Engineering Optimization: Theory and Practice, Fourth Edition

506 Geometric Programming

The pumping cost is given by (300Q^2 /D^5 ) Find the optimal size of the pipe and the.

amount of fluid handled for minimum overall cost.

SOLUTION

f (D, Q)= 100 D^1 Q^0   0 + 5 D^2 Q^0 + 02 D^0 Q−^1 + 003 D−^5 Q^2 (E 1 )

Here we can see that

c 1 = 001 , c 2 = 05 , c 3 = 02 , c 4 = 003

(

a 11 a 12 a 13 a 14

a 21 a 22 a 23 a 24

)

=

(

1 2 0− 5

0 0 −1 2

)

The orthogonality and normality conditions are given by







1 2 0− 5

0 0−1 2

1 1 1 1

















 1

 2

 3

 4











=







0

0

1







SinceN>(n+ 1 ), these equations do not yield the requiredj(j = 1 to 4)directly.

But any three of thej’s can be expressed in terms of the remaining one. Hence by

solving for 1 ,  2 , and 3 in terms of 4 , we obtain

 1 = 2 − 11  4

 2 = 8  4 − 1 (E 2 )

 3 = 2  4

The dual problem can now be written as

Maximizev( 1 ,  2 ,  3 ,  4 )

=

(

c 1

 1

) 1 (

c 2

 2

) 2 (

c 3

 3

) 3 (

c 4

 4

) 4

=

(

100

2 − 11  4

) 2 1 − 1  4 (

50

8  4 − 1

) 8  4 − 1 (

20

2  4

) 2  4 (

300

 4

) 4

Since the maximization ofvis equivalent to the maximization of lnv, we will maximize

lnvfor convenience. Thus

lnv=( 2 − 11  4 ) ln 100[ −ln( 2 − 11  4 )]+( 8  4 − 1 )

×[ln 50−ln( 8  4 − 1 )]+ 2  4 [ln 20 −ln( 2  4 )]

+ 4 [ln 300 −ln( 4 )]