Engineering Optimization: Theory and Practice, Fourth Edition

(Martin Jones) #1
8.10 Geometric Programming with Mixed Inequality Constraints 519

The constraints are given by (see Table 8.2)


∑N^0

j= 1

λ 0 j= 1

∑m

k= 0

∑Nk

j= 1

σkakijλkj= 0 , i= 1 , 2 ,... , n

∑Nk

j= 1

λkj≥ 0 , k= 1 , 2 ,... , m

that is,


λ 01 +λ 02 +λ 03 = 1
σ 0 a 011 λ 01 +σ 0 a 012 λ 02 +σ 0 a 013 λ 03 +σ 1 a 111 λ 11 +σ 1 a 112 λ 12 +σ 2 a 211 λ 21 = 0

σ 0 a 021 λ 01 +σ 0 a 022 λ 02 +σ 0 a 023 λ 03 +σ 1 a 121 λ 11 +σ 1 a 122 λ 12 +σ 2 a 221 λ 21 = 0
σ 0 a 031 λ 01 +σ 0 a 032 λ 02 +σ 0 a 033 λ 03 +σ 1 a 131 λ 11 +σ 1 a 132 λ 12 +σ 2 a 231 λ 21 = 0
σ 0 a 041 λ 01 +σ 0 a 042 λ 02 +σ 0 a 043 λ 03 +σ 1 a 141 λ 11 +σ 1 a 142 λ 12 +σ 2 a 241 λ 21 = 0
λ 11 +λ 12 ≥ 0
λ 21 ≥ 0

thatis,


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 2 )

λ 11 +λ 12 ≥ 0

λ 21 ≥ 0

Since Eqs. (E 2 ) re same as Eqs. (Ea 3 ) f the preceding example, the equality constraintso
can be used to expressλ 02 ,λ 03 ,λ 11 ,λ 12 , andλ 21 in terms ofλ 01 as


λ 02 = 8 λ 01 − 4

λ 03 = − 9 λ 01 + 5
λ 11 = 6 λ 01 − 3
λ 12 = 4 λ 01 − 2

λ 21 = 22 λ 01 − 21

(E 3 )
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