Question

ANSWER PART B, I HAVE PART A.

MATLAB ONLY!! WRITE IN COMPUTER SO I CAN COPY PASTE!! The student must use the following case provided and apply most of the commands leared in the previous modules. You should use the command folder that you must be expanding with each module. Case (Gilat, 2015): The concentration of a drug in the body Cp can be modeled by the equation: Doka (e-ketek) Va (ka-k) Where Do is the dosage administrated (mg). Ve is the volume of distribution (L), k, is the absorption rate constant (h ), k, is the elimination rate constant (h ), and t is the time (h) since the drug was administered. For a certain drug, the following quantities are given Dc 150 mg, V = 50 L, ke=0.4h, ka = 1.6 -1 Create a program editor that use while-loop prints in a single graph the function of sine a) A single dose is administered at t= 0. Calculate and plot Cp vs t for 10 hours. Hint: to get a smooth graph use 0.1 as time interval b) A fist dose is administered at t = 0, and subsequently four more doses are administered at intervals of 4 hours (ie at 4, 8, 12 and 16). Calculate and plot Cp vs t for 24 hours. Hint: use continue command in this case. c) Build the flowchart that represent the program.

Answer 1

EXPLANATION:

Here the concentration values(Cp) along with time interval for 0,4,8,12,16 are also displayed along with the plot of Cp vs t for 24 Hours.

continue statement is used to omit formulated(as per question) Cp values for interval (t) 20 and 24 or for the value of i 6 and 7.

here set(gca,'XTick',t); command is used to give x axis with intervals 0, 4, 8, 12, 16, 20 and 24

clc
clear all
close all
Dg=150;
Vd=50;
ke=0.4;
ka=1.6;
t=0:4:24;
i=1;
t1=0;
fprintf('Concentration of Drug in body:\n');
for i=1:length(t)
if(i>5)
% as at regular interval of 4hours, 4 more does are administered
% after first dose given at t=0
% So, total 5 doses are given at t=0,4,8,12,16 time interval
% after that no does is given.So,concentration of drug is taken for
%last 2 intervals of t=20 ,24 as 0.Because final plot will be taken
%as t(0:4:24) vs cp.So for last 2 intervals plot requires value for
%cp.
Cp(i)=0.0;
% as alrady all doses are given,so continue statement move the flow
%control to for loop and increnent the value of i.
continue;
end
if(i<=5)
% insert dose concentration details in array for interval
% t=0,4,8,12,16
Cp(i)=(Dg/Vd)*(ka/(ka-ke))*(exp(-ke*t(i))-exp(-ka*t(i)));
% print t and Cp value
fprintf('t = %d : %f \n ',t(i),Cp(i));
end

end
% plot graph for t=24 hours and Cp with blue line plotting
t=0:4:24;
plot(t,Cp, 'blue')
%x axis labeling
xlabel('t(interval)-->')
%y axis labeling
ylabel('Cp(Concentration)-->')
% set x axis values from 0 to 24 at regular interval of 4
set(gca,'XTick',t);
%x graph title
title('Concentration(Cp) and time interval(t) plot ')
fprintf('\n');

SCREEN SHOT

clc clear all close all Dg=150; Vd=50; ke=0.4; ka=1.6; t=0:4:24; i=1; t1=0; fprintf('Concentration of Drug in body:\n'); for i=1:length(t) if(i>5) as at regular interval of 4hours, 4 more does are administered $ after first dose given at t=0 So, total 5 doses are given at t=0,4,8,12,16 time interval * after that no does is given.So, concentration of drug is taken for Blast 2 intervals of t=20 ,24 as 0.Because final plot will be taken as t(0:4:24) vs cp. So for last 2 intervals plot requires value for scp. Cp (i)=0.0; * as alrady all doses are given, so continue statement move the flow control to for loop and increnent the value of i. continue; end

if(i<=5) * insert dose concentration details in array for interval t=0,4,8,12,16 Cp (1) = (Dg/Va) * (ka/ (ka-ke)) * (exp(-ke*t (1))-exp(-ka*t (1))); print t and Cp value fprintf('t = $d : 3f \n',t(i), Cp (i)); end - end plot graph for t=24 hours and Cp with blue line plotting t=0: 4:24; plot(t, Cp, 'blue') tx axis labeling xlabel('t(interval)-->') ty axis labeling ylabel('Cp (Concentration) -->') set x axis values from 0 to 24 at regular interval of 4 set (gca, 'XTick',t); fx graph title title('Concentration (Cp) and time interval(t) plot ') fprintf('\n');

OUTPUT

Concentration of Drug in body: t = 0: 0.000000 t = 4 : 0.800940 t = 8: 0.163038 t = 12 : 0.032919 t = 16 : 0.006646

- OX Figure 1 File Edit View Insert Tools Desktop Window Help OHSAS E DO Concentration(Cp) and time interval(t) plot Cp(Concentration)--> o 4 8 16 20 24 12 t(interval)-->