Homework 3 (HW 3) Solutions, m=79.9/100, s=22.2 / Due: Mon. 20 September 2010
CE361 HW 3 Fall 2010 - 5 -
HIGHWAY DESIGN FOR PERFORMANCE
· You will be permitted to submit this HW with as many as three other CE361 students. (signatures and printed names at the top of the front page)
· Identify each problem by its number and name, be clear, be concise, cite your sources, attach documentation (if appropriate), and let your methodology be known.
· “FTE” = Fundamentals of Transportation Engineering, the textbook for CE361.
1. Poisson models for natural events. A list of Atlantic hurricanes at http://weather.unisys.com/hurricane/atlantic/.A. (5 points) For the years 1995-2009, what was the average “arrival rate” for Atlantic storms?
= 14.46 storms/yr. See spreadsheet table at right.
B. (10 points) Calculate P(n) for the range of events per year that appear in the website’s archives for 1995-2009. If you use a spreadsheet, show one P(n) calculation by hand.
Range of n is 8n28. See spreadsheet table at right. Sample calculation:
(2.24) P(n) = ;
P(17) = = 0.077964
C. (5 points) What is the probability that the number of Atlantic storms in 2010 will exceed 18?
P(n>18) = 1 – = 1 – 0.8551 = 0.1449.
2. (10 points) Time between events. Problem 2.48 in FTE.
Crash rate, l = 37 collisions/year or 37/52 = 0.711 collisions per week.
Solve for P(n>1) using Equation (2.24) and Time interval t = 1 week. Start with
= 0.491. Then Pr (n1) = 1 – P (0) = 1 – 0.492 = 0.5093
3. Level of service on rural 2-lane highway. ADT is 8954 vehicles; the K factor is 0.111. V = 0.111*8954 = 994vph. 8 county roads that cross this segment of SR361 + 25 driveways: (2*8)+25=41 access points. WB shoulders are 6 feet wide, EB shoulders are only 5.3 feet wide: average shoulder width = 0.5*(6+5.3)=5.65ft. Early one morning, the field-measured speed was 55.0 mph, when V = 82 veh/hr, six of which were trucks, with no buses or RVs. To save space, the entries in the Figure 3.4 worksheet are shown in tabular form on the next page.
A. (10 points) Average travel speed (ATS). Equation 3.1 in the first iteration is ==0.943
Equation 3.2 in the first iteration is
=1288 pc/hr. The calculations associated with free-flow speed measurements are as follows. Equation 3.1 = 0.965. Equation 3.4 is == 55.7 mph. Because 1288 pc/h does not lie in the same range in the Exhibits as V=994, a second iteration was necessary. The result in the 2nd iteration is vp = 1179 pc/h, which takes us back to the “600-1200 pc/h” rows in the exhibits. An endless cycle is possible. Using 1179 pc/h in Equation 3.5 produces ATS = 45.7 mph. If 1288 pc/h is used, ATS = 44.9 mph. Unfortunately, 45.7 mph lies in the LOS C range of Table 3.2, while 44.9 mph lies in the LOS D range.
B. (10 points) Percent time spent following (PTSF). Equation 3.1 in the first iteration is = 0.968. Equation 3.2 is = 1241 pc/h. Because 1241 pc/h does not lie in the same range in the Exhibits as V=994, a second iteration was necessary. The second iteration results in vp = 1129 pc/h. This would put us back in the original range of values, and an endless cycle is likely. If we use vp = 1129 pc/h in the subsequent calculations, the PTSF value of 68.4 mph lies in the LOS D range of Table 3.2. If we use vp = 1241 pc/h in the subsequent calculations, the PTSF = 62.9 mph, which lies in the LOS C range of Table 3.2.
