The green points represent data collected after the infusion is stopped. Notice they approximate a straight line on the semi-log graph.
Extrapolation of this line back to time = 0 gives Cp(0)** = 7.5 mg/L. The value of Cp from the line at time = 6 hr gives Cp(6) = 2.3 mg/L. The elimination rate constant can be calculated from the ln slope of this line: kel = ln(Cp(0)-Cp(6))/6 = (ln(7.5) - ln(2.3))/6 = 0.197 hr-1. NOTE: The term Cp(0)** is the intercept of the green line on the y axis and does not equal Dose/V. It is the value of Cp from the line at time zero.
Extrapolation to time = T = 0.75 hr (end of the infusion) gives Cp(T) = 7.5 x e-0.197*0.75 = 6.47 mg/L.
Rearranging 
(Equation 46) allows calculation of V from: V = (k0/kel*Cp(T)) * (1 - e-kel*T) = (200/0.197 x 6.47) x (1 - e-0.197 x 0.75) = 21.6 L.
| Parameter | Value | Units |
| kel | 0.197 | hr-1 |
| V | 21.6 | L |