Multiple-Dosage Regimens

m
After single-dose drug

.co
ot
administration, the plasma drug

sp
og
l
.b
level rises above and then falls

ist
ac
below the minimum effective
r m
ha
P
concentration (MEC), resulting in
ch
en
stB

a decline in therapeutic effect.

La

 To maintain prolonged therapeutic activity, many

drugs are given in a multiple-dosage regimen.
Multiple-Dosage Regimens
 The plasma levels of drugs given in multiple doses

m
must be maintained within the narrow limits of

.co
ot
the therapeutic window ( CP above the MEC and

sp
og
l
below the MTC) to achieve optimal clinical

.b
ist
ac
effectiveness.
r m
ha
P
ch
en
stB

 Dosage regimen is established for drug to provide

La

the correct plasma level without excessive

fluctuation and drug accumulation outside the
therapeutic window.
 MSC = Maximum Safety Concentration
MEC = Minimum Effective Concentration

TOXICITY

m
.co
ot
sp
og
l
.b
ist
MSC

ac
Cp

m
(mg/ml)

r
ha
THERAPEUTIC WINDOW
SUCCESS P
ch
MEC
en
stB
La

FAILURE

Time (h)
 Multiple-Dosage Regimens
Criteria for optimum dosage regimen:

m
.co
ot
The plasma levels of drug given must be

sp
I.

og
l
maintained within the therapeutic window.

.b
ist
ac
 Ex. The therapeutic range of theophylline is 10-20µg/L.

r m
ha
P
So, the best is to maintain the CP around 15µg/L.
ch
en
stB
La

II. Should be convenient to the patient

 It is difficult to take I.V. injection every ½ hour or one
tablet every 2 hour, this lead to poor compliance.
Multiple-Dosage Regimens
Factors affecting the design of dosage regimen:

m
.co
ot
sp
(1) The size of the drug dose.

og
l
.b
ist
(2) The frequency of drug administration (τ) (i.e.,

ac
r m
ha
the time interval between doses).
ch
P
en
stB
La
Superposition Principle
 The superposition principle can be used when all the
PK processes are linear.

m
.co
ot
sp
og
l
 That is when distribution, metabolism, and excretion

.b
ist
ac
(DME) processes are linear or first order.

r m
ha
P
ch
en
stB

 Thus, concentrations after multiple doses can be

La

calculated by adding together the concentrations from

each dose. Also, doubling the dose will result in the
concentrations at each time doubling.
 principle of superposition

m
.co
ot
sp
og
 The basic assumptions are

l
.b
ist
ac
r m
 (1) that the drug is eliminated by first-order kinetics
ha
P
ch
and
en
stB
La

 (2) that the pharmacokinetics of the drug after a single

dose (first dose) are not altered after taking multiple
doses.
 Drug Accumulation
 When a drug is administered at regular intervals

m
over a prolonged period, the rise and fall of drug

.co
ot
concentration in blood will be determined by the

sp
og
l
relationship between the half-life of elimination

.b
ist
ac
and the time interval between doses.
r m
ha
P
ch
en
stB

 If the drug amount administered in each dose has

La

been eliminated before the next dose is applied,

repeated intake at constant intervals will result in
similar plasma levels.
 Drug Accumulation
 If intake occurs before the preceding dose has been

m
eliminated completely, the next dose will add on to

.co
ot
the residual amount still present in the body, i.e., the

sp
og
l
drug accumulates.

.b
ist
ac
r m
ha
 The shorter the dosing interval relative to the
P
ch
en

elimination half-life, the larger will be the residual

stB
La

amount of drug to which the next dose is added and

the more extensively will the drug accumulate in the
body.
 Drug Accumulation
 At a given dosing frequency, the drug does not

m
accumulate infinitely and a steady state (Css) or

.co
ot
accumulation equilibrium is eventually

sp
og
l
reached.

.b
ist
ac
r m
ha
P
ch
 This is so because the activity of elimination
en
stB

processes is concentration dependent.

La

 The higher the drug concentration rises, the

greater is the amount eliminated per unit of time.
La
stB
en
ch
Pha
rm
ac
ist
.b
log
sp
ot
.co
m
 Drug Accumulation
if two successive doses are omitted, the plasma level will
drop below the therapeutic range and a longer period

m
will be required to regain the desired plasma level.

