Multiple dosing IV bolu

1. Multiple dosing: intravenous bolus
administration

2. Multiple IV Bolus Dose
Administration
 Objectives:
 1) To understand drug accumulation
after repeated dose administration
 2) To recognize and use the
integrated equations used to
describe plasma concentrations
versus time after multiple IV. doses
 3) To calculate appropriate multiple
dose drug regimen
2

3. Multiple-Dosage Regimens:
 After single-dose drug administration, the
plasma drug level rises above and then
falls below the minimum effective
concentration (MEC)  decline in
therapeutic effect.
 To maintain prolonged therapeutic
activity, many drugs are given in a
multiple-dosage regimen.
 The plasma levels of drugs given in
multiple doses must be maintained within
the narrow limits of the therapeutic
window
3

4.  Ideally, a dosage regimen is established for
each drug to provide the correct plasma level
without excessive fluctuation and drug
accumulation outside the therapeutic
window.
 Some drugs that have a narrow therapeutic
range (eg, digoxin and phenytoin) require
definition of the therapeutic minimum and
maximum nontoxic plasma concentrations
(MEC and MTC).
4

5. Multiple dosing calculations using
Superposition
Let:
Dose 1  Conc. 1
and:
Dose 2  Conc. 2
then the response system behaves according to
the superposition principle if:
Dose 1 +Dose 2  Conc. 1 + Conc. 2
and in that case the response system is a linear
response system
5

6. Multiple dosing calculations using
Superposition
 A patient is to be given 100 mg of a
drug intravenously. Assuming that K
= 0.10 hr-1 and a V = 15 L,
estimate the following:
1. The half life
hr
hr
K
t 93
.
6
1
.
0
)
2
ln(
)
2
ln(
2
/
1 


6

7. Multiple dosing calculations using
Superposition
2. The concentration 2 hrs after the dose
3. The concentration 10 hrs after the
dose
mg/L
5.46
D
)
2
( )
2
(


 

 t
K
e
V
t
C
mg/L
2.45
D
)
10
( )
10
(


 

 t
K
e
V
t
C
7

8. Multiple dosing calculations using
Superposition
4. The concentration 18 hrs after the
dose
mg/L
10
.
1
D
)
18
( )
18
(


 

 t
K
e
V
t
C
8

9. 9
0
1
2
3
4
5
6
7
0 8 16 24 32
Conc.
(mg/L)
Time (hr)

10. Multiple dosing calculations using
Superposition
5. Assuming that 100 mg of the drug is
administered every 8 hrs, estimate
the concentration 2 hrs after the third
dose using the values calculated in
parts 2-4. What property of the linear
systems did you use to answer this
question?
10

11. 0
2
4
6
8
10
12
0 8 16 24 32
Conc.
(mg/L)
Time (hr)
Conc. After the first dose
)
(
1 t
C
11

12. 0
2
4
6
8
10
12
0 8 16 24 32
Conc.
(mg/L)
Time (hr)
Conc. After the second dose
)
(
2 t
C
12

13. 0
2
4
6
8
10
12
0 8 16 24 32
Conc.
(mg/L)
Time (hr)
Conc. After the third dose
)
(
3 t
C
13

14. 0
2
4
6
8
10
12
0 8 16 24 32
Conc.
(mg/L)
Time (hr)
Total Conc.
)
'
(
3 t
Cn
14

15. 0
2
4
6
8
10
12
0 8 16 24 32
Conc.
(mg/L)
Time (hr)
)
2
(
)
10
(
)
18
(
)
2
'
( 3
2
1
3 






 t
C
t
C
t
C
t
Cn
t = 2 hrs after third dose
= 10 hrs after second dose
= 18 hrs after first dose
15

16. Multiple dosing calculations using
Superposition
5. Assuming that 100 mg of the drug is
administered every 8 hrs, estimate the
concentration 2 hrs after the third dose
using the values calculated in parts 2-4.
What property of the linear systems did
you use to answer this question?
mg/L
01
.
9
)
2
'
(
46
.
5
45
.
2
10
.
1
)
2
'
(
)
2
(
)
10
(
)
18
(
)
2
'
(
3
3
3
2
1
3
















t
C
t
C
t
C
t
C
t
C
t
C
n
n
n
16

17. 1
7

18. 1
8

19. 1
9

20. 2
0

21. Multiple dosing calculations using
Superposition
 The principle of superposition assumes
that early doses of drug do not affect the
pharmacokinetics of subsequent doses.
 Therefore, the blood levels after the
second, third, or nth dose will overlay or
superimpose the blood level attained after
the (n – 1)th dose

22. Multiple administration every 4 hrs
Dose
Number
Time
(hr)
Dose 1 Dose 2 Dose 3 Dose 4 Total
1 0 0 0
1 21.0 21.0
3 19.8 19.8
2 4 16.9 0 16.9
5 14.3 21.0 35.3
7 10.1 19.8 29.9
3 8 8.50 16.9 0 25.4
9 7.15 14.3 21.0 42.5
11 5.06 10.1 19.8 35.0
4 12 4.25 8.50 16.9 0 29.7
13 3.58 7.15 14.3 21.0 46.0
15 2.53 5.06 10.1 19.8 37.5

