Model

Elastic or provisioned

The same architecture is obviously right for a development environment and obviously wrong for a platform with continuous traffic.

elastic wins ⟺ d \times k < 1, cold start bills \frac{t _{warm} + t _{job}}{t _{job}}

$d$: Duty cycle; the fraction of time you are active
$k$: Elastic premium; the premium the elastic rate carries over the provisioned rate

Trade-off: Cost as a substitutable resource

How it behaves

One input on the horizontal axis, the others held fixed.

k = 2k = 3

Data table

Elastic cost ÷ provisioned cost, by duty cycle
Duty cycle d	series	Elastic ÷ provisioned
0.01	k = 2	0.02×
0.022	k = 2	0.045×
0.035	k = 2	0.07×
0.047	k = 2	0.094×
0.06	k = 2	0.119×
0.072	k = 2	0.144×
0.084	k = 2	0.168×
0.097	k = 2	0.193×
0.109	k = 2	0.218×
0.121	k = 2	0.243×
0.134	k = 2	0.268×
0.146	k = 2	0.292×
0.159	k = 2	0.317×
0.171	k = 2	0.342×
0.183	k = 2	0.366×
0.196	k = 2	0.391×
0.208	k = 2	0.416×
0.22	k = 2	0.441×
0.233	k = 2	0.466×
0.245	k = 2	0.49×
0.258	k = 2	0.515×
0.27	k = 2	0.54×
0.282	k = 2	0.565×
0.295	k = 2	0.589×
0.307	k = 2	0.614×
0.319	k = 2	0.639×
0.332	k = 2	0.663×
0.344	k = 2	0.688×
0.356	k = 2	0.713×
0.369	k = 2	0.738×
0.381	k = 2	0.762×
0.394	k = 2	0.787×
0.406	k = 2	0.812×
0.418	k = 2	0.837×
0.431	k = 2	0.861×
0.443	k = 2	0.886×
0.456	k = 2	0.911×
0.468	k = 2	0.936×
0.48	k = 2	0.96×
0.493	k = 2	0.985×
0.505	k = 2	1.01×
0.517	k = 2	1.03×
0.53	k = 2	1.06×
0.542	k = 2	1.08×
0.554	k = 2	1.11×
0.567	k = 2	1.13×
0.579	k = 2	1.16×
0.592	k = 2	1.18×
0.604	k = 2	1.21×
0.616	k = 2	1.23×
0.629	k = 2	1.26×
0.641	k = 2	1.28×
0.653	k = 2	1.31×
0.666	k = 2	1.33×
0.678	k = 2	1.36×
0.691	k = 2	1.38×
0.703	k = 2	1.41×
0.715	k = 2	1.43×
0.728	k = 2	1.46×
0.74	k = 2	1.48×
0.752	k = 2	1.5×
0.765	k = 2	1.53×
0.777	k = 2	1.55×
0.79	k = 2	1.58×
0.802	k = 2	1.6×
0.814	k = 2	1.63×
0.827	k = 2	1.65×
0.839	k = 2	1.68×
0.851	k = 2	1.7×
0.864	k = 2	1.73×
0.876	k = 2	1.75×
0.889	k = 2	1.78×
0.901	k = 2	1.8×
0.913	k = 2	1.83×
0.926	k = 2	1.85×
0.938	k = 2	1.88×
0.95	k = 2	1.9×
0.963	k = 2	1.93×
0.975	k = 2	1.95×
0.988	k = 2	1.98×
1	k = 2	2×
0.01	k = 3	0.03×
0.022	k = 3	0.067×
0.035	k = 3	0.104×
0.047	k = 3	0.141×
0.06	k = 3	0.179×
0.072	k = 3	0.216×
0.084	k = 3	0.253×
0.097	k = 3	0.29×
0.109	k = 3	0.327×
0.121	k = 3	0.364×
0.134	k = 3	0.401×
0.146	k = 3	0.438×
0.159	k = 3	0.476×
0.171	k = 3	0.513×
0.183	k = 3	0.55×
0.196	k = 3	0.587×
0.208	k = 3	0.624×
0.22	k = 3	0.661×
0.233	k = 3	0.698×
0.245	k = 3	0.735×
0.258	k = 3	0.772×
0.27	k = 3	0.81×
0.282	k = 3	0.847×
0.295	k = 3	0.884×
0.307	k = 3	0.921×
0.319	k = 3	0.958×
0.332	k = 3	0.995×
0.344	k = 3	1.03×
0.356	k = 3	1.07×
0.369	k = 3	1.11×
0.381	k = 3	1.14×
0.394	k = 3	1.18×
0.406	k = 3	1.22×
0.418	k = 3	1.26×
0.431	k = 3	1.29×
0.443	k = 3	1.33×
0.456	k = 3	1.37×
0.468	k = 3	1.4×
0.48	k = 3	1.44×
0.493	k = 3	1.48×
0.505	k = 3	1.52×
0.517	k = 3	1.55×
0.53	k = 3	1.59×
0.542	k = 3	1.63×
0.554	k = 3	1.66×
0.567	k = 3	1.7×
0.579	k = 3	1.74×
0.592	k = 3	1.77×
0.604	k = 3	1.81×
0.616	k = 3	1.85×
0.629	k = 3	1.89×
0.641	k = 3	1.92×
0.653	k = 3	1.96×
0.666	k = 3	2×
0.678	k = 3	2.03×
0.691	k = 3	2.07×
0.703	k = 3	2.11×
0.715	k = 3	2.15×
0.728	k = 3	2.18×
0.74	k = 3	2.22×
0.752	k = 3	2.26×
0.765	k = 3	2.29×
0.777	k = 3	2.33×
0.79	k = 3	2.37×
0.802	k = 3	2.41×
0.814	k = 3	2.44×
0.827	k = 3	2.48×
0.839	k = 3	2.52×
0.851	k = 3	2.55×
0.864	k = 3	2.59×
0.876	k = 3	2.63×
0.889	k = 3	2.67×
0.901	k = 3	2.7×
0.913	k = 3	2.74×
0.926	k = 3	2.78×
0.938	k = 3	2.81×
0.95	k = 3	2.85×
0.963	k = 3	2.89×
0.975	k = 3	2.93×
0.988	k = 3	2.96×
1	k = 3	3×

Worked examples, recomputed

Every number the mechanisms quote for this relationship is recomputed from it, so none of them disagrees with it.

Where it appears

Each mechanism states it in its own names. The general symbol each name stands for is shown with it.

Cost Mechanics
What is metered, and what is the smallest increment you are charged for?
$d \times k < 1$
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