Guide · Biopharmaceutics & Pharmacokinetics
Pharmacokinetic Parameters: AUC, Clearance, Half-Life and Bioavailability
Work out kₑ, half-life, volume of distribution, AUC, clearance and bioavailability from concentration–time data, with a full worked IV-bolus example.
Pharmacokinetics describes what the body does to a drug: how quickly it is absorbed, how widely it spreads, and how fast it is removed. A handful of parameters — the elimination rate constant, half-life, volume of distribution, clearance, area under the curve and bioavailability — summarise all of that in numbers you can calculate from a set of plasma concentrations.
This guide works with the simplest useful model: a one-compartment body and linear (first-order) kinetics, where the rate of elimination is proportional to the concentration. It is the model behind most biopharmaceutics practicals and exam questions in the Pharm-D syllabus, and the starting point for every more complex model.
We first define each parameter, then analyse one illustrative IV-bolus data set from start to finish, then cover bioavailability and repeated dosing. The data and drug names are invented for teaching. Nothing here is a dosing recommendation for a real patient or a real medicine.
Prepared by the PharmaWallah student team. Verify results with your textbook or instructor. Report an error · Editorial policy
Learning objectives
After working through this guide you should be able to:
Key terms
- Plasma concentration–time curve
- A graph of drug concentration in plasma (C) against time after a dose. Its shape shows absorption, distribution and elimination.
- First-order (linear) kinetics
- Elimination in which a constant fraction of the drug is removed per unit time, so the amount removed per hour falls as the concentration falls.
- Elimination rate constant (kₑ)
- The fraction of drug in the body eliminated per unit time, in units of reciprocal time (h⁻¹). It is the negative slope of ln C against time.
- Half-life (t½)
- The time for the plasma concentration (or amount in the body) to fall by half. In first-order kinetics it is constant: t½ = 0.693/kₑ.
- C₀
- The concentration at time zero after an IV bolus, found by extrapolating the ln C line back to t = 0. It is not measured directly.
- Apparent volume of distribution (Vd)
- The volume the dose would occupy if it were everywhere at the plasma concentration. It is a proportionality constant, not a real anatomical volume, and can exceed body water.
- Clearance (CL)
- The volume of plasma completely cleared of drug per unit time (L/h or mL/min). The rate of elimination equals CL × C.
- AUC
- Area under the plasma concentration–time curve (mg·h/L). AUC₀–t runs to the last sample; AUC₀–∞ adds the extrapolated tail. It reflects total exposure to the drug.
- Bioavailability (F)
- The fraction of an administered dose that reaches the systemic circulation unchanged. By definition F = 1 for an IV dose.
- Steady state (Css)
- The condition during repeated dosing or a constant infusion when the rate of drug in equals the rate of drug out, so the concentration profile repeats from one dose to the next.
- Accumulation factor (R)
- How much higher the steady-state concentrations are than those after the first dose, for doses given every τ hours: R = 1/(1 − e^(−kₑτ)).
The concentration–time curve, kₑ and half-life
After an IV bolus into a one-compartment body, the whole dose is in the plasma at once and concentration falls exponentially. Plotted on ordinary axes the curve bends; plotted as the natural logarithm of concentration against time it becomes a straight line. That straight line is why we take logarithms: a straight line has a slope and an intercept we can calculate by linear regression.
C = C₀ · e^(−kₑt) so ln C = ln C₀ − kₑ · t
- C
- plasma concentration at time t (mg/L)
- C₀
- concentration extrapolated to t = 0 (mg/L)
- kₑ
- elimination rate constant (h⁻¹)
- t
- time after the dose (h)
The slope of ln C against t is −kₑ, and the intercept is ln C₀.
t½ = ln 2 / kₑ = 0.693 / kₑ
Half-life and kₑ carry the same time unit: kₑ in h⁻¹ gives t½ in hours.
Many textbooks and older practical manuals plot log₁₀ C on semi-log paper instead of ln C. The line is still straight, but its slope is −kₑ/2.303, because ln x = 2.303 × log₁₀ x. If you read a slope from a log₁₀ plot, multiply it by 2.303 before calling it kₑ.
Use the terminal points
After an oral dose, or when a drug distributes slowly, the early points do not lie on the elimination line. Fit only the straight terminal portion of the ln C plot. In a true one-compartment IV bolus, all points should lie close to one line.
Volume of distribution, AUC and clearance
Once you know C₀, the volume of distribution for an IV bolus follows directly, because at time zero the whole dose is in the body and none has yet been eliminated.
Vd = Dose(IV) / C₀
- Dose(IV)
- intravenous bolus dose (mg)
- C₀
- back-extrapolated concentration at t = 0 (mg/L)
Express dose and concentration in the same mass unit: mg with mg/L gives L. Note that 1 µg/mL = 1 mg/L.
