Dosing Regimen DesignMultiple Dosing

Myrna Y. Munar, Pharm.D., BCPSAssociate Professor of Pharmacy

ObjectivesThrough the preparation for and participation in this lecture, a successful student should be able to:

Use the Multiple Dosing Function to convert a single dose equation to a multiple dose equation.Use the accumulation index to convert a single dose equation to a steady-state equation and predict the extent of drug accumulation for a given dosage regimen.List the factors that determine if a drug concentration will stay within a given therapeutic target window.Develop and alter a dosage regimen under given conditions.

Multiple Dosing

Most drugs are given more than once ∴ we need to be able to understand and predict what will happen when a dose of drug is given repeatedly at constant intervals

Bolus Dosing

0

20

40

60

80

100

120

140

0 20 40 60 80 100 120 140

Time (hrs)

Con

cent

ratio

n (m

g/L)

Principle of SuperpositionLo

g Pl

asm

a D

rug

Conc

Time

Cmax1

Cmax2

Cmin2

Cmin1 Cmin1

Cmax1

+

Cmax2

Etc for“n” doses

Tau

Recall,

After “n” number of doses, a pattern emerges and the “n”numbers of equations can be simplified to….

...)1(1max2max

1max1max2max

1max1min

1min1max2max

etceCC

eCCC

eCC

CCC

kTau

kTau

kTau

−

−

−

+=

+=

=

+=

Before Steady State: Multiple Dosing Function (MDF)

MDF is used to convert a single dose equation to a multiple dose equation

MDFee

n k

k=−−

− ⋅ ⋅

− ⋅

11

τ

τ

Bolus Dosing

0

20

40

60

80

100

120

140

0 20 40 60 80 100 120 140

Time (hrs)

Con

cent

ratio

n (m

g/L)

Using MDF

Fast Forward from first dose to any dose after that

Concentration after single dose X MDF = Concentration at that time at “n” dosesEquation for single dose x MDF = Equation at “n” doses

Recall

e-kt = fraction of drug remaining at time “t”∴ e-kτ = fraction of drug remaining at the end of the dosing interval1-e-kt = fraction of drug removed at time “t”∴ 1-e-kτ = fraction of drug removed at the end of the dosing interval

Review: Cp=Coe-kt

m=ΔY / Δ Xm=Y2-Y1/X2-X1

Y=mX+b

-k= Δ lnC / Δ t-k=lnC2-lnC1/t2-t1lnCp=-kt+Co

Antilog:Cp=Coe-kt

X axis Time

Y axisLogC

b=y-intercept Co=y-intercept

m=slope -k=slope

Using MDF

If given by instantaneous input the highest concentration occurs at t=0.

ktt eCC −= 0

DVDoseFC ⋅

=∴ 0

ktk

kn

Dt e

ee

VDoseFC −

⋅−

⋅⋅−

⎟⎟⎠

⎞⎜⎜⎝

⎛−−⋅

= τ

τ

11

At “n” doses

CF Dose

Vee

etD

n k

kkt=

⋅ −−

⎛⎝⎜

⎞⎠⎟

− ⋅ ⋅

− ⋅−1

1

τ

τ

Multiple Dosing Function

Bolus Dosing

0

20

40

60

80

100

120

140

0 20 40 60 80 100 120 140

Time (hrs)

Con

cent

ratio

n (m

g/L)

SS is reached

Using MDF to Determine Steady State Concentrations

⎟⎟⎠

⎞⎜⎜⎝

⎛−−

= ⋅−

⋅⋅−

τ

τ

k

kn

eeMDF

11

When n=large number e-nkτ approaches 0

ττ kkn eeMDF −⋅−=⇑ −

≅⎟⎠⎞

⎜⎝⎛

−−

≅1

11

01

At Steady State= Accumulation Index/Factor

Reac k=

− −

11 τ

Relates the concentrations in the dosing interval at steady state to the values after a single dose

Accumulation

used to “fast forward” from first dose concentrations to steady-state concentrations

