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TAMU PETE 301 - Numerical Methods for Engineers Ch. 11 Solutions

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1 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. CHAPTER 11 11.1 (a) First, the decomposition is implemented as e2 = 0.4/0.8 = 0.5 f2 = 0.8 0.5)(0.4) = 0.6 e3 = 0.4/0.6 = 0.66667 f3 = 0.8 0.66667)(0.4) = 0.53333 Transformed system is 0.8 0.4 00.5 0.6 0.40 0.66667 0.53333 which is decomposed as 100[] 0.5 1 00 0.66667 1L 0.8 0.4 0[] 0 0.6 0.40 0 0.53333U The right hand side becomes r1 = 41 r2 = 25 0.5)(41) = 45.5 r3 = 105 0.66667)45.5 = 135.3333 which can be used in conjunction with the [U] matrix to perform back substitution and obtain the solution x3 = 135.3333/0.53333 = 253.75 x2 = (45.5 – (–0.4)253.75)/0.6 = 245 x1 = (41 0.4)245)/0.8 = 173.75 (b) The first iteration can be implemented as 2141 0.441 0.4(0)51.250.8 0.8xx  13225 0.4 0.425 0.4(51.25) 0.4(0)56.8750.8 0.8xxx  23105 0.4105 0.4(56.875)159.68750.8 0.8xx  Second iteration: 141 0.4(56.875)79.68750.8x 225 0.4(79.6875) 0.4(159.6875)150.93750.8x 3105 0.4(150.9375)206.71880.8x2 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. The error estimates can be computed as ,179.6875 51.25100% 35.69%79.6875a ,2150.9375 56.875100% 62.32%150.9375a ,3206.7188 159.6875100% 22.75%206.7188a The remainder of the calculation proceeds until all the errors fall below the stopping criterion of 5%. The entire computation can be summarized as iteration unknown value amaximum a1 x1 51.25 100.00% x2 56.875 100.00% x3 159.6875 100.00% 100.00% 2 x1 79.6875 35.69% x2 150.9375 62.32% x3 206.7188 22.75% 62.32% 3 x1 126.7188 37.11% x2 197.9688 23.76% x3 230.2344 10.21% 37.11% 4 x1 150.2344 15.65% x2 221.4844 10.62% x3 241.9922 4.86% 15.65% 5 x1 161.9922 7.26% x2 233.2422 5.04% x3 247.8711 2.37% 7.26% 6 x1 167.8711 3.50% x2 239.1211 2.46% x3 250.8105 1.17% 3.50% Thus, after 6 iterations, the maximum error is 3.5% and we arrive at the result: x1 = 167.8711, x2 = 239.1211 and x3 = 250.8105. 11.2 As in Example 11.1, the LU decomposition is 10.49 1[]0.645 10.717 1L 2.04 11.550 1[]1.395 11.323U To compute the first column of the inverse 10[]{}00LD Solving this gives3 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 10.490196{}0.3162960.226775D Back substitution, [U]{X} = {D}, can then be implemented to give to first column of the inverse 0.7558410.541916{}0.3496670.171406X For the second column 01[]{}00LD which leads to 0.5419161.105509{}0.7133220.349667X For the third column 00[]{}10LD which leads to 541916.0105509.1713322.0349667.0}{X For the fourth column 00[]{}01LD which leads to 0.1714060.349667{}0.5419160.755841X4 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. Therefore, the matrix inverse is 10.755841 0.541916 0.349667 0.1714060.541916 1.105509 0.713322 0.349667[]0.349667 0.713322 1.105509 0.5419160.171406 0.349667 0.541916 0.755841A 11.3 First, the decomposition is implemented as e2 = 0.020875/2.01475 = 0.01036 f2 = 2.014534 e3 = 0.01036 f3 = 2.014534 e4 = 0.01036 f4 = 2.014534 Transformed system is 2.01475 0.028750.01036 2.014534 0.028750.01036 2.014534 0.028750.01036 2.014534 which is decomposed as 10.01036 1[]0.01036 10.01036 1L 2.01475 0.028752.014534 0.02875[]2.014534 0.028752.014534U Forward substitution yields r1 = 4.175 r2 = 0.043258 r3 = 0.000448 r4 = 2.087505 Back substitution x4 = 1.036222 x3 = 0.01096 x2 = 0.021586 x1 = 2.072441 11.4 Here’s a MATLAB script to verify the results of Example 11.2, clc L=[2.4495 0 0;6.1237 4.1833 0;22.454 20.917 6.1101]; L*L' When this script is run, the results are (note that the discrepancies are due to round-off errors): A = 6.0001 15.0000 55.00115 PROPRIETARY MATERIAL. © The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it


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