compartmental modelling: applications€¦ · consider the most general two-compartment ‘open’...
TRANSCRIPT
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Compartmental modelling: Applications
Dr Phil Arundel
Hon Prof; University of Warwick
talk presented at Systems Pharmacology School, 24th March 2014
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Mathematical transforms have been developed over
centuries for the purpose of reducing effort - Fourier
Analysis and the Laplace Transform to name but two.
Some are so familiar that we forget they are there,
e.g. logarithms. They are so useful that new ideas
follow on; for example fractional roots, digital filters
and convolution/deconvolution.
Abstract
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Go round it
If you want to turn right on a bike what is the first
thing you do?
Turn the handlebars a bit to the left
This starts you falling to the right, putting your centre
of gravity to the right
Now you turn slightly to the right and continue round
the curve, matching your speed to the angle of the
bike to the ground.
This is counter-intuitive, lucky you didn’t realise that
as a youngster.
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Go round it
Short story
Two dogs see a bone in the next-door garden which is fenced-off.
Dog1 charges at the fence (closing in on the target) but fails to get over it, he sits down and barks. He is barking “this problem has not changed even though I am closer……………(sound familiar?)…”
Dog2 sets off along the fence, away from the target (a dumb move according to Dog1 who really considers it counter-intuitive…) looking for a way through, …round, …over at the lowest point of the… fence. He finds it and traverses the fence and reaches the bone.
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Go round it: the principle
Logarithms (Napier 1614)
Division is very difficult, digital computers can’t do it (analog computers can)
Subtraction is not easy, digital computers can just do it. Fortunately logarithms allow us to do division by arithmetic.
Process to divide two numbers: • Take their logarithms (away from target)
• Subtract them (cross the fence)
• Take the anti-log (move to the target)
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Go round it: the principle
Laplace transform (LT)
Algebra can be done by humans but solving
simultaneous differential equations is difficult.
Fortunately the LT allow us to solve linear
differential equations using algebra.
Process:
• Take their LT (away from target)
• Solve equations by algebra (cross the fence)
• Take the inverse LT (move to the target)
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Introduction
The human body is so complex that all mathematical equations relating to it
are approximations (models) of varying accuracy and relevance.
When compounds are injected into the body via the blood stream they can
undergo a number of possible sequences, here are two.
1. They may stay mostly in the blood, not transferring to other parts of the
body, being eliminated from the blood by the liver and/or kidneys.
1-compartment model (open)
2. They may transfer readily into tissues; e.g. muscle and then be eliminated
by the liver/kidneys.
2-compartment model (open)
Transfers are governed by the Law of Mass Action
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The law of mass action
This states that
“the rate at which a process occurs is dependent on the amount of
reactant present”.
It was first announced by Guldberg and Waage on 11th March 1864,
in Norway.
Xkdt
dXv *
For pharmacokinetics this means that the transfer rates into
and out of compartments are related to the mass remaining.
The concept of drug concentration (the most readily
measurable variable) is accommodated by identifying a value
for the volume of the ‘blood’ compartment.
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Modelling
Consider the most general two-compartment ‘open’ model for intravenous (IV) input
into compartment 1. Using LT, the transfer function for this system will be derived
and then, by convolution, the solution for other forms of input will be simply produced.
1x 2xoutk
ink
elk
This system has a
compartment ‘like’ the
blood into which a drug is
added and from which it
is removed, and a
compartment ‘like’ the
rest of the body with
which it can exchange.
INPUT
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Differential equations
We can write differential equations for the two compartments,
and solve for x1 the mass in compartment 1.
)1....(............ 2101 XkXkkXsX inoutel
)2....(............0 212 XkXksX inout
211 xkxkk
dt
dxinoutel
212 xkxk
dt
dxinout
Laplace transformed, with transforms of initial values included
1x 2xoutk
ink
elk
INPUT
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Manipulating the LT equations: analysing compartment 1
By rearranging Eq.2
in
out
ks
sXksX
)(.)( 12
Substituting in Eq.1 in
outinoutel
ks
sXkkXsXkks
)(..)(. 101
)(...)(. 101 sXkkXkssXkskks outinininoutel
)3.........(...........)(..)( 012 XkssXkkskkks inelinoutinel
0121 .)(.. XkssXss in
where we shall assume 21
)).((
).()(
21
01
ss
XkssX in
We need to take the LT inverse
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Finding the Inverse by partial fractions
This is done using either the cover-up rule or full manipulation.
tintin exk
exk
tx 21 ..)( 012
20
21
11
Where it can be shown that 21 ink
tintin exk
exk
tx 21 ..)( 021
20
21
11
Typically 21 .10
in a practical case
I.V.
