Chapter 6 VBA Program to Implement the Newton-Raphs on Algorithm and Solve Example

6.16 Here is a VBA program to implement the Newton-Raphson algorithm and solve Example 6.3. Option Explicit Sub NewtRaph() Dim imax As Integer, iter As Integer Dim x0 As Single, es As Single, ea As Single x0 = 0# es = 0.01 imax = 20 MsgBox "Root: " & NewtR(x0, es, imax, iter, ea) MsgBox "Iterations: " & iter MsgBox "Estimated error: " & ea End Sub Function df(x) df = -Exp(-x) - 1# End Function Function f(x) f = Exp(-x) - x End Function Function NewtR(x0, es, imax, iter, ea) Dim xr As Single, xrold As Single xr = x0 iter = 0 Do xrold = xr xr = xr - f(xr) / df(xr) iter = iter + 1 If (xr <> 0) Then ea = Abs((xr - xrold) / xr) * 100 End If If ea < es Or iter >= imax Then Exit Do Loop NewtR = xr End Function It’s application yields a root of 0.5671433 after 4 iterations. The approximate error at this point is 2.1×10−5 %. 6.17 Here is a VBA program to implement the secant algorithm and solve Example 6.6. Option Explicit Sub SecMain() Dim imax As Integer, iter As Integer Dim x0 As Single, x1 As Single, xr As Single Dim es As Single, ea As Single x0 = 0 x1 = 1 es = 0.01 imax = 20 MsgBox "Root: " & Secant(x0, x1, xr, es, imax, iter, ea) MsgBox "Iterations: " & iter MsgBox "Estimated error: " & ea End Sub Function f(x) f = Exp(-x) - x End Function Function Secant(x0, x1, xr, es, imax, iter, ea) xr = x1 iter = 0 Do xr = x1 - f(x1) * (x0 - x1) / (f(x0) - f(x1)) iter = iter + 1 If (xr <> 0) Then ea = Abs((xr - x1) / xr) * 100 End If If ea < es Or iter >= imax Then Exit Do x0 = x1 x1 = xr Loop Secant = xr End Function It’s application yields a root of 0.5671433 after 4 iterations. The approximate error at this point is 4.77×10−3 %. 6.18 Here is a VBA program to implement the modified secant algorithm and solve Example 6.8. Option Explicit Sub SecMod() Dim imax As Integer, iter As Integer Dim x As Single, es As Single, ea As Single x = 1 es = 0.01 imax = 20 MsgBox "Root: " & ModSecant(x, es, imax, iter, ea) MsgBox "Iterations: " & iter MsgBox "Estimated error: " & ea End Sub Function f(x) f = Exp(-x) - x End Function Function ModSecant(x, es, imax, iter, ea) Dim xr As Single, xrold As Single, fr As Single Const del As Single = 0.01 xr = x iter = 0 Do xrold = xr fr = f(xr) xr = xr - fr * del * xr / (f(xr + del * xr) - fr) iter = iter + 1 If (xr <> 0) Then ea = Abs((xr - xrold) / xr) * 100 End If If ea < es Or iter >= imax Then Exit Do Loop ModSecant = xr End Function It’s application yields a root of 0.5671433 after 4 iterations. The approximate error at this point is 3.15×10−5 %. 6.19 Here is a VBA program to implement the 2 equation Newton-Raphson method and solve Example 6.10. Option Explicit Sub NewtRaphSyst() Dim Dim Dim Dim imax As Integer, x0 As Single, y0 xr As Single, yr es As Single, ea x0 = 1.5 y0 = 3.5 es = 0.01 imax = 20 iter As Integer As Single As Single As Single Call NR2Eqs(x0, y0, xr, yr, es, imax, iter, ea) MsgBox "x, y = " & xr & ", " & yr MsgBox "Iterations: " & iter MsgBox "Estimated error: " & ea End Sub Sub NR2Eqs(x0, y0, xr, yr, es, imax, iter, ea) Dim J As Single, eay As Single iter = 0 Do J = dudx(x0, y0) * dvdy(x0, y0) - dudy(x0, y0) * dvdx(x0, y0) xr = x0 - (u(x0, y0) * dvdy(x0, y0) - v(x0, y0) * dudy(x0, y0)) / J yr = y0 - (v(x0, y0) * dudx(x0, y0) - u(x0, y0) * dvdx(x0, y0)) / J iter = iter + 1 If (xr <> 0) Then ea = Abs((xr - x0) / xr) * 100 End If If (xr <> 0) Then eay = Abs((yr - y0) / yr) * 100 End If If eay > ea Then ea = eay If ea < es Or iter >= imax Then Exit Do x0 = xr y0 = yr Loop End Sub Function u(x, y) u = x ^ 2 + x * y - 10 End Function Function v(x, y) v = y + 3 * x * y ^ 2 - 57 End Function Function dudx(x, y) dudx = 2 * x + y End Function Function dudy(x, y) dudy = x End Function Function dvdx(x, y) dvdx = 3 * y ^ 2 End Function Function dvdy(x, y) dvdy = 1 + 6 * x * y End Function It’s application yields roots of x = 2 and y = 3 after 4 iterations. The approximate error at this point is 1.59×10−5 %. 6.20 The program from Prob. 6.19 can be set up to solve Prob. 6.11, by changing the functions to Function u(x, y) u = y + x ^ 2 - 0.5 - x End Function Function v(x, y) v = x ^ 2 - 5 * x * y - y End Function Function dudx(x, y) dudx = 2 * x - 1 End Function Function dudy(x, y) dudy = 1 End Function Function dvdx(x, y) dvdx = 2 * x ^ 2 - 5 * y End Function Function dvdy(x, y) dvdy = -5 * x End Function Using a stopping criterion of 0.01%, the program yields x = 1.233318 and y = 0.212245 after 7 iterations with an approximate error of 2.2×10−4. The program from Prob. 6.19 can be set up to solve Prob. 6.12, by changing the functions to Function u(x, y) u = (x - 4) ^ 2 + (y - 4) ^ 2 - 4 End Function Function v(x, y) v = x ^ 2 + y ^ 2 - 16 End Function Function dudx(x, y) dudx = 2 * (x - 4) End Function Function dudy(x, y) dudy = 2 * (y - 4) End Function Function dvdx(x, y) dvdx = 2 * x End Function Function dvdy(x, y) dvdy = 2 * y End Function Using a stopping criterion of 0.01% and initial guesses of 2 and 3.5, the program yields x = 2.0888542 and y = 3.411438 after 3 iterations with an approximate error of 9.8×10−4. Using a stopping criterion of 0.01% and initial guesses of 3.5 and 2, the program yields x = 3.411438 and y = 2.0888542 after 3 iterations with an approximate error of 9.8×10−4. 6.21 x= a x2 = a f ( x) = x 2 − a = 0 f ' ( x) = 2 x Substitute into Newton Raphson formula (Eq. 6.6), x2 − a x= x− 2x Combining terms gives x= 2 x ( x) − x 2 + a x 2 + a / x = 2 2 6.22 SOLUTION: ( f ( x ) = tanh x 2 − 9 [ ( ) )] f ' ( x ) = sech x − 9 ( 2 x ) x o = 3. 1 xi +1 = x i − 2 2 f ( x) f '(x) iteration 1 2 3 4 xi+1 2.9753 3.2267 2.5774 7.9865 The solution diverges from its real root of x = 3. Due to the concavity of the slope, the next iteration will always diverge. The sketch should resemble figure 6.6(a). 6.23 SOLUTION: 4 3 2 f ( x ) = 0.0074 x − 0.284 x + 3.355 x − 12.183 x + 5 f ' ( x) = 0.0296 x 3 − 0 .852 x 2 + 6.71 x − 12.183 xi +1 = x i − i 1 2 3 f ( xi ) f ' ( xi ) xi+1 9.0767 -4.01014 -3.2726 The solution converged on another root. The partial solutions for each iteration intersected the x-axis along its tangent path beyond a different root, resulting in convergence elsewhere. 6.24 SOLUTION: f(x) = ± 16 − ( x + 1) 2 + 2 xi+1 = x i − f ( xi )( xi −1 − xi ) f ( xi −1 ) − f ( xi ) 1st iteration xi −1 = 0.5 ⇒ f ( xi −1 ) = −1.708 xi = 3 ⇒ f ( xi ) = 2 2(0.5 − 3) xi +1 = 3 − = 1.6516 ( −1.708 − 2) 2nd iteration xi = 1.6516 ⇒ f ( xi ) = −0.9948 xi −1 = 0.5 ⇒ f ( xi −1 ) = −1.46 xi +1 = 1.6516 − − 0.9948( 0.5 − 1 .6516 ) = 4.1142 (−1 .46 − −0.9948) The solution diverges because the secant created by the two x-values yields a solution outside the functions domain.