Light timing at a road intersection with a steady rate of vehide arrivais. A light is to b...

Light timing at a road intersection with a steady rate of vehide arrivais. A light is to be timed at an intersection of two one-way roads for the period of the moming rush hour. The intersection is relativeiy isolated from other décisions being made in the highway network. The length of a light cycle has already been set at three minutes. There are 40 light cycles in the morning rush hour. Of this cycle time, a certain amount is lost to the movement of vehicles. That is, when the light turns from red to green in each of the two directions, there is a delay in vehicles entering the intersection. Generaliy, this total loss time is approximated by the length of the amber phase. We wiil use 20 seconds as the loss time. A survey of traffic on the two roads during the morning rush hour has given ar¬rivai rates at the intersection through time. (See the accompanying figure.)

The light cycle is the standard green-amber-red sequence. The décision variables for light timing are

The diagram of light timing shows that green time, direction 1, corresponds closely to red time, direction 2, and green time, direction 2, corresponds closely to red time, direction 1.

Thus, the sum of green time in the two directions is the total usable time in the light cycle. That is,

Not all the green time in each direction is always needed. Sometimes, all vehicles in line in a particular direction can be cleared in less than the available green time, leaving a slack green time when the light is green when no vehicles are available to clear. Thus, we need four new variables:

The sum of these variables is equal to the usable time in the light cycle,

and this equation replaces the previous equation.

An intersection has a clearance rate in each direction, the maximum rate at which vehicles move through the intersection in each direction, once flow “gets up to speed.”

Let:

r1 = vehicles per minute that can clear in direction 1, and

r2 = vehicles per minute that can clear in direction 2.

The vehicle movement through the intersection in direction 1 is the product of the clearance rate and used green time r1 xt. The movement in direction 2 is by similar convention . Assume that both directions start with no vehicles in line waiting at the time of the first green phase.

Let:

S1t = vehicle storage at the end of the tth light cycle in direction 1, unknown, and

S2t = vehicle storage at the end of the tth light cycle in direction 2, unknown

Write the storage/inventory equations for directions 1 and 2 that calculate S1t and S2t from the data given and the decisions made on green time in each direction.

Using the storage equations, show how to minimize the maximum line length (storage of vehicles) in any direction through the morning rush period.