.co
ot
sp
og
l
.b
ist
ac
r m
ha
P
ch
en
stB
La
 Drug Accumulation
 When the daily dose is given in several divided doses,
the mean plasma level shows little fluctuation.

m
.co
ot
sp
og
l
.b
ist
ac
r m
ha
P
ch
en
stB
La
 Drug Accumulation
 The time required to reach steady state accumulation
during multiple constant dosing depends on the rate

m
.co
of elimination.

ot
sp
og
l
.b
ist
 The steady-state plasma level declines to a new value

ac
r m
corresponding to the new rate of elimination.
ha
P
ch
en
stB

 When elimination is impaired (e.g., in progressive

La

renal insufficiency), the mean plasma level of renally

eliminated drugs rises and may enter a toxic
concentration range.
La
stB
en
ch
Pha
rm
ac
ist
.b
log
sp
ot
.co
m
 Drug Accumulation
 The 2nd dose is taken before the 1st dose is
eliminated.

m
.co
ot
sp
 Subsequent doses are then taken following the

og
l
.b
ist
identical interval (.

ac
r m
ha
P
ch
 Accumulation of the drug up to a steady-state is
en
stB

seen where drug intake = drug elimination.

La

 If dose (D) and  are properly selected, blood levels

will rise and fall between peak (Cmax) and (Cmin)
within the therapeutic range (TR):
Drug Accumulation

m
.co
ot
sp
og
l
.b
ist
ac
r m
ha
P
ch
en
stB
La

Dose: 100 mg every t1/2

Repetitive Intravenous Injections
 The maximum amount of drug in the body following
a single rapid IV injection is equal to the dose of the

m
.co
drug.

ot
sp
og
l
 For a one-compartment open model, the drug will be

.b
ist
ac
eliminated according to first-order kinetics.
r m
ha
P  kt
DB  D0e
ch
en
stB
La

 If τ is equal to the dosage interval, then the amount

of drug remaining in the body after several hours can
be determined with:  k
DB  D0e
Repetitive Intravenous Injections
 The fraction (f) of the dose remaining in the body

m
is related to the elimination constant (k) and the

.co
ot
dosage interval (τ) as follows:

sp
og
l
.b
ist
DB
 e  k

ac
f 

r m
ha
D0 ch
P
en
stB
La

 If τ is large, f will be smaller because DB (the

amount of drug remaining in the body) is smaller.
Repetitive Intravenous Injections
 If the dose D0= 100mg is given by rapid injection

m
every t1/2 , then τ= t1/2 .

.co
ot
 After the 1st t1/2 the DB= 50mg.

sp
og
l
.b
 After the 2nd dose the DB= 150mg.

ist
ac
m
 After the 2nd t1/2 the DB= 75mg.

r
ha
P
 After the 3rd t1/2 the DB= 175mg,
ch
en
stB

and so on.
La

you will reach equilibrium at which:

Dmax =200mg and Dmin =100mg
Note that: D  D  D0
ss
max
ss
min
 mg Drug in Body
after Each Dose

m
.co
ot
sp
og
mg Drug in Body

l
.b
after each t1/2 as

ist
ac
well as Mg Drug

r m
ha
Eliminated per t1/2
P
ch
en
stB
La

Thus, the absolute amount of drug eliminated per

unit time increases with increasing body load, which
is exactly what characterizes a 1st order process.
This is the sole reason why a steady-state is
reached, i.e. Drug Going Out = Drug Coming In.
Example
 A patient receives 1000 mg every 6 hours by

m
.co
repetitive IV injection of an antibiotic with an

ot
sp
elimination half-life of 3 hours. Assume the drug is

og
l
.b
ist
distributed according to a one-compartment model

ac
m
and the volume of distribution is 20 L.
r
ha
P
ch
en
stB

Find the maximum and minimum amount of drug in

La

a.
the body.
b. Determine the maximum and minimum plasma
concentration of the drug.
 solution
 We must first find the value of k from the t ½ .
0.693 0.693
k   0.231hr 1
t1/ 2 3

m
.co
 The time interval(τ) is equal to 6 hours:

ot
sp
og
l
f  e  ( 0.231)(6)  0.25

.b
ist
ac
r m
ha
P
 In this example, 1000 mg of drug is given intravenously, so the
ch
en

amount of drug in the body is immediately increased by 1000

stB
La

mg.

