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 {{Elizabeth_Polidan}}   {{Elizabeth_Polidan}} 
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 +  Plots from running Matlab 
   
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Revision as of 02:59, 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
Plots from running Matlab

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
