Ct = I1/[ln (A10/d)] + I3 d (5+b+c) + A9 d
where Ct = Ct [(1+ PVFm&rAm&r)(A1np+ A3L)]
I3 = I2 + (1+PVFm&rAm&r) A5
A9 = (1+PVFm&rAm&r) A4L.
Minimizing Ct with respect to d is, of course, equivalent to minimizing the
original total cost function Ct. Therefore, we can work with Ct for convenience. If
we neglect the first term, which represents the cost of heat losses, we have a
geometric programming problem (Papalambros and Wilde 1988) with zero degrees
of difficulty. Without specifying the parameter values, we see from inspection that
the weights of the two remaining terms will be
w1 = 1/(6 + b + c) and w2 = (5 + b + c)/(6 + b + c).
With heat losses neglected, at the optimum pipe diameter the variable costs
associated with pumping are 1/(6 + b + c) of the total variable costs. The variable
costs attributable to pipe capital and maintenance costs are the remaining portion.
Here, the variable costs represent that portion which is a function of our decision
variable, the pipe diameter. Also note that the pumping costs include the variable
portion of the capital cost of the pumps and the maintenance associated with that
portion, as well as the pumping energy costs.
Considering a more specific case, if the values of parameters b and c found in the
example given for eq 2-11 (b = 0.152, c = 0.0568) are used, we find the following
values for the weights
w1 = 16.4%
w2 = 83.6%.
These results vary very little over the range of values found for b and c in
Appendix A. Thus, when heat losses are neglected, we find this very simple solution
is applicable in most cases.
Once values for the remaining parameters are known, the pipe diameter is found
by using the equations given above and the two terms of the objective function
d = [(5 + b + c)(I3/A9)][1/(6+b+c)].
It should be noted that this solution obtained using geometric programming theory
also could have been easily obtained using classical differential methods, as used
later. The advantage of the geometric programming method is that it ensures that a
global rather than a local minimum has been found. Differential methods only ensure
a local extremum and require the evaluation of second order terms to determine the
nature of the extremum, i.e., maximum or minimum.
To arrive at this simple expression for the pipe diameter, we have neglected the
heat losses. Because the cost of heat losses will always be greater than zero, we have
constructed a lower bounding problem for our original problem by neglecting this
I3 d(5+b+c) + A9 d ≤ I1/[ln(A10/d)] + I3 d(5+b+c) + A9 d.
The cost of any design that includes heat losses can never be less than the same