Engineering Optimization: Theory and Practice, Fourth Edition

514 Geometric Programming

a 021 λ 01 +a 022 λ 02 +a 023 λ 03 +a 121 λ 11 = 0

a 031 λ 01 +a 032 λ 02 +a 033 λ 03 +a 131 λ 11 = 0 (E 2 )

λ 0 j≥ 0 , j= 1 , 2 , 3

λ 11 ≥ 0

In this problem,c 01 = 0, 2 c 02 = 0, 4 c 03 = 0, 8 c 11 = , 8 a 011 = , 1 a 021 = , 0 a 031 = , 1

a 012 ,= 0 a 022 = , 1 a 032 = , 1 a 013 = , 1 a 023 = , 1 a 033 = , 0 a 111 = − 1 ,a 121 = − 1 , and

a 131 = − 1. Hence Eqs. (E 1 ) nd (Ea 2 ) ecomeb

v(λ)=

[

20

λ 01

(λ 01 +λ 02 +λ 03 )

]λ 01 [

40

λ 02

(λ 01 +λ 02 +λ 03 )

]λ 02

×

[

80

λ 03

(λ 01 +λ 02 +λ 03 )

]λ 03 (

8

λ 11

λ 11

)λ 11

(E 3 )

subjectto

λ 01 +λ 02 +λ 03 = 1

λ 01 +λ 03 −λ 11 = 0

λ 02 +λ 03 −λ 11 = 0 (E 4 )

λ 01 +λ 02 −λ 11 = 0

λ 01 ≥ 0 , λ 02 ≥ 0 , λ 03 ≥ 0 , λ 11 ≥ 0

The four linear equations in Eq. (E 4 ) ield the unique solutiony

λ∗ 01 =λ∗ 02 =λ∗ 03 =^13 , λ∗ 11 =^23

Thus the maximum value ofvor the minimum value ofx 0 is given by

v∗=x∗ 0 0 =( 6 )^1 /^3 ( 201 )^1 /^3 ( 402 )^1 /^3 ( 8 )^2 /^3

= [( 60 )^3 ]^1 /^3 ( 8 )^1 /^3 ( 8 )^2 /^3 = 0 ( 6 )( 8 )= 480

The values of the design variables can be obtained by applying Eqs. (8.62) and (8.63)

as

λ∗ 01 =

c 01 (x∗ 1 )^11 a^0 (x 2 ∗)^21 a^0 (x 3 ∗)a^310

x∗ 0

1

3 =

20 (x 1 ∗)(x 3 ∗)

480

=

x 1 ∗x 3 ∗

24

(E 5 )

λ∗ 02 =

c 02 (x∗ 1 )^12 a^0 (x 2 ∗)^22 a^0 (x 3 ∗)a^320

x∗ 0

1

3 =

40 (x 2 ∗)(x 3 ∗)

480

=

x 2 ∗x 3 ∗

12

(E 6 )