Translating chemical equations from article, results differ (Matlab) - matlab

I've been trying to translate a set of chemical equations to MATLAB code, to be able to solve for different chemical species. I have the approximate solution (as it's from a graph) but after entering all the data and checking multiple times I still haven't been able to find what is wrong. I'm wondering what is going wrong and if anyone could please help me out. The source for the graph/equation is the article at this link: The chemistry of co-injected BOE. The graph I want to reproduce later on is figure 2 in the paper, see the image below:
Now the results I get for 10cc, 40cc and 90cc are respectively:
HF 43%, H2F2 48%, F- 3%, HF2- 6% in comparison ~28%, 63%, 2%, 7% (10cc).
HF 35%, H2F2 33%, F- 14%, HF2- 18% in comparison ~24%, 44%, 6%, 26% (40cc).
HF 21%, H2F2 12%, F- 37%, HF2- 30% in comparison ~18%, 23%, 20%, 45% (90cc).
The script is the following:
clc;
clear all;
%Units to be used
%Volume is in CC also cm^3, 1 litre is 1000 CC, 1 cc = 1 ml
%density is in g/cm^3
%weigth percentages are in fractions of 0 to 1
%Molecular weight is in g/mol
% pts=10; %number of points for linear spacing
%weight percentages of NH4OH and HF
xhf=0.49;
xnh3=0.28;
%H2O
Vh2o=1800;
dh2o=1.00; %0.997 at 25C when rounded 1
mh2o=18.02;
%HF values
Vhf=100;
dhf49=1.15;
dhf=dh2o+(dhf49-dh2o)*xhf/0.49; %# 25C
Mhf=20.01;
nhf=mols(Vhf,dhf,xhf,Mhf);
%NH4OH (NH3) values
% Vnh3=linspace(0.1*Vhf,1.9*Vhf,pts);
Vnh3=10;
dnh3=0.9; %for ~20-31% #~20-25C
Mnh3=17.03; %The wt% of NH4OH actually refers to the wt% of NH3 dissolved in H2O
nnh3=mols(Vnh3,dnh3,xnh3,Mnh3);
if max(nnh3)>=nhf
error(['There are more mols NH4OH,',num2str(max(nnh3)),', than mols HF,',num2str(nhf),'.'])
end
%% Calculations for species
Vt=(Vhf+Vh2o+Vnh3)/1000; %litre
A=nhf/Vt; %mol/l
B=nnh3/Vt; %mol/l
syms HF F H2F2 HF2 NH3 NH4 H OH
eq2= H*F/HF==6.85*10^(-4);
eq3= NH3*H/NH4==6.31*10^(-10);
eq4= H*OH==10^(-14);
eq5= HF2/(HF*F)==3.963;
eq6= H2F2/(HF^2)==2.7;
eq7= H+NH4==OH+F+HF2;
eq8= HF+F+2*H2F2+2*HF2==A;
eq9= NH3+NH4==B;
eqns=[eq2,eq5,eq6,eq8,eq4,eq3,eq9,eq7];
varias=[HF, F, H2F2, HF2, NH3, NH4, H, OH];
assume(HF> 0 & F>= 0 & H2F2>= 0 & HF2>= 0& NH3>= 0 & NH4>= 0 & H>= 0 & OH>= 0)
[HF, F, H2F2, HF2, NH3, NH4, H, OH]=vpasolve(eqns,varias);% [0 max([A,B])])
totalHF=double(HF)+double(F)+double(H2F2)+double(HF2);
HFf=double(HF)/totalHF %fraction of species for HF
H2F2f=double(H2F2)/totalHF %fraction of species for H2F2
Ff=double(F)/totalHF %fraction of species for F-
HF2f=double(HF2)/totalHF %fraction of species for HF2-
an extra function needed is called mols.m
%%%% amount of mol, Vol=volume, d=density, pwt=%weight, M=molecularweight
function mol=mols(Vol, d, pwt, M)
mol=(Vol*d*pwt)/M;
end
The equations being used from the article are in the image below:
(HF)2 is H2F2 in my script


So appears the issue wasn't so much with Matlab, had some help in that area as well.
Final solution and updated Matlab code can be found here:
https://chemistry.stackexchange.com/questions/98306/why-do-my-equilibrium-calculations-on-this-hf-nh4oh-buffer-system-not-match-thos

