Source: https://guthlabs.ai/docs/instruments/caffeine-half-life-level-tracker/implementation/

instrument implementation / implementation-contract

# Caffeine Half-Life / Level Tracker implementation

A source-bound implementation contract for Caffeine Half-Life / Level Tracker; deployment and executable qualification remain separate.

Status: **VERIFIED_CLEAN**Lifecycle: **implementation-contract**

## Binding contract

This record describes what an executable surface must bind and validate. It does not claim that a public calculator, package, or production route is deployed.

**Record owner**: Guth Instruments

**Lifecycle**: verified-clean

**Evidence ID**: t6-069

**Effect**: Computation or source lookup only

## Input and validation requirements

A zero half-life produces an undefined division by zero, and evaluating the literal formula before an intake produces an extrapolated value outside the physical model domain. Inputs at or below the validity boundary should be rejected or explicitly flagged.

## Required corrections

No binding erratum is attached to this unique-name record. Implementers must still preserve the declared units, domain, and failure behavior.

## Inspectable interface fixture

The sample below is evidence for interface and validation tests, not a production implementation.

````
{
  "inputs": {
    "doses": [
      {
        "time_hr": 0,
        "dose_mg": 100
      }
    ],
    "half_life_hr": 5,
    "sample_times_hr": [
      0,
      5,
      10
    ],
    "threshold_mg": 10
  },
  "expected_outputs": {
    "sampled_points": [
      {
        "t_hr": 0,
        "y_mg": 100
      },
      {
        "t_hr": 5,
        "y_mg": 50
      },
      {
        "t_hr": 10,
        "y_mg": 25
      }
    ],
    "time_to_below_threshold_hr": 16.6096404744,
    "curve_features": {
      "finite_roots": [],
      "horizontal_asymptote_mg": 0,
      "dose_jump_at_hr": [
        0
      ]
    }
  },
  "working": "Use C(t)=100·2^(-(t−0)/5). At t=0: exponent=0, so C=100·2^0=100 mg. At t=5: exponent=−1, so C=100·(1/2)=50 mg. At t=10: exponent=−2, so C=100·(1/4)=25 mg. For the 10 mg crossing: 100·2^(−t/5)=10; 2^(−t/5)=0.1; −t/5=log2(0.1)=−3.32192809489; therefore t=5·3.32192809489=16.6096404744 hr. The curve remains positive for every finite t, so it has no finite root and approaches 0 mg."
}
````

## Other records

[01**Overview**](<https://guthlabs.ai/docs/instruments/caffeine-half-life-level-tracker/>)[02**Mechanism**](<https://guthlabs.ai/docs/instruments/caffeine-half-life-level-tracker/mechanism/>)[04**Qualification**](<https://guthlabs.ai/docs/instruments/caffeine-half-life-level-tracker/qualification/>)

## Sources and evidence

These pointers bind the page to evidence. They do not grant authority or collapse verification into deployment.

- `VERIFIED_LEDGER.json`

- `evidence-tranche6/tranche6-results.json`