HCM 2000 LOS for 2-lane roadsValue / Input variable and description
SR361 / Site ID
12 / Lane width
5.65 / Avg shoulder width
9.2 / Segment length
rolling / terrain
994 / 2-way hourly volume
0.60 / Major direction split, d(maj)
0.88 / PHF (rural default = 0.88)
0.066 / P(T) trucks and buses
0.006 / P(R) rec vehs
20 / % no passing zones
41 / Nr of access points
4.46 / Access pts/mi
Average Travel Speed / Iter. 2
0.93 / f(G) Exhibit 20-7 / 0.99
1.9 / E(T) Exhibit 20-9 / 1.5
1.1 / E(R) Exhibit 20-9 / 1.1
0.943 / f(HV) Equation 3.1 / 0.967
1287 / v(p) Equation 3.2 / 1179
No / v(p) > 3200? / No
772 / d(maj)*v(p) > 1700? Eqn 3.3 / 707
No / v(p) in same range? / Yes
(If NO, iterate)
Free-Flow Speed from Field Measurement
55.0 / S(FM) field measured speed
82 / V(f) observed volume
0.073 / P(T)
0.00 / P(R)
1.50 / E(T) Exhibit 20-9
1.10 / E(R) Exhibit 20-9
0.965 / f(HV) Equation 3.1
55.7 / FFS free-flow speed Eqn 3.4
0.83 / f(np) no-passing Exhibit 20-11
45.7 / ATS Equation 3.5
/ Percent Time Spent Following / Iter. 2
0.94 / f(G) Exhibit 20-8 / 1.00
1.5 / E(T) Exhibit 20-10 / 1.0
1.0 / E(R) Exhibit 20-10 / 1.0
0.968 / f(HV) Equation 3.1 / 1.000
1241 / v(p) Equation 3.2 / 1129
No / v(p) > 3200? / No
745 / d(maj)*v(p) > 1700? Eqn 3.3 / 678
No / v(p) in same range? / No
(If NO, iterate)
62.9 / Base PTSF Equation 3.7
5.46 / f(d/np) Exhibit 20-12
68.4 / PTSF Equation 3.8
C. (5 points) In both Parts A and B, using the second iteration vp values (1179 and 1129) for are LOS C. If the 1st (and 3rd?) iteration vp values (1288 and 1241) are used, the level of service for the peak hour is LOS D. LOS D is OK as a “conservative” solution, but the LOS C solution is also acceptable, as long as the correct procedures were followed.
4. I-96 Incident and queueing analysis. For the first ten minutes after the truck's mishap, the ramp’s service rate for detouring traffic was approximately 325 vph. After ten minutes, the ramp’s service rate for detouring traffic was 650 vph. At exactly 1:00PM (t=63 minutes), NB I-96 flow rate became 1550 vph:
A. (10 points) Draw a queueing diagram that shows the buildup and dissipation of the queue.
B. (10 points) The queue dissipates when the departure curve reaches the arrival curve at CV = 2383 cumulative vehicles. The departure curve has the equation CV = 628 + (t-63). The arrival curve has the equation CV = t. The two curves cross when 628 + 60(t-63) = 25.83 t. t = = 92.2 minutes.
C. (10 points) The longest vehicle queue occurs at t = 63 minutes (1:00PM), when 1628 vehicles have arrived and 628 vehicles have been served. Therefore, 1000 vehicles (the length of the vertical dashed line) are in the queue at t = 63 minutes. The longest vehicle delay also occurs at t = 63 minutes (1:00PM), and applied to the 628th vehicle to arrive. The 628th vehicle arrived at t = 60 = 24.3 minutes. It left the queue at t = 63 minutes, after having been in the queue 38.7 minutes (the length of the horizontal dashed line).
5. Queueing at drive-up window. The first Taco Terrace server window has a mean service rate of = 1.56 vehs/min. The second server window has a mean service rate of = 0.99 vehs/min. The expected l is 50 drive-up customers per hour, or 0.833 vehs/min. Because both interarrival times and departure time patterns are exponentially (Poisson) distributed, the windows are both M/M/1.
A. (8 points) Average time in the 2-window system = t1 + t2. Use Equation 3.20: . = 1.37 min.; = 6.25 min. t1 + t2. = 1.37 + 6.25 = 7.62 min. The service at the first window, on the average, does not change the arrival pattern from the first to the second window.
B. (7 points) The number of car lengths needed between the two server windows equals the average queue length at the second window. = 0.841. Equation 3.18 = 4.46 vehs. Five vehicle lengths are needed.