170
Chapter 3
From Eq. (3.71) Example 3.10
4 terms in the original expression
3 variables
DD = zero
This means the equation can be solved as an algebra problem. However, if
DD>zero this becomes an interation problem for a geometric programming
optimization routine. The next example contains constraints and is more
difficult.
EXAMPLE 3.11. Look at an example from [3.12] expande d from
the original text which outlines a method to formulate other problems.
Find the minimum area of an open cylindrical tank Example 3.6 with
volume no less than 1 unit. The radius is r and the height is h.
go(x) = ~rr: + 2nrh
area of tank
(3.79)
gl(x) = ~r2h > 1
constant
Degree difficulty = T-(N + 1) = 3-(2 + 1)
Let u~ =~r 2 and U 2 = 2nrh Eq. (3.70) substituted into Eq. (3.79)
or
> (~r2"~ ~’ {2r~rh’] h
(3.80)
1
In the volume constraint divide by ~ so
1
1 >_~
or
1
gl =g~-~_< 1
1
or placing g~ in a similar form as g0 with u I =~ or 1 ~g~
l a [~j ~
(3.81)
Chapter 3
From Eq. (3.71) Example 3.10
4 terms in the original expression
3 variables
DD = zero
This means the equation can be solved as an algebra problem. However, if
DD>zero this becomes an interation problem for a geometric programming
optimization routine. The next example contains constraints and is more
difficult.
EXAMPLE 3.11. Look at an example from [3.12] expande d from
the original text which outlines a method to formulate other problems.
Find the minimum area of an open cylindrical tank Example 3.6 with
volume no less than 1 unit. The radius is r and the height is h.
go(x) = ~rr: + 2nrh
area of tank
(3.79)
gl(x) = ~r2h > 1
constant
Degree difficulty = T-(N + 1) = 3-(2 + 1)
Let u~ =~r 2 and U 2 = 2nrh Eq. (3.70) substituted into Eq. (3.79)
or
> (~r2"~ ~’ {2r~rh’] h
(3.80)
1
In the volume constraint divide by ~ so
1
1 >_~
or
1
gl =g~-~_< 1
1
or placing g~ in a similar form as g0 with u I =~ or 1 ~g~
l a [~j ~
(3.81)