Using Rac

Fast Forward to steady-stateConcentration for single dose X Rac = Concentration at steady-stateEquation for single dose x Rac = Equation at Steady-State

Determinants of Accumulation

Frequency of Administration (τ) relative toHalf-life (t1/2)

t12

τ OR

1kτ

ExampleGiven t1/2 = 6 hrs & Tau = 6 hrs, what happens when:

Tau=t1/2

Tau>t1/2; eg. Tau is 5 times greater than t1/2

Tau<t1/2; eg. Tau is 1/3 of t1/2

Calculate Rac

Reac k=

− −

11 τ

ExcelApplication of Accumulation Index (Rac)

Prediction t1/2 (hrs) Tau (hrs) k RacTau = t1/2 6 6 0.1155 2.000294Tau 5 x > t1/2 6 30 0.1155 1.032283Tau 1/3 t1/2 6 2 0.1155 4.848237

τkac eR

tk

kt

−−=

==

11

693.0;693.0

2/12/1

Application of RacPrediction: τ relative to t1/2

When τ=t1/2: 2When τ>t1/2: No or less accumulationWhen τ<t1/2: More accumulation

C

tTau

C

tTau

Example: t1/2 =6 hours

If τ is 6 hours Rac=2, CSS 2x first doseIf τ is 30 hrs(~ >5 times t1/2) Rac →1 no accumulation; all of the drug will be eliminated from the first dose before the second is given.If τ is 2 hours (1/3 t1/2) the Rac would be 4.8. Conc, CSS ~ 5x first dose

Application of RacCalculation of Loading Dose

LD= MD*Rac

τ=t1/2 Rac=2 ∴ LD=MD x2τ>t1/2 Rac→1 ∴ LD= MDτ<t1/2 Rac→large number ∴large LD

Example – Drug X

Loading Dose (LD) can be calculated using Rac, where LD = MD*Rac

Drug X with t1/2 < Taut1/2 = 1 hrTau = 8 hrsTo achieve target Css, we must give MD = 750 mg/day OR 250 mg q 8 hours

Reac k=

− −

11 τ

Drug X: t1/2<τ

· MD 750 mg/day OR 250 mg q 8 hrsτ=8 hourst1/2 = 1 hr

onaccumulatiNo

0.11

1R

11R

8*693.0ac

ac

∴

=−

=

−=

−

−

e

e kτ

Drug X

Time to steady-state ~ 5 hoursNo need for loading dose LD= MD*Rac=250 mg (normal MD)

Example – Drug Y

Loading Dose (LD) can be calculated using Rac, where LD = MD*Rac

Drug Y with t1/2 = Taut1/2 = 8 hrTau = 8 hrsTo achieve target Css, we must give MD = 750 mg/day OR 250 mg q 8 hours

Reac k=

− −

11 τ

Drug Y: t1/2=τ

· MD 750 mg/day or 250 mg q 8 hrsτ=8 hourst1/2 = 8 hr

0.21

1R

11R

8*087.0ac

ac

=−

=

−=

−

−

e

e kτ

Drug Y

Time to steady-state ~ 40 hoursLD= MD*Rac=500 mg

Example – Drug ZLoading Dose (LD) can be calculated using Rac, where LD = MD*Rac

Drug Z with t1/2 > Taut1/2 = 24 hrTau = 8 hrsTo achieve target Css, we must give MD = 750 mg/day OR 250 mg q 8 hours

