Download C ss,av

Multiple Dosing:
Intermittent or multiple dose
regimen
100 mg q. t1/2 via i.v. bolus
Amount in Body [mg]
200
150
100
50
Time [t1/2]:
1
2
3
4
5
6
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

AN,max & AN,min
N= 1
2
3
4
 NK E
AN ,max
5
6
7
1 e
 NK E


F  Dose  Ass ,max 1  e
 K E
1 e
AN ,min  AN ,max e
 K E
Not AN,min = AN,max - F•Dose
Why?
Css,max & Css,min
C ss ,max
F  Dose

 K E

V 1  e
Css,min  Css,max e
 K E
 Css,max
F  Dose

V
Average amount of drug in the
body at steady state
At steady state, Rate In = Rate Out
FDose/ = KE Ass,av
Ass ,av
F  Dose 1.44  F  Dose  t1 / 2


K E

1/KE = t1/2/ln 2 = 1.44 t1/2
Average steady-state plasma
concentration
F  Dose

C ss,av
 CL  C s ,av
F  Dose

CL  
AUC
C ss ,av 
Equal Areas
AUC0   AUC
AUC0 

Css,av
Concept applies to all routes of
administration and it is independent of
absorption rate.
Dosing rate from AUC0-
Given AUC after a single dose, D, the maintenance
dose, DM , is:
C ss ,av
DM 
D
AUC0 
Where D produced the AUC0-, Css,av is the desired
steady-state plasma concentration, and
desired dosing interval.
 is the
Example: 50 mg p.o. dose
2 3
Cp
mg/L
2.4
1.8
0 2

AUC 0   2 1  0  1
2
 2  3 2  1  2.5

AUC1  2 
1
3  2.4 



1 2
4
8 AUC 2   2  4  2  5.4
Time [h]
2.4  1.8 
8

AUC 4   2 8  4  8.4

AUC8  Clast  K E  1.8  0.3  6
4
AUC = 1+2.5+5.4+8.4+6 = 23.3 mg•h/L
Example, continued
Calculate a dosing regimen for this drug that would
provide an average steady-state plasma
concentration of 15 mg/L.
C ss ,av
15mg / L 6h 
50mg 
DM 
D
AUC0
23.3mg  h / L
DM = 193 mg
Dosing regimen: 200 mg q. 6 h.
Example, continued - 2
There is a problem with this approach ??
Peak and trough concentrations are unknown.
Css,max
Css,min
Another example - digitoxin
t1/2 = 6 days; usual DR is 0.1 mg/day
Assuming rapid and complete absorption of digitoxin,
What would be the
average steady-state body
level?
Maximum and minimum
plateau values?
Is there accumulation of digitoxin?
How long to reach steady state?
Ass,av = 1.44 F Dose t1/2/
= (1.44)(1)(0.1mg/d)(6d)/(1d)
= 0.864 mg
Ass,max = 0.1/(1-e-(0.116)(1))
= 0.909 mg
Ass,min = 0.909 – 0.1
= 0.809 mg
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
KE
AI

KE
FI

When
 = t1/2
AI = FI = 2
Absorption Rate influence on
Rate of Accumulation
v
ka
4.0
CL
3.5
3.0
2.5
2.0
ka=0.5.pw o
1.5
ka=0.01.pw o
1.0
KE = 0.1
0.5
0.0
0
20
40
60
80
100
Time
 NK E
 Nka
AN ,max
ka e
KEe
 f ss  1 

Ass , max
ka  K E ka  K E
120
 NK E
 Nka
AN ,max
ka e
KEe
 f ss  1 

Ass ,max
ka  K E ka  K E
When ka >> KE, control is by drug t1/2:
f ss  1  e
 NKE
When ka << KE, control is by absorption t1/2:
f ss  1  e
 Nka
Loading Dose (LD) = Ass,max  F
Whether a LD is needed depends upon:
•Accumulation Index
•Therapeutic Index
•Drug t1/2
•Patient Need
Dosing Regimen Design
OBJECTIVE: Maintain Cp within the therapeutic window.
Cp
Time
Dosing Regimen Design
APPROACH: Calculate
max and DM,max.
Cu
Cp
max
Time
 max
Cl
Cl  Cu e
 K E max
Cu 

ln

Cl 
 Cu 


 1.44 t1 / 2 ln 
KE
 Cl 
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
The Dosing Rate (DR) is DM,max 
DR 
DM ,max
 max
max
Time

V F Cu  Cl  K EV F Cu  Cl 


ln Cu Cl  / K E
ln Cu Cl 
DR 
DM ,max
 max

V F Cu  Cl  K EV F Cu  Cl 


ln Cu Cl  / K E
ln Cu Cl 
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.
Cu
DR = (CL/F)Css,av
max
Cl
Average Concentration Approach
1. Choose the average to maintain:
Css,av = (Cu - Cl)/ln (Cu/Cl)
2. Choose :


max
;usually 4, 6, 8, 12, 24 h
3. Calculate DR:
DR = (CL/F)Css,av
4. Calculate DM:
DM = DR•
Example
V = 35L
KE = 0.143 h-1
t1/2 = 4.85 h
CL = 5L/h F = 0.80
Cu = 10 mg/L
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) = 36.25 mg/h
DM = (36.25 mg/h)(8 h) = 290 mg  300 mg
Dosing Regimen: 300 mg q 8 h
Peak Concentration Approach
1. Choose the peak concentration to maintain.
2. Choose :


max
;usually 4, 6, 8, 12, 24 h
3. Calculate DM:
DM = (V•Cpeak/F)(1 - e-KE)
from:
C ss ,max
F  Dose

 K E

V 1  e
Example
V = 35L
KE = 0.143 h-1
t1/2 = 4.85 h
CL = 5L/h F = 0.80
Cu = 10 mg/L
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
Check
C ss ,max
F  Dose

 K E

V 1  e
C ss ,min  C ss ,max
F  Dose

V
Css,max = [(0.8)(200)/35]/(1 – e-(0.143)(6)) = 7.93 mg/l
Css,min = 7.93 - (0.8)(200)/35 = 3.4 mg/L
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.
Assessment of PK parameters
CL:
CL/F = (DM/)/Css,av and Css,av = AUCss,/
Relative F:
  DM
FB  C ss ,av, B   

 
FA  C ss ,av, A   DM
  
 
 A 

 
 B 
CLR:
CLR = (Ae,ss/Css,av) where Ae,ss is the amount of
drug excreted in the urine over one .