CHAPTER 14 14.1 The elevation can be determined as f (0.8, 1.2) = 2(0.8)1.2 + 1.5(1.2) − 1.25(0.8) 2 − 2(1.2) 2 + 5 = 5.04 The partial derivatives can be evaluated, ∂f = 2 y − 2.5 x = 2(1.2) − 2.5(0.8) = 0.4 ∂x ∂f = 2 x + 1.5 − 4 y = 2(0.8) + 1.5 − 4(1.2) = −1.7 ∂y which can be used to determine the gradient as ∇ f = 0.4i – 1.7j. This corresponds to the direction θ = tan–1(–1.7/0.4) = –1.3397 radians (= –76.76o). This vector can be sketched on a topographical map of the function as shown below: 2 1.6 1.2 0.8 0.4 0

0

0.4

0.8

1.2

1.6

2

The slope in this direction can be computed as 0.4 2 + ( −1.7) 2 = 1.746 14.2 The partial derivatives can be evaluated, ∂f = 2 x = 2( 2) = 4 ∂x ∂f = 4 y = 4( 2) = 8 ∂y The angle in the direction of h is 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.

2 3 θ = tan −1 = 0.9828 radians (= 56.31°) 2 The directional derivative can be computed as g ' (0) = 4 cos(0.9828) + 8 sin(0.9828) = 8.875 14.3 (a) 3 y 2 + 2 ye xy ∇f = xy 6 xy + 2 xe

2 y 2 e xy H = xy xy 6 y + 2 xye + 2e

6 y + 2 xye xy + 2e xy 6 x + 2 x 2 e xy

(b) 4 x ∇f = 2 y 2 z

4 0 0 H = 0 2 0 0 0 2

(c) x 2 ∇f = x 2

2x + 3y 2 + 3xy + 2 y 3x + 4 y + 3xy + 2 y 2

− 2 x 2 − 6 xy − 5 y 2 − 3x 2 − 8 xy − 6 y 2 − 3x 2 − 8 xy − 6 y 2 − 5 x 2 − 12 xy − 8 y 2 H= 2 2 2 x + 3 xy + 2 y

(

)

14.4 The partial derivatives can be evaluated, ∂f = −3x + 2.25 y ∂x ∂f = 2.25 x − 4 y + 1.75 ∂y These can be set to zero to generate the following simultaneous equations 3x − 2.25 y = 0 − 2.25 x + 4 y = 1.75 which can be solved for x = 0.567568 and y = 0.756757, which is the optimal solution. 14.5 The partial derivatives can be evaluated at the initial guesses, x = 1 and y = 1, ∂f = −3x + 2.25 y = −3(1) + 2.25(1) = −0.75 ∂x ∂f = 2.25 x − 4 y + 1.75 = 2.25(1) − 4(1) + 1.75 = 0 ∂y 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.

3

Therefore, the search direction is –0.75i. f (1 − 0.75h, 1) = 0.5 + 0.5625h − 0.84375h 2 This can be differentiated and set equal to zero and solved for h* = 0.33333. Therefore, the result for the first iteration is x = 1 – 0.75(0.3333) = 0.75 and y = 1 + 0(0.3333) = 1. For the second iteration, the partial derivatives can be evaluated as, ∂f = −3(0.75) + 2.25(1) = 0 ∂x ∂f = 2.25(0.75) − 4(1) + 1.75 = −0.5625 ∂y Therefore, the search direction is –0.5625j. f (0.75, 1 − 0.5625h) = 0.59375 + 0.316406h − 0.63281h 2 This can be differentiated and set equal to zero and solved for h* = 0.25. Therefore, the result for the second iteration is x = 0.75 + 0(0.25) = 0.75 and y = 1 + (–0.5625)0.25 = 0.859375. 1.2 1

0

1 2

0.8

max

0.6 0.4 0.2 0

0

0.2

0.4

0.6

0.8

1

1.2

14.6 The partial derivatives can be evaluated at the initial guesses, x = 1 and y = 1, ∂f = 2( x − 3) = 2(1 − 3) = −4 ∂x ∂f = 2( y − 2) = 2(1 − 2) = −2 ∂y

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4 f (1 − 4h, 1 − 2h) = (1 − 4h − 3) 2 + (1 − 2h − 2) 2 g ( h) = ( −4h − 2) 2 + ( −2h − 1) 2 Setting g′ (h) = 0 gives h* = –0.5. Therefore, x = 1 –4(–0.5) = 3 y = 1 –2(–0.5) = 2 Thus, for this special case, the approach converges on the correct answer after a single iteration. This occurs because the function is spherical as shown below. Thus, the gradient for any guess points directly at the solution. 5 4.2 3.4 2.6

1

4

3.2

2.4

1.6

0.8

0

1.8

14.7 The partial derivatives can be evaluated at the initial guesses, x = 0 and y = 0, ∂f = 4 + 2 x − 8 x 3 + 2 y = 4 + 2(0) − 8(0) 3 + 2(0) = 4 ∂x ∂f = 2 + 2 x − 6 y = 2 + 2(0) − 6(0) = 2 ∂y f (0 + 4h, 0 + 2h) = 20h + 20h 2 − 512h 4 g ' ( h) = 20 + 40h − 2048h 3 The root of this equation can be determined by bisection. Using initial guesses of h = 0 and 1 yields a root of h* = 0.244 after 13 iterations with εa = 0.05%. iteration 1

hl 0.00000

hu 1.00000

hr 0.50000

g(hl) 20.000

g(hr) -216.000

g(hl)*g(hr) -4320.00

εa 100.00%

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5 2 3 4 5 6 7 8 9 10 11 12 13

