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Quiz 2
12/14/2005
Problem 1
(a) The longest path on the tree is 13 units.
Therefore, for the 3center problem,13/6 is
the smallest maximum distance we can hope for.
This can indeed be achieved by placing
three centers on the BC edge: one is a distance of 7/6 from B; the second is a distance of
(7/6) + 2(13/6) = 33/6 = 5.5 from B; and the third is a distance of (33/6) + 2(13/6) = 59/6
from B (or 1/6 to the left of C). This is also the unique optimal solution to the 3center
problem.
(b) Now we also have to worry about the “geometry” of the network.
Some thought will
convince you that the minimum maximum distance we can hope for is 2 units. To
achieve this we must definitely place one facility at C. The other three facilities must be
placed to the left of the point which is 8 units away from B on BC (or 2 units away from
C on BC).
For example one possible solution would place one of the three other facilities
at B, a second facility 4 units away from B on BC, and the third facility 7 units away
from B on BC.
(c) We are now looking for the 1absolutemedian of the network, because all we want to
do is minimize the expected travel time from the facility to a random demand.
The easy way to find this 1median is by using the majority theorem, thus requiring that
we must have as much length to the “left” of the facility as to its “right”.
(Length is the
same as weight in this case, since the demands are distributed uniformly on the edges of
the network.) Then, by inspection, the optimal location is at the point which is 6 units
away from B on BC. (Note that we then have 8 units of length to the left and 8 to the
right.)
A more formal way is by assuming that the optimal location is at a point which is
x
units
away from B on BC.
We then have for the expected travel distance:
x
x
2
1
10
−
x
10
−
x
4
x
2
−
12
x
+
95
E
[
D
]
=
⋅
+
⋅
(
x
+
)
+
⋅
+
⋅
(11
−
x
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 Fall '06
 hansman

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