Related

Solve for 3 variables with a quite complex equation in Matlab

syms omega j theta sigma
k=[90 94 98 102 106 110]; % Strike prices
putprices=[.1606 .2806 .9285 2.8801 6.1523 10.0147]; % Put prices at each strike
mustar = log(100)-sigma.^2/2-omega.*(exp(-theta)-1); % risk-neutral mean
Es1j=exp(mustar-theta.*j+sigma.^2./2); % Expected value of stock price at state j 1 period from now
x1=(log(k)-(mustar-theta.*j))./sigma; % the first normcdf value
x2=(log(k)-(mustar-theta.*j)-sigma.^2)./sigma; % the second normcdf value
qpkandj = k.*normcdf(x1)-Es1j.*normcdf(x2); % combination of normcdfs and strike prices and expected value of stock price 1 peroid from now
qpk=exp(-omega).*omega.^j./factorial(j).*qpkandj; % Poisson formula
eqns = symsum(qpk,j,0,inf)== putprices % Continuation of Poisson formula
assume(omega > 0)
assume(theta, 'real')
assume(sigma, 'real')
S = solve(eqns, [omega theta sigma], 'Real', true, 'ReturnConditions', true)
S.omega
S.theta
S.sigma
Trying to work on a Poisson mixture model to find the values of the three parameters (omega, theta, and sigma) that match the Poisson equation's outputs to the putprices in the third line. Whenever I run this, I get a very, very long conditions list and I tried to put it all into the assume condition and it made no difference. Anyone have any ideas what's going on or how I could manage to make this work?

Matlab will print the condition list because you don't add a semicolon. To avoid print the condition list, just add semicolon like this:
eqns = symsum(qpk,j,0,inf) == putprices; % Continuation of Poisson formula

linear combination of curves to match a single curve with integer constraints

I have a set of vectors (curves) which I would like to match to a single curve. The issue isnt only finding a linear combination of the set of curves which will most closely match the single curve (this can be done with least squares Ax = B). I need to be able to add constraints, for example limiting the number of curves used in the fitting to a particular number, or that the curves lie next to each other. These constraints would be found in mixed integer linear programming optimization.
I have started by using lsqlin which allows constraints and have been able to limit the variable to be > 0.0, but in terms of adding further constraints I am at a loss. Is there a way to add integer constraints to least squares, or alternatively is there a way to solve this with a MILP?
any help in the right direction much appreciated!
Edit: Based on the suggestion by ErwinKalvelagen I am attempting to use CPLEX and its quadtratic solvers, however until now I have not managed to get it working. I have created a minimal 'notworking' example and have uploaded the data here and code here below. The issue is that matlabs LS solver lsqlin is able to solve, however CPLEX cplexlsqnonneglin returns CPLEX Error 5002: %s is not convex for the same problem.
function [ ] = minWorkingLSexample( )
%MINWORKINGLSEXAMPLE for LS with matlab and CPLEX
%matlab is able to solve the least squares, CPLEX returns error:
% Error using cplexlsqnonneglin
% CPLEX Error 5002: %s is not convex.
%
%
% Error in Backscatter_Transform_excel2_readMut_LINPROG_CPLEX (line 203)
% cplexlsqnonneglin (C,d);
%
load('C_n_d_2.mat')
lb = zeros(size(C,2),1);
options = optimoptions('lsqlin','Algorithm','trust-region-reflective');
[fact2,resnorm,residual,exitflag,output] = ...
lsqlin(C,d,[],[],[],[],lb,[],[],options);
%% CPLEX
ctype = cellstr(repmat('C',1,size(C,2)));
options = cplexoptimset;
options.Display = 'on';
[fact3, resnorm, residual, exitflag, output] = ...
cplexlsqnonneglin (C,d);
end

