1.
WENTWORTH INSTITUTE OF TECHNOLOGY
Cleaning the Great Lakes
Raed Alsulami, Mark Falcone, and Simon Filippenko
MATH 2500­5
Professor Mel Henriksen
12 April, 2016

2. Alsulami, Falcone, Filippenko 1
The basis of this project is a study being conducted on the Great Lakes in an attempt to
improve the quality and purity of the lakes. The objective of this team is to estimate the time it
would take to reduce the pollution in each of the Great Lakes. This is based on the assumption
that all water coming from outside the Great Lakes is clean and pure. As this water flows into
the Lakes, that new cleaner water will flow from Lake to Lake, and slowly reduce the
concentration of pollution in each Lake. If each Lake is treated as a separate tank or container,
compartmental analysis can be used to help form equations to evaluate. Once these equations
have been formed, tools such as Excel and MATLAB can be used to determine the time it would
take to clean the Great Lakes.
Part 1:​ The purpose of these equations is to calculate, in years, the time it would take to drain
all of the water out of each lake. They were found using the volumes of each lake along with the
outflow rates given in the figure shown above. The volume of each lake was divided by the
outflow rate per year to yield the time.
Lake Superior:
610 mi 74 years2 3
× 1 year
15 mi3 = 1
Lake Michigan:
062 mi 7.95 years1 3
× 1 year
38 mi3 = 2
Lake Huron:
35 mi 3.75 years9 3
× 1 year
68 mi3 = 1
Lake Erie:
28 mi .51 years1 3
× 1 year
85 mi3 = 1

3. Alsulami, Falcone, Filippenko 2
Lake Ontario:
54 mi .58 years3 3
× 1 year
99 mi3 = 3
Part 2: ​The Great Lakes can all be represented by their own tank with only clean water going in
and polluted water going out. An equation can be found for the change in pollutant over time,
and time can be found if a pollution level is plugged in. The levels of pollutant provided are 50%
and 5% for each lake, so the time at which this concentration would be found must be solved
for.
Lake Superior: (S)
S(0) 100% )(2, 10 mi ) , 10 slugs = ( mi3
slugs
6 3
= 2 6
S(t ) 50% )(2, 10mi ) slugs 1 = ( mi3
slugs
6 3
= 2
2,610
(t ) 5% )(2, 10 mi ) slugsS 2 = ( mi3
slugs
6 3
= 20
2,610
Equation: S dt
ds
= −15
2610
Separate the above equation to get:
dt S
ds
= −15
2610
Integrate the equation to get:
dt∫
S
ds
=∫
−15
2610
n(S) t l + C1 = −15
2610 + C2
n(S) t l = −15
2610 + C3
Use the exponential­log relationship to get S alone:
(t) C S = e t−15
2610
4
Solve using the initial condition at t=0:
(0) C , S(0) 610 S = e (0)−15
2610
4 = 2
610 2 = C4
Plug this value into the general equation
(t) 610e S = 2 t−15
2610
This equation can be used to solve for the time at which the pollution is 50% and 5%.
Calculations for 50%:
(t ) 610e S 1 = 2 t−15
2610 1
610e 2
2610
= 2 t−15
2610 1
n(0.5) n(e ) l = l t−15
2610 1
.693 − t − 0 = 15
2610 1
20 years 218 days t1 = 1 +
The same process can be done to obtain a value for 5% pollution:
or 5% pollution, S(t ) F 2 = 20
2610
(t ) , 10e S 2 = 2 6 − t15
3190 2

4. Alsulami, Falcone, Filippenko 3
, 10e 20
2,610
= 2 6 − t15
2610 2
)n( ) n( l 1
20 = l e− t15
2,610 2
.995 t − 2 = − 15
2,610 2
21 years 47 days t2 = 5 +
The following table shows the time it would take for the pollution content to reach the desired
levels in each lake, that is 50% pollution (at t1) and 5% pollution (at t2).
(Y ears) t1 (extra days) t1 (Y ears) t2 (extra days) t2
Superior 120 222 521 95
Michigan 19 135 83 263
Huron 9 193 41 69
Erie 1 15 4 186
Ontario 2 175 10 259
Part 3: ​The water from several of the lakes flows directly into another lake, so it’s important to
consider this when analyzing the pollution content of each lake at any given time. To get a
complete model for the pollution in the lakes as a function of time, we can use scalar equations
that were setup to analyze the previous multi­compartmental model. The assumptions made to
create this model are that all lakes have the same initial concentration of pollutant and all
incoming water other than what is coming from another lake is completely clean.
i.) Each lake will be assigned a letter to represent it, as it will become increasingly difficult to
create a matrix equation without properly assigned variables and the organization required to
separate the lakes and their respective equations from one another. For the sake of simplicity,
each lake will be assigned the first letter of its name as a variable.
Lake Superior (S):
)( ) )S( yr
−15mi3 Sslug
2610mi3 = ( −15
2610 yr
slugs
Lake Michigan (M):
)( ) )M( yr
−38mi3 Mslug
1062mi3 = ( −38
1062 yr
slugs
Lake Huron (H):

