

Line 1: 
Line 1: 
 {{Elizabeth_Polidan}}   {{Elizabeth_Polidan}} 
 +  
 +  [[Media:Assignment4.pdfAnswers to questions in A second chemostat assignment]] 
   
 { class="wikitable" border="1"   { class="wikitable" border="1" 
Revision as of 00:55, 8 February 2013
Elizabeth Polidan
BIOL 398.03 / MATH 388
 Loyola Marymount University
Elizabeth Polidan Home
Course Home
Answers to questions in A second chemostat assignment

Figure 1
Original script
t0 = 0; % initial time
t1 = 100; % final time
c0 = 0; % initial nutrient conc
N0 = 30;
c20 = 0;
S0 = [c0;N0;c20];
q = 0.15
u = 120
r = 1.0
K = 5
V = 0.5
u2 = 60


Figure 2
t0 = 0; % initial time
t1 = 100; % final time
c0 = 0; % initial nutrient conc
N0 = 30; c20 = 0;
S0 = [c0;N0;c20];
q = 0.15
u = 30
r = 1.0
K = 5
V = 0.5
u2 = 60


Figure 3
t0 = 0; % initial time
t1 = 100; % final time
c0 = 0; % initial nutrient conc
N0 = 30;
c20 = 0;
S0 = [c0;N0;c20];
q = 0.15
u = 120
r = 1.0
K = 5
V = 0.25
u2 = 60


Figure 4
t0 = 0; % initial time
t1 = 100; % final time
c0 = 0; % initial nutrient conc
N0 = 30;
c20 = 0;
S0 = [c0;N0;c20];
q = 0.30
u = 120
r = 1.0
K = 5
V = 0.5
u2 = 60


Figure 5
t0 = 0; % initial time
t1 = 100; % final time
c0 = 0; % initial nutrient conc
N0 = 30;
c20 = 0;
S0 = [c0;N0;c20];
q = 0.45
u = 120
r = 1.0
K = 5
V = 0.5
u2 = 60


Figure 6
t0 = 0; % initial time
t1 = 100; % final time
c0 = 0; % initial nutrient conc
N0 = 30;
c20 = 0;
S0 = [c0;N0;c20];
q = 0.15
u = 120
r = 1.0
K = 5
V = 0.5
u2 = 30