0.00000 0.00000 0.12500 0.18750 0.21875 0.23438 0.24219 0.24219 0.24219 0.24316 0.24365 0.24390

0.50000 0.25000 0.25000 0.25000 0.25000 0.25000 0.25000 0.24609 0.24414 0.24414 0.24414 0.24414

0.25000 0.12500 0.18750 0.21875 0.23438 0.24219 0.24609 0.24414 0.24316 0.24365 0.24390 0.24402

20.000 20.000 21.000 14.000 7.313 3.008 0.595 0.595 0.595 0.280 0.122 0.043

-2.000 21.000 14.000 7.313 3.008 0.595 -0.680 -0.037 0.280 0.122 0.043 0.003

-40.00 420.00 294.00 102.38 21.99 1.79 -0.40 -0.02 0.17 0.03 0.01 0.00

100.00% 100.00% 33.33% 14.29% 6.67% 3.23% 1.59% 0.80% 0.40% 0.20% 0.10% 0.05%

Therefore, x = 0 + 4(0.244) = 0.976 y = 0 + 2(0.244) = 0.488 14.8

∂f = −8 + 2 x − 2 y ∂x ∂f = 12 + 8 y − 2 x ∂y

At x = y = 0, ∂f = −8 ∂x ∂f = 12 ∂y f (0 − 8h,0 + 12h) = g (h) g ( h) = 832h 2 + 208h At g′ (h) = 0, h* = −0.125. Therefore, x = 0 − 8(−0.125) = 1 y = 0 + 12(−0.125) = −1.5 14.9 The following code implements the random search algorithm in VBA. It is set up to solve Prob. 14.7. Option Explicit

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6 Sub RandSearch() Dim n As Long Dim xmin As Double, xmax As Dim maxf As Double, maxx As xmin = -2: xmax = 2: ymin = n = InputBox("n=") Call RndSrch(n, xmin, xmax, MsgBox maxf MsgBox maxx MsgBox maxy End Sub

Double, ymin As Double, ymax As Double Double, maxy As Double -2: ymax = 2 ymin, ymax, maxy, maxx, maxf)

Sub RndSrch(n, xmin, xmax, ymin, ymax, maxy, maxx, maxf) Dim j As Long Dim x As Double, y As Double, fn As Double maxf = -1000000000# For j = 1 To n x = xmin + (xmax - xmin) * Rnd y = ymin + (ymax - ymin) * Rnd fn = f(x, y) If fn > maxf Then maxf = fn maxx = x maxy = y End If Next j End Sub Function f(x, y) f = 4 * x + 2 * y + x ^ 2 - 2 * x ^ 4 + 2 * x * y - 3 * y ^ 2 End Function

The result of running this program for different number of iterations yields the results in the following table. We have also included to exact result. n 1000 10000 100000 1000000 10000000 truth

f(x, y) 4.31429 4.34185 4.34338 4.34397 4.34401 4.34401

x 1.011782 0.961359 0.964238 0.965868 0.967520 0.967580

y 0.613747 0.678104 0.641633 0.656765 0.655715 0.655860

14.10 The following code implements the grid search algorithm in VBA: Option Explicit Sub GridSearch() Dim nx As Long, ny As Long Dim xmin As Double, xmax As Dim maxf As Double, maxx As xmin = -2: xmax = 2: ymin = nx = 1000 ny = 1000 Call GridSrch(nx, ny, xmin, MsgBox maxf MsgBox maxx MsgBox maxy

Double, ymin As Double, ymax As Double Double, maxy As Double 1: ymax = 3 xmax, ymin, ymax, maxy, maxx, maxf)

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7 End Sub Sub GridSrch(nx, ny, xmin, xmax, ymin, ymax, maxy, maxx, maxf) Dim i As Long, j As Long Dim x As Double, y As Double, fn As Double Dim xinc As Double, yinc As Double xinc = (xmax - xmin) / nx yinc = (ymax - ymin) / ny maxf = -1000000000# x = xmin For i = 0 To nx y = ymin For j = 0 To ny fn = f(x, y) If fn > maxf Then maxf = fn maxx = x maxy = y End If y = y + yinc Next j x = x + xinc Next i End Sub Function f(x, y) f = y - x - 2 * x ^ 2 - 2 * x * y - y ^ 2 End Function

14.11 f ( x, y ) = 6 x 2 y − 9 y 2 − 8 x 2 ∂f = 12 xy − 16 x ⇒ 12( 2)(4) − 16( 4) = 32 ∂x ∂f = 6 x 2 − 18 y ⇒ 6(4) 2 − 18(2) = 60 ∂y ∇f = 32iˆ + 60 ˆj ∂f ∂f f x o + h, y o + h = f (4 + 32h,2 + 60h) ∂x ∂y = 6(4 + 32h) 2 ( 2 + 60h) − 9( 2 + 60h) 2 − 8( 4 + 32h) 2 g ( x) = 368,640h 3 + 63,856h 2 + 4,624h + 28 14.12 f ( x, y ) = 2 x 3 y 2 − 7 yx + x 2 + 3 y

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8 ∂f = 6 x 2 y 2 − 7 y + 2 x ⇒ 6(1)(1) − 7(1) + 2(1) = 1 ∂x ∂f = 4 x 3 y − 7 x + 3 ⇒ 4(1)(1) − 7(1) + 3 = 0 ∂x ∇f = 1iˆ + 0 ˆj f ( xo +

∂f ∂f h, y o + h) = f (1 + h, 1 + 0h) ∂x ∂y = 2(1 + h) 3 (1) 2 − 7(1 + h)(1) + (1 + h) 2 + 3(1)

g ( x) = 2h 3 + 7 h 2 + h − 1

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