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KA6423 P57600 Assignment 1
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  • 1. Page 1 of 5 Intelligent Urban Transport Management System Assignment 1Name Muhammad bin RamlanMatrix No. P57600Subject KA 6423Session 2012/2013Lecturer Prof Ir Dr RizaAtiq O.K. Rahmat
  • 2. Page 2 of 5QuestionBelow is an illustration of two signalised intersections which are to be optimised.Based on the given data, you are required to determine optimum common cycletime, green time split for each intersection and offset time. 95 204 102 89 200 92 356 225 521 490 Queue length = 15 cars 80 95 180 190 80 180 90 95 152 95 400m
  • 3. Page 3 of 5Answer JUNCTION 1 Number Of Saturation Flow Per Lane Saturation Flow Actual Flows Flow / Saturation Flow Green Time Phase (pcu/Hr) (pcu/Hr) (pcu/Hr) Lanes Ration Split 1 2 1800 3600 1,005 0.28 0.477 2 2 1800 3600 401 0.11 0.190 3 2 1800 3600 350 0.10 0.166 4 2 1800 3600 350 0.10 0.166 Jumlah, Y 0.59 1.000 Cycle Time Co = 1.5 L + 1.5 L + 5 5 L=4*4= 16 seconds [lost time per phase 4 seconds] Co = 1-Y Co = (1.5 x 16 + 5) / ( 1 - 0.59) = 69.88 Take Co = 70 seconds Effective Green Time = 70 -L = 54 seconds
  • 4. Page 4 of 5Green Time SplitPhase Green Time Round-Up To 1 25.769 26 Take amber time = 3 seconds 2 10.282 11 Take all red time = 1 seconds 3 8.974 9 Total of all amber and red time = 16 sec 4 8.974 9 total 54 55 Actual CycleTime = 71 secJUNCTION 2 Number Of Saturation Flow Per Lane Saturation Flow Actual Flows Flow / Saturation Flow Green TimePhase (pcu/Hr) (pcu/Hr) (pcu/Hr) Lanes Ration Split 1 2 1800 3600 857 0.24 0.449 2 2 1800 3600 381 0.11 0.200 3 2 1800 3600 327 0.09 0.171 4 2 1800 3600 342 0.10 0.179 Jumlah, Y 0.53 1.000 Co = 1-YCycle Time 1.5 L + 5 Co = L=4*4= 16 seconds [lost time per phase 4 seconds] 1-Y Co = (1.5 x 16 + 5) / ( 1 - 0.59) = 61.67
  • 5. Page 5 of 5 Take Co = 62 seconds Effective Green Time = 62 -L = 46 secondsGreen time split Phase Green Time Round Round-Up To 1 20.67 21 Teke amber time = 3 seconds 2 9.19 10 Take all red time = 1 seconds 3 7.89 8 Total of all amber and red time = 16 sec 4 8.25 9 total 46 48 Actual CycleTime = 64 secOFFSET TIME L=400 meter S=10m/s Q=15 kenderaan Loss time= 2 second= 400/10- (15(2)+2)= 40-32= 8# So, Offset Junction 2 is8 Second