walk onto the scale at the same time On a given day. the naturalist records in the logbook, Observation 1 Goose 1 on scale scale transducer indicates 12 3 kilos. Observation 2 Goose 1 and Goose 2 on scale scale transducer indicates 26 3 kilos. Observation 3 Goose 2 on scale scale transducer indicates 13 1 kilos. We wish to infer from this data a good estimate for the weights of each of the geese. In what follows we shall denote by W1 the weight of Goose 1 in kilos and by W2 the weight of. Goose 2 in kilos we further denote by W the 2 1 vector. We further de ne the 3 1 vector of scale readings, where again the units are kilos The naturalist s three observations above can be summarized as. where B is a matrix and E is a 3 1 noise vector We shall assume that the noise vector E in. kilos relates to transducer error and does not depend on the weight on the scale. i 8 points The matrix B is given by,a 26 3 26 3,c 26 3 26 3 2. ii 4 points The matrix equation system of linear equations BW S has. a a unique solution for W,b no solution for W,c many solutions for W. iii 8 points The weight vector W, minimizes the norm of the residual squared IS BW I2 over all vectors W. satis es the linear system of equations,Question 5 20 points. The average skin friction coe cient nondimensional and positive C f for a turbulent boundary. layer depends on the Reynolds number nondimensional and positive Re as. where and are unknown constants i e which do not depend on Re Note that is a positive. multiplicative factor and is the exponent of Re The above form would lead to a nonlinear. regression problem We thus de ne y log C f and x log Re such that now y may be expressed. as y Ymodel x with,Ymodel x 0 1 x 10, for 0 and 1 independent of x Here log refers to the natural logarithm with corresponding. Matlab built in function log This form you will agree is amenable to linear regression analysis. We shall assume that the model of equation 10 is bias free there exists a unique true. 0true 1true T such that equation 10 exactly predicts log C f. We further provide m 1 arrays xpts values of log Re and Y values of log C f corresponding to. noisy measurements xpts i Y i 1 i m for m 52 We assume that the measurements. are given by,Y i Ymodel xpts i true E i 1 i m, where the noise E i satis es our assumptions N1 N2 and N3 normal zero mean homoscedastic. with standard deviation uncorrelated in x,We now run the script. begin script,X ones m 1 xpts,betahat X Y,end script. i 5 points Based on the regression analysis above our estimate for is. a betahat 1,b exp betahat 1,c log betahat 1,d betahat 1. where exp is the Matlab built in exponential function. Hint Consider the log transformation of equation 9 to equation 10. Note that prior to running the script the Matlab workspace contains only xpts and Y. ii 5 points Based on the regression analysis above our estimate for is. a betahat 2,b exp betahat 2,c log betahat 2,d betahat 2. Hint Consider the log transformation of equation 9 to equation 10. iii 5 points We now calculate based on the data provided joint 0 95 con dence level con dence. intervals for 0true and 1true I0joint and I1joint respectively We obtain for I1joint the con dence. interval for 1true the interval 0 1560 0 1262 With con dence level 0 95 we may then. conclude that over the range of Re covered by the experimental data. a C f increases with increasing Re,b C f decreases with increasing Re. c C f is independent of Re,d C f is always negative. iv 5 points The length of the con dence interval of Question 5 iii is deemed too large and. hence the accuracy of our 1true estimate potentially too low To remedy the situation you. a increase the number of measurements from m 52 to mnew m and then re perform. the regression analysis, b retain the m 52 measurements but replace the model of equation 10 which has. n 2 terms by a new model which has n 52 terms,Ymodel x j x j 0 1 x 51 x51. and then re perform the regression analysis,Appendix A. Matlab Script for Question 2 iii,begin script,B zeros 1000 5. w zeros 5 1,end script, where max is the Matlab built in function which returns the maximum of a vector. Answer Key,Q1 i c 4 4,ii d w u v,Q3 i b 1 1,ii a no solution for. v a 0 0 3333 1 1 0,Q4 i b 1 1,ii b no solution for W. Q5 i b exp betahat 1,ii a betahat 2,iii b C f decreases with increasing Re. iv a increase the number of measurements,MIT OpenCourseWare. http ocw mit edu, 2 086 Numerical Computation for Mechanical Engineers. For information about citing these materials or our Terms of Use visit http ocw mit edu terms.

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