# USER EQUILIBRIUM—EFFECT OF DECREASED TRAFFIC

Two routes connect an origin and a destination, and the flow is 15,000 veh/h. Route 1 has a performance function tl = 4 + 3×1, and route 2 has a function of t2 = b + 6×2, with the x’s expressed in thousands of vehicles per hour and the t’s in minutes. The user- equilibrium flow on route 1 is 9780 veh/h and it is know that both routes are used. First, determine the free-flow travel time on route 2 (the parameter b in route 2’s performance function) and equilibrium travel times. Second, if population declines reduce the number of travelers at the origin, and the total origin–destination flow is reduced to 7000 veh/h, determine user-equilibrium travel times and flows.

SOLUTION Note: Open boxes in equations “ ” are to be completed by the reader

First, using the known traffic flow (9780 veh/h) on route 1, the route-equilibrium travel time can be determined by finding the travel time on route 1 as,

tl = 4 + 3 × = . Because it is known that both routes are used and user equilibrium exists, t1 = t2. It is also know from the conservation of flow that q = x1 + x2, so,

x2 = q x1 = = . With t2 known (since t1 = t2 at user equilibrium) and x2 known, the free-flow travel time on route 2 (the b parameter in route 2’s performance function) can be determined as,

b = t2 6×2 = 6 × = 2.02 min. For the second part of the problem, with the origin to destination flow reduced to 7000 veh/h, we first check that both routes are used (using performance functions), t1(7) = min, t2(0) = min

t1(0) = min, t2(7) = min. Because t1(7) > t2(0) and t2(7) > t1(0), both routes are used. Now, setting route travel times equal and substituting performance functions gives

1 24 3 2.02 6x x+ = +

From conservation of flow, x2 = − , so that

4 + 3×1 = 2.02 + 6( − )

Solving gives x1 = and x2 = − = , or x1 = 4,447 veh/h and x2 =

2,553 veh/h.

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