AUC is usually calculated from the data with the linear trapezoidal rule: the area between each pair of consecutive samples is treated as a trapezium. For an IV bolus the first trapezium starts at C₀. Because sampling stops before the drug is fully gone, the remaining tail is added by extrapolation from the last measured concentration.
AUC(tᵢ→tᵢ₊₁) = (tᵢ₊₁ − tᵢ) × (Cᵢ + Cᵢ₊₁) / 2 AUC₀–∞ = AUC₀–t + C_last / kₑ
- C_last
- last measured concentration (mg/L)
- kₑ
- terminal elimination rate constant (h⁻¹)
Clearance links the dose to the exposure. For a single IV dose, every milligram given is eventually cleared, and the total cleared equals CL × AUC₀–∞. In the one-compartment model clearance is also the product of kₑ and Vd, which shows why a drug with a large volume of distribution is eliminated slowly unless its clearance is also high.
CL = Dose(IV) / AUC₀–∞ = kₑ · Vd
L/h × 1000 ÷ 60 gives mL/min. Clearance is a volume per time, never an amount per time.
Worked example: analysing an IV-bolus data set end to end
Drug X (illustrative) is given as a single 500 mg IV bolus. Plasma samples give the concentrations below. We will find kₑ, t½, C₀, Vd, AUC₀–∞ and CL.
| Time (h) | C (mg/L) | ln C |
|---|---|---|
| 0.5 | 22.6 | 3.1179 |
| 1 | 20.5 | 3.0204 |
| 2 | 16.8 | 2.8214 |
| 4 | 11.2 | 2.4159 |
| 6 | 7.6 | 2.0281 |
| 8 | 5.0 | 1.6094 |
| 12 | 2.3 | 0.8329 |
Worked example
Step 1 — kₑ, t½ and C₀ from the regression of ln C on time
Fit a straight line to ln C against t for all seven points, then find kₑ, the half-life and C₀.
Take the natural log of each concentration (third column of the table).
Least-squares regression of ln C on t gives slope = −0.19928 h⁻¹ and intercept = 3.21743, with R² = 0.99993, so the points lie on a straight line.
kₑ = −slope = 0.1993 h⁻¹ (about 19.9% of the drug in the body is eliminated per hour at any moment).
t½ = 0.693 / 0.1993 = 3.48 h.
C₀ = e^(intercept) = e^3.21743 = 24.96 mg/L.
kₑ = 0.199 h⁻¹, t½ = 3.48 h, C₀ = 25.0 mg/L.
Using ln 2 = 0.693147 instead of 0.693 changes t½ only in the fourth decimal place (3.4783 versus 3.4775 h).
Worked example
Step 2 — Volume of distribution
Use the dose and C₀ from Step 1.
Vd = Dose / C₀ = 500 mg ÷ 24.96 mg/L.
Vd = 20.03 L.
Vd ≈ 20.0 L.
Worked example
Step 3 — AUC by the trapezoidal rule plus extrapolation
Calculate AUC₀–12 starting from C₀ at t = 0, then add the tail to infinity.
0–0.5 h: 0.5 × (24.96 + 22.6)/2 = 11.89 mg·h/L.
0.5–1 h: 0.5 × (22.6 + 20.5)/2 = 10.78.
1–2 h: 1 × (20.5 + 16.8)/2 = 18.65.
2–4 h: 2 × (16.8 + 11.2)/2 = 28.00.
4–6 h: 2 × (11.2 + 7.6)/2 = 18.80.
6–8 h: 2 × (7.6 + 5.0)/2 = 12.60.
8–12 h: 4 × (5.0 + 2.3)/2 = 14.60.
AUC₀–12 = sum of the trapezia = 115.32 mg·h/L.
Extrapolated tail = C_last / kₑ = 2.3 ÷ 0.1993 = 11.54 mg·h/L.
AUC₀–∞ = 115.32 + 11.54 = 126.86 mg·h/L. The tail is 9.1% of the total.
AUC₀–∞ ≈ 126.9 mg·h/L.
Worked example
Step 4 — Clearance, two ways
Calculate CL from Dose/AUC₀–∞ and check it against kₑ × Vd.
CL = Dose / AUC₀–∞ = 500 ÷ 126.86 = 3.94 L/h.
Check: CL = kₑ × Vd = 0.1993 × 20.03 = 3.99 L/h.
In mL/min: 3.94 × 1000 ÷ 60 ≈ 66 mL/min.
CL ≈ 3.94 L/h (model check 3.99 L/h).
The two values differ by about 1%. Straight trapezia lie slightly above a curve that bends downwards, so the linear trapezoidal rule slightly overestimates AUC during a decline (here 126.9 against the model's C₀/kₑ = 125.3), and CL from Dose/AUC comes out slightly lower. Wider sampling intervals make the gap larger.