5. Alsulami, Falcone, Filippenko 4
)( ) )( ) )( ) ( )M )S )H]( yr
−38mi3 Mslug
1062mi3 + ( yr
15mi3 Sslug
2610mi3 − ( yr
68mi3 Hslug
935mi3 = [ 38
1062 + ( 15
2610 − ( 68
935 yr
slugs
Lake Erie (E):
)( ) )( ) ( )H )E]( yr
68mi3 Hslug
935mi3 − ( yr
85mi3 Eslug
354mi3 = [ 68
935 − ( 85
354 yr
slugs
Lake Ontario (O):
)( ) )( ) ( )E )O]( yr
85mi3 Eslug
128mi3 − ( yr
99mi3 Oslug
354mi3 = [ 85
128 − ( 99
354 yr
slugs
ii.) The equations in part i.) yield the following 5x5 matrix:
iii.) These equations were then plugged into the following MATLAB code, similar to what was
demonstrated in class, to obtain Eigenvalues and Eigenvectors.
The order of the matrix is as follows: Superior, Michigan, Huron, Erie, Ontario.
A=[­15/2610,0,0,0,0;0,­38/1062,0,0,0;15/2610,38/1062,­68/935,0,0;0,0,68/935,­85/128,0;0,0,0,8
5/128,­99/354];
p=poly(A);
evalues=roots(p);
I=eye(5,5);
r1=evalues(1);
u1=null(A­r1*I);
r2=evalues(2);
u2=null(A­r2*I);
r3=evalues(3);
u3=null(A­r3*I);
r4=evalues(4);
u4=null(A­r4*I);
r5=evalues(5);
u5=null(A­r5*I);
The results are as follows:

6. Alsulami, Falcone, Filippenko 5
r1: ­0.6641
r2:­0.2797
r3:­0.0727
r4:­0.0358
r5:­0.0057
u1=[0,0,0,­0.5010,0.8655]
u2=[0,0,0,0,­1]
u3=[0,0,­0.9241,­0.1137,­0.3647]
u4=[0,0.6995,0.6775,0.0784,0.2135]
u5=[­0.9960,0,­0.0855,­0.0094,­0.0229]
When the roots are real and distinct, use the equation:
where “u” represents the eigenvector, r represents the eigenvalue, and C is anCeu rt →
unknown constant.
Superior:
.9960(C )e − 0 5
(−0.0057t)
Michigan:
.6995(C )e 0 4
(−0.0358t)
Huron:
.9241(C )e .6775(C )e .0855(C )e − 0 3
(−0.0727t) + 0 4
(−0.0358t) − 0 5
(−0.0057t)
Erie:
.5010(C )e − .1137(C )e .0784(C )e .0094(C )e − 0 1
(−0.6641t) + 0 3
(−0.0727t) + 0 4
(−0.0358t) − 0 5
(−0.0057t)
Ontario:
.8655(C )e (C )e .3647(C )e .2135(C )e 0229(C )e 0 1
(−0.6641t) − 1 2
(−0.2797t) − 0 3
(−0.0727t) + 0 4
(−0.0358t) − . 5
(−0.0057t)
The C values can be found by setting t=0, and using initial conditions.
Superior
(t) − .9960(C )e S = 0 5
(−0.0057t)
The volume of Lake Superior at t=0 was determined to be S(0)=2,610 : mi3
(0) − .9960(C )e , 10 S = 0 5
(−0.0057(0)) = 2 6

