Solution Chapter 10

# Solution Chapter 10 - 1 Traffic Engineering, 4th Edition...

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Unformatted text preview: 1 Traffic Engineering, 4th Edition Roess, R.P., Prassas, E.P., and McShane. W.R. Solutions to Problems in Chapter 10 Problem 10‐1 Parts a and b: To plot a frequency distribution curve and a cumulative frequency distribution curve, a frequency distribution table must be constructed from the data given. Table 1 provides this. Then: • The Frequency Distribution Curve (FDC) is plotted as the % Vehicles in Group vs. the Middle Speed of the group. • The Cumulative Frequency Distribution Curve (CFDC) is plotted as the Cum % Vehicles in Group vs. the Upper Speed of the speed group. • The median speed is the 50th percentile speed from the CFDC. • The modal speed is estimated as the peak of the FDC. • The pace is the 10‐mi/h increment in speed that captures the highest percentage of observed speeds compared to any other 10‐mi/h increment. • The percent vehicles in the pace are found by finding the percentile speed representing the boundaries of the pace, and subtracting their values. Figure 1 shows the two curves, and the solution for the values called for in Part b of the question. From Figure 1: Mode = 42.0 mi/h Median = 42.0 mi/h Pace = 37.5 – 47.5 mi/h % Veh in Pace = 72% ‐ 24% = 48% 2 Table 1: Frequency Distribution Table for Speed Data Speed Group (mi/h) Low High Speed Speed 15 20 20 25 25 30 30 35 35 40 40 45 45 50 50 55 55 60 60 65 Middle Speed, S (mi/h) 17.5 22.5 27.5 32.5 37.5 42.5 47.5 52.5 57.5 62.5 Observed Freq n 0 4 9 18 35 42 32 20 9 0 169 Percent Freq (%) 0.00% 2.37% 5.33% 10.65% 20.71% 24.85% 18.93% 11.83% 5.33% 0.00% 100.00% Cum % Freq (%) 0.00% 2.37% 7.69% 18.34% 39.05% 63.91% 82.84% 94.67% 100.00% 100.00% n*S 0.00 90.00 247.50 585.00 1,312.50 1,785.00 1,520.00 1,050.00 517.50 0.00 7,107.50 n*S2 0.00 2,025.00 6,806.25 19,012.50 49,218.75 75,862.50 72,200.00 55,125.00 29,756.25 0.00 310,006.25 30.00% 25.00% Percent Frequency 20.00% 15.00% 10.00% 5.00% 0.00% 10 20 30 40 42 Middle Speed, mi/h 50 60 70 100.00% 90.00% Cumulative Percent Frequency 80.00% 70.00% 60.00% 50.00% 40.00% 30.00% 20.00% 10.00% 0.00% 10 20 30 40 42 24% 72% 50 60 70 Upper Speed of Group, mi/h Figure 1: Frequency and Cumulative Frequency Distribution Curves 3 Part c: The mean speed of the distribution is computed using Equation 10‐3; the standard deviation is computed using Equation 10‐5. Both use column totals from the Frequency Distribution Table. ∑ ni Si 7107.50 = = 42.06 mi / h x= i N 169 2 2 2 ∑ ni Si − N x = 310,006.25 − 169 * 42.06 = 11,037.88 = 65.7 = 8.11 mi / h s= N −1 169 − 1 168 Part d: The standard error of the mean, E, is computed as: s 8.11 8.11 E= = = = 0.624 13 N 169 Then: 95% Confidence : μ = x ± 1.96 E = 42.06 ± (1.96 * 0.624) = 42.06 ± 1.22 ( 40.84 to 43.28 mi / h) 99.7% Confidence : μ = x ± 3 E = 42.06 ± (3 * 0.624) = 42.06 ± 1.87 ( 40.19 to 43.93 mi / h) Part e: Sample size is estimated using Equation 10‐11 for 95% confidence: 3.84 s 2 3.84 * 8.11 n= = = 38.9, say 40 samples e 0 .8 Part f: This question is answered by conducting a Chi‐Squared Goodness of Fit test. In this test, actual observed frequency values are compared to theoretical frequencies that would have been observed if the distribution were “Normal.” Table 2 shows the test computations. 