I could reproduce the Cplex problem. Here is a workaround. Instead of solving the first model, use a model that is less nonlinear:
The second model solves fine with Cplex. The problem is somewhat of a tolerance/numeric issue. For the second model we have a much more well-behaved Q matrix (a diagonal). Essentially we moved some of the complexity from the objective into linear constraints.
You should now see something like:
Tried aggregator 1 time.
QP Presolve eliminated 1 rows and 1 columns.
Reduced QP has 401 rows, 443 columns, and 17201 nonzeros.
Reduced QP objective Q matrix has 401 nonzeros.
Presolve time = 0.02 sec. (1.21 ticks)
Parallel mode: using up to 8 threads for barrier.
Number of nonzeros in lower triangle of A*A' = 80200
Using Approximate Minimum Degree ordering
Total time for automatic ordering = 0.00 sec. (3.57 ticks)
Summary statistics for Cholesky factor:
Threads = 8
Rows in Factor = 401
Integer space required = 401
Total non-zeros in factor = 80601
Total FP ops to factor = 21574201
Itn Primal Obj Dual Obj Prim Inf Upper Inf Dual Inf
0 3.3391791e-01 -3.3391791e-01 9.70e+03 0.00e+00 4.20e+04
1 9.6533667e+02 -3.0509942e+03 1.21e-12 0.00e+00 1.71e-11
2 6.4361775e+01 -3.6729243e+02 3.08e-13 0.00e+00 1.71e-11
3 2.2399862e+01 -6.8231454e+01 1.14e-13 0.00e+00 3.75e-12
4 6.8012056e+00 -2.0011575e+01 2.45e-13 0.00e+00 1.04e-12
5 3.3548410e+00 -1.9547176e+00 1.18e-13 0.00e+00 3.55e-13
6 1.9866256e+00 6.0981384e-01 5.55e-13 0.00e+00 1.86e-13
7 1.4271894e+00 1.0119284e+00 2.82e-12 0.00e+00 1.15e-13
8 1.1434804e+00 1.1081026e+00 6.93e-12 0.00e+00 1.09e-13
9 1.1163905e+00 1.1149752e+00 5.89e-12 0.00e+00 1.14e-13
10 1.1153877e+00 1.1153509e+00 2.52e-11 0.00e+00 9.71e-14
11 1.1153611e+00 1.1153602e+00 2.10e-11 0.00e+00 8.69e-14
12 1.1153604e+00 1.1153604e+00 1.10e-11 0.00e+00 8.96e-14
Barrier time = 0.17 sec. (38.31 ticks)
Total time on 8 threads = 0.17 sec. (38.31 ticks)
QP status(1): optimal
Cplex Time: 0.17sec (det. 38.31 ticks)
Optimal solution found.
Objective : 1.115360
See here for some details.
Update: In Matlab this becomes:

Matlab forecasting with autoregressive exogenous modell

i have a file, which is the energy consumption of a house.
every 10 minute one value (watt):
10:00 123
10:10 125
10:20 0
...
It means each day have 144 value (Rows).
i want to forecast the energy of the next day with ARX an ARMAX programm. i did write ARX code in Matlab. but i can't forecast the next day. My code take the last 5 consumption and forecast the 6th one. How can i forecast the nex 144 value ( = the day after)
% ARX Process----------------------------
L=length(u_in)
u_in_ID=u_in;% Input data used for Identification
u_in_vfy=u_in;% Input data used for verification
y_out_ID=y_out;% Output data used for Identification
y_out_vfy=y_out;%Output data used for verification
m=5; %Parameter to be used to generate order of delay for Input, Output and Error
n=length(y_out_ID)-m;
I=eye(n,1)+1;
I(1)=I(1)-1;
A=I; % Initialize Matrix A
Y=y_out_ID((m+1):end); % Defining Y vector
length(Y)
na=1;
% Put output delay 1 to m-na in A matrix
for k=1:1:m-na
A=[A y_out_ID((m-k+1):(end-k))];
end
% Put "Current Input -- mth delayed Input" to Matrix A
for p=1:1:m
k=p-1;
A=[A u_in_ID((m-k+1):(end-k))];
end
A(:,1)=[]; % Delete 1st column of Matrix A, which was used to Initialize it
parsol=inv(A'*A)*A'*Y;
BB=A*parsol;
% Generate Identified Output vector based on previous
% outputs, current and previous Inputs and Parameters solved by Least
% square method
n=length(y_out_vfy)-m;
I=eye(n,1)+1;
I(1)=I(1)-1;
A=I;
for k=1:1:m-na
A=[A y_out_vfy((m-k+1):(end-k))];
end
for p=1:1:m
k=p-1;
A=[A u_in_vfy((m-k+1):(end-k))];
end
A(:,1)=[]; % Delete 1st column of Matrix A, which was used to Initialize it
A;
y_out_sysID=A*parsol;
can anyone help me?

