1. Multiple Dosing: Intermittent or multiple dose regimen
Dosing Regimen Design
Multiple Dosing:
Intermittent or multiple dose regimen

2. 100 mg q. t1/2 via i.v. bolus 200 150 Amount in Body [mg] 100 50
Time [t1/2]:

3. Principle: 1 dose lost per 
At steady state, Rate In = Rate Out
F•Dose = Ass,max - Ass,min
Ass,min = Ass,maxe-KE
F•Dose = Ass,max (1 - e-KE)
Ass,max = F•Dose / (1 - e-KE)
Ass,min = Ass,max - F•Dose

4. AN,max & AN,min
N =
Not AN,min = AN,max - F•Dose Why?

5. Css,max & Css,min

6. Average amount of drug in the body at steady state
At steady state, Rate In = Rate Out
FDose/ = KE Ass,av
1/KE = t1/2/ln 2 = t1/2

7. Average steady-state plasma concentration

8. AUC
Equal Areas

9. Css,av
Concept applies to all routes of administration and it is independent of absorption rate.

10. Dosing rate from AUC0-
Given AUC after a single dose, D, the maintenance dose, DM , is:
Where D produced the AUC0-, Css,av is the desired steady-state plasma concentration, and  is the desired dosing interval.

11. Example: 50 mg p.o. dose AUC = 1+2.5+5.4+8.4+6 = 23.3 mg•h/L Cp
mg/L
Time [h]
AUC = = 23.3 mg•h/L

12. Example, continued
Calculate a dosing regimen for this drug that would provide an average steady-state plasma concentration of 15 mg/L.
DM = 193 mg
Dosing regimen: 200 mg q. 6 h.

13. Example, continued - 2 There is a problem with this approach ??
Peak and trough concentrations are unknown.
Css,max
Css,min

14. Another example - digitoxin
t1/2 = 6 days; usual DR is 0.1 mg/day
Assuming rapid and complete absorption of digitoxin,
Ass,av = 1.44 F Dose t1/2/
= (1.44)(1)(0.1mg/d)(6d)/(1d)
= mg
What would be the average steady-state body level?
Maximum and minimum plateau values?
Ass,max = 0.1/(1-e-(0.116)(1))
= mg
Is there accumulation of digitoxin?
Ass,min = – 0.1
= mg
How long to reach steady state?

15. Rate of Accumulation, AI, FI
The rate of accumulation depends on the half-life of the drug: 3.3 x t1/2 gives 90% of the steady-state level.
Accumulation Index (AI): Ass,max/F DM = (1 – e-KE)-1
Fluctuation Index (FI): Ass,max/Ass,min = e+KE AI FI KE 
When  = t1/2
AI = FI = 2

16. Absorption Rate influence on Rate of Accumulationv CL ka
Absorption Rate influence on Rate of Accumulation
KE = 0.1

17. When ka >> KE, control is by drug t1/2:
When ka << KE, control is by absorption t1/2:

18. Loading Dose (LD) = Ass,max  F Whether a LD is needed depends upon:
Accumulation Index
Therapeutic Index
Drug t1/2
Patient Need

19. Dosing Regimen Design
OBJECTIVE: Maintain Cp within the therapeutic window. Cp
Time

20. Dosing Regimen Design max APPROACH: Calculate max and DM,max. Cu Cp
Time Cu Cl
max

21. DM,max and Dosing Rate
From the principle that one dose is lost over a dosing interval at steady state:
DM,max = (V/F)(Cu - Cl) Cp
Time
The Dosing Rate (DR) is DM,max  max

22. KEV = CL
(Cu - Cl)/ln(Cu/Cl) = Css,av = logarithmic average of Cu and Cl.
The log average is the concentration at the midpoint of the dosing interval; it’s less than the arithmetic average.
DR = (CL/F)Css,av

23. Average Concentration Approach
Choose the average to maintain:
Css,av = (Cu - Cl)/ln (Cu/Cl)
Choose :
  max ;usually 4, 6, 8, 12, 24 h
Calculate DR:
DR = (CL/F)Css,av
Calculate DM:
DM = DR•

24. Example max = (1.44)(4.85)[ln (10/3)] = 8.41 h
V = 35L
KE = h-1
t1/2 = 4.85 h
Cu = 10 mg/L
CL = 5L/h
F = 0.80
Cl = 3 mg/L
Css,av = (10 – 3)/ln (10/3) = 5.8 mg/L
max = (1.44)(4.85)[ln (10/3)] = 8.41 h
Choose  < max: 8 h
DR = (5 L/h)(5.8 mg/L)/(0.8) = mg/h
DM = (36.25 mg/h)(8 h) = 290 mg  300 mg
Dosing Regimen: 300 mg q 8 h

25. Peak Concentration Approach
Choose the peak concentration to maintain.
Choose :
  max ;usually 4, 6, 8, 12, 24 h
Calculate DM:
DM = (V•Cpeak/F)(1 - e-KE)
from:

26. Example max = (1.44)(4.85)[ln (10/3)] = 8.41 h
V = 35L
KE = h-1
t1/2 = 4.85 h
Cu = 10 mg/L
CL = 5L/h
F = 0.80
Cl = 3 mg/L
Cpeak = 8 mg/L
max = (1.44)(4.85)[ln (10/3)] = 8.41 h
Choose  < max: 6 h
 set to 6 h so that Css,min > Cl
DM = [(35 L)(8 mg/L)/0.8](1 – e-(0.143)(6)) = 202 mg
Dosing Regimen: 200 mg q 6 h

27. Check Css,max = [(0.8)(200)/35]/(1 – e-(0.143)(6)) = 7.93 mg/l
Css,min = (0.8)(200)/35 = 3.4 mg/L

28. Rationale for controlled release dosage forms
Compliance vs. fluctuation: when the dosing interval is less than 8 h, compliance drops.
For short half-life drugs, either  must be small (2, 3, 4, 6 h), or the fluctuation must be quite large, when conventional dosage forms are used.
Use of controlled release permits long  while maintaining low fluctuation.
Not generally of value for drugs with long half lives (> 12 h). Due to extra expense, they should not be recommended.

29. Assessment of PK parameters
CL:
CL/F = (DM/)/Css,av and Css,av = AUCss,/
Relative F:
CLR:
CLR = (Ae,ss/ x Css,av) where Ae,ss is the amount of drug excreted in the urine over one .