ENGR62/MS&E111
Spring 2006
Introduction to Optimization
June 5, 2006
Prof. Amin Saberi
Homework Assignment 3: Solutions
Problem 1
(a) The given data correspond to the following graph (not to scale):
o
A
B
C
D
E
d
40
50
60
10
20
50
70
10
80
60
55
40
Figure 1: Graph for problem 1
where each line represents
two
arcs
1
in opposite directions (note, however, that not all of
these arcs are necessary to find the shortest path since, for instance, we will never come
back to the origin after leaving it). Thus, we have a graph where:
V
=
{
o,A,B,C,D,E,d
}
(1)
E
=
{
(
o,A
)
,
(
o,B
)
,
(
o,C
)
,
(
A,B
)
,
(
B,C
)
,...
}
(2)
(b) The shortest path can be found by solving the following LP:
min
∑
(
i,j
)
∈
E
c
ij
f
ij
subject to
∑
{
k

(
j,k
)
∈
E
}
f
jk
−
∑
{
i

(
i,j
)
∈
E
}
f
ij
= 0
∀
j
∈
V
\{
o,d
}
∑
{
o

(
j,o
)
∈
E
}
f
ok
−
∑
{
i

(
i,o
)
∈
E
}
f
io
= 1
∑
{
d

(
j,d
)
∈
E
}
f
dk
−
∑
{
i

(
i,d
)
∈
E
}
f
id
=
−
1
0
≤
f
ij
≤
1
∀
(
i,j
)
∈
E
(3)
where the
c
ij
correspond to the distances given in the graph above.
(c) The shortest path, as computed in Excel, is given by (
o,A,B,E,D,d
) with a corresponding
distance of 160 miles (see the spreadsheet on the course webpage
2
).
1
Note that the roads can alternatively be interpreted as being unidirectional.
This is ok, and results in a
shortest path of 165.
2
Note: This spreadsheet shows how formulate network flow problems in Excel using a different method than
you may be used to. It is simply another way you can formulate these problems in Excel.
1
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(d) If the numbers in the table are interpreted as times rather than distances then the process
of finding the minimumtimepath is identical to the one shown above. Thus, the shortest
time path is (
o,A,B,E,D,d
) with a corresponding time of 160.
(e) If the numbers in the graph instead correspond to flow capacities, then we can find the
maximum flow between o and d by appending an edge, (
d,o
), to the graph above and
solving the following LP:
max
f
do
subject to
∑
{
k

(
j,k
)
∈
E
′
}
f
jk
−
∑
{
i

(
i,j
)
∈
E
′
}
f
ij
= 0
∀
j
∈
V
0
≤
f
ij
≤
u
ij
∀
(
i,j
)
∈
E
(4)
where the
u
ij
are the “capacities” given in the graph above and
E
′
=
E
∪ {
(
d,o
)
}
as
discussed in the notes.
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 Spring '06
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 shortest path, Flow network, following graph, following LP, cij xij, cij correspond

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