Absolute and relative bioavailability
Because clearance does not depend on the route, the AUC produced by a dose is proportional to the amount that actually reaches the circulation. Comparing AUCs therefore compares the fractions absorbed — provided each AUC is first divided by its own dose, and AUC₀–∞ is used for both.
F(absolute) = (AUC_oral × Dose_IV) / (AUC_IV × Dose_oral)
Equivalent to (AUC_oral/Dose_oral) ÷ (AUC_IV/Dose_IV). It assumes the same clearance on both occasions, usually ensured by a crossover study in the same subjects.
F(relative) = (AUC_test / Dose_test) ÷ (AUC_reference / Dose_reference)
Compares two non-IV products, for example a new tablet against an oral suspension. It can exceed 1 (100%) if the test product is better absorbed than the reference.
Worked example
Absolute bioavailability of an oral tablet
Drug X (illustrative): a 50 mg IV dose gives AUC₀–∞ = 12.5 mg·h/L; a 100 mg oral tablet in the same subjects gives AUC₀–∞ = 17.0 mg·h/L. Find F.
F = (AUC_oral × Dose_IV) / (AUC_IV × Dose_oral).
F = (17.0 × 50) / (12.5 × 100) = 850 / 1250.
F = 0.68.
F = 0.68, i.e. 68% of the oral dose reaches the systemic circulation.
Without dose normalisation, 17.0 ÷ 12.5 would wrongly give 1.36.
Worked example
Relative bioavailability of a test tablet
A 250 mg test tablet gives AUC₀–∞ = 40.2 mg·h/L; a 200 mg reference suspension gives 36.0 mg·h/L (illustrative). Find the relative bioavailability.
Dose-normalised test AUC = 40.2 ÷ 250 = 0.1608 h/L.
Dose-normalised reference AUC = 36.0 ÷ 200 = 0.1800 h/L.
F(relative) = 0.1608 ÷ 0.1800 = 0.893.
F(relative) ≈ 0.89 (89.3%).
Repeated dosing: accumulation and time to steady state
If a dose is given again before the previous one has been eliminated, drug accumulates. It does not accumulate forever: as the amount in the body rises, so does the amount eliminated per interval, until elimination over one interval equals one dose. That balance is steady state.
How long steady state takes depends only on the half-life, not on the dose or the interval. After n half-lives of constant dosing the fraction of steady state reached is 1 − (½)ⁿ.
| Half-lives elapsed | Fraction of steady state |
|---|---|
| 1 | 50% |
| 2 | 75% |
| 3 | 87.5% |
| 3.32 | 90% |
| 4 | 93.8% |
| 5 | 96.9% |
| 7 | 99.2% |
That is why steady state is said to take about 4–5 half-lives: by then the concentrations are within roughly 3–6% of their final values.
R = 1 / (1 − e^(−kₑτ))
- R
- accumulation factor (steady-state level ÷ first-dose level at the same time after a dose)
- τ
- dosing interval (h)
If τ equals one half-life, e^(−kₑτ) = ½ and R = 2. A shorter interval relative to t½ means more accumulation.
Worked example
Accumulation for a 12-hourly regimen
Drug Y (illustrative) has t½ = 8 h and is given every 12 h. Find the accumulation factor and the time to reach 90% of steady state.
kₑ = 0.693 ÷ 8 = 0.0866 h⁻¹.
kₑτ = 0.0866 × 12 = 1.0395.
e^(−1.0395) = 0.3536, so R = 1 / (1 − 0.3536) = 1.547.
90% of steady state takes 3.32 half-lives = 3.32 × 8 = 26.6 h.
R ≈ 1.55; about 27 h to reach 90% of steady state (about 40 h for 5 half-lives).
Loading and maintenance doses — the concepts
The same parameters explain the two dose types students meet in therapeutics. A loading dose fills the volume of distribution up to a target concentration straight away, so the patient does not wait 4–5 half-lives. A maintenance dose replaces what clearance removes, so the target is held.
Loading dose = Css(target) × Vd / F Maintenance dose rate = CL × Css(target) / F
The maintenance dose for an interval τ is (CL × Css × τ) / F. For an IV route F = 1.
Worked example
Illustrative arithmetic only
Drug Z is entirely fictional: target average Css 10 mg/L, Vd 40 L, CL 4 L/h, oral F = 0.8. Show how the formulas combine.
Loading dose = 10 mg/L × 40 L ÷ 0.8 = 500 mg.
Maintenance dose rate = 4 L/h × 10 mg/L ÷ 0.8 = 50 mg/h.
Over a 12-hour interval: 50 mg/h × 12 h = 600 mg.
Loading dose 500 mg; maintenance 50 mg/h (600 mg per 12 h) — for this fictional drug only.
