Engineering Optimization: Theory and Practice, Fourth Edition

516 Geometric Programming

a 142 ,= 1 a 211 = , 1 a 221 = , 1 a 231 = , and 0 a 241 = 0,Eqs. (E 1 ) ecomeb

Maximizev(λ)=

[

c 01

λ 01

(λ 01 +λ 02 +λ 03 )

]λ 01 [

c 02

λ 02

(λ 01 +λ 02 +λ 03 )

]λ 02

×

[

c 03

λ 03

(λ 01 +λ 02 +λ 03 )

]λ 03 [

c 11

λ 11

(λ 11 +λ 12 )

]λ 11

×

[

c 12

λ 12

(λ 11 +λ 12 )

]λ 12 (

c 21

λ 21

λ 21

)λ 21

subjectto

λ 01 +λ 02 +λ 03 = 1

a 011 λ 01 +a 012 λ 02 +a 013 λ 03 +a 111 λ 11 +a 112 λ 12 +a 211 λ 21 = 0

a 021 λ 01 +a 022 λ 02 +a 023 λ 03 +a 121 λ 11 +a 122 λ 12 +a 221 λ 21 = 0

a 031 λ 01 +a 032 λ 02 +a 033 λ 03 +a 131 λ 11 +a 132 λ 12 +a 231 λ 21 = 0

a 041 λ 01 +a 042 λ 02 +a 043 λ 03 +a 141 λ 11 +a 142 λ 12 +a 241 λ 21 = 0

λ 11 +λ 12 ≥ 0

λ 21 ≥ 0

or

Maximizev(λ)=

(

1

λ 01

)λ 10 (

2

λ 02

)λ 20 (

10

λ 03

)λ 03 [

3

λ 11

(λ 11 +λ 12 )

]λ 11

×

[

4

λ 12

(λ 11 +λ 12 )

]λ 12

( 5 )λ^21 (E 2 )

subjectto

λ 01 +λ 02 +λ 03 = 1

λ 01 −λ 02 +λ 03 −λ 11 +λ 21 = 0

2 λ 01 − 3 λ 02 +λ 21 = 0

−λ 01 +λ 03 +λ 11 +λ 12 = 0

λ 02 − 2 λ 11 +λ 12 = 0

(E 3 )

λ 11 +λ 12 ≥ 0

λ 21 ≥ 0

Equations(E 3 ) an be used to express any five of thec λ’s in terms of the remaining

one as follows: Equations (E 3 ) an be rewritten asc

λ 02 +λ 03 = 1 −λ 01 (E 4 )

λ 02 −λ 03 +λ 11 −λ 21 =λ 01 (E 5 )

3 λ 02 −λ 21 = 2 λ 01 (E 6 )