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# homework7 - CIV ENG 376 Homework#7 Due on Wendesday Nov 23...

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CIV ENG 376: Homework #7 Due on Wendesday, Nov. 23 Instructor: Marco Peng Chen

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Peng Chen CIV ENG 376 (Instructor: Marco ): Homework #7 Problem 1: Textbook 7.17 First, determine the critical ﬂow ratio for each phase. ( v 1 s 1 ) crit = max { 245 1750 , 230 1725 } = 245 1750 = 0 . 140 ( v 2 s 2 ) crit = max { 975 3350 , 1030 3400 } = 1030 3400 = 0 . 303 ( v 3 s 3 ) crit = max { 255 1725 , 235 1750 } = 255 1725 = 0 . 148 ( v 4 s 4 ) crit = max { 225 1700 , 215 1750 } = 225 1700 = 0 . 132 Thus, the sum of the ﬂow ratios for the critical lane groups is C = 4 X i =1 ( v i s i ) crit = 0 . 140 + 0 . 303 + 0 . 148 + 0 . 132 = 0 . 723 Problem 2: Textbook 7.20 According to the minimum cycle length formula, we can calculate the minimum cycle length as follows. C = L 1 - 4 i =1 ( v i s i ) crit X c = 5 × 4 1 - 0 . 225+0 . 175+0 . 200+0 . 150 0 . 85 = 170 s In order that the lane group v c ratios are equalized, the phase eﬀective green times are allocated as follows. g 1 = ( v 1 s 1 ) crit X c · C = 0 . 225 0 . 85 × 170 = 45 s g 2 = ( v 2 s 2 ) crit X c · C = 0 . 175 0 . 85 × 170 = 35 s g 3 = ( v 3 s 3 ) crit X c · C = 0 . 200 0 . 85 × 170 = 40 s g 4 = ( v 4 s 4 ) crit X c · C = 0 . 150 0 . 85 × 170 = 30 s Problem 3 According to the ﬂow ratios of the lane group, we can take advantage of the overlapping phase. The phase
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homework7 - CIV ENG 376 Homework#7 Due on Wendesday Nov 23...

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