1. Intravenous Infusion Previous rates of administration were instantaneous IV bolus and first order absorption. As a rate (mg/hr) a first order rate constantly declines even though a constant percentage of what remains is handled.

2. Intravenous Infusion A first order rate (ka) such as 0.693 hr -1, will peak at about 2.6 hours when the T½ = 4 hours. If we deal with ka as we would K, since ka = 0.693, then T½ ka = 1 hour. This means that following a 400 mg dose: TimeAm’t Unabs 1 200 2 100 3 50 4 25 5 12.5 6 6.25 25 mg/hr 100 mg/hr 6.25 mg/hr

3. Intravenous Infusion A zero order rate is constant (mg/hr). With a 400 mg dose, a rate of 200mg/hr results in all drug being delivered in 2 hours Time Am’t Delivered hr (mg) 1 200 2 200 3 0 25 mg/hr 100 mg/hr 6.25 mg/hr Zero-Order Rate 200 mg/hr Tmax = 2 hours The infusion rate is identified as k 0 with units of am’t/time.

4. Intravenous Infusion-continuous If a drug is infused at a constant rate (4 mg/hr) for a long period of time, concentrations will rise and approach a constant concentration (C SS ). At that concentration (C SS ), the rate of elimination of the drug equals the infusion rate. IN = OUT C SS = k 0 /VK Zero-Order Rate 4 mg/hr Approaching maximum concentration of 7.3 mg/L

5. Intravenous Infusion-continuous NumberPercent Of of C SS T½ 150.0 275.0 387.5 493.75 596.875 698.438 799.219 Zero-Order Rate 4 mg/hr Approaching maximum concentration of 7.3 mg/L The rate at which the steady-state concentration is approached is proportional to the half-life.

6. Intravenous Infusion Many drugs are administered by intermittent intravenous infusions, where the time of the infusion is short and SS is not achieved. 15 min -  -lactams 30 min – aminoglycosides 1 hr vancomycin 4 hour amphotericin. For these drugs the infusion rate (mg/hr) is: k 0 = dose / infusion time IV Intermittent Infusion Dose = 400 mg Infusion Time = 2 hours Infusion Rate (k 0 ) = 200 mg/hr

7. Intravenous Infusion During the infusion period the concentrations increase. The equation predicting these concentrations is: C SS = (k 0 /KV) *(1-e -kt ) This has rough similarity to the oral dosing equation IV Intermittent Infusion Dose = 160 mg Infusion Time = 4 hours Infusion Rate (k 0 ) = 40 mg/hr Vol = 10 L; T½ = 2hr C SS = 8.66 mg/L Dose Clearance

8. Intravenous Infusion During the infusion period the concentrations increase. The equation predicting these concentrations is: C SS = (k 0 /KV) *(1-e -kt ) Here 160 mg is infused over 4 hours to a patient with a V of 10 L and a half-life of 2 hours IV Intermittent Infusion Dose = 160 mg Infusion Time = 4 hours Infusion Rate (k 0 ) = 40 mg/hr Vol = 10 L; T½ = 2hr C SS = 8.66 mg/L The concentration at the end of the infusion (4 hr) is: C SS = (k 0 /KV) *(1-e -kt ) = (40/(0.3465 * 10))*(1- -0.3465*4 ) = (11.54)(1-0.25) = 8.66 mg/L

9. Intravenous Infusion During the infusion period the concentrations increase. The equation predicting these concentrations is: C SS = (k 0 /KV) *(1-e -kt ) Here 160 mg is infused over 4 hours to a patient with a V of 10 L and a half-life of 2 hours IV Intermittent Infusion Dose = 160 mg Infusion Time = 4 hours Infusion Rate (k 0 ) = 40 mg/hr Vol = 10 L; T½ = 2hr C SS = 8.66 mg/L What is the Cl in this patient? Cl = KV = 0.3465 * 10 = 3.465 L/hr If the infusion is continuous C SS = (k 0 /KV) *(1-e -kt ) = (k 0 /KV) Cl can be obtained from k 0 & C SS

10. Intravenous Infusion As soon as the infusion is complete, the concentrations will begin to decline. If distribution is instantaneous This equation calculates the concentration from the start of the infusion to the end of the infusion at time t C SS = (k 0 /KV) *(1-e -kt ) This equation calculates the concentration from the END of the infusion on and so t is the duration of infusion and t’ is the time from the end of the infusion C SS = [(k 0 /KV) *(1-e -kt )]e- Kt’

11. Intravenous Infusion This equation calculates the concentration from the END of the infusion on and so t is the duration of infusion and t’ is the time from the end of the infusion C SS = [(k 0 /KV) *(1-e -Kt )]e -Kt’ To calculate the concentration at 6 hours from the Start of a 4 hour infusion, t will be 4 hours and t’ is the time from the end of the infusion to the time of interest – 2 hours. C SS = [(k 0 /KV) *(1-e -kt )]e -Kt’ = [(11.54)(1-0.25)]e -0.3465*2 = [8.66]*0.500 = 4.33 mg/L 86428642

12. Intravenous Infusion This equation will calculate the concentrations during the infusion but requires that the time t’ is zero. Example: Calculate the concentration @ 2 hours C SS = [(k 0 /KV) *(1-e -Kt )] e -Kt’ = (11.54)*(1-e -0.3465*2 ) e -0.3465*0 = [(11.54)(1-0.50)] x 1 = [11.54*0.500] x 1 = 5.77 mg/L 86428642

13. Intravenous Infusion Question A patient is given an intravenous infusion of 200 mg over 4 hours. Six and 8 hrs after the infusion began samples were taken and the concentrations measured. TimeConc (hr)(mg/L) 63.333 81.667 1.What loading dose would achieve the end of infusion concentration immediately? 2.Would a loading dose be considered necessary for this drug in this patient? 86428642

