5.7 Computational Cost - Control Co-Design

Count comparable work¶

Computational cost must include all model, derivative, optimization, recovery, and validation work:

T_{total}=T_{model}+T_{deriv}+T_{solve}+T_{restart}+T_{validation}.

One-pass sequential design is usually cheapest because it solves two restricted problems once. Iterated sequential design repeats those solves. Nested CCD multiplies outer evaluations by inner control solves, although warm starts and parallel candidate evaluations can help. Simultaneous CCD solves a larger sparse NLP and often trades memory for fewer repeated integrations.

Runtime alone is not meaningful. A method using loose defects or failed inner solves is not cheaper if its result must be rejected.

Activity 5.7: compare computational work fairly¶

Activity 5.7: compare computational work fairly

Workflow	Major solves	Time per solve	Derivative time	Restarts	Validation
One-pass baseline	2	18 s	6 s	0	20 s
Iterated baseline	10	18 s	20 s	0	20 s
Nested CCD	36 inner solves	12 s	95 s	2 at 30 s	20 s
Simultaneous CCD	1 sparse NLP	390 s	70 s	1 at 160 s	20 s

Compute total serial time for each workflow.
If 12 nested inner solves can run concurrently on four workers with perfect load balance, recompute nested wall time assuming the remaining 24 solves are serial.
Compute CCD runtime ratios relative to the one-pass baseline.
The iterated baseline fails final feasibility and needs a 90 s repair solve. Update its cost.
State which cost and accuracy quantities should appear in a fair publication table.