Attempted to access sym(67); index out of bounds because numel(sym)=2

I am receiving the following error message:
Attempted to access sym(67); index out of bounds because numel(sym)=2.
I have been working on this for three days. I looked for similar error, but it didn't help. My code is below:
filename='DriveCyclesCP.xlsx';
V=xlsread('DriveCyclesCP.xlsx',2,'C9:C774'); % Get the velocity values, they are in an array V.
N=length(V); % Find out how many readings
mass = 1700 ; % Vehicle mass+ two 70 kg passengers.
area_Cd = 0.75; % Frontal area in square metres
Crr=0.009; %rolling resistance
g=9.8; % gravity acceleration
T=774; %UDDS cycle time duration
V_ave = 21.5; % UDDS avearage speed im m/s
rd=0.3; % Effective tire radius
Qhv =12.22; % E85 low Heating value in kWh/kg
Vd = 2.189; % engine size in L
md=0.801; % mass density of Ethanol
mf =Vd*md; % mf is the fuel mass consumed per cycle
Per = zeros(1,N); % engine power for each point of the drive cycle
a = zeros(1,N); % acceleration
SFC = zeros(1,N); % specific fuel consumption
Wc = zeros (1,N); % mass flow rate
nf = zeros (1,N); %fuel efficiency
Pm = zeros (1,N); % motor power
Pt = zeros (1,N);
Te =zeros (1,N); % Engine Troque
Tt = zeros (1,N);
Tm =zeros (1,N);
we =zeros (1,N); % Engine rot speed
wt = zeros (1,N);
wm =zeros (1,N);
S =zeros (1,8);
int (sym ('C'));
for C=1:N
a(C)=V(C+1)-V(C);
Pt(C)= V(C)*(mass*g*Crr + (0.5*area_Cd*1.202*(V(C))^2) + mass*a(C))/1000;
Per(C)=(mass*g*Crr +0.5*area_Cd*1.202*(V(C))^2 +mass*g*0.03)/1000*0.85;% e
syms Te(C) Tt(C) Tm(C) wt(C) we(C) wm(C) k1 k2
S = solve( Pm(C)==Pt(C) - Per(C), Tt(C)*wt(C)== Pt(C), Tt(C)*wt(C)== Te(C)*we(C) + Tm(C)*wm(C), wt(C)==we(C)/k1, wt(C)==wm(C)/k2, Pm(C)==wm(C) *Tm(C), Per(C)==we(C) *Te(C), Tt == k1*Te + k2*Tm );
end

The problem is on the line
int (sym ('C'));
You have defined sym to be a matrix with 2 entries somewhere (either earlier in the code or in a previous mfile), thus it treats sym as a matrix instead of a function. Thus when Matlab gets to the statement sym('C') it first converts the character 'C' to its ASCII integer representation (this just happens to be the number 67), then it tries to calculate sym(67) which is impossible as sym only has 2 elements.
Thus you have to stop sym from being a matrix (variable) and let it be a function again. There are two ways to solve this, either you can start you file with the statement clear;, this will remove all variables in memory, which might not be what you want; or you can use a function instead of script, as this hides all variables that have been defined previously and prevents this sort of error.
Note the line numel(X) is a way to measure how many elements are in X. Thus numel(sym)=2 means that sym has 2 elements.
P.S. There is an error in the lines (notice that I only taken some of the lines of you code)
N=length(V); % Find out how many readings
for C=1:N
a(C)=V(C+1)-V(C);
end
When C becomes equal to N, then V(C+1) will generate an error.