Reac k=

− −

11 τ

Drug Z: t1/2>τ

· MD 750 mg/day or 250 mg q 8 hrsτ=24 hourst1/2 = 8 hr

8.41

1R

11R

8*029.0ac

ac

=−

=

−=

−

−

e

e kτ

Drug Z

Time to steady-state ~ 5 days

LD= MD*Rac=1200 mg

ExcelCalculation of LD Time to SS in hrs and days:Prediction t1/2 (hrs) Tau (hrs) k Rac 3*t1/2 (hrs) 5*t1/2 (hrs) MD (mg) LD (mg)t1/2 < Tau 1 8 0.693 1.003926 3 5 250 250.9816t1/2 = Tau 8 8 0.086625 2.000294 24 40 250 500.0736t1/2 > Tau 24 8 0.028875 4.848237 72 120 250 1212.059

ac

kac

RMDLDe

R

tk

kt

*1

1

693.0;693.0

2/12/1

=−

=

==

− τ

Maintenance of Concentrations within TR

CssCss e k

,max,min

= −

1τ

•Css,avg

•Fluctuation within a dosing interval (τ) determined by:

−τ

-t1/2

-rate of drug input

•Peak to Trough Ratio (P:T)

Peak to Trough Ratio: Cmax

⎟⎠⎞

⎜⎝⎛

−⋅

=

−

⋅=

− τke11

VDoseFC

statesteadyat Therefore,VDoseFC

Dmaxss,

Dmax

Peak to Trough Ratio: Cmin

τke−⋅= maxCss,minCss,

Peak to Trough Ratio

τ

ττ

τ

-k

kk

D

kD

e

eeV

DoseFeV

DoseF

11

11

1

minCss,maxCss,

=

⋅⎟⎠⎞

⎜⎝⎛

−⋅

⎟⎠⎞

⎜⎝⎛

−⋅

=−

−

−

Peak to trough ratio depends upon t1/2 and dose interval

Peak to Trough Ratio

• τ=t1/2 P:T =2• τ>t1/2 P:T will be large ∴large

fluctuations• τ<t1/2 P:T will be small, as τ →0 there

is no fluctuation like constant infusion

Example: t1/2 =6 hours

• τ=t1/2, eg. τ =6 hrs, P:T=2, Css,max will be twice Css,min

• τ>t1/2, eg. τ =24 hrs, P:T=16, Css,max will be 16x Css,min

• τ<t1/2, eg. τ =2 hrs, P:T=1.3 , Css,max will be 1/3 larger than Css,min

τkeTratioP −=

1:

Rate of Drug Input

PO administration results in a slower rate of input Peak conc occur later and are smaller P:T is also smallerEquations become complex, but more accurate

Average Concentration at Steady-State

CAUC

ss avgss

,( )= −0 τ

τ

time

lnco

nc

Css,avg

tautau

Determinants of Css,avg

CF Dose

V eess t

Dk

kt( ) =

⋅−

⎛⎝⎜

⎞⎠⎟−

−11 τ

C AUCF Dose

V eess t ss

Dk

kt( ) ,( )

00

0

11

τ

τ τ

τ

∫ ∫= =⋅

−⎛⎝⎜

⎞⎠⎟− −

−

AUCF Dose

V kF DoseV k

F DoseCLss

D D( )0

1− =

⋅ ⎛⎝⎜

⎞⎠⎟ =

⋅⋅

=⋅

τ

Determinants of Css,avg

CAUC

F DoseCL

FDose

CLss avgss

,( )= =

⋅

=⋅

−0 τ

τ ττ

•F

•Dose/τ

•CL

Dosage Regimen Design

Dose Rate (Dose/τ) Rate of drug administered over timeDaily Dose (ie. 1 G Q8, Dose/τ=3 G/d)

Dose Interval (τ)How often givenHow small the pieces are

Examples

Dose/τ τ Regimen1 G/day 6 hours 250 mg Q61 G/day 12 hours 500 mg Q122 G/day 6 hours 500 mg Q62 G/day 12 hours 1 G Q12

When to alter Dose/τ

Is Css,avg in acceptable range?