23. 2
3

24. 2
4

25. Multiple IV bolus administration
 Concentration after n doses:
where r:
n: number of doses, T: dosing interval






 
 )
(
D t
K
n e
V
r
C
KT
nKT
e
e
r 




1
1

26. Multiple IV bolus administration: useful
equations
 Maximum concentration after n doses:
 Maximum concentration at steady state:







V
R
CSS
D
max







V
r
CN
D
max

27. Multiple IV bolus administration: useful
equations
 Minimum concentration after n doses:
 Minimum concentration at steady state:







 KT
e
V
R
CSS
D
min







 KT
e
V
r
CN
D
min

28. Multiple IV bolus administration
 Concentration at steady state:
where R is the accumulation ratio:
T: dosing interval






 
 )
(
D t
K
ss e
V
R
C
KT
e
R 


1
1

29. Conc time profile:

30. The AUC during a dosing interval at
steady state is equal to the total AUC
following a single dose (For linear PK)

31. AUC for a single dose is:
As explained in the previous slide,
Multiple IV bolus administration: useful
equations
 Average concentration at steady state:



 0
.dt
C
C
SS
average
SS

KVd
X
Caverage
SS
0





0
τ
0
SS dose).dt
C(single
.dt
C
KVd
X0
0
dose).dt
C(single 



32. Predicting average Css using single
dose data

33. Time to reach steady state conc.
 The time required to reach to a certain
fraction of the steady-state level is given
by:
 Time required to achieve steady-state
depends on the half-life and is
independent of the rate of dosing and the
clearance
 To get to 95% of the steady-state: 5 half-
lives are needed
 To get to 99% of the steady-state: 7 half-
lives are needed
)
1
ln(
44
.
1 5
.
0 fss
t
n 




34. Different doses regimen have the same
average steady state conc: The same dosing
rate (Dose/ T)

35. Multiple IV bolus dosing compared to
IV infusion
Multiple IV
bolus
IV infusion

36. Multiple IV bolus dosing compared to
IV infusion
 For IV infusion:
 For multiple IV bolus (dosing rate =
dose/ dosing interval):
 The steady-concentration depends
on the rate of dosing and the
clearance
clearance
rate
Dosing
0


KVd
K
Caverage
SS
clearance
rate
dosing
0



KVd
X
Caverage
SS

37. Example 1
 To a patient 250 mg penicillin with t½ of
1 h and Vd of 25 L is administered every
6 h intravenously
1. Estimate Cmax, Cmin and Cav at steady
state
2. Has the objective of maintaining
concentration above minimum inhibitory
concentration (4 mg/L) been achieved in
this therapy? Elaborate!
3. How long did it take to reach 95% of Css?
4. Is the idea of giving a bolus dose to
achieve Css in a shorter time feasible
with regard to this drug?

38. Example 1
016
.
1
1
1
1
1
6
*
693
.
0




 

e
e
R KT
1
0.5
hr
0.693
1
0.693
t
0.693
K 



mg/L
10.16
25
250
1.016
V
D
R
Cmax
SS















mg/L
0.16
e
25
250
1.016
e
V
D
R
C *6
0.693
KT
min
SS
















 

mg/L
2.41
6
*
25
*
0.693
250
KVdτ
X
C 0
average
SS




39. Example 1
 Drug concentration cannot be maintained above
the MIC if it is being administered every 6 h (6 x
t½). Because almost 98% of the dose is out of
the body at the time of the next administration.
However, conventionally penicillins are given
q.i.d. and it is known that they are effective.
Therefore, there is no need for keeping the
concentration above MIC during the entire
therapy.

 4.3 hrs are needed to get to 95% of Css (i.e.
Css was obtained as a result of the first dose)
hr
4.3
0.95)
ln(1
*
1.44
fss)
ln(1
t
1.44
*
1
0.5
nτ
nτ








40. Example 1
 The steady-state is achieved very
rapidly (after the first dose). Since
there is no need for accumulation,
there is little justification for giving
a loading dose.

41. Example 2
 A patient is receiving 1000 mg of
sulfamethoxazole iv every 12 hours
for the treatment of severe gram-
negative infection. At steady state
the maximum and minimum serum
sulfamethoxazole concentrations
were 81.5 mg/L and 40 mg/L,
respectively. Estimate the values of
K and VD

42. Example 2
T
C
C
t
t
C
C
K SS
SS )
ln(
)
ln(
)
ln(
)
ln( min
max
1
2
2
1 




1
hr
0.059
12
ln(40)
ln(81.5)
K 



97
.
1
1
1
1
1
12
*
059
.
0




 

e
e
R KT
L
24.2
81.5
1000
1.97
C
D
R
V
V
D
R
C
max
max
SS
SS
























