Equation 13 · The Economics of Codex: Tokens, Sandboxes, Review Time, and Software Throughput

What does this equation mean?

m^*(x)=\arg\max_m \left[p_m(x)V(x)-C_m(x)-H_m(x)-R_m(x)\right],

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Inputs and operationsargmax_m [p_m(x)V(x)-C_m(x)-H_m(x)-R_m(x)]

Result or conditionm^*(x)

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This equation states an equality: the expressions on both sides have the same value under the article’s assumptions. Read the equation part by part below; each part has a contextual explanation and a link to its mathematical background.

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m^*

Symbol m^*

$m^*$ is part of the quantity the equation computes from the expression on the right.

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x

Symbol x

x is an argument of the function-like quantity on the left; its role is set by that function’s stated inputs.

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m

Symbol m

m is part of the quantity the equation computes from the expression on the right.

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p_m

Symbol p_m

$p_m$ is one of the signed contributions combined to compute the quantity on the left.

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V

Symbol V

V is one of the signed contributions combined to compute the quantity on the left.

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C_m

Symbol C_m

$C_m$ is one of the signed contributions combined to compute the quantity on the left.

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H_m

Symbol H_m

human supervision and $R_m$ expected residual risk.

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R_m

Symbol R_m

$R_m$ is one of the signed contributions combined to compute the quantity on the left.

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=

The expressions on both sides represent the same quantity under the stated assumptions.

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subscript

The lower label selects a particular version, component, or indexed member of the quantity. For example, x₀ and xₜ can be values at different positions.

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How to interpret it

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What the article says around this equation

For task features x , choose model and reasoning configuration m to maximize $m^*(x)=\arg\max_m \left[p_m(x)V(x)-C_m(x)-H_m(x)-R_m(x)\right]$ . where $H_m$ is human supervision and $R_m$ expected residual risk. The cheapest model can cost more if it creates repeated attempts or review. The strongest model can be wasteful on deterministic extraction.

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Sources cited in the article section

[2] Models ↗

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