What inputs does the estimator require?

It requires a positive measured peak Cmax, a positive half-life, and a positive time τ from the peak to the next dose.

What units are used for the result?

Cmax, Cmin, and the peak-to-trough drop are reported in mg/L. Half-life, τ, and k use hours and h⁻¹.

What does the fraction remaining mean?

It is the proportion of the measured peak concentration expected to remain at the trough.

Can this tool replace clinical judgment?

No. It is a pharmacokinetic estimation tool and is not medical advice.

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Trough Level Estimator (Cmin = Cmax · e^(-kτ))

Predict the trough concentration Cmin from a measured peak concentration Cmax during therapeutic drug monitoring (TDM): Cmin = Cmax · e^(-kτ), where k = ln2 / t½ and τ is the time from the peak to the next dose. Inputs are the measured peak, the elimination half-life, and the dosing interval. Also reports the fraction of the peak remaining at the trough and the peak-to-trough drop. This tool takes a measured Cmax (the TDM workflow), unlike drug-half-life-dose (which derives C0 from Dose/Vd) and dosing-interval-designer (which outputs a dimensionless peak/trough ratio, not an absolute concentration). Valid for one-compartment first-order elimination with the peak measured post-distribution. Not medical advice.

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2 examples

Aminoglycoside: Cmax=15 mg/L, t½=8 h, τ=8 h → Cmin = 7.5 mg/L

k = ln2/8 = 0.0866 h⁻¹; e^(-k·8) = e^(-ln2) = 0.5; Cmin = 15 × 0.5 = 7.5 mg/L. Half the peak remains after one half-life.

{"result":"Cmin = 7.5 mg/L (50% of peak 15 mg/L remaining; drop 7.5 mg/L). Method: Cmin = Cmax · e^(-kτ).","metadata":{"input":{"cmax":15,"halfLife":8,"tau":8},"result":{"cmin":7.5,"cminUnit":"mg/L","k":0.0866,"kUnit":"h⁻¹","fractionRemaining":0.5,"peakToTroughDrop":7.5,"method":"Trough from peak: Cmin = Cmax · e^(-kτ), k = ln2/t½","model":"One-compartment first-order elimination, peak measured post-distribution. Not medical advice."}}}

Long interval: Cmax=15 mg/L, t½=8 h, τ=12 h → Cmin = 5.3033 mg/L

k = ln2/8 = 0.0866 h⁻¹; e^(-k·12) = e^(-1.5·ln2) = 0.3536; Cmin = 15 × 0.3536 = 5.3033 mg/L. A longer interval (1.5 half-lives) gives a deeper trough.

{"result":"Cmin = 5.3033 mg/L (35.36% of peak 15 mg/L remaining; drop 9.6967 mg/L). Method: Cmin = Cmax · e^(-kτ).","metadata":{"input":{"cmax":15,"halfLife":8,"tau":12},"result":{"cmin":5.3033,"cminUnit":"mg/L","k":0.0866,"kUnit":"h⁻¹","fractionRemaining":0.3536,"peakToTroughDrop":9.6967,"method":"Trough from peak: Cmin = Cmax · e^(-kτ), k = ln2/t½","model":"One-compartment first-order elimination, peak measured post-distribution. Not medical advice."}}}

Inputs

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4 options

SettingsAdjust formats, ranges, numbers, and modes.4

Measured peak Cmax (mg/L)numberRequiredMeasured peak plasma concentration (post-distribution). Must be positive.Half-life t½ (h)numberRequiredElimination half-life. Must be positive.Time from peak to next dose τ (h)numberRequiredDosing interval — time from the measured peak to the next dose. Must be positive.Decimal PlacesnumberOptional

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Trough Level Estimator (Cmin = Cmax · e^(-kτ))

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1. One half-life from peak to next dose

2. Longer interval produces a deeper trough

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