Vol. 10 No 5 - Pi Mu Epsilon

BENNINK, RED LIGHT, GREEN LIGHT 36103 0300 01 0202 02 03 00 0101 - 01 02 03 00 (5)0 00 01 02 03This pattern can be extended indefinitely in all four directions. Thus onecan come up with a traffic signal schedule which maximizes unimpeded trafficflow for a lattice of intersections.Computer SimulationI thought it would be fun to simulate traffic flow in a city byimplementing my lattice of intersections and traffic signals schemes on acomputer. Figure 7 shows position-versus-time data extracted from a typical run.In this particular run, four cars were driving north along a street with a 4-intersection signal pattern similar to that shown in Figure 6. One can see smoothcurves indicating periods of acceleration or braking, as well as flat regions wherecars are waiting at a red light. As one would expect. Figure 7 is reminiscent ofFigure 1.To assess the effectiveness of the 4-intersection pattern suggested by mymodel, I decided to compare the average speed of cars in a city with the 4-intersection pattern with that of cars in a city with random signal schedules (SeeFigure 8). The average speed was calculated by taking the total distance traveledby all the cars in the city divided by the length of time the simulation was run(chosen to be 10 periods). As one can see, when there are few cars on the road theaverage speed in the "planned" city (upper line) is quite close to the given speedlimit of 30 mihr (dashed line). Cars in the "random" city (lower line) averagelittle more than half the speed limit. As the traffic becomes heavier, however, theplanned signal scheme seems to be less and less effective, to the point that itproduces results hardly better than those of traffic signals given arbitrary phases.This result surprised me; I expected that the 4-intersection pattern would besignificantly better up to much heavier traffic loads. Apparently, only a shortcaravan ofvehicles are able to pass uninhibited through green lights at successiveintersections.and H respectively. Show that the areas of sectors OBH and OGC areequal.See the accompanying figure.I. Solution by Robert Downes, Mountain Lakes High School, Plainfield,New Jersey.Place the ellipse in the Cartesian plane so that we have B(b, O), C(O, c),E(e, O), and F(f, 0). The equation of the ellipse then becomesSince we were given OE + OF = OB 2, from which we get e + f =b 2, then we find that H(f, celb) and G(e, cfib). The areas of sectors OBHand OCG, K(0BH) and K(OCG), are given byand11. Solution by Skidmore College Problem Group, Saratoga Springs,New York.Place the ellipse in the Cartesian plane (as in Solution I above). Thelinear transforrnationf, given by fix, y) = (xlb, ylc), sends the ellipse to the