Engineering Optimization: Theory and Practice, Fourth Edition

(Martin Jones) #1
Answers to Selected Problems 801

CHAPTER 13


13.1Before:X 1 =


{

17

13

}

,X 2 =

{

15

22

}

;After:X 1 =

{

23

22

}

, X 2 =

{

9

13

}

13.3(a) 9, (b) 10, (c) 11 13.4 10 13.6(i j )=(i 4 ) 13.8x∗= 2


1 3.9x 1 ( 2 )= 2. 8297 , x 2 ( 2 )= 1. 9345 , x 3 ( 2 )= 1. 6362 , x 4 ( 2 )= 1. 1887


13.12Number of copies of strings 1, 2, 3, 4, 5, 6, 7 are 0, 0, 1, 2, 5, 2, 2, respectively


13.14String length= 37


CHAPTER 14


14.1c∗ 1 = 0. 04 , c∗ 2 = 0. 81


14.3(a){ 0. 001165 , 0. 002329 , 0. 03949 ,− 0. 05635 },


(b){ 0. 0009705 , 0. 001941 , 0. 05273 ,− 0. 084102 },


(c) { 0. 0009704 , 0. 001941 , 0. 05265 ,− 0. 08395 }


14.5


{

∂yi
∂x 1

}

={− 0. 000582 ,− 0. 001165 ,− 0. 002329 , 0. 002329 }

14.7

{

∂yi
∂x 3

}

={ 0. 4693 × 10 −^7 , 4770. 9 × 10 −^7 , − 0. 027948 , 0. 027947 }

14.9(a)


{

0. 000125

0. 000458

}

(b)

{

− 0. 000229

− 0. 000229

}

,

{

0. 0

0. 000333

}

(c)


{

− 275

0

}

,

{

0

200

}

14.11

∂λ 1
∂A 2

= 2. 28840 ,

∂λ 2
∂A 2

= 64. 8649 ,

∂Y 1

∂A 2

=

{

− 0. 312639 × 10 −^21

0 91666. 3 × 10 −^6

}

,

∂Y 2

∂A 2

=

{

0. 698492 × 10 −^8

0 83790. 8 × 10 −^2

}

1 4.15

∂ω 1
∂D

= − 1. 584664 ,

∂ω 2
∂D

= − 2. 744719

14.16y∗= 3 ,A∗ 1 = 0. 316228 × 10 −^7 , A∗ 2 = 0. 948683 × 10 −^7 , f∗= 0. 6 × 10 −^6


14.18y∗= 0. 25 , A 1 ∗= 1. 0 ,A∗ 2 = 1. 0 ,f∗= 34. 7565


14.20 X∗= { 0. 7635 , 1. 0540 }, f∗= 871 .5670 withf= 0. 625 f 1 + 0611. 0 f 2


14.21 X∗= { 0. 8 , 1. 1 }, F∗= 3. 1267


14.22 X∗= { 0. 75 , 1. 25 }

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