dosing regimen design multiple dosing - dr ted...
Post on 22-Apr-2018
230 Views
Preview:
TRANSCRIPT
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
top related