Getting unexpected results while using ode45

I am trying to solve a system of differential equations by writing code in Matlab. I am posting on this forum, hoping that someone might be able to help me in some way.
I have a system of 10 coupled differential equations. It is a vector-host epidemic model, which captures the transmission of a disease between human population and insect population. Since it is a simple system of differential equations, I am using solvers (ode45) for non-stiff problem type.
There are 10 differential equations, each representing 10 different state variables. There are two functions which have the same system of 10 coupled ODEs. One is called NoEffects_derivative_6_15_2012.m which contains the original system of ODEs. The other function is called OnlyLethal_derivative_6_15_2012.m which contains the same system of ODEs with an increased withdrawal rate starting at time, gamma=32 %days and that withdrawal rate decays exponentially with time.
I use ode45 to solve both the systems, using the same initial conditions. Time vector is also the same for both systems, going from t0 to tfinal. The vector tspan contains the time values going from t0 to tfinal, each with a increment of 0.25 days, making a total of 157 time values.
The solution values are stored in matrices ye0 and yeL. Both these matrices contain 157 rows and 10 columns (for the 10 state variable values). When I compare the value of the 10th state variable, for the time=tfinal, in the matrix ye0 and yeL by plotting the difference, I find it to be becoming negative for some time values. (using the command: plot(te0,ye0(:,10)-yeL(:,10))). This is not expected. For all time values from t0 till tfinal, the value of the 10 state variable, should be greater, as it is the solution obtained from a system of ODEs which did not have an increased withdrawal rate applied to it.
I am told that there is a bug in my matlab code. I am not sure how to find out that bug. Or maybe the solver in matlab I am using (ode45) is not efficient and does give this kind of problem. Can anyone help.
I have tried ode23 and ode113 as well, and yet get the same problem. The figure (2), shows a curve which becomes negative for time values 32 and 34 and this is showing a result which is not expected. This curve should have a positive value throughout, for all time values. Is there any other forum anyone can suggest ?
Here is the main script file:
clear memory; clear all;
global Nc capitalambda muh lambdah del1 del2 p eta alpha1 alpha2 muv lambdav global dims Q t0 tfinal gamma Ct0 b1 b2 Ct0r b3 H C m_tilda betaHV bitesPERlanding IC global tspan Hs Cs betaVH k landingARRAY muARRAY
Nhh=33898857; Nvv=2*Nhh; Nc=21571585; g=354; % number of public health centers in Bihar state %Fix human parameters capitalambda= 1547.02; muh=0.000046142; lambdah= 0.07; del1=0.001331871263014; del2=0.000288658; p=0.24; eta=0.0083; alpha1=0.044; alpha2=0.0217; %Fix vector parameters muv=0.071428; % UNIT:2.13 SANDFLIES DEAD/SAND FLY/MONTH, SOURCE: MUBAYI ET AL., 2010 lambdav=0.05; % UNIT:1.5 TRANSMISSIONS/MONTH, SOURCE: MUBAYI ET AL., 2010
Ct0=0.054;b1=0.0260;b2=0.0610; Ct0r=0.63;b3=0.0130;
dimsH=6; % AS THERE ARE FIVE HUMAN COMPARTMENTS dimsV=3; % AS THERE ARE TWO VECTOR COMPARTMENTS dims=dimsH+dimsV; % THE TOTAL NUMBER OF COMPARTMENTS OR DIFFERENTIAL EQUATIONS
gamma=32; % spraying is done of 1st feb of the year
Q=0.2554; H=7933615; C=5392890;
m_tilda=100000; % assumed value 6.5, later I will have to get it for sand flies or mosquitoes betaHV=66.67/1000000; % estimated value from the short technical report sent by Anuj bitesPERlanding=lambdah/(m_tilda*betaHV); betaVH=lambdav/bitesPERlanding; IC=zeros(dims+1,1); % CREATES A MATRIX WITH DIMS+1 ROWS AND 1 COLUMN WITH ALL ELEMENTS AS ZEROES
t0=1; tfinal=40; for j=t0:1:(tfinal*4-4) tspan(1)= t0; tspan(j+1)= tspan(j)+0.25; end clear j;
% INITIAL CONDITION OF HUMAN COMPARTMENTS q1=0.8; q2=0.02; q3=0.0005; q4=0.0015; IC(1,1) = q1*Nhh; IC(2,1) = q2*Nhh; IC(3,1) = q3*Nhh; IC(4,1) = q4*Nhh; IC(5,1) = (1-q1-q2-q3-q4)*Nhh; IC(6,1) = Nhh; % INTIAL CONDITIONS OF THE VECTOR COMPARTMENTS IC(7,1) = 0.95*Nvv; %80 PERCENT OF TOTAL ARE ASSUMED AS SUSCEPTIBLE VECTORS IC(8,1) = 0.05*Nvv; %20 PRECENT OF TOTAL ARE ASSUMED AS INFECTED VECTORS IC(9,1) = Nvv; IC(10,1)=0;
