CEE434-Exam1-Solution
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University of Illinois, Urbana Champaign *
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Course
434
Subject
Industrial Engineering
Date
Jan 9, 2024
Type
docx
Pages
4
Uploaded by hdaggett001
Solution: CEE 434 – Exam 1 (2023 Fall)
Problem 1
1.1
Formulation of the primary model:
z
=
¿
η
1
x
1
+
η
2
x
2
+
η
3
x
3
max
¿
s .t .
{
L
1
x
1
+
L
2
x
2
+
L
3
x
3
≤L
W
1
x
1
+
W
2
x
2
+
W
3
x
3
≤ R
P
1
x
1
+
P
2
x
2
+
P
3
x
3
≤TPL
x
i
≥
0,
i
=
1,2,3
Decision variables: x
1
, x
❑
2
, x
3
, the number of families willing to move to each community.
Data needed: η
1
,η
2
,η
3
, L
1
, L
2
, L
3
,L,W
1
,W
2
,W
3
,R ,P
1
,P
2
, P
3
,TPL
1.2
Check the shadow price (marginal value) for each resource. If the shadow price of a resource is larger than zero (implying the constraint is binding), then that resource will affect the solution. 1.3
Check the reduced cost for the community that is zero. The value of the reduced cost represents
the willingness level that needs to be improved for that community to have a certain size in the
solution.
1.4
Formulation of the dual model:
w
=
¿
L y
1
+
Ry
2
+
TPL y
3
min
¿
s .t .
{
L
1
y
1
+
W
1
y
2
+
P
1
y
3
≥η
1
L
2
y
1
+
W
2
y
2
+
P
2
y
3
≥η
2
L
3
y
1
+
W
3
y
2
+
P
3
y
3
≥η
3
Objective function: minimize the total opportunity cost of the resources needed for the new communities (from the perspective of society).
Constraints: the revenue (the willingness to move to the new communities) cannot exceed the total value of consumed resources.
Problem 2
Model formulation:
max z
=
a
1
x
1
−
b
1
x
1
2
+
a
2
x
2
−
b
2
x
2
2
+
a
3
x
3
−
b
3
x
3
2
s .t .
{
L
1
x
1
+
L
2
x
2
+
L
3
x
3
=
L
W
1
x
1
+
W
2
x
2
+
W
3
x
3
≤ R
P
1
x
1
+
P
2
x
2
+
P
3
x
3
≤TPL
x
i
≥
0,
i
=
1,2,3
Standard format:
min z
'
=−
a
1
x
1
+
b
1
x
1
2
−
a
2
x
2
+
b
2
x
2
2
−
a
3
x
3
+
b
3
x
3
2
s .t .
{
L
1
x
1
+
L
2
x
2
+
L
3
x
3
−
L
=
0
… λ
W
1
x
1
+
W
2
x
2
+
W
3
x
3
−
R≤
0
… μ
w
P
1
x
1
+
P
2
x
2
+
P
3
x
3
−
TPL≤
0
…μ
p
x
i
≥
0,
i
=
1,2,3
Lagrangian function:
L
(
x
1
,x
2
, x
3
, λ,μ
)
=−
a
1
x
1
+
b
1
x
1
2
−
a
2
x
2
+
b
2
x
2
2
−
a
3
x
3
+
b
3
x
3
2
+
λ
(
L
1
x
1
+
L
2
x
2
+
L
3
x
3
−
L
)
+
μ
w
(
W
1
x
1
+
W
2
x
2
+
W
3
x
3
−
R
)
+
μ
p
(
P
1
x
1
+
P
2
x
2
+
P
3
x
3
−
TPL
)
KKT conditions:
Optimality:
∂ L
∂x
1
=−
a
1
+
2
b
1
+
λ L
1
+
μ
w
W
1
+
μ
p
P
1
=
0
∂ L
∂x
2
=−
a
2
+
2
b
2
+
λ L
2
+
μ
w
W
2
+
μ
p
P
2
=
0
∂ L
∂x
3
=−
a
3
+
2
b
3
+
λ L
3
+
μ
w
W
3
+
μ
p
P
3
=
0
Feasibility:
L
1
x
1
+
L
2
x
2
+
L
3
x
3
−
L
=
0
W
1
x
1
+
W
2
x
2
+
W
3
x
3
−
R≤
0
P
1
x
1
+
P
2
x
2
+
P
3
x
3
−
TPL ≤
0
Complementary slackness:
μ
w
(
W
1
x
1
+
W
2
x
2
+
W
3
x
3
−
R
)
=
0
μ
p
(
P
1
x
1
+
P
2
x
2
+
P
3
x
3
−
TPL
)
=
0
Nonnegativity:
μ
w
,μ
p
≥
0
Problem 3
Model formulation:
z
=
¿
η
1
x
1
+
η
2
x
2
+
η
3
x
3
max
¿
min p
=
P
1
x
1
+
P
2
x
2
+
P
3
x
3
s .t .
{
L
1
x
1
+
L
2
x
2
+
L
3
x
3
≤ L
W
1
x
1
+
W
2
x
2
+
W
3
x
3
≤R
x
i
≥
0,
i
=
1,2,3
Payoff table:
Willingness (
z
)
Pollutant (
p
)
max z
min p
In order to obtain the values, we need to set each one of the objectives as the single objective and solve the problems with the same set of constraints, respectively.
Problem 4
(Values in
bold were to be filled by students)
String
x
f(x)
f(x)/∑f(x)
0101
5
125
0.044
0010
2
8
0.003
1010
10
1000
0.350
1100
12
1728
0.604
∑f(x) = 2861
Pair
Mating Pool
Crossover Location
Generated String
1
1100
1 | 100
1100
1100
1 | 100
1100
2
1010
10
| 10
1000
1100
11
| 00
1110
String
x
f(x)
f(x)/∑f(x)
1100
12
1728
0.257
1100
12
1728
0.257
1000
8
512
0.076
1110
14
2744
0.409
∑f(x) = 6712
Problem 5
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