Engineering Optimization: Theory and Practice, Fourth Edition

534 Geometric Programming

or

1. 75

√

900 +h^2

dh

≤ 1 (E 2 )

It can be seen that the functions in Eqs. (E 1 ) nd (Ea 2 ) re not posynomials, due to thea

presence of the term

√

900 +h^2. The functions can be converted to posynomials by

introducing a new variableyas

y=

√

900 +h^2 or y^2 = 009 +h^2

and a new constraint as

900 +h^2

y^2

≤ 1 (E 3 )

Thus the optimization problem can be stated, withx 1 =y, x 2 = h,andx 3 = das

design variables, as

Minimizef= 0. 188 yd (E 4 )

subjectto

1. 75 yh−^1 d−^1 ≤ 1 (E 5 )

900 y−^2 +y−^2 h^2 ≤ 1 (E 6 )

For this zero-degree-of-difficulty problem, the associated dual problem can be stated

as

Maximizev(λ 01 , λ 11 , λ 21 , λ 22 )

=

(

0. 188

λ 01

)λ 01 (

1. 75

λ 11

)λ 11 (

900

λ 21

)λ 21 (

1

λ 22

)λ 22

(λ 21 +λ 22 )λ^21 +λ^22 (E 7 )

subjectto

λ 01 = 1 (E 8 )

λ 01 +λ 11 − 2 λ 21 − 2 λ 22 = 0 (E 9 )

−λ 11 + 2 λ 22 = 0 (E 10 )

λ 01 −λ 11 = 0 (E 11 )

The solution of Eqs. (E 8 ) o (Et 11 ) ivesg λ∗ 01 = 1 ,λ∗ 11 = 1 ,λ∗ 21 =^12 , andλ∗ 22 =^12. Thus

the maximum value ofvand the minimum value off is given by

v∗=

(

0. 188

1

) 1

( 51. 7 )^1

(

900

0. 5

) 0. 5 (

1

0. 5

) 0. 5

( 0. 5 + 0. 5 )^0.^5 +^0.^5 = 91. 8 =f∗