Hs=2000000; Cs=3000000; k=1; landingARRAY=zeros(tfinal*50,2); muARRAY=zeros(tfinal*50,2);
[te0 ye0]=ode45(#NoEffects_derivative_6_15_2012,tspan,IC); [teL yeL]=ode45(#OnlyLethal_derivative_6_15_2012,tspan,IC);
figure(1) subplot(4,3,1); plot(te0,ye0(:,1),'b-',teL,yeL(:,1),'r-'); xlabel('time'); ylabel('S'); legend('susceptible humans'); subplot(4,3,2); plot(te0,ye0(:,2),'b-',teL,yeL(:,2),'r-'); xlabel('time'); ylabel('I'); legend('Infectious Cases'); subplot(4,3,3); plot(te0,ye0(:,3),'b-',teL,yeL(:,3),'r-'); xlabel('time'); ylabel('G'); legend('Cases in Govt. Clinics'); subplot(4,3,4); plot(te0,ye0(:,4),'b-',teL,yeL(:,4),'r-'); xlabel('time'); ylabel('T'); legend('Cases in Private Clinics'); subplot(4,3,5); plot(te0,ye0(:,5),'b-',teL,yeL(:,5),'r-'); xlabel('time'); ylabel('R'); legend('Recovered Cases');
subplot(4,3,6);plot(te0,ye0(:,6),'b-',teL,yeL(:,6),'r-'); hold on; plot(teL,capitalambda/muh); xlabel('time'); ylabel('Nh'); legend('Nh versus time');hold off;
subplot(4,3,7); plot(te0,ye0(:,7),'b-',teL,yeL(:,7),'r-'); xlabel('time'); ylabel('X'); legend('Susceptible Vectors');
subplot(4,3,8); plot(te0,ye0(:,8),'b-',teL,yeL(:,8),'r-'); xlabel('time'); ylabel('Z'); legend('Infected Vectors');
subplot(4,3,9); plot(te0,ye0(:,9),'b-',teL,yeL(:,9),'r-'); xlabel('time'); ylabel('Nv'); legend('Nv versus time');
subplot(4,3,10);plot(te0,ye0(:,10),'b-',teL,yeL(:,10),'r-'); xlabel('time'); ylabel('FS'); legend('Total number of human infections');
figure(2) plot(te0,ye0(:,10)-yeL(:,10)); xlabel('time'); ylabel('FS(without intervention)-FS(with lethal effect)'); legend('Diff. bet. VL cases with and w/o intervention:ode45');
The function file: NoEffects_derivative_6_15_2012
function dx = NoEffects_derivative_6_15_2012( t , x )
global Nc capitalambda muh del1 del2 p eta alpha1 alpha2 muv global dims m_tilda betaHV bitesPERlanding betaVH
dx = zeros(dims+1,1); % t % dx
dx(1,1) = capitalambda-(m_tilda)*bitesPERlanding*betaHV*x(1,1)*x(8,1)/(x(7,1)+x(8,1))-muh*x(1,1);
dx(2,1) = (m_tilda)*bitesPERlanding*betaHV*x(1,1)*x(8,1)/(x(7,1)+x(8,1))-(del1+eta+muh)*x(2,1);
dx(3,1) = p*eta*x(2,1)-(del2+alpha1+muh)*x(3,1);
dx(4,1) = (1-p)*eta*x(2,1)-(del2+alpha2+muh)*x(4,1);
dx(5,1) = alpha1*x(3,1)+alpha2*x(4,1)-muh*x(5,1);
dx(6,1) = capitalambda -del1*x(2,1)-del2*x(3,1)-del2*x(4,1)-muh*x(6,1);
dx(7,1) = muv*(x(7,1)+x(8,1))-bitesPERlanding*betaVH*x(7,1)*x(2,1)/(x(6,1)+Nc)-muv*x(7,1);
%dx(8,1) = lambdav*x(7,1)*x(2,1)/(x(6,1)+Nc)-muvIOFt(t)*x(8,1);
dx(8,1) = bitesPERlanding*betaVH*x(7,1)*x(2,1)/(x(6,1)+Nc)-muv*x(8,1);
dx(9,1) = (muv-muv)*x(9,1);
dx(10,1) = (m_tilda)*bitesPERlanding*betaHV*x(1,1)*x(8,1)/x(9,1);
The function file: OnlyLethal_derivative_6_15_2012
function dx=OnlyLethal_derivative_6_15_2012(t,x)
global Nc capitalambda muh del1 del2 p eta alpha1 alpha2 muv global dims m_tilda betaHV bitesPERlanding betaVH k muARRAY
dx=zeros(dims+1,1);
% the below code saves some values into the second column of the two arrays % t muARRAY(k,1)=t; muARRAY(k,2)=artificialdeathrate1(t); k=k+1;
dx(1,1)= capitalambda-(m_tilda)*bitesPERlanding*betaHV*x(1,1)*x(8,1)/(x(7,1)+x(8,1))-muh*x(1,1);
dx(2,1)= (m_tilda)*bitesPERlanding*betaHV*x(1,1)*x(8,1)/(x(7,1)+x(8,1))-(del1+eta+muh)*x(2,1);
dx(3,1)=p*eta*x(2,1)-(del2+alpha1+muh)*x(3,1);
dx(4,1)=(1-p)*eta*x(2,1)-(del2+alpha2+muh)*x(4,1);
dx(5,1)=alpha1*x(3,1)+alpha2*x(4,1)-muh*x(5,1);
dx(6,1)=capitalambda -del1*x(2,1)-del2*( x(3,1)+x(4,1) ) - muh*x(6,1);
dx(7,1)=muv*( x(7,1)+x(8,1) )- bitesPERlanding*betaVH*x(7,1)*x(2,1)/(x(6,1)+Nc) - (artificialdeathrate1(t) + muv)*x(7,1);
dx(8,1)= bitesPERlanding*betaVH*x(7,1)*x(2,1)/(x(6,1)+Nc)-(artificialdeathrate1(t) + muv)*x(8,1);
dx(9,1)= -artificialdeathrate1(t) * x(9,1);
dx(10,1)= (m_tilda)*bitesPERlanding*betaHV*x(1,1)*x(8,1)/x(9,1);
The function file: artificialdeathrate1
function art1=artificialdeathrate1(t)
global Q Hs H Cs C
art1= Q*Hs*iOFt(t)/H + (1-Q)*Cs*oOFt(t)/C ;
The function file: iOFt
function i = iOFt(t)
global gamma tfinal Ct0 b1
if t>=gamma && t<=tfinal
i = Ct0*exp(-b1*(t-gamma));
else
i =0;
end
The function file: oOFt
function o = oOFt(t)
global gamma Ct0 b2 tfinal
if (t>=gamma && t<=tfinal)
o = Ct0*exp(-b2*(t-gamma));
else
o = 0;
end

