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

Dosing Regimen Design

Multiple Dosing:Intermittent or multiple dose

regimen

100 mg q. t1/2 via i.v. bolus

Time [t1/2]: 1 2 3 4 5 6

Am

ou

nt

in B

od

y

[mg

]

50

100

150

200

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

E

E

ENK

ssK

NK

N eADoseFee

A

1

11

max,max,

N = 1 2 3 4 5 6 7

EKNN eAA max,min,

Not AN,min = AN,max - F•Dose Why?

Css,max & Css,min

EKss eVDoseF

C

1max,

V

DoseFCeCC ss

Kssss

E

max,max,min,

Average amount of drug in the body at steady state

At steady state, Rate In = Rate Out

FDose/ = KE

Ass,av

2/1

,

44.1 tDoseF

KDoseF

AE

avss

1/KE = t1/2/ln 2 = 1.44 t1/2

Average steady-state plasma concentration

avsCCLDoseF

,

CLDoseF

C avss,

AUC

Equal Areas

AUCAUC 0

0

,

AUCC avss

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:

DAUC

CD avssM

0

,

Where D produced the AUC0-, Css,av is the desired steady-state plasma concentration,

and is the desired dosing interval.

Example: 50 mg p.o. dose

Cp

mg/L

1 2 4 8Time [h]

2 3 2.4 1.8 1012

201

0

AUC

5.2122

322

1

AUC

4.5242

4.234

2

AUC

4.8482

8.14.28

4

AUC

63.08.18

Elast KCAUC

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.

mgLhmghLmg

DAUC

CD avssM 50

/3.236/15

0

,

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?

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 mgHow long to reach steady state?

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

KE

When = t1/2

AI = FI = 2

Absorption Rate influence on Rate of Accumulation

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

0 20 40 60 80 100 120

Time

ka=0.5.pwo

ka=0.01.pwo

v

CL

ka

KE = 0.1

Ea

NKa

Ea

NkE

ssss

N

Kk

ek

Kk

eKf

A

A Ea

1max,

max,

Ea

NKa

Ea

NkE

ssss

N

Kk

ek

Kk

eKf

A

A Ea

1max,

max,

When ka >> KE, control is by drug t1/2:

When ka << KE, control is by absorption t1/2:

ENKss ef 1

aNkss ef 1

Loading Dose (LD)

Whether a LD is needed depends upon:

•Accumulation Index

•Therapeutic Index

•Drug t1/2

•Patient Need

= Ass,max F

Dosing Regimen Design

OBJECTIVE: Maintain Cp within the therapeutic window.

Cp

Time

Dosing Regimen DesignAPPROACH: Calculate max and DM,max.

Cp

Time

Cu

Clmax

l

u

E

l

u

Kul

C

Ct

K

CC

eCC E

ln44.1ln

2/1max

max

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)

The Dosing Rate (DR) is DM,max max

lu

luE

Elu

luM

CC

CCFVK

KCC

CCFVDDR

ln/lnmax

max,

Cp

Time

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

lu

luE

Elu

luM

CC

CCFVK

KCC

CCFVDDR

ln/lnmax

max,

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 hCu = 10 mg/L

CL = 5L/h

F = 0.80 Cl = 3 mg/L

Css,av = (10 – 3)/ln (10/3) = 5.8 mg/Lmax = (1.44)(4.85)[ln (10/3)] = 8.41 h Choose < max: 8

hDR = (5 L/h)(5.8 mg/L)/(0.8) = 36.25 mg/hDM = (36.25 mg/h)(8 h) = 290 mg 300 mgDosing 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: EKss eVDoseF

C

1max,

Example

V = 35L KE = 0.143 h-1 t1/2 = 4.85 hCu = 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

Check EKss eVDoseF

C

1max,

VDoseF

CC ssss

max,min,

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:

B

M

A

M

Aavss

Bavss

A

B

D

D

C

C

F

F

,,

,,

CLR:

CLR = (Ae,ss/ x Css,av) where Ae,ss is the amount of drug excreted in the urine over one .