Newton-Raphson Method of Solving a Nonlinear Equation- More ...

Chapter 03.04Newton-Raphson Method of Solving a NonlinearEquation – More ExamplesIndustrial EngineeringExample 1You are working for a start-up computer assembly company and have been asked todetermine the minimum number of computers that the shop will have to sell to make a profit.The equation that gives the minimum number of computers n to be sold after considering thetotal costs and the total sales isf n40n1.5 875n 35000 0Use the Newton-Raphson method of finding roots of equations to find the minimum numberof computers that need to be sold to make a profit. Conduct three iterations to estimate theroot of the above equations. Find the absolute relative approximate error at the end of eachiteration and the number of significant digits at least correct at the end of each iteration.Solutionf n40n1.5 875n 35000 00.5f n60n 875Let us take the initial guess of the root of n 0f as n 50 .Iteration 1The estimate of the root isf n0n1 n0f n01.54050 8755035000 50 0. 56050 8755392.1 50 450.74 50 11.963 61.963The absolute relative approximate error aat the end of Iteration 1 is0 03.04.1

03.04.2 Chapter 03.04n1 n0a 100n161.963 5010061.963 19.307%The number of significant digits at least correct is 0, as you need an absolute relativeapproximate error of less than 5% for one significant digit to be correct in your result.Iteration 2The estimate of the root isf n1n2 n1f n11.540 61.963 875 61.963 61.9630. 560 61.963 292.45 61.963 402.70 61.963 0.72623 62.689The absolute relative approximate error 875 35000aat the end of Iteration 2 isn2 n1a 100n262.689 61.96310062.689 1.1585%The number of significant digits at least correct is 1, because the absolute relativeapproximate error is less than 5% .Iteration 3The estimate of the root isf n2n3 n2f n21.562.689 87562.6890.6062.689 87540 62.68951.0031 62.689 399.943 62.689 2.508010 62.692The absolute relative approximate error 35000aat the end of Iteration 3 is

Newton-Raphson Method-More Examples: Industrial Engineering 03.04.3n3 n2a 100n362.692 62.689 a10062.6923 4.000610%Hence the number of significant digits at least correct is given by the largest value ofwhich2ma 0 .51032m4.000610 0.51032m8 .001110 103log8.001110 2 m3m 2 log8.001110 4. 0968Som 4The number of significant digits at least correct in the estimated root 62.692 is 4.mforNONLINEAR EQUATIONSTopic Newton-Raphson Method-More ExamplesSummary Examples of Newton-Raphson MethodMajor Industrial EngineeringAuthors Autar KawDate August 7, 2009Web Site http://numericalmethods.eng.usf.edu