function [t,CA,CB,y] = equilibrium(DoseA,DoseB,tend,P)
%DoseA=IV bouls dose of drug A.
%DoseB=IV bolus dose of drug B.
%tend=tast time point for simulations.
%P=vector of model parameters (see below).
%t=column vector of times for soultions.
%y=length(t)x5 matrix of solutions:
%  y(:,1)=CAtot
%  y(:,2)=ATA
%  y(:,3)=CBtot
%  y(:,4)=ATB
%  y(:,5)=Rtot
%CA=column vector of free drug A concentrations.
%CB=column vector of free drug B concentrations.

% Model parameters.
Vc   =P(1);
kelA =P(2);
kelB =P(3);
ktpA =P(4);
ktpB =P(5);
kptA =P(6);
kptB =P(7);
kintA=P(8);
kintB=P(9);
KDA  =P(10);
KDB  =P(11);
kdeg =P(12);
Rtot0=P(13);
CA0  =P(14);
CB0  =P(15);

% Model secondary parameters

RCA0=Rtot0*CA0/KDA/(1+CA0/KDA+CB0/KDB);
RCB0=Rtot0*CB0/KDB/(1+CA0/KDA+CB0/KDB);
CAtot0=CA0+RCA0;
CBtot0=CB0+RCB0;
InA0=kelA*CA0+kintA*RCA0;
InB0=kelB*CB0+kintB*RCB0;
ATA0=kptA*Vc*CA0/ktpA;
ATB0=kptB*Vc*CB0/ktpB;
ksyn=kdeg*(Rtot0-RCA0-RCB0)+kintA*RCA0+kintB*RCB0;

% Model initial conditions;

IC=[DoseA/Vc+CAtot0;ATA0;DoseB/Vc+CBtot0;ATB0;Rtot0];

% solving model ODEs over time interval [0, tend].

RelTol=10^(-8);% default 10^(-3)
AbsTol=10^(-8);% default 10^(-6)
options = odeset('RelTol',RelTol,'AbsTol',AbsTol);

[t,y]=ode45(@model,[0:0.1:tend],IC,options);

% Calculating CA and CB based on the model variables:CAtot,CBtot, Rtot.
CA=t;
CB=t;
for i=1:length(t)
CAtot=y(i,1);
ATA  =y(i,2);
CBtot=y(i,3);
ATB  =y(i,4);
Rtot =y(i,5);
[CA(i),CB(i)] = agonism_roots(Rtot,CAtot,CBtot,KDA,KDB);
end

% Model ODEs.
function [dydt] = model(t,y);
    
CAtot=y(1);
ATA  =y(2);
CBtot=y(3);
ATB  =y(4);
Rtot =y(5);
[CA,CB] = equilibrium_roots(Rtot,CAtot,CBtot,KDA,KDB);
dydt=[InA0-(kelA+kptA)*CA+ktpA*ATA/Vc-kintA*(CAtot-CA);
      kptA*Vc*CA-ktpA*ATA;
      InB0-(kelB+kptB)*CB+ktpB*ATB/Vc-kintB*(CBtot-CB);
      kptB*Vc*CB-ktpB*ATB;
      ksyn-kdeg*Rtot-(kintA-kdeg)*(CAtot-CA)-(kintB-kdeg)*(CBtot-CB)];
end
% Endo of function agonism
end

function [ CA,CB ] = equilibrium_roots(Rtot,CtotA,CtotB, KDA,KDB )
%   This function solves the equilibrium equations using roots function 
%   to find the only solution z to the equlibrium polynomial such that
%   0 < z & z < 1 & z < aA+aB.
aA=CtotA/Rtot;
aB=CtotB/Rtot;
kA=KDA/Rtot;
kB=KDB/Rtot;

if (kA == kB)
% root of z^2+b*z+c=0
 b=-(1+aA+aB+kA);
 c=aA+aB;
 p=roots([1,b,c]);
 if (0<p(1) & p(1)<1 & p(1)<aA+aB)
  z=p(1);
 end
 if (0<p(2) & p(2)<1 & p(2)<aA+aB)
  z=p(2);
 end
else
% root of z^3+b*z^2+c*z+d=0
 b=-(2+kA+kB+aA+aB);
 c=1+2*aA+2*aB+kA+kB+kB*aA+kA*aB+kA*kB;
 d=-(kB*aA+kA*aB+aA+aB);
 p=roots([1,b,c,d]);
 if (0<p(1) & p(1)<1 & p(1)<aA+aB)
  z=p(1);
 end
 if (0<p(2) & p(2)<1 & p(2)<aA+aB)
  z=p(2);
 end
 if (0<p(3) & p(3)<1 & p(3)<aA+aB)
  z=p(3);
 end
end
CA=CtotA*KDA/(KDA+Rtot*(1-z));
CB=CtotB*KDB/(KDB+Rtot*(1-z));

end