1
10
100
1000
10000
0 2 4 6 8
time (hr)
Co
nc (
ng
/ml)
Cp act
Cp pred
Search line
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The ‘Mass’ v ‘Time’ curve
We can write the equation in the following form, tt BeAextx 21.)( 01
where,
21
2
21
1
inin
kBand
kA
Note A + B = 1
I.V.
1
10
100
1000
10000
0 2 4 6 8
time (hr)
Co
nc (
ng
/ml)
Cp act
Cp pred
Search line
I.V.
1
10
100
1000
10000
0 2 4 6 8
time (hr)
Co
nc (
ng
/ml)
Cp act
Cp pred
Search line
In this example x0 = 2500, A = 0.94 and B = 0.06
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To analyse compartment 2 we eliminate
X1(t) using Eq.2, so from before
Manipulating the LT equations: analysing compartment 2
out
in
k
ksXX
.21
212 ..0 XkXksX inout
then from Eq.3
012 .)(..)( XkssXkkskkks inelinoutinel
022 ..)(. XkkkskkksX outelinoutinel
The underlined section is like before, “characteristic” of the differential
equation pair, so the roots
(eigenvalues) are unchanged. Again by partial fractions 21,
ttout eexktx 21.)(
21
02
0)(, 2 txt
0)(,0 2 txt
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The transfer function
What have we achieved using the LT to study the 2-
compartment model with an instantaneous (bolus) input into
compartment 1?
)).((
).()(
21
01
ss
XkssX in
In engineering terms X1(s) is the impulse
response to the instantaneous input X0, where
)).((
)()(
210
1
ss
ks
X
sX in1x 2x
outk
ink
elk
INPUT
is the TRANSFER FUNCTION for the 2-
compartment open model.
One great advantage of the LT is the ease with
which new inputs can analysed. This will be shown
next.
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Changing the input
Consider the case of a ZERO ORDER input, of rate “k0”.
It has the LT, k0/s found as follows
k0
time
s
ke
s
kdtekkL stst 0
00
000 .
The pharmaceutical equivalent to the zero order input is an INFUSION.
In the LT domain, once we know the transfer function, the LT for a new
form of input is obtained by simply multiplying (convolution), as follows,
LT for an INFUSION:-
s
k
ss
kssX in 0
21
1 .)).((
)()(
This can be solved in two stages by Partial Fractions.
Note convolution in the time domain is much more complicated.
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LT analysis of an infusion
)(
1.
.
)(
1.
.)(
221
20
121
101
ss
kk
ss
kksX inin
)(
11.
.
)(
11.
.
2212
20
1211
10
ss
kk
ss
kk inin
tt ekBekAtx 21 11.)(2
0
1
01
Effortless!! For an infusion to time T
TT ekBekAtx 21 11.)(2
0
1
01
Now switch off the infusion at
t = T, a “stopped infusion”.
time
k0
T
Solving it this way preserves the symmetry of the terms, …second
stage of partial fractions
First stage partial
fractions
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Simplifying the analysis
The best way to tackle this, is by superposition.
The rate of input depicted above can be
expressed in two stages.
Firstly the initial phase starts from zero with a
rate of k0, which it maintains throughout, while
the second infusion is offset by T and has a
‘negative’ infusion rate (-k0).
time
k0
T
-k0
The time shift of T is produced by means of
an exponential term in the ‘s’ domain,
Tse
termsBss
ekksX
Ts
in '')(
..
)(121
101
time
k0
T
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continued
termsBekAekAtx Ttt ''11.)( )(1
0
1
01
11
termsBeekA tTt ''. 11 )(1
0
It is important to learn to recognise equalities like,
for t > T
TTtteee 111 .