If your working code is even remotely as messy as the code you posted, then that should IMHO the first thing you should address.
I cleaned up iOFt, oOFt a bit for you, since those were quite easy to handle. I tried my best at NoEffects_derivative_6_15_2012. What I'd personally change to your code is using decent indexes. You have 10 variables, there is no way that if you let your code rest for a few weeks or months, that you will remember what state 7 is for example. So instead of using (7,1), you might want to rewrite your ODE either using verbose names and then retrieving/storing them in the x and dx vectors. Or use indexes that make it clear what is happening.
E.g.
function ODE(t,x)
insectsInfected = x(1);
humansInfected = x(2);
%etc
dInsectsInfected = %some function of the rest
dHumansInfected = %some function of the rest
% etc
dx = [dInsectsInfected; dHumansInfected; ...];
or
function ODE(t,x)
iInsectsInfected = 1;
iHumansInfected = 2;
%etc
dx(iInsectsInfected) = %some function of x(i...)
dx(iHumansInfected) = %some function of x(i...)
%etc
When you don't do such things, you might end up using x(6,1) instead of e.g. x(3,1) in some formulas and it might take you hours to spot such a thing. If you use verbose names, it takes a bit longer to type, but it makes debugging a lot easier and if you understand your equations, it should be more obvious when such an error happens.
Also, don't hesitate to put spaces inside your formulas, it makes reading much easier. If you have some sub-expressions that are meaningful (e.g. if (1-p)*eta*x(2,1) is the number of insects that are dying of the disease, just put it in a variable dyingInsects and use that everywhere it occurs). If you align your assignments (as I've done above), this might add to code that is easier to read and understand.
With regard to the ODE solver, if you are sure your implementation is correct, I'd also try a solver for stiff problems (unless you are absolutely sure you don't have a stiff system).