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Page 1 of 5 Intelligent Urban Transport Management System Assignment 1 Name Muhammad bin Ramlan Matrix No. P57600 Subject KA 6423 Session 2012/2013 Lecturer Prof Ir Dr RizaAtiq O.K. Rahmat

KA6423 P57600 Assignment 1

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Page 1: KA6423 P57600 Assignment 1

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Intelligent Urban Transport Management System

Assignment 1

Name Muhammad bin Ramlan Matrix No. P57600 Subject KA 6423 Session 2012/2013 Lecturer Prof Ir Dr RizaAtiq O.K. Rahmat

Page 2: KA6423 P57600 Assignment 1

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Question

Below is an illustration of two signalised intersections which are to be optimised.

Based on the given data, you are required to determine optimum common cycle

time, green time split for each intersection and offset time.

356

521

225

490

95

190

80

18095 152 95 80 180 90

95 204 102 89 200 92

400m

Queue length = 15 cars

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Answer

JUNCTION 1

Phase Number Of Lanes

Saturation Flow Per Lane (pcu/Hr)

Saturation Flow (pcu/Hr)

Actual Flows (pcu/Hr)

Flow / Saturation Flow Ration

Green Time 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

L = 4 * 4 = 16 seconds [lost time per phase 4 seconds]

Co = (1.5 x 16 + 5) / ( 1 - 0.59) = 69.88

Take Co = 70 seconds

Effective Green Time = 70 - L = 54 seconds

Co =

1.5 L + 5 Co =

1.5 L + 5

1 - Y

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Green Time Split

Phase 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 sec

JUNCTION 2

Phase Number Of Lanes

Saturation Flow Per Lane (pcu/Hr)

Saturation Flow (pcu/Hr)

Actual Flows (pcu/Hr)

Flow / Saturation Flow Ration

Green Time 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

Cycle Time

L = 4 * 4 = 16 seconds [lost time per phase 4 seconds]

Co = (1.5 x 16 + 5) / ( 1 - 0.59) = 61.67

1 - Y Co =

Co =

1.5 L + 5

1 - Y

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Take Co =

Effective Green Time =

Green time split

Phase Green Time Round

1 20.67 212 9.19 103 7.89

4 8.25

total 46 48

OFFSET TIME

= 400/10- (15(2)+2) = 40-32 = 8

# So, Offset Junction

62 seconds

62 - L = 46 seconds

Round-Up To

21 Teke amber time =

10 Take all red time =

8 Total of all amber and red time =

9

48 Actual CycleTime = 64

L=400 meter S=10m/s Q=15 kenderaan Loss time= 2 second

Junction 2 is8 Second

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3 seconds

1 seconds

16 sec

sec