Engineering Optimization: Theory and Practice, Fourth Edition

Answers to Selected Problems

CHAPTER 1

1.1Min.f= 5 xA− 08 xB+ 601 xC+ 51 xD, 0. 05 xA+ 0. 05 xB+ 0. 1 xC+

0. 15 xD≤ 0001 , 0. 1 xA + 0. 15 xB+ 0. 2 xC+ 0. 05 xD≤ 0002 , 0. 05 xA+ 0. 1 xB+

0. 1 xC+ 0. 15 xD≤ 5001 , xA≥ 0005 , xB≥ 0 ,xC≥ 0 , xD≥ 0004

1.2(a) X∗= { 0. 65 , 0. 53521 } (b) X∗= { 0. 9 , 2. 5 }

(c) X∗= { 0. 65 , 0. 53521 } 1.5x∗ 1 =x 2 ∗= 003

1.9(a)R∗ 1 = 4. 472 , R 2 ∗= 2. 236 (b)R 1 ∗= 3. 536 , R 2 ∗= 3. 536

(c) R∗ 1 = 6. 67 , R∗ 2 = 3. 33

1.11(a)y 1 = nl x 1 , y 2 = nl x 2 , nl f= 2 y 1 + 3 y 2

(b) f= 10 y^2 x^2 ,x 1 = 01 y^2 , ln(log 10 f ) =ln(log 10 x 1 ) n+l x 2

1.14xij= if city 1 jis visited immediately after cityi, and=0 otherwise.

Find{xij} o minimizet f=

∑n

i= 1

∑n

j= 1

dijxijsubject to

∑n

i= 1

xij= 1 (i= 1 , 2 ,... , n),

i=jand

∑n

j= 1

xij= 1 (i= 1 , 2 ,... , n), j=i

1.19Min.f=ρlbd,

Py

bd

+

6 Pxl

bd^2

≤σy,

Py

bd

+

6 Pxl

bd^2

≤

π^2 Ed^2

48 l^2

, b≥ 0. 5 , b≤ 2 d.

1.25Max.f=^23 tm+^35 td, tm+td≤ 04 , td≥ 1. 25 tm, 0 ≤tm≤ 42 , 0 ≤td≤ 0. 2

1.29Min.f=π x 3 [x^21 − (x 1 −x 2 )^2 ]+^43 π[x^31 − (x 1 −x 4 )^3 ] π x, 3 (x 1 −x 2 )^2 +

4

3 π(x^1 −x^4 )

(^3) − 4 , 619 , 606 ≤ 0 , x
2 −
pR 0
S.e+ 0. 4 p
≤ 0 , x 4 −
pR 0
S.e+ 0. 8 p

≤ 0

CHAPTER 2

2.1r∗= R 2 .3x∗= 1. 5 (inflection point)

2.5x= −1 (not min, not max),x=2 (min) 2.9d=

(

D^5

8 f l

) 1 / 4

2.10 3 5.36 m 2.11(a) 79. 28 ◦ (b) 0 .911 from end of stroke

Engineering Optimization: Theory and Practice, Fourth Edition Singiresu S. Rao 795
Copyright © 2009 by John Wiley & Sons, Inc.