Compartmental modelling: Applications

Dr Phil Arundel

Hon Prof; University of Warwick

talk presented at Systems Pharmacology School, 24th March 2014

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

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.

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.

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)

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)

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

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.

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

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

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

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

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

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

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.

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.

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

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

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

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.

Representation of “Bolus” and “Stopped Infusion”

BOLUS

STOPPED

INFUSION

0 T time

H

It behaves as if its parameters have changed

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

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 ,,

Time shifting

00

.).().()( deefdefLT sTsTs

0

).( defe ssT

)(. sFe sT

)().( TtfTtuL )(. sFesT

So

Where )()( tfLsF T

0

0 t

t

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

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

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?

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.

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.