Numerical Methods with Worked Examples: Matlab Edition
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This book is for students following an introductory course in numerical methods, numerical techniques or numerical analysis. It introduces MATLAB as a computing environment for experimenting with numerical methods. It approaches the subject from a pragmatic viewpoint; theory is kept at a minimum commensurate with comprehensive coverage of the subject and it contains abundant worked examples which provide easy understanding through a clear and concise theoretical treatment. This edition places even greater emphasis on ‘learning by doing’ than the previous edition.
Fully documented MATLAB code for the numerical methods described in the book will be available as supplementary material to the book on http://extras.springer.com
obtain the rest of the solution values we make use of the values that we already know. Substituting the value of x7 into (2.11) we have 3x6 + 8 = 17 and so x6 = 3. Now that we know x6 and x7 we can reduce (2.10) to an equation involving x5 only, to give x5 + 15 − 2 = 14, and so x5 = 1. Similarly we can determine x4 from (2.9) using 3x4 − 2 + 6 + 6 = 2, to give x4 = −2 23 . Finally, to complete the solution we substitute the known values for x5 and x6 into (2.8) to obtain 2x3 − 10 + 9 = 5 and
original coefficient matrix. We have from (2.71) x1 = 1− 6 1 6 = 0.1389 2.8 Gauss–Seidel Iteration 39 Table 2.4 Gauss–Seidel iteration, results Step x1 x2 x3 x4 x5 x6 ··· x10 1 0.1667 0.1667 0.1667 0.1667 0.1667 0.1667 ··· 0.1667 2 0.1389 0.1157 0.1196 0.1190 0.1191 0.1190 ··· 0.1468 3 0.1474 0.1222 0.1265 0.1257 0.1259 0.1258 ··· 0.1465 4 0.1463 0.1212 0.1255 0.1248 0.1249 0.1249 ··· 0.1464 5 0.1465 0.1213 0.1256 0.1249 0.1250 0.1250 ···
nonlinear equation. We start with an interval containing a root and divide it into a left and a right half. We decide which of the two halves contains a root and proceed with a further division of that half. We do this repeatedly until the interval containing a root is sufficiently narrowed down to meet the level of accuracy required. If we take the mid-point of the latest interval to be an approximation to a root, we can say that this is accurate to within ± half the width of the interval.
shown below. Noting that the decision variable d is already in single digit binary form, represent each of the problem variables, x1 , x2 and x3 as strings of four binary digits. u = [ 1 2 4 8 ]; % Use the continuation mark . . . for multi-line input A = [ 5*u 12*u 12*u 0; 10*u 7*u 16*u 0; 12*u 15*u 10*u 0; . . . 0*u 1*u 0*u 16; 0*u 0*u 1*u −16 ]; b = [ 120; 140; 180; 16; 0 ]; f = [ −200*u −400*u −550*u 0 ]; x = bintprog( f, A, b ) Deduce that the factory should make the switch and that retail
two variables and subject to more than one equality constraint. Table 8.14 has the details. The advantage of the method is that it generally avoids the problems of large σ values which are often found when using the simple penalty function method on the function to be minimised rather than its Lagrangian. A comparison of Tables 8.9 and 8.13 show how the multiplier penalty function method may avoid the problem of having to use very large σ values. Summary In this chapter we have looked at ways of