14. Intravenous Infusion Question A patient is given an intravenous infusion of 200 mg over 4 hours. Six and 8 hrs after the infusion began samples were taken and the concentrations measured. TimeConc (hr)(mg/L) 63.333 81.667 [ ] end of infusion requires K. T½ = 2 hours;K = 0.3465 hr-1 C 4 = 3.333 e Kt = 3.333 e (0.3465 * 2) = 6.667 mg/L 86428642 86428642

15. Intravenous Infusion Question A patient is given an intravenous infusion of 200 mg over 4 hours. Six and 8 hrs after the infusion began samples were taken and the concentrations measured. TimeConc (hr)(mg/L) 63.333 81.667 Calculate Volume: C 4 = [(k 0 /KV) *(1-e -Kt )]e -Kt’ V = [(k 0 /KC 4 ) *(1-e -Kt )]e -Kt’ t’= 0; t=4; C4 = 6.667; k 0 =50 mg/hr V = [(50/(0.3465*6.667) *(1-e -0.3465*4 ] = 16.23 L 86428642 Volume based on this concentration 86428642

16. Intravenous Infusion Question A patient is given an intravenous infusion of 200 mg over 4 hours. Six and 8 hrs after the infusion began samples were taken and the concentrations measured. TimeConc (hr)(mg/L) 63.333 81.667 1.What loading dose would achieve the end of infusion concentration immediately? Load = V * C SS = 16.23 L * 6.667 mg/L = 108.21 mg 86428642 86428642

17. Intravenous Infusion Question A patient is given an intravenous infusion of 200 mg over 4 hours. Six and 8 hrs after the infusion began samples were taken and the concentrations measured. TimeConc (hr)(mg/L) 63.333 81.667 1.What loading dose would achieve the end of infusion concentration immediately? 2.Would a loading dose be considered necessary for this drug in this patient? 86428642 86428642

18. Intravenous Infusion Question A patient is given an intravenous infusion of 200 mg over 4 hours. Six and 8 hrs after the infusion began samples were taken and the concentrations measured. Would a loading dose be considered necessary for this drug in this patient? How long does it take to get to steady state? 3.3 (90%) – 5 (96.8) T½ With a T½ of 2 hours …between 6.6 and 10 hours. Bolus?? How rapid is response required? Does a 108.2 mg bolus achieve C SS ? 86428642 86428642

19. Intravenous Infusion Question If this patient receives a continuous infusion (> 24 hr) what is the steady-state concentration?

20. Intravenous Infusion Question If this patient receives a continuous infusion (> 24 hr) what is the steady-state concentration? C SS = [(k 0 /KV) *(1-e -Kt )] When t is large e -Kt becomes Small and equation becomes C SS = [(k 0 /KV)] = 50 (0.3465 x 16.23) = 8.89 mg/L What bolus loading dose would achieve this true steady-state?

21. Intravenous Infusion Question If this patient receives a continuous infusion (> 24 hr) what is the steady-state concentration? C SS = [(k 0 /KV) *(1-e -Kt )] When t is large e -Kt becomes Small and equation becomes C SS = [(k 0 /KV)] = 50 (0.3465 x 16.23) = 8.89 mg/L Load = Css V = (8.89 mg/L)(16.23 L) = 144.3 mg

22. Intravenous Infusion Question If this patient receives a bolus of 144.3 mg and simultaneously receives an intravenous infusion of 50 mg/hr, what will the concentration-time profile look like? … what concentrations are attained?

23. Intravenous Infusion Question If this patient receives a bolus of 144.3 mg and simultaneously receives an intravenous infusion of 50 mg/hr, what will the concentration-time profile look like? … what concentrations are attained? What would the initial concentration of the bolus be? C 0 = Dose / V = 144.3 mg / 16.23 L = 8.89 mg/L

24. Intravenous Infusion Question If this patient receives a bolus of 144.3 mg and simultaneously receives an intravenous infusion of 50 mg/hr, what will the concentration-time profile look like? … what concentrations are attained? What would the concentration -time profile for the bolus look like? T½ = 742742

25. Intravenous Infusion Question If this patient receives a bolus of 144.3 mg and simultaneously receives an intravenous infusion of 50 mg/hr, what will the concentration-time profile look like? … what concentrations are attained? What would the concentration -time profile for the bolus look like? T½ = 2 hours 742742 Should the conc.–time profile from the bolus and the simultaneous infusion just be the sum of both profiles?

26. Intravenous Infusion Question TimeBolusInfusion Total (hr)(mg/L) (mg/L) (mg/L) 0.08.891 0.00 8.891 0.57.477 1.414 8.891 1.06.287 2.604 8.891 1.55.287 3.604 8.891 2.04.446 4.445 8.891 3.03.144 5.747 8.891 4.02.223 6.668 8.891 6.01.112 7.779 8.891 8.00.556 8.335 8.891 12.00.139 8.752 8.891 16.00.035 8.856 8.891 24.00.002 8.889 8.891 742742

27. Intravenous Infusion Question TimeBolusInfusion Total (hr)(mg/L) (mg/L) (mg/L) 0.08.891 0.00 8.891 0.57.477 1.414 8.891 1.06.287 2.604 8.891 1.55.287 3.604 8.891 2.04.446 4.445 8.891 3.03.144 5.747 8.891 4.02.223 6.668 8.891 6.01.112 7.779 8.891 8.00.556 8.335 8.891 12.00.139 8.752 8.891 16.00.035 8.856 8.891 24.00.002 8.889 8.891 742742 When bolus and infusion are exactly matched concentration starts at true steady state and remains unchanged

28. Intravenous Infusion Question What happens if the bolus is larger or smaller? 742742

29. Don’t let this be you on Exam Day.