4 Table 2: Chi‐Square Goodness of Fit Test Speed Group (mi/h) High Low Speed Speed ∞ 60 60 55 55 50 50 45 45 40 40 35 35 30 30 25 25 20 20 15 TOTAL Observed Freq n 0 9 20 32 42 35 18 9 4 0 169 Upper Limit "zd" ∞ 2.21208385 1.59556104 0.86806412 0.36251541 -0.2540074 -0.8705302 -1.487053 -2.1035758 -2.7200986 Prob z<zd Tab 7.3 1.0000 0.9864 0.9432 0.8078 0.6406 0.4013 0.1922 0.0681 0.0188 0.0033 Prob of of Being in Group 0.0136 0.0432 0.1354 0.1672 0.2393 0.2091 0.1241 0.0493 0.0155 0.0033 1.0000 Theoretical Freq f 2.2984 7.3008 22.8826 28.2568 40.4417 35.3379 20.9729 8.3317 2.6195 0.5577 169 Combined Group f 9.5992 22.8826 28.2568 40.4417 35.3379 20.9729 11.5089 n 9 20 32 42 35 18 13 0.0374 0.3631 0.4959 0.0600 0.0032 0.4214 0.1932 Group χ2 After all speed groups are combined to insure that all values of f ≥ 5, the result is a Chi‐Square value of 1.5743, with 7 – 3 = 4 degrees of freedom. From text Table 7.11, the probability of a value this high or higher is between 0.75 (for X2 = 1.923) and 0.90 (for X2 = 1.064). Interpolating: Xd2 Prob (X2 ≥ Xd2) 1.064 0.90 1.574 ? 1.923 0.75 ⎛ 1.923 − 1.574 ⎞ ? = 0.75 + (0.90 − 0.75) * ⎜ ⎟ = 0.75 + (0.15 * 0.406) = 0.811 or 81.1% ⎝ 1.923 − 1.064 ⎠ In order to reject the hypothesis that the data and the Normal Distribution are the same, the Prob (X2 ≥ Xd2) would have to be less than 5%. Therefore, the hypothesis is confirmed. The data may be considered to be normally distributed. Problem 10‐2 a. To determine whether or not the observed reduction in speeds was significant, a Normal Approximation Test must be conducted: 169 169 1.5743 sy = zd (x − x ) − 0 = 43.5 − 40.8 = = 1 2 s12 s s2 + = N1 N 2 sd 4.82 5.32 + = 0.192 + 0.260 = 0.452 = 0.672 120 108 0.672 2.7 = 4.02 0.672 From text Table 7.3, the Prob (zd ≤ 4.02) = 0.9999. Thus, the difference is highly significant. 5 b. The standard error of the mean for the after sample is: E= s = 5.3 = 5.3 = 0.515 10.39 N 108 Therefore, it is 95% probable that the true mean speed of the distribution lies between 40.8 + (1.96*0.515) and 40.8 – (1.96*0.515) or 39.8 to 41.8 mi/h. As the target speed of 40 mi/h is in this range, it may be considered to have been achieved. Problem 10.3 Note: This solution assumes that there is one lane being observed. Table 3: Summary of Queue Counts Clock Cycle Time Number 9:00 1 9:01 2 9:02 3 9:03 4 9.04 5 9:05 6 9:06 7 9.07 8 9:08 9 9:09 10 9:10 11 9:11 12 9:12 13 9:13 14 9:14 15 Sum Total for All Time Periods: +0 s 3 1 4 2 0 2 4 5 2 0 1 1 2 2 4 33 No. of Vehicles in Queue at: +15 s +30 s 4 2 2 3 3 3 3 3 1 2 1 1 3 3 5 6 3 4 3 2 2 3 0 1 2 1 3 2 3 3 38 39 +45 s 4 3 4 4 3 2 3 4 3 2 1 0 2 2 3 40 150 ∑V qi = 150 Then: TQ ⎛ ∑ V qi ⎜ i = ⎜I s * ⎜ VT ⎜ ⎝ ⎞ ⎟ 150 ⎞ ⎛ ⎟ * 0.90 = ⎜15 * ⎟ * 0.90 = 4.66 s / veh ⎜ ⎟ ⎟ 435 ⎠ ⎝ ⎟ ⎠ 6 VSLC = FVS = VSTOP 305 = = 20.3 vehs / cycle N c * N L 15 *1 VSTOP 305 = = 0.701 VT 435 d = TQ + (FVS * CF ) Where: CF = +2 (Text Table 10.6, FFS = 35 mi/h, VSLC = 20.3) d = 4.66 + (0.701 * 2) = 6.062 s / veh Then: Problem 10‐4 For 95% confidence: 3.84 s 2 n= e2 Table 4 executes this equation for tolerance levels (e) of 2 min, 5 min, and 10 min with base standard deviations (s) of 5 min, 10 min, and 15 min. Table 4: Required Number of Samples Tolerance (min) 2 5 10 Standard Deviations (min) 5 10 15 24 96 216 4 15 35 1 4 9 Problem 10‐5 Table 5 uses the problem data to determine the average travel time and average running time in each section shown. These values are used to compute the average travel speed and the average running speed. Note that data for the 1st segment must be added as follows: Cumulative Section Length = 0.50 mi; Cumulative Travel Time = 1.0 min; Delay = 0 s; No. of Stops = 0. Then, using the data, Segment Lengths must be established, as well as travel times in each segment (converted to seconds). As an example, select the segment between checkpoints 3 and 4: 7 Section Length (mi) = 3.50 – 2.25 = 1.25 mi Travel Time = 7 min, 30 s – 4 min, 50 s = 450 s – 290 s = 160 s. Running Time = 160 – 25 = 135 s. Average Travel Speed = (1.25/160)*3600 = 28.1 mi/h. Average Running Speed = (1.25/135)*3600 = 33.3 mi/h Table 5: Average Travel Speed and Average Running Speed Section Length (mi) 0.50 0.50 1.25 1.25 0.50 0.25 0.75 Travel Time (s) 60 65 165 160 100 77 87 Running Time (s) 60 55 135 135 58 30 73 Travel Speed (mi/h) 30.0 27.7 27.3 28.1 18.0 11.7 31.0 Running Speed (mi/h) 30.0 32.7 33.3 33.3 31.0 30.0 37.0 1 2 3 4 5 6 7 40 35 Average Speed (mi/h) 30 25 20 15 10 5 0 1 2 3 4 5 6 7 Section Number Travel Speed Running Speed Figure 2: Average Travel Speed and Average Running Speed 8 The answer to Part b depends upon which measure is to have a tolerance of 3 mi/h – average travel speed or average running speed? We also need an estimate of the standard deviation of travel speed and/or running speed. The standard deviations are computed in Table 6 as: s= ∑ (x i − x) 2 N −1 Table 6: Computation of Standard Deviations Segment 1 2 3 4 5 6 7 Total Mean STD Travel Speed (mi/h) 30 27.7 27.3 28.1 18.0 11.7 31.0 173.8 24.8 Running Speed (mi/h) 30 32.7 33.3 33.3 31.0 30.0 37.0 227.4 32.5 (TS-Mean)2 (RS-Mean)2 26.72 8.19 5.96 10.85 46.65 172.71 38.49 6.19 0.06 0.71 0.71 2.11 6.19 20.24 Then: 3.84 s 2 n= e2 3.84 * 6.42 2 n ATS = = 17.6, say 18 runs 32 3.84 * 3.37 2 n ARS = = 4.8, say 5 runs 32 If each car makes 5 runs, 3 cars would be needed to estimate ATS; only 1 would be needed to estimate ARS. 6.42 3.37 ...
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