PROPT Drug Displacement Problem - Revision history

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	[[File:drugDisplacement_01.png]]		[[File:drugDisplacement_01.png]]

			[[Category:PROPT Examples]]

Mbot at 12:07, 9 November 2011

2011-11-09T12:07:30Z

Mbot at 13:04, 8 November 2011

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	<pre>		<pre>
	Problem type appears to be: lpcon		Problem type appears to be: lpcon
	Time for symbolic processing: 0.~~43888~~ seconds		Time for symbolic processing: 0.41796 seconds
	Starting numeric solver		Starting numeric solver
	===== * * * =================================================================== * * *		===== * * * =================================================================== * * *

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	{{Part Of Manual\|title=the PROPT Manual\|link=[[PROPT \|PROPT Manual]]}}		{{Part Of Manual\|title=the PROPT Manual\|link=[[PROPT\|PROPT Manual]]}}

	ITERATIVE DYNAMIC PROGRAMMING, REIN LUUS		ITERATIVE DYNAMIC PROGRAMMING, REIN LUUS
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	title('Drug Displacement control');		title('Drug Displacement control');
	</source>		</source>

			[[File:drugDisplacement_01.png]]

Mbot: Created page with "{{Part Of Manual|title=the PROPT Manual|link=PROPT Manual}} ITERATIVE DYNAMIC PROGRAMMING, REIN LUUS 12.4.3 Example 3: The desired level of two drugs, warfarin and..."

2011-11-02T15:29:06Z

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{{Part Of Manual|title=the PROPT Manual|link=[[PROPT |PROPT Manual]]}}

ITERATIVE DYNAMIC PROGRAMMING, REIN LUUS

12.4.3 Example 3: The desired level of two drugs, warfarin and phenylbutazone, must be reached in a patients bloodstream in minimum time.

CHAPMAN & HALL/CRC Monographs and Surveys in Pure and Applied Mathematics

==Problem Formulation==

Find u over t in [0; t ] to minimize

<math> J = t_F </math>

subject to:

<math> \frac{dx_1}{dt} = g_1*(g_4*(0.02-x_1)+46.4*x_1*(u-2*x_2)) </math>
<math> \frac{dx_2}{dt} = g_1*(g_3*(u-2*x_2)+46.4*(0.02-x_1)) </math>
<math> g_2 = 1+0.2*(x_1+x_2) </math>
<math> g_3 = g_2^2+232+46.4*x_2 </math>
<math> g_4 = g_2^2+232+46.4*x_1 </math>
<math> g_1 = \frac{g_2^2}{g_3*g_4-2152.96*x_1*x_2} </math>
<math> 0 <= u <= 8 </math>

x1 is the concentration of warfarin, and x2 of phenylbutazone. The initial and final condition are:

<math> x_0 = [0.02 \ 0] </math>
<math> x_{t_f} = [0.02 \ 2.00] </math>

<source lang="matlab">
% Copyright (c) 2007-2008 by Tomlab Optimization Inc.
</source>

==Problem setup==

<source lang="matlab">
toms t
toms t_f
p = tomPhase('p', t, 0, t_f, 50);
setPhase(p);

tomStates x1 x2
tomControls u

% Initial guess
x0 = {t_f == 300
icollocate({
x1 == 0.02; x2 == 2*t/t_f})
collocate(u == 8-8*t/t_f)};

% Box constraints
cbox = { 1 <= t_f <= 500
0 <= collocate(u) <= 8};

% Boundary constraints
cbnd = {initial({x1 == 0.02; x2 == 0})
final({x1 == 0.02; x2 == 2})};

% General variables
g2 = 1+0.2*(x1+x2);
g3 = g2.^2+232+46.4*x2;
g4 = g2.^2+232+46.4*x1;
g1 = g2.^2./(g3.*g4-2152.96*x1.*x2);

% ODEs and path constraints
ceq = collocate({
dot(x1) == g1.*(g4.*(0.02-x1)+46.4*x1.*(u-2*x2))
dot(x2) == g1.*(g3.*(u-2*x2)+46.4*(0.02-x1))});
</source>

==Solve the problem==

<source lang="matlab">
options = struct;
options.name = 'Drug Displacement';
% Objective is first parameter
solution = ezsolve(t_f, {cbox, cbnd, ceq}, x0, options);
t = subs(collocate(t),solution);
u = subs(collocate(u),solution);
</source>

<pre>
Problem type appears to be: lpcon
Time for symbolic processing: 0.43888 seconds
Starting numeric solver
===== * * * =================================================================== * * *
TOMLAB - TOMLAB Development license 999007. Valid to 2011-12-31
=====================================================================================
Problem: --- 1: Drug Displacement f_k 221.333418113505790000
sum(|constr|) 0.000000061270786413
f(x_k) + sum(|constr|) 221.333418174776570000
f(x_0) 300.000000000000000000

Solver: snopt. EXIT=0. INFORM=1.
SNOPT 7.2-5 NLP code
Optimality conditions satisfied

FuncEv 1 ConstrEv 14 ConJacEv 14 Iter 10 MinorIter 256
CPU time: 0.046800 sec. Elapsed time: 0.048000 sec.

</pre>

==Plot result==

<source lang="matlab">
figure(1)
plot(t,u,'+-');
legend('u');
title('Drug Displacement control');
</source>

← Older revision		Revision as of 07:07, 9 November 2011
Line 12:		Line 12:

	<math> J = t_F </math>		<math> J = t_F </math>


	subject to:		subject to:

	<math> \frac{dx_1}{dt} = g_1(g_4(0.02-x_1)+46.4x_1(u-2*x_2)) </math>		<math> \frac{dx_1}{dt} = g_1(g_4(0.02-x_1)+46.4x_1(u-2*x_2)) </math>

	<math> \frac{dx_2}{dt} = g_1(g_3(u-2x_2)+46.4(0.02-x_1)) </math>		<math> \frac{dx_2}{dt} = g_1(g_3(u-2x_2)+46.4(0.02-x_1)) </math>

	<math> g_2 = 1+0.2*(x_1+x_2) </math>		<math> g_2 = 1+0.2*(x_1+x_2) </math>

	<math> g_3 = g_2^2+232+46.4*x_2 </math>		<math> g_3 = g_2^2+232+46.4*x_2 </math>

	<math> g_4 = g_2^2+232+46.4*x_1 </math>		<math> g_4 = g_2^2+232+46.4*x_1 </math>

	<math> g_1 = \frac{g_2^2}{g_3g_4-2152.96x_1*x_2} </math>		<math> g_1 = \frac{g_2^2}{g_3g_4-2152.96x_1*x_2} </math>

	<math> 0 <= u <= 8 </math>		<math> 0 <= u <= 8 </math>


	x1 is the concentration of warfarin, and x2 of phenylbutazone. The initial and final condition are:		x1 is the concentration of warfarin, and x2 of phenylbutazone. The initial and final condition are:

	<math> x_0 = [0.02 \ 0] </math>		<math> x_0 = [0.02 \ 0] </math>

	<math> x_{t_f} = [0.02 \ 2.00] </math>		<math> x_{t_f} = [0.02 \ 2.00] </math>


	<source lang="matlab">		<source lang="matlab">