Has there been a change inFCL

The only thing we can control is Dose/τ

Out DrugIn Drug

CL

DoseFavgCss, =

⋅= τ

Determinants of P:T

CssCss e k

,max,min

= −

1τ k

CLVD

=

•Half-life; k

•CL

•V

•The only thing we control is τ

When to alter τ

CssCss e k

,max,min

= −

1τ

•Is P:T clinically acceptable?

kCLVD

=

•Has there been a change in

-CL

-V

•The only thing we control is τ

Examples

Increase Dose Rate

1

10

0 2 4 6 8

Time (hrs)

Con

cent

ratio

n (m

g/L)

250mg Q6500mg Q6

Css,min=3.0

Css,min=1.5

Css,max=5.6

Css,max=2.8

Css,avg=2.1

Css,avg=4.2

P:T=1.9

P:T=1.9

Determinants of Css,avg

CAUC

F DoseCL

FDose

CLss avgss

,( )= =

⋅

=⋅

−0 τ

τ ττ

•F

•Dose/τ; increase Dose/τ, increase Css,avg

•CL

Increase Dose Interval

0 2 4 6 8 10 12 14Time (hrs)

Con

cent

ratio

n (m

g/L)

250mg Q6

500mgQ12

Css,max=3.6

Css,max=2.77

Css,,min=1.52

Css,min=1.08

Css,avg=2.1

P:T=3.3

P:T=1.9

Determinants of P:TCssCss e k

,max,min

= −

1τ k

CLVD

=

•CL•VD

•Only thing we can control is Tau (τ)•P:T too small, increase dose interval•i.e. give less often eg. q 6h to q 12h

Decreased Volume

1

10

0 1 2 3 4 5 6 7

Time (hrs)

Con

cent

ratio

n (m

g/L)

250mgQ6

Decreased VCss,max=3.6

Css,max=2.8

Css,min=1.5

Css,min=1.1

Css,avg=2.1

P:T=3.3

P:T=1.9

Determinants of Css,avg

CAUC

F DoseCL

FDose

CLss avgss

,( )= =

⋅

=⋅

−0 τ

τ ττ

•F

•Dose/τ

•CL

Determinants of P:T

CssCss e k

,max,min

= −

1τ

kCLVD

=

•CL•VD; decreased VD will increase k (steeperslope) and increase P:T ratio

•Tau (τ)

Decreased Clearance

1

10

0 1 2 3 4 5 6 7

Time (hrs)

Con

cent

ratio

n (m

g/L)

250mg Q6 Decreased CL

Css,max=2.8

Css,max=4.8

Css,min=3.6

Css,min=1.5

Css,avg=4.2

Css,avg=2.1

P:T=1.3

P:T=1.9

Determinants of Css,avg

CAUC

F DoseCL

FDose

CLss avgss

,( )= =

⋅

=⋅

−0 τ

τ ττ

•F

•Dose/τ

•CL; decreased CL will increase Css,avg

Determinants of P:T

CssCss e k

,max,min

= −

1τ k

CLVD

=

•CL; decreased CL, will decrease k (flatterslope), and decrease P:T ratio

•VD

•Tau (τ)

When to change Dose Rate

Is Css,avg in the target range?

Yes No change in DR

No

Is Css,avg too high or too low?

Too High

↓ DR

Too Low

↑DR

P:T fluctuation

Conc.

Time

ShorterTau

LongerTau

When to change Dose Interval

Is P:T clinically acceptable?

Yes No change in DI

No

Is P:T too great or too small?

Too Great ↓ DI

(give more often)

Too Small

↑DI (Give less often)

P:T fluctuation

Conc.

Time

ShorterTau

LongerTau

P:T ratio (fluctuation) too small?

Conc

Time

Next doseTau

Peak

Trough

P:T ratio (fluctuation) too small?

Action is to increase or lengthen the dosing interval (tau)

Result is to increase P:T ratio

Conc

Time Next dose

Increase TauPeak

Trough

P:T ratio (fluctuation) too large?

Conc

Time

Next dose

Tau

Peak

Trough

P:T ratio (fluctuation) too large?

Reflects faster ksteeper slope

Reflects shorter t1/2

Action is to decrease or shorten the dosing interval (tau)

Result is to decrease P:T ratio

Conc

Time

Next dose

Tau

Peak

Trough