7. Alsulami, Falcone, Filippenko 6
Isolate :C5
− 620.48 C5 = 2
Michigan
(t) .6995(C )e M = 0 4
(−0.0358t)
=1062(0) .6995(C )e M = 0 4
(−0.0358(0))
518.23 C4 = 1
Huron:
(t) − .9241(C )e .6775(C )e .0855(C )e H = 0 3
(−0.0727t) + 0 4
(−0.0358t) − 0 5
(−0.0057t)
(0) − .9241(C )e .6775(C )e .0855(C )e 35 H = 0 3
(−0.0727(0)) + 0 4
(−0.0358(0)) − 0 5
(−0.0057(0)) = 9
To obtain the value for , substitute the and values from the two previous equations.C3 C4 C5
(0) − .9241(C )e .6775(1518.23)e .0855(− 620.48)e 35 H = 0 3
(−0.0727(0)) + 0 (−0.0358(0)) − 0 2 (−0.0057(0)) = 9
(0) − .9241(C )e 028.60 24.05 35 H = 0 3
(−0.0727(0)) + 1 + 2 = 9
43.74 C3 = 3
Erie​:
(t) − .5010(C )e − .1137(C )e .0784(C )e .0094(C )e E = 0 1
(−0.6641t) + 0 3
(−0.0727t) + 0 4
(−0.0358t) − 0 5
(−0.0057t)
(0) − .5010(C )e − .1137(C )e .0784(C )e .0094(C )e E = 0 1
(−0.6641(0)) + 0 3
(−0.0727(0)) + 0 4
(−0.0358(0)) − 0 5
(−0.0057(0))
(0) − .5010(C )e − .1137(343.74)e .0784(1518.23)e E = 0 1
(−0.6641(0)) + 0 (−0.0727(0)) + 0 (−0.0358(0))
.0094(− 620.48)e 28 − 0 2 (−0.0057(0)) = 1
− 6.75 C1 = 4

8. Alsulami, Falcone, Filippenko 7
Ontario​:
(t) .8655(C )e (C )e .3647(C )e .2135(C )e 0229(C )e O = 0 1
(−0.6641t) − 1 2
(−0.2797t) − 0 3
(−0.0727t) + 0 4
(−0.0358t) − . 5
(−0.0057t)
54(continued from line above) = 3
(0) .8655(C )e (C )e .3647(C )e O = 0 1
(−0.6641(0)) − 1 2
(−0.2797(0)) − 0 3
(−0.0727(0))
.2135(C )e 0229(C )e + 0 4
(−0.0358(0)) − . 5
(−0.0057t)
(0) .8655(− 6.75)e (C )e .3647(343.74)e O = 0 4 (−0.6641(0)) − 1 2
(−0.2797(0)) − 0 (−0.0727(0))
.2135(1518.23)e 0229(− 620.48)e + 0 (−0.0358(0)) − . 2 (−0.0057(0))
− 35.673 C2 = 1
iv.) Write simplified equation for each lake:
(t) 609.998e S = 2 −0.0057t
(t) 062.002e M = 1 −0.0358t
(t) − 17.65e 028.601e 24.051e H = 3 (−0.0727t) + 1 (−0.0358t) + 2 (−0.0057t)
(t) 3.422e 9.083e 19.029e 4.633e E = 2 (−0.6641t) − 3 (−0.0727t) + 1 (−0.0358t) + 2 (−0.0057t)
(t) − 0.462e 35.673e 25.36e 24.14e 0.009e O = 4 (−0.6641(t)) + 1 (−0.2797(t)) − 1 (−0.0727(t)) + 3 (−0.0358(t)) − 6 (−0.0057(t))
v.) Use MATLAB to solve each equation for the time at which the pollution level is at 50% its
initial level. As for an equation, use: (S equation used as example)
609.998e 2
2610
= 2 −0.0057t → 2
1
= e−0.0057t
This yields the following MATLAB code to solve this equation:
syms t
S = solve(0.5 == exp(­0.0057*t));
vpa(S,6)
This yields a value of t=121.605 years for Lake Superior, or 121 years and 221 days.
Running this same script but replacing the equation inside the solve() parentheses with the
equations displayed in part iv.) and substituting V/2 for 50% and V/20 for 5% if V is the volume
of the lake in question.. This yields the numbers displayed in the following table.

9. Alsulami, Falcone, Filippenko 8
(time for 50% in years) t1 (time for 5% in years) t2
Superior 121.605 525.567
Michigan 19.362 83.680
Huron 33.482 275.127
Erie 24.750 237.127
Ontario 21.574 215.63
When these results are compared to the results displayed in the table in part 2, it’s clear
that in modelling Lake Superior and Lake Michigan, these two methods for modelling match up
closely. However, as more time passes, the results from Part 2 become more inaccurate. In this
context, the results for 50% pollution are close, but the results for 5% show a significant
deviation. It is also notable that the first method appears to be increasingly inaccurate as we get
toward the end of the system. That is to say that, the more compartments the water passes
through the less accurate the method in Part 2 becomes. This makes using this method with
exponentials important for evaluating lakes at the end of the chain like Erie and Ontario.
vi.) Use MATLAB to solve each equation for the time at which the pollution level is at 5% its
initial level. The procedure is the same as part v.) except in this case, V/20 is used rather than
V/2, where V represents the volume of the lake. The same script as in part v.) can be used to
make this calculation, with some minor changes. The script is as follows:
syms t;
S = solve(0.05 == exp(­0.0057*t));
vpa(S,6)
This yields a value of t=525.567 years for Lake Superior, or 525 years and 207 days.
The table above depicts the values for this part as well. It is important to note that, as mentioned
above, this method is a great deal more accurate than the simplified version used in Part 2. This
means that, while the previous method may work well enough on Lake Superior and Lake
Michigan which have mostly or entirely pure water flowing into them, it does not allow for much
accuracy when evaluating a lake like Ontario that is receiving polluted water from other lakes.