 At the end of the dosage interval (ie, before the next dose), the
amount of drug remaining in the body is 25% of the amount of
drug present just after the previous dose, because f = 0.25.
solution
 if the value of f is known, a table can be
constructed relating the fraction of the dose in the

m
body before and after rapid IV injection.

.co
ot
sp
og
l
.b
ist
ac
r m
ha
P
ch
en
stB
La
Repetitive Intravenous Injections
 The maximum amount of drug in the body is 1333 mg.
 The minimum amount of drug in the body is 333 mg.

m
.co
 The difference between the maximum and minimum values,

ot
sp
D0, will always equal the injected dose.

og
l
.b
ist
Dmax  Dmin  D0

ac
r m
1333  333  1000 mg
ha
P
ch
en


 Dmax
stB

can also be calculated directly by:

La

 D0
Dmax 
1 f
 1000
Dmax   1333 mg
1  0.25

Dmin  1333  1000  333mg
Repetitive Intravenous Injections
 The average amount of drug in the body at steady state:
 FD0
D 

m
k
av

.co
ot
sp
FD01.44t1/ 2

og

Dav 

l
.b


ist
ac
 Any equations can be used for repetitive dosing at constant

r m
ha
time intervals and for any route of administration as long as
P
ch
en

elimination occurs from the central compartment

stB

(1)(1000 )(1.44)(3)
La


D 
av  720 mg
6
 Since the drug in the body declines exponentially (ie, first-
order drug elimination), the value D∞av is not the arithmetic
mean of D ∞max and D ∞min.
Repetitive Intravenous Injections
 To determine the concentration of drug in the body
after multiple doses:

m
.co
ot

C Po

sp
 Dmax 
 

og
Cmax 0r Cmax

l
1  e  k

.b
VD

ist
ac
m

C Po e  kt
r
Dmin
ha
 
Cmin  0r Cmin
P 
ch
VD 1  e  k
en
stB


La

 Dav  FD0
C  0r C 
VD k
av av
VD
 For this example, the values for C ∞max, C ∞min, and
C ∞av are 66.7, 16.7, and 36.1 μg/mL, respectively.
Repetitive Intravenous Injections
 The C ∞av is equal to the AUC for a dosage interval

m
at steady state divided by the dosage interval .

.co
ot
AUC tt

sp
2

og



l
Cav 1

.b


ist
ac
rm
AUC  t2
ha

P
FD0

FD0
ch
t1
en

ClT kVD
stB
La
Repetitive Intravenous Injections
 To know the plasma drug concentration at any time

m
after the administration of n doses of drug:

.co
ot
 nk

D0 1  e  kt

sp
Where: n is the number of doses given.

og
CP   e

l
.b
t is the time after the nth dose.

ist
 k
VD  1  e
ac

m
r
Pha
ch
en

 At steady state, e–nk approaches zero and equation

stB
La

reduces to:
D0  1   kt
C    k e
VD  1  e
P

where CP∞ is the steady-state drug concentration at time t after the dose.
Example
 The patient in the previous example received 1000 mg of
an antibiotic every 6 hours by repetitive IV injection.

m
The drug has an apparent volume of distribution of 20 L

.co
ot
and elimination half-life of 3 hours. Calculate:

sp
og
l
.b
ist
ac
a) The plasma drug concentration Cp at 3 hours after the

r m
ha
second dose. ch
P
en

b) The steady-state plasma drug concentration C∞p at 3

stB
La

hours after the last dose

c) C∞max
d) C∞min
e) CSS.
 Solution
a. The Cp at 3 hours after the second dose ( n = 2, t = 3 hrs)

m
.co
1000  1  e  ( 2)(0.231)(6)  ( 0.231)(3)

ot
sp
CP   e
 ( 0.231)(6 ) 
 31.3mg / l

og
20  1  e

l


.b
ist
ac
m
r
ha
b. The C∞p at 3 hours after the last dose. Because steady
ch
P
state is reached:
en
stB

