Six-tenths rule on log–log plots — what is the slope? Using the six-tenths factor rule for scaling equipment costs with capacity, if you plot cost vs capacity on a log–log graph, what slope do you expect?
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A0.1
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B0.6
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C0.2
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D0.8
Answer
Correct Answer: 0.6
Explanation
Introduction / Context: Early-stage cost estimation often relies on capacity scaling laws: Cost_2 = Cost_1 * (Capacity_2 / Capacity_1)^n. The exponent n captures economies of scale. The classic six-tenths rule sets n ≈ 0.6 for many types of process equipment.
Given Data / Assumptions:
- Cost ∝ Capacity^n with n ≈ 0.6.
- Plot is log(Cost) vs log(Capacity).
Concept / Approach: On a log–log plot, a power law appears as a straight line with slope equal to the exponent. Therefore, using the six-tenths factor rule, the slope equals 0.6. Deviations from 0.6 occur for specific equipment classes and time periods, but 0.6 remains a widely taught heuristic.
Step-by-Step Solution:
Write Cost = k * Capacity^n.Take logs: log(Cost) = log(k) + n * log(Capacity).Identify slope on log–log axes: slope = n = 0.6.Verification / Alternative check: Any two points (C1, Cap1) and (C2, Cap2) on the line satisfy n = log(C2/C1) / log(Cap2/Cap1). Using data generated with n = 0.6 reproduces the same slope.
Why Other Options Are Wrong:
0.1 or 0.2 — far too low; would imply minimal economies of scale.0.8 — possible for some classes but not the six-tenths rule.Common Pitfalls: Applying n = 0.6 blindly to all equipment; always sanity-check against vendor quotes and technology specifics.
Final Answer: 0.6