10. Alsulami, Falcone, Filippenko 9
vii.) On a single graph, display the mass of pollutant vs. time for each lake (assuming 1 slug per
cubic mile because no initial concentration is given).
The graph above displays the mass of pollutant in each lake as a function of time. This
gives a good visual representation of how the pollution in each lake would decay over time, and
gives a visual representation of approximately when each lake would reach the goal of having 0
pollution. These functions are exponential, which means some of them may never truly reach
zero pollution mass, which means the best that can be done is to find a near­zero value, and this
graph can be used to find such a value for each lake. This graph was generated using the
MATLAB code on the following page.
s=@(t)2609.998*exp(­0.0057*t);
m=@(t)1062.002*exp(­0.0358*t);
h=@(t)­317.650*exp(­0.0727*t)+1028.601*exp(­0.0358*t)+224.051*exp(­0.0057*t);
e=@(t)23.422*exp(­0.6641*t)­39.083*exp(­0.0727*t)+119.029*exp(­0.0358*t);
o=@(t)(­40.462*exp(­.6641*t)+(135.673*exp(­0.2797*t))­(125.362*exp(­0.0727*t))+(324.142*
exp(­0.0358*t))+(60.009*exp(­0.0057*t)));
tval=0:700;

11. Alsulami, Falcone, Filippenko 10
plot(tval,s(tval),'Magenta');
hold on;
plot(tval,m(tval),'Blue');
hold on;
plot(tval,h(tval),'Red');
hold on;
plot(tval,e(tval),'Green');
hold on;
plot(tval,o(tval),'Black');
title('Mass of Pollutant vs. Time');
xlabel('Time (Years)');
ylabel('Mass of Pollutant (Slugs)');
legend('s(t)','m(t)','h(t)','e(t)','o(t)');
viii.) On another graph, plot the concentration vs. time for each lake.
The graph above displays the concentration of pollution in each lake as a function of
time. This could prove very valuable in the event that the goal of this exercise is to get each lake
below a certain percentage of pollution, or to decrease the pollution by a certain percentage. This
was gained by dividing the constant that accompanies each exponential by the volume of the lake

12. Alsulami, Falcone, Filippenko 11
in question, and creating a graph of those modified functions. This was created using the
following code in MATLAB.
s=@(t)(2609.998./2610)*exp(­0.0057*t);
m=@(t)(1062.002./1062)*exp(­0.0358*t);
h=@(t)(­317.650./935)*exp(­0.0727*t)+(1028.601./935)*exp(­0.0358*t)+(224.051./935)*exp(­0
.0057*t);
e=@(t)(23.422./128)*exp(­0.6641*t)­(39.083./128)*exp(­0.0727*t)+(119.029./128)*exp(­0.0358
*t);
o=@(t)(­40.462./354)*exp(­.6641*t)+(135.673./354)*exp(­0.2797*t)­(125.362./354)*exp(­0.072
7*t)+(324.142./354)*exp(­0.0358*t)+(60.009./354)*exp(­0.0057*t);
tval=0:700;
plot(tval,s(tval),'Magenta');
hold on;
plot(tval,m(tval),'Blue');
hold on;
plot(tval,h(tval),'Red');
hold on;
plot(tval,e(tval),'Green');
hold on;
plot(tval,o(tval),'Black');
title('Concentration of Pollution vs. Time');
xlabel('Time (Years)');
ylabel('Concentration of Pollution');
legend('Superior','Michigan','Huron','Erie','Ontario');
Conclusion:
While there are many ways to model a problem like the flow of pollution through, and
the cleaning of the Great Lakes, it is clear that some ways are better than others. Throughout this
project, multiple options were explored, and it seems apparent which ones would be more
practical for certain situations, and which situations these more effective equations would be
unnecessary in. This provides useful insight into the different options to be considered when
modelling, and when each option might be best applied for the sake of both effectiveness and
efficiency. While the method used in Part 3 is the most complete method, and will likely prove to
be the most accurate, it is also the most time consuming option, and has by far and wide the most
opportunities for error during the calculations.