Homework 4 / Due: Friday 23 September 2005
TRANSPORTATION PLANNING AND DEMAND MODELING
Dear Consultant:
The steady (and sometimes rapid) growth in Mythaca County makes it important that you can demonstrate basic travel demand modeling capabilities. Please complete the exercises below completely and clearly. You may work in a group of CE361 students not to exceed three in size. If the HW is submitted by more than one student, the signatures of those students must appear at the top of the front page of the materials submitted.
1. Trip Generation. A developer proposes to build a subdivision called Sleepy Manor with 1528 homes on 1/5-acre lots. The developer says that every household will have 4 members and two vehicles.
A. (10 points) ITE Method. Use FTE Figures 4.4 to estimate the number of vehicle trip ends that will be generated by Sleepy Manor on an average weekday.
Note: ITE yields vehicle trips.
Fig. 4.4(a) Dwelling Units: T = 9.57 * 1528 DU = 14,623 by average rate;
Ln(T) = 0.920 * Ln(1528) + 2.707 = 9.452 à T = 12,736 by fitted curve equation.
Fig. 4.4(b) Persons: T = 2.55 * 1528 * 4 = 15,586 by average rate;
Ln(T) = 0.909 * Ln(1528*4) + 1.519 = 9.44367 à T = 12,628 by fitted curve equation.
Fig. 4.4(c) Vehicles: T = 6.02 * 1528 * 2 = 18,397 by average rate;
Ln(T) = 0.994 * Ln(1528*2) + 1.811 = 9.7877 à T = 17,814 by fitted curve equation.
Fig. 4.4(d) Acres: T = 26.04 * 1528 * 1/5 = 7,958 by average rate; no fitted curve equation.
B. (5 points) Regression. Use FTE Equation 4.3 to estimate the number of trips that will be generated by Sleepy Manor on an average weekday.
(4.3) T = 0.69 + (1.39*pers/HH) + (1.94*vehs/HH) = 0.69 + (1.39*4) + (1.94*2) = 0.69 + 5.56 + 3.88 = 10.13 trips/HH; 1528 HHs * 10.13 = 15,479 trips/day. It is not explicit in FTE, but these are person trips, not vehicle trips. Person trips can be converted to vehicle trips by dividing by auto occupancy (e.g., 1.29 persons/vehicle on FTE p. 227): T = 15,479/1.29 = 11,999 vehicle trips.
C. (5 points) Cross-classification. Use FTE Table 4.2 to estimate the number of trips that will be generated by Sleepy Manor on an average weekday.
Every HH is predicted to have 4 persons and 2 vehicles. The rate in Table 4.2 for this category of HH is 13.3. 1528 * 13.3 = 20,322 person trips per day. T =20,322 person trips/1.29 = 15,753 vehicle trips
D. (5 points) Recommendation. Of all the values of T computed in Parts A-C above, which value do you believe is the best estimate? Why?
In Part A, dwelling units and acres are the most reliable inputs. Persons are a better basis for tripmaking, but the predicted household size of 4 may not be so reliable – perhaps too high?
In Part B, T=12,.209 vehicle trips T is lower than four of the five T’s found in Part A.
In Part C, T=15,753 vehicle trips is higher than four of the five T’s found in Part A.
The median value of T=14,623 from Part A is based on a database defined for a residential neighborhood. Furthermore, this T=14,623 value is based on Dwelling Units, not the predictions of number of persons or vehicles associated with those households. It is better to choose a single T value and justify it than to take the average of all T values, some of which may be suspect.
2. (20 points) Trip Distribution by Gravity Model. FTE Problem 4.11.
Sample calculation for F24: (4.8) 1000*= 1000*12-2.8 = 0.951
The revised version of FTE Table 4.9 is shown below.
P = / 901 / from Zone 2a = / 1000 / b = / -2.8
(1) / (2) / (3) / (4) / (5) / (6) / (7)
Zone j / Aj / t2j / F2j / AjF2j / AFj/∑(AF) / T2j
1 / 4909 / 35 / 0.047 / 233.1 / 0.026 / 24
2 / 774 / 5 / 11.038 / 8543.3 / 0.962 / 867
3 / 174 / 20 / 0.228 / 39.6 / 0.004 / 4
4 / 69 / 12 / 0.951 / 65.6 / 0.007 / 7
Total / 5926 / 8881.7 / 1.000 / 901
3. (20 points) Mode Choice. FTE Problem 4.18(a).
For mode A: UA = -0.12 – (0.025*5) – (0.032*0) – (0.015*20) – (0.002*100) = -0.745
For mode B: UB = -0.56 – (0.025*10) – (0.032*15) – (0.015*40) – (0.002*50) = -1.990
eA = e-0.745 = 0.474; eB = e-1.99 = 0.137; PA = = 0.776; PA = = 0.224
Trips by mode A = 0.776 * 5000 = 3880; Trips by mode B = 0.224 * 5000 = 1120.
4. Trip Assignment. Carlos Andre, a new member of the County Highway Department, has just learned that the State DOT plans to expand a 5.75-mile section of nearby I-25 from 4 lanes to 6 lanes. The project with proceed in two phases.