1000   ( 0.231)(3)
La

 1
CP    ( 0.231)(6 ) 
e  33.3mg / l
20  1  e 

 1000 / 20
c. The C∞max is: C   ( 0.231)(6 )
 66.7mg / l
1 e
max
Solution
d. The C∞min may be estimated as the drug concentration

m
.co
after the dosage interval , or just before the next dose.

ot
sp
   kt  ( 0.231)(6)

og
C C e  66.7e  16.7mg / l

l
.b
ist
min max

ac
r m
e. The CSS is estimated by: ha
P
ch
en
stB

1000
  36.1mg / l
La

CSS
(0.231)( 20)(6)
Because the drug is given by IV bolus injections, F = 1

C∞av is represented as CSS in some references.

Multiple Oral Dose Administration
 To start the plasma concentration achieved
following a single oral dose can be given by:

m
.co
ot
sp
Fka D0
 

og
 kt kat
CP  e e

l
.b
VD ka  k 

ist
ac
r m
ha
P
ch
 This can be converted to an equation describing
en
stB

plasma concentration at any time following:

La

Fka D0  1  e   nk
 1 e  nka
 kat 
CP    e  kt
   e 
VD k a  k    1  e 
 k
 1  e  k a



Multiple Oral Dose Administration
 The plasma concentration versus time curve described by
this equation is similar to the IV curve in that there is
accumulation of the drug in the body to some plateau

m
.co
level and the plasma concentrations fluctuate between a

ot
sp
minimum and a maximum value.

og
l
.b
ist
ac
m

for Multiple Oral Doses

Plot of Cp Versus Time
r
ha
P
ch
en
stB
La
Multiple Oral Dose Administration
 The Cpmax value could be calculated at the time t =

m
tpeak after many doses, but it is complicated by the

.co
ot
need to determine the value for tpeak.

sp
og
l
.b
ist
ac
m
 However Cpmin can be more easily determined:
r
ha
P
ch
en
stB
La

 k a FD0  1   k
C   e
VD k a  k   1  e  k
min

Multiple Oral Dose Administration
 if we assume that the subsequent doses are given after
the plasma concentration has peaked and eka•τ is close

m
to zero. That is the next dose is given after the

.co
ot
sp
absorption phase is complete.

og
l
.b
ist
ac
Plot Cp Versus Time after a Single Dose

rm
showing Possible Time of Second Dose
Pha
ch
en

 Cpmin then becomes:

stB
La

 k
Fka D0  e 
C P min    k 
VD k a  k  1  e 
Multiple Oral Dose Administration
 if we assume that ka >> k then (ka - k) is approximately
equal to ka and ka/(ka - k) is approximately equal to one.

m
 k
 

.co
FD0 e


ot
CP min   k 

sp
og
VD 1  e 

l
.b
ist
ac
m
 Equation above is an even more extreme simplification.
r
ha
P
ch
 It can be very useful if we don't know the ka value but we
en
stB

can assume that absorption is reasonably fast.

La

 It will tend to give concentrations that are lower than those

obtained with the full equation (previous eq.). Thus any
estimated fluctuation between Cpmin and Cpmax will be
overestimated using the simplified equation.
 Multiple Oral Dose
Administration

 Dav  FD0
C  
 FD0  1   kt P av 0r Cav
VD k
C    k 
e VD

1 e
max

m
VD 

.co
ot
sp
k a FD0    k

og
 1

l
   k 

.b
Cmin e
VD k a  k   1  e

ist
ac

r m
ha
P
ch
2.3 ka
en

t max  log
stB

ka  k
La

tp 
1 
k a 1  e  k 
ka  k
ln
k 1 e 
 k a

Multiple Oral Dose Administration
 C∞av ( ) is the average plasma concentration during the
dosing interval at steady state.

m
.co
ot
sp
 This term is defined as the area under the plasma

og
l
.b
concentration versus time curve during the dosing interval at

ist
ac
steady state divided by the dosing interval.