Not for real dosing
These formulas teach how the parameters interact. Real doses are chosen from validated, drug-specific references and product information, adjusted for the patient's renal and hepatic function, age, weight and interacting drugs, and confirmed by clinical judgement and, where appropriate, drug-level monitoring. Population values of Vd, CL and F vary widely between patients.
Common mistakes
Practice questions
Try each one before opening the answer.
Q1.A drug has kₑ = 0.25 h⁻¹. What is its half-life?
Show answer and working
t½ = 0.693 / kₑ.
t½ = 0.693 ÷ 0.25 = 2.772 h.
Answer: t½ ≈ 2.77 h
Q2.A student plots log₁₀ C against time and finds a slope of −0.0652 h⁻¹. Find kₑ and t½.
Show answer and working
kₑ = −slope × 2.303 = 0.0652 × 2.303 = 0.15016 h⁻¹.
t½ = 0.693 ÷ 0.15016 = 4.62 h.
Answer: kₑ ≈ 0.150 h⁻¹; t½ ≈ 4.62 h
Q3.A 300 mg IV bolus gives a back-extrapolated C₀ of 7.5 µg/mL. Calculate Vd.
Show answer and working
Convert: 7.5 µg/mL = 7.5 mg/L.
Vd = Dose / C₀ = 300 mg ÷ 7.5 mg/L = 40 L.
Answer: Vd = 40 L
Q4.After a 200 mg IV bolus, concentrations are 10, 8.0, 6.4 and 4.1 mg/L at 0, 1, 2 and 4 h, and kₑ = 0.22 h⁻¹. Calculate AUC₀–∞ and CL.
Show answer and working
0–1 h: 1 × (10 + 8.0)/2 = 9.0 mg·h/L.
1–2 h: 1 × (8.0 + 6.4)/2 = 7.2 mg·h/L.
2–4 h: 2 × (6.4 + 4.1)/2 = 10.5 mg·h/L.
AUC₀–4 = 9.0 + 7.2 + 10.5 = 26.7 mg·h/L.
Tail = C_last / kₑ = 4.1 ÷ 0.22 = 18.64 mg·h/L.
AUC₀–∞ = 26.7 + 18.64 = 45.34 mg·h/L.
CL = Dose / AUC₀–∞ = 200 ÷ 45.34 = 4.41 L/h.
Answer: AUC₀–∞ ≈ 45.3 mg·h/L; CL ≈ 4.41 L/h
Q5.A 25 mg IV dose gives AUC₀–∞ = 5.0 mg·h/L and a 50 mg oral dose gives AUC₀–∞ = 7.2 mg·h/L in the same subjects. Calculate the absolute bioavailability.
Show answer and working
F = (AUC_oral × Dose_IV) / (AUC_IV × Dose_oral).
F = (7.2 × 25) / (5.0 × 50) = 180 / 250 = 0.72.
Answer: F = 0.72 (72%)
Q6.A drug with t½ = 6 h is given every 6 h. What is the accumulation factor? What would it be if it were given every 12 h instead?
Show answer and working
τ = 6 h is one half-life, so e^(−kₑτ) = 0.5 and R = 1 / (1 − 0.5) = 2.0.
τ = 12 h is two half-lives, so e^(−kₑτ) = 0.25 and R = 1 / (1 − 0.25) = 1.33.
Answer: R = 2.0 every 6 h; R ≈ 1.33 every 12 h
Q7.A drug has kₑ = 0.1 h⁻¹ and Vd = 35 L. Calculate its clearance in L/h and mL/min.
Show answer and working
CL = kₑ × Vd = 0.1 × 35 = 3.5 L/h.
3.5 L/h × 1000 mL/L ÷ 60 min/h = 58.3 mL/min.
Answer: CL = 3.5 L/h ≈ 58.3 mL/min
Q8.A drug with t½ = 10 h is started as a constant infusion with no loading dose. How long until it reaches 90% of steady state?
Show answer and working
Fraction reached = 1 − (½)ⁿ; set it to 0.90, so (½)ⁿ = 0.10.
n = ln 10 / ln 2 = 3.32 half-lives.
Time = 3.32 × 10 h = 33.2 h.
Answer: About 33 h
Limitations and safe use
Practise with the calculators
References
- Ducharme MP, Shargel L, editors. Shargel and Yu's Applied Biopharmaceutics & Pharmacokinetics. 8th ed. New York: McGraw Hill; 2022.
- Rowland and Tozer's Clinical Pharmacokinetics and Pharmacodynamics: Concepts and Applications. 5th ed. Philadelphia: Wolters Kluwer; 2019.
- Gibaldi M, Perrier D. Pharmacokinetics. 2nd ed. New York: Marcel Dekker; 1982.
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