)(
termsBeekAtx TtT ''.1.)( )(1
01
11
for t > T
When T is large termsBek
AtxTt
''.)()(
1
01
1
which looks a bit like a ‘time displaced’ bolus equation
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Comparing “bolus” and “stopped infusion”
For large values of T, and t > T the two equations are
Consider the case when
)(
2
0)(
1
01
21 ..)(TtTt
ek
Bek
Atx
tteBxeAxtx 21 ..)( 001
HBA
xk 21
00 ;1
t = 60 for the bolus and ( t -T) = 60 for the “stopped infusion”,
then for the bolus; and for the infusion;
).()(6060
121 BeAeHtx
Typically
60
2
60
1
121 ..)(
e
Be
Atx
21 .10 The influence of B is now much greater in the second equation.
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Representation of “Bolus” and “Stopped Infusion”
BOLUS
STOPPED
INFUSION
0 T time
H
It behaves as if its parameters have changed
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Significance
In Engineering we are used to systems where the parameters
remain fixed. For some PHYSIOLOGICAL systems, parameters
apparently vary depending on the history of the system.
Applying FT analysis to biological systems opens up large areas of new work
This relatively simple analysis led to an important paper in
Pharmacokinetics about 20 years ago.
Hughes MA et al (1992); Anesthesiology, vol 76, p334-341
There must be several more that remain to be discovered.
Examples
Deconvolution & convolution: when doses are not straight into the blood
Matrix exponential for linear systems
Similar eigenvalues and model reduction
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The unit step function, u(t)
Important when manipulating LT expressions
0,0
0,1)(
t
ttu
Tt
TtTtu
,0
,1)(now shift along the t-axis
0
).().().( dtTtfTtudtTtfT
It’s useful when changing limits
0
0 T t
t
Finding the LT of a function shifted along the time axis.
For the previous function which has been time-shifted, the LT is by definition,
0
).().( dteTtfTtu st
T
stdteTtf ).(
Using a change of variable, let ddtTtTt ,,
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Time shifting
00
.).().()( deefdefLT sTsTs
0
).( defe ssT
)(. sFe sT
)().( TtfTtuL )(. sFesT
So
Where )()( tfLsF T
0
0 t
t
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Making use of the ‘time shift’ facility in FT analysis
What if we delay the instantaneous dose
by T in the 2-compartment model.
Previously the transfer function was
)).((
)(
21
ss
ks in
Now the transfer function becomes )).((
).(
21
ss
kse insT
for t > T and a unit dose )()(
121)(
TtTtBeAetx
This is useful when we consider multiple, evenly spaced dosing
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Multiple dosing
Consider the case where we have 3 equal doses, the first at t = 0, the
second at t = T and the third at t = 2T. Remember that the first dose will
still be having an effect when t > T and the second when t > 2T. So the
combined situation will be for t > 2T,
"")()2()(
1111 termsBAeAeAetx
TtTtt
"".1 )2(2. 111 termsBeeeA TtTT
This can be rearranged to form a geometric progression whose sum can
be found,
thenehIfT1
h
hhhee
TT
1
111
322.11
As 11 T
eh and setting 12
Tek
)()(
121
1
1
1
1)(
timetimee
kBe
hAtx
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Numerical example
If
then
Say A = 0.5 and B = 0.5 then for multiple doses to steady state,
1,24,02.02.0 021 xTand
)()(
121
381.0992.0)(
timetimee
Be
Atx
)()(
121 3.15.0)(
timetimeeetx
The sum of the coefficients is now 0.5 + 1.3 = 1.8.
For a single dose the sum is only 0.5 + 0.5 = 1.0 So for
multiple dosing to steady state the sum of the coefficients
is 80% higher, what does this mean in practice?
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Graphically
AUC
AUC
0 T 2T 3T 4T
1
1.8
The drug is said to accumulate in the body. The shaded area in GREEN
eventually becomes equal in one interval to the whole BLUE area, out to infinity.
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Beware
Model complexity should be related to the number of data points. For example extracting more than 3 or 4 PK parameters from 8 data points would be dubious.
A model (equation) is only useful while it continues to accommodate new data without major repairs.
A model only proves itself by making predictions (preferably beyond the known region) which are then successfully validated.