r m
ha
P
ch
en

Plot of Cp versus Time after

stB

Multiple Oral Administration

La

showing AUC at Steady State

Multiple Oral Dose Administration

AUC tt 2

Cav  1

m
.co
ot
AUC 

sp
FD0 FD0
 
t2

og
l
.b
t1
ClT kVD

ist
ac
m
FD0
r
ha

C  P
kVD
ch
av
en
stB
La

 Note that the AUC during one dosing interval at

steady state is the same as the AUC from zero to
infinity after one single dose.
Loading Dose
 Since extravascular doses require time for absorption into
the plasma to occur, therapeutic effects are delayed until

m
.co
sufficient plasma concentrations are achieved.

ot
sp
og
l
.b
ist
 To reduce the onset time of the drug a loading or initial dose

ac
m
of drug is given.
r
ha
P
ch
en
stB

 The main objective of the loading dose is to achieve desired

La

plasma concentrations, C∞av, as quickly as possible.

 A maintenance dose is given to maintain C∞av and steady

state so that the therapeutic effect is also maintained.
 Single dose –
Loading dose
7

6
Therapeutic
Plasma Concentration

5 level

m
.co
ot
sp
4

og
l
.b
ist
3

ac
m
Repeated doses –
r
2 ha
P Maintenance dose
ch
en
stB

1
La

0
0 5 10 15 20 25 30

Time
Loading Dose
 The time required for the drug to accumulate to a
steady-state plasma level is dependent mainly on its

m
.co
elimination half-life.

ot
sp
og
l
.b
ist
 The time needed to reach 90% of C∞av is

ac
m
approximately 3.3 half-lives, and the time required to
r
ha
P
ch
reach 99% of C∞av is equal to approximately 6.6 half-
en
stB

lives.
La

 For a drug with a half-life of 4 hours, it will take

approximately 13 and 26 hours to reach 90% and 99%
of C ∞ av, respectively.
La
stB
en
ch
Pha
rm
ac
ist
.b
log
sp
ot
.co
m
 Loading doses

Cp

m
.co
ot
sp
Maintenance doses

og
l
.b
ist
ac
r m
ha
P
ch
en
stB
La

time

e.g. Tetracycline t1/2 = 8 hours

500mg loading dose followed by 250mg every 8 hours

 How long does it take to reach steady-state?

This depends entirely on the t1/2 of the drug:

Half-lives of the drug % of steady-state

1 50
2 75

m
.co
3 88

ot
4 93 PRACTICAL STEADY-STATE

sp
og
7 99

l
.b
ist
Thus, it takes 4 t1/2s for all drugs!!

ac
r m
ha
Ex. t1/2(hr)
P time to steady-state (hr)
ch
en
stB

antibiotics 1 4
La

acetaminophen 2 8
propranolol 4 16
carbamazepine 15 60 or - 3 days
digoxin, diazepam 40 160 or - 1 week
phenobarbital 100 400 or - 2 weeks
chloroquine 9 days 36 days or - 1 month
Multiple Oral Dose Administration
 The relationship between loading dose and
maintenance dose and thus drug accumulation

m
.co
during multiple dose administration can be

ot
sp
studied by looking at the ratio between the

og
l
.b
ist
minimum concentration at steady state and the

ac
r m
concentration one dosing interval, τ, after the first
ha
P
ch
dose. [Assuming e-ka • τ is close to zero].
en
stB
La

 k
Fka D0  e 

VD k a  k  1  e  k 
C P min 1 1
  
C P1 Fka D0  k
e 1  e  k
1  f
VD k a  k 
Multiple Oral Dose Administration
 This turns out to be the same equation as for the
IV bolus. Therefore we can calculate a loading dose

m
.co
just as we did for an IV multiple dose regimen.

ot
sp
og
l
.b
Maintenanc e Dose

ist
Loading Dose 

ac
1 f
r m
ha
P
ch
en
stB

 This equation holds if each dose is given after

La

the absorption phase of the previous dose is

complete.
 Example
 With F = 1.0; V = 30 liter; t1/2 = 6 hours or kel = 0.693/6 =
0.116 hr-1, calculate the dose given every 12 hours that will

m
.co
achieve an average plasma concentration of 15 mg/L.

