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A metropolitan area lies along the two banks of a river – or the two shores of a bay as in the SF Bay Area. This body of water is W km wide and L km long. Studies show that F cars will travel across...

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A metropolitan area lies along the two banks of a river – or the two shores of a bay as in the SF Bay Area. This body of water is W km wide and L km long. Studies show that F cars
will travel across the water each day if bridges with enough capacity are constructed to carry the
traffic. We estimate that the daily interest society must pay on the bonds issued to pay for a
bridge with a capacity to carry f cars per day is C = (a + bf)W dollars per day. We also assume
that the origins and destinations of the F trips are uniformly and independently distributed along
the two shores and that everyone uses the bridge that involves the shortest detour. These detours
cost commuters . dollars per vehicle-km, including time wasted and out of pocket expenses. Let
n be the number of bridges constructed between the two shores (and note that bridges cannot be
constructed at the northern or southern edge of the metropolitan area).
Assume that the n bridges are equidistant, that a trip requires a detour if its origin and destination are between the same two consecutive bridges or between a bridge and the
northern or southern edge of the metropolitan area, and that the average length of a detour
is 1/3 of the separation between bridges, L/(n+1).
Write a formula for the number of bridges that
minimizes the sum of the construction and detour costs. (This is not necessarily the
solution that society would choose based on the political process, but it should be a rough
approximation.)
Answered Same Day Dec 22, 2021

Solution

Robert answered on Dec 22 2021
127 Votes
Sol.
It is given in the problem that in the whole bay area there are originated a total of
F trips per day. As the origins and destinations of the F trips are uniformly and independently distributed along
the two shores we can assume that there is the
same number of trips that originated and terminated in each area between two successive

idges (an area is shown in figure). If n is the number of
idges we have n+1 areas in
each side of the
idge, which means a total of 2(n+1)...
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