If t1/2 = 4 h and dosing interval τ = 24 h, the accumulation factor R is approximately:

• A. 1.02
• B. 1.50
• C. 1.25
• D. 1.10

Answer: (A)

Repeated dosing with first-order elimination leads to accumulation that depends on how long you interval between doses (τ) is compared with the drug’s half-life. The accumulation factor, R, tells you how much higher the steady-state concentrations are compared with a single-dose scenario. It’s calculated from k, the elimination rate constant, using R = 1 / (1 − e^(−kτ)), where k = ln 2 / t1/2.

Here, t1/2 is 4 hours, so k ≈ 0.693 / 4 = 0.173 h⁻¹. With τ = 24 hours, the term e^(−kτ) = e^(−0.173 × 24) ≈ e^(−4.16) ≈ 0.016. Thus R ≈ 1 / (1 − 0.016) ≈ 1 / 0.984 ≈ 1.02.

So the accumulation is small: steady-state concentrations are about 2% higher than after a single dose because the long dosing interval (about six half-lives) allows most of the drug to be cleared before the next dose.