ot
sp
og
l
.b
ist
Solution
ac
r m
ha
P
ch
en

FD0
stB


C 
La

kVD
av


Cav kVD 15  0.116  30  12
D0    624 mg
F 1.0
Multiple Oral Dose Administration
 We could now calculate the loading dose

f  e  k  e  ( 0.166)(12)  0.25

m
.co
ot
sp
og
l
Maintenanc e Dose 624

.b
Loading Dose    832 mg

ist
ac
1 f 1  0.25
r m
ha
P
ch
en
stB
La
 To get some idea of the fluctuations in plasma
concentration we could calculate the Cpmin value.
 Assuming that ka >> k and that e-kaτ --> 0, using
Equation:

m
.co
 k
FD0  e  1.0  624  0.25 

ot
    6.93mg / L

sp
CP min   k  

og
1  e  1  0.25 

l
VD 30

.b
ist
ac
r m
ha
Therefore the plasmach
P
concentration would probably
en
stB

fluctuate between 7 and 23 mg/L

La

(very approximate) with an

average concentration of about 15
mg/L.
 Multiple Oral Dose Administration
 As an alternative we could give half the dose, 312
mg, every 6 hours to achieve:

m
 e  k  1.0  312  0.5 

.co
FD0
   

ot
CP min   k   10 .4 mg / L

sp
   

og
VD 1 e  30 1 0.5

l
.b
ist
ac
r m
 The C∞av would be the same
ha
P
ch

1 312
en

FD0
stB


Cav    15mg / L
La

kVD 0.116  30  6
 Thus the plasma concentration would fluctuate
between about 10.4 to 20 with an average of 15
mg/L.
Example: Superposition Principle

 Calculate drug concentration at 24 hours after the first

dose of 200 mg. The second dose of 300 mg was given

m
.co
at 6 hours and the third dose of 100 mg at 18 hours.

ot
sp
og
The apparent volume of distribution is 15 L and the

l
.b
ist
elimination rate constant is 0.15 hr-1.

ac
r m
ha
P
ch
en
stB
La
Superposition Principle

m
.co
ot
sp
og
l
.b
ist
ac
r m
ha
P
ch
en
stB
La

 This method involved calculating the contribution

from each dose at 24 hours after the first dose.
Superposition Principle
 This result is shown graphically in in the following
figure:

m
.co
ot
sp
og
l
.b
ist
ac
r m
ha
P
ch
en
stB
La

Drug Concentration after Three IV Bolus Doses

Non-uniform dosing intervals
 The calculations we have looked at consider that the

m
.co
dosing intervals are quite uniform, however,

ot
sp
commonly this ideal situation is not adhered to

og
l
.b
ist
completely.

ac
r m
ha
P
ch
 Dosing three times a day may be interpreted as with
en
stB

meals, the plasma concentration may then look like

La

the plot in Figure:

 Cp versus Time during Dosing at
8 am, 1 pm, and 7 pm

m
.co
ot
sp
og
l
.b
ist
ac
r m
ha
P
ch
en
stB

• The ratio between Cpmax and Cpmin is seven fold (8.2/1.1 = 7.45) in
La

this example.

• This regimen may be acceptable if :

1) The drug has a wide therapeutic index
2) There is no therapeutic disadvantage to low overnight plasma
concentrations, e.g., analgesic of patient stays asleep.
Homework # 5
 An adult male patient (46 years old, 81 kg) was given
orally 250 mg of tetracycline hydrochloride every 8 hours
for 2 weeks. From the literature, tetracycline

m
.co
hydrochloride is about 75% bioavailable and has an

ot
sp
apparent volume of distribution of 1.5 L/kg. The

og
l
.b
elimination half-life is about 10 hours. The absorption rate

ist
ac
constant is 0.9 hr–1. From this information, calculate:
r m
ha
P
ch
en
stB

a) Cmax after the first dose.

La

b) Cmin after the first dose.

c) Plasma drug concentration Cp at 4 hours after the 7th dose.
d) Maximum plasma drug concentration at steady-state C∞max.
e) Minimum plasma drug concentration at steady-state C∞min.
f) Average plasma drug concentration at steady-state C∞av.