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Equation 15 · Part 1 · The Economics, Energy, and Physical Limits of Tool Protocols and the Model Context Protocol

Symbol n_max

nmax⁡=⌊C−Tsystem−Treservetˉtool⌋.n_{\max} = \left\lfloor \frac{C - T_{\mathrm{system}} - T_{\mathrm{reserve}}}{\bar{t}_{\mathrm{tool}}} \right\rfloor.
nmax⁡n_{\max}

What this part means

the maximum number of tools a deployment can nominally offer before it must start pruning or retrieving.

Its job in the formula

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

Where the article explains it

If every tool schema costs on average tˉtool\bar{t}_{\mathrm{tool}} tokens and all of them are injected in full on every turn, the maximum number of tools a deployment can nominally offer before it must start pruning or retrieving is

The passage around this formula

Put the three costs together and a deployment question comes into focus: given a fixed context window, how many tools can actually be offered before something has to give? A simple ceiling model makes the trade-off explicit. Let C be the model’s usable context window, TsystemT_{\mathrm{system}} the fixed cost of operator instructions, and TreserveT_{\mathrm{reserve}} a reserved allowance for the user’s message and the model’s own answer. If every tool schema costs on average tˉtool\bar{t}_{\mathrm{tool}} tokens and all of them are injected in full on every turn, the maximum number of tools a deployment can nominally offer before it must start pruning or retrieving is nmax⁡=⌊C−Tsystem−Treservetˉtool⌋n_{\max} = \left\lfloor \frac{C - T_{\mathrm{system}} - T_{\mathrm{reserve}}}{\bar{t}_{\mathrm{tool}}} \right\rfloor. The assumption doing the…

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