§ Phase 1 will close the two SB lanes and give NB and SB traffic each one lane on the NB side of the median, while the two SB lanes are expanded to three lanes.
§ Phase 2 will use the three new SB lanes for two-way traffic (2 lanes in the direction of heavier flow) while the two old NB lanes are expanded to three lanes.
Carlos wonders if the delay suffered by drivers during the construction phases can be justified by the eventual saving in travel time after the expanded I-25 is completed. You are asked to conduct a study for the AM peak hour, in which NB flow is 1598 vph and SB flow is 3379 vph. Each lane on I-25 has LOS “E” capacity 2000 vph. In your analysis, use FTE Equation 4.13 with a = 0.15 and b = 4.0, and free-flow speed is 72 mph. Remember that “C” in Equation 4.13 is LOS “C” capacity. If you use a spreadsheet to facilitate your calculations, show sample calculations by hand to demonstrate your expertise or allow someone else to check your work.
A. (10 points) Current conditions. What are the AM peak hour values for NB travel time, NB VHT, SB travel time, and SB VHT for the current 4-lane section of I-25?
(4.13) tNB = with *60 min/hr = 4.79 minutes. tNB = 4.848 minutes. Note that LOS “E” capacity has been converted to LOS “C” capacity by using the 0.75 factor (FTE p. 230). NB VHT = (4.85 min * 1598 vph)/60 min per hr = 129.16 VHT in the AM peak hour. In the SB direction, only the flow rate changes: = 5.946 minutes; SB VHT = (3379 * 5.946)/60 = 334.9 VHT in the AM peak hour.
In the spreadsheet summary below, the small differences are due to round-off error.
B. (10 points) Phase 1 conditions. What will be the AM peak hour values for NB travel time, NB VHT, SB travel time, and SB VHT during Phase 1? Phase 1 will last about 6 months, during which there will be about 130 AM peak hours. Be clear about how many lanes NB and SB traffic will have.
In Phase 1, there is only one lane in each direction. Only the number of lanes changes in the (4.13) calculations: tNB = =5.715; NB VHT = (1598 * 5.715)/60 = 152.2 VHT in the AM peak hour.
tSB = =23.29; SB VHT = (3379 * 23.29)/60 = 1311.7 VHT in the AM peak hour.
C. (10 points) Phase 2 conditions. What will be the AM peak hour values for NB travel time, NB VHT, SB travel time, and SB VHT during Phase 2? Phase 2 will last about 6 months, during which there will be about 130 AM peak hours. Be clear about how many lanes NB and SB traffic will have.
Now the heavier SB traffic can use two lanes and the lighter NB traffic can use one lane. The NB 1-lane calculations match those in Part B for Phase 1. The SB 2-lane calculations match those in Part A for Current Conditions.
D. (10 points) 6-lane conditions. What will the AM peak hour values for NB travel time, NB VHT, SB travel time, and SB VHT for the completed 6-lane section of I-25?
At last, there are lanes in each direction. Only the number of lanes changes in the (4.13) calculations: tNB = =4.801; NB VHT = (1598 * 4.801)/60 = 127.88 VHT in the AM peak hour. tSB = =5.018;
SB VHT = (3379 * 5.018)/60 = 282.62 VHT in the AM peak hour.
E. (10 points) Is it worth it? How much extra delay to combined NB + SB traffic was caused by Phases 1 and 2? During an average AM peak hour, how much VHT is saved by users of the 6-lane section of I-25, when compared with the old 4-lane section? How many AM peak periods will be needed to compensate AM peak period drivers for the construction delays?
Extra delay in 260 AM peak hours = (130*(152.28+1312.17))+(130*(152.28+335.00))-(260*(129.16+335.00) = 133,038.3 VHT during Phases 1 and 2.
Time saved each AM peak hr on 6-lane section of I-25 = (129.16+335.00)-(127.92+282.72) = 53.51 VHT.
Number of AM peak hrs to recoup the construction delays = 133,038.3/53.51 = 2486.2. 2486.2/260 = 9.56 years.
AM peak on I-25 / Year 0 / Phase 1 / Phase 2 / 6 lanes5.75 / mi / 6 mos. / 6 mos.
free-flow speed (mph) = / 72 / 72 / 72 / 72
free-flow time (min) = / 4.79 / 4.79 / 4.79 / 4.79
a = / 0.15 / 0.15 / 0.15 / 0.15
b = / 4.00 / 4.00 / 4.00 / 4.00
LOS E Capac NB = / 4000 / 2000 / 2000 / 6000
LOS E Capac SB = / 4000 / 2000 / 4000 / 6000
q NB = / 1598 / 1598 / 1598 / 1598
q SB = / 3379 / 3379 / 3379 / 3379
time NB = / 4.85 / 5.72 / 5.72 / 4.80
time SB = / 5.95 / 23.30 / 5.95 / 5.02
VHT NB = / 129.16 / 152.28 / 152.28 / 127.92
VHT SB = / 335.00 / 1312.17 / 335.00 / 282.72
chg in VHT ea day = / 1000.29 / 23.12 / -53.51
chg in VHT, 130 days = / 130038.3 / 3005.1 / -6956.8
chg in VHT, tot proj = / 133043.4
days to recoup constru delay = / 2486.2