Section 5.3.1: Optimisation

Choices in Process Design


Continuous parameter trade offs

Example: pipeline for constant flow rate.

Capital cost $\propto$ pipe diameter, D - maybe should be D0.6 or D0.9.


Operating cost $\propto$ D-4.75 - assumes friction factor $\propto$ Re-0.25.


Typically, capital cost versus operating cost.

The Optimisation problem

f(x) - the objective function
x - the decision variable
constraints: $ x \geq x_L $
  $ x \leq x_U $

Constraints define feasible region.

Minimise f(x) with respect to (w.r.t.) x, subject to (s.t.) $ x \geq x_L $, $ x \leq x_U $

Example: (use D instead of x) minimise f(D) = 2.0 D + 0.4 + 0.5 D-4.75 (s.t.) $ D \geq 0.1$, $D \leq 2.5 $.

If expression available for derivative..

Solve

\begin{displaymath}\frac{df}{dx} \equiv f^{'}(x) = 0 \end{displaymath}

for stationary point(s). Try to get all SPs.

Nature of each SP at xsp is given by sign of second derivative f''(x):

+ implies local minimum
- implies local maximum
0 don't know

Example: $f^{'}(D) = 2.0 - 4.75 \times 0.5 D^{-5.75} $
implies $ D_{sp} = \left[ \frac{2.0}{4.75 \times 0.5}\right]^{\frac{-1}{5.75}} = 1.030 $ m.
and $ f^{''}(D_{sp}) = + 5.75 \times 4.75 \times 0.5 D_{sp}^{-6.75} =
\frac{ 5.75 \times 4.75 \times 0.5}{1.030^{6.75}} > 0 $

Hence local minimum.

Global minimum

Check



Beware of discontinuities Simple theory assumes that both f(x) and f'(x) are continuous. If not, global minimum may exist at or near a discontinuity in f or f'.



Maximisation

Profit, throughput, reactor conversion,...

`Maximise f(x)' is equivalent to `Minimise -f(x)'.

Without loss of generality, a computer code can treat all problems as minimisations.


Next - Section 5.3.2: Non-gradient Methods

Return to Section 5.3 Index

Return to Section 5 Index

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Last modified: Tue Aug 25 10:29:53 BST