8.4 Solution Using Differential Calculus 497These equations, in the case of problems with a zero degree of difficulty, give a unique
solution tow 1 , w 2 ,... , wn. Oncewiare found, the desired solution can be obtained as
x∗i =ewi, i= 1 , 2 ,... , n (8.19)In a general geometric programming problem with a nonnegative degree of difficulty,
N≥n+1, and hence Eqs. (8.18) denoteNequations innunknowns. By choosing
anynlinearly independent equations, we obtain a set of solutionswiand hencexi∗.
The solution of an unconstrained geometric programming problem is illustrated
with the help of the following zero-degree-of-difficulty example [8.1].
Example 8.1 It has been decided to shift grain from a warehouse to a factory in an
open rectangular box of lengthx 1 meters, widthx 2 meters, and heightx 3 meters. The
bottom, sides, and the ends of the box cost, respectively, $80, $10, and $20/m^2. It costs
$1 for each round trip of the box. Assuming that the box will have no salvage value,
find the minimum cost of transporting 80 m^3 of grain.
SOLUTION The total cost of transportation is given by
total cost=cost of box+cost of transportation=(cost of sides+cost of bottom+cost of ends of the box)
+(number of round trips required for transporting the grain×cost of each round trip)f (X)=[( 2 x 1 x 3 ) 01 +(x 1 x 2 ) 08 +( 2 x 2 x 3 ) 0] 2 +[
80
x 1 x 2 x 3( 1 )
]
=$
(
80 x 1 x 2 + 04 x 2 x 3 + 02 x 1 x 3 +80
x 1 x 2 x 3)
(E 1 )
wherex 1 , x 2 , andx 3 indicate the dimensions of the box, as shown in Fig. 8.1. By
comparing Eq. (E 1 ) ith the general posynomial of Eq. (8.1), we obtainw
c 1 = 08 , c 2 = 04 , c 3 = 02 , c 4 = 08
a 11 a 12 a 13 a 14
a 21 a 22 a 23 a 24
a 31 a 32 a 33 a 34
=
1 0 1− 1
1 1 0 − 1
0 1 1 − 1
The orthogonality and normality conditions are given by