1 - Purdue University



|CE361 Introduction to Transportation Engineering |Posted: Sat. 18 September 2010 |

|Homework 4 Solutions |Due: Wed. 29 September 2010 |

TRANSPORTATION PLANNING AND DEMAND MODELING

• You may work in a group of CE361 students not to exceed four in size.

• Signatures (and printed names) at the top of the front page of the materials submitted.

• Label each problem (see bold font below).

1. Site-based Trip Generation. Mythaca Community College is planning a downtown campus. It plans to occupy 41,900 sq ft of floor space in a newly-constructed building. Expected enrollment is 1033 students. MCC will employ 74 faculty and staff at its downtown location. How many vehicle trip ends per average weekday does the ITE Trip Generation report indicate will be caused by the new MCC campus?

A. (10 pts) Show all your calculations. Summarize your results in a format like the table at the end of FTE Example 4.1 on page 194.

T = 1033 stus * 1.20 trip ends/stu = 1240 trip ends by avg trip rate.

Ln(T) = 0.89*Ln(1033) + 1.24 = (0.89*6.940) + 1.24 = 6.1768 + 1.24 = 7.4168;

e7,4168= 1664.

T = 74 employees *15.55 trips ends/empl = 1151 trip ends

T = (11.27*74) + 3163.13 = 833.98 + 3163.13 = 3997 trip ends

T = 41.9 GFA * 27.49 trip ends/GFA = 1152 trip ends

No fitted curve given for GFA.

|X |units |Avg Trip Rate |Fitted Curve |

|1033 |stus |1240 |1664 |

|74 |empl |1151 |3997 |

|41.9 |GFA |1152 |No curve given |

B. (5 pts) What should bother you about using the ITE Trip Generation data for Part A?

• In each of the three ITE plots, the MCC value is well below the minimum value in the ITE data. The MCC data are “out of range”.

• For employees, the fitted curve and average curve diverge significantly for high and low values if X.

• ITE data are for vehicle trips. No mention or consideration is given for use of public transit instead of private automobile.

C. (5 pts) Which single value of T would you adopt? Why?

There is a cluster of similar values -- 1240, 1151, and 1152 –based on Average Trip Rate that seems to form a consensus from three different X variables. Any of these values –- or T=1200 – would be satisfactory. The fitted curve values are erratic (or missing). With such small values of X, we ought to use the Average Trip rate, whose curve goes through the origin (0,0).

2. Trip Generation by Regression. Assume that a study area can be modeled using four TAZs. The auto ownership and employment data for these zones are given in Table 2.

|Table 2 |

|TAZ |Cars in zone |Jobs in zone |P(i) |A(j) |A’(j) |

|1 |1499 |0 |7666 |0 |0 |

|2 |2099 |1642 |10720 |2381 |7294 |

|3 |2719 |2039 |13876 |2957 |9057 |

|4 |0 |3590 |36.03 |5206 |15947 |

| |6317 |7271 |32298 |10543 |32298 |

A. (15 points) Use FTE equations 4.2 and 4.5 to compute Total Productions and HBW Attractions for each zone. Summarize your results using the format of FTE Table 4.6.

(4.2) P(1) = 36.03 + (5.09 * 1499) = 7666, etc.

(4.5) A(2) = 1.45 * 1642 = 2381, etc.

Note P(4) = 36.03, even though X = 0 cars in zone. Seems strange, but trips can be produced without a car.

B. (5 points) Balance the P and A values for each zone as described at the start of FTE Section 4.3.3. Show the revised values in a new column in the table you created in Part A.

Total P/Total A = 32298/10543 = 3.063, so multiply each A(j) by 3.063 to balance the A’s against the P’s. A’(3) = 2957*3.063 = 9058, etc. (9058 in table above is from spreadsheet calc.)

3. (20 points) Trip Distribution by Gravity Model. Regardless of your results in Problem 1, use P(5)=1000 to represent the number of trips produced in Zone 5 by the CC campus. Use A(1)=579, A(2)=731, A(3)=738, and A(4)=898 in this problem. Using the Tanner Function, the format of FTE Table 4.11, and the auto travel times t(5,1)= 8, t(5,2)= 11, t(5,3)= 9, and t(5,4)= 18, determine how many trips produced by MCC in Zone 5 will be attracted to each of the other four zones. Use a = 1.0, b = 1.4, and c = -0.22 in the Tanner Function. Show your intermediate results in the format of FTE Tables 4.9-4.11. Show the calculations needed for the j=3 row.

j=3, A(3)=738, t(5,3)=9,

(4.9) F5,3= 1.0*[pic]=91.4e(-0.22*9)=21.674*0.1381 = 2.993,

A(3)F(5,3)= 738*2.993= 2208.5, AF(3)/sum(AF) = 2208.5/6709.8 = 0.329,

T(5,3) = P(5)*0.329 = 1000 * 0.329 = 329.

|P= |1000 | from Zone 5 | | |  |

|a = |1 |b = |1.40 |c = |-0.22 |  |

|  | | | | | |  |

|(1) |(2) |(3) |(4) |(5) |(6) |(7) |

|Zone j |A(j) |t(5j) |F(5j) |A(j)F(5j) |AF(j)/sum(AF) |T(5j) |

|  | | | | | |  |

|1 |579 |8 |3.162 |1830.8 |0.273 |273 |

|2 |731 |11 |2.552 |1865.8 |0.278 |278 |

|3 |738 |9 |2.993 |2208.5 |0.329 |329 |

|4 |898 |18 |1.090 |804.7 |0.120 |120 |

|  |---------- | | |---------- |---------- |---------- |

|  |2946 |  |  |6709.8 |1.000 |1000 |

4. (20 points) Mode Choice. By which mode do the students and employees of MCC get to/from the downtown campus? The MCC hired a consulting firm last year to develop a mode choice model to explain how commuters choose between driving to campus and taking the bus. The consultants found that the only measurable factors that matter are total travel time (TTT) and out-of-pocket costs (OOPC). Their proposed utility function was

Vm = -0.39*TTTm –0.80*OOPCm. MCC has a contract with Mythaca Bus Company that permits MCC students and employees to ride MBC buses for free if they show their MCC ID. Parking at the campus costs $5.00. Bus service between MCC and Zone 4 takes 27 minutes of in-vehicle travel time. Add three minutes wait time at the start of the bus trip. What proportion of MCC tripmakers will use bus to/from Zone 4? Show your calculations for Vm and Pm. Use auto travel time from Problem 3.

Vbus = (-0.39*(27+3)) + (-0.80*0) = -11.70; eV(bus)= e-11.70=0.00000829

Vauto = (-0.39*18)+ (-0.80*5) = -11.02; eV(auto)= e-11.02=0.00001637

(4.11) Pbus = [pic]= [pic]= [pic]= 0.336

(4.11) Pauto = [pic]= [pic]= [pic]= 0.664

Check: Pbus + Pauto = 0.336 + 0.664 = 1.000.

5. Trip Assignment. There are two ways to drive from Zone 4 to the MCC downtown campus. Route A is the “direct” (shortest distance) route on arterial streets. Route B is longer by distance, but it includes part of a freeway that offers higher speeds and capacity. Route A has a free-flow travel time of 18 minutes, C(A) = 772 vph, a = 0.25, and b = 5.8. Route B has a free-flow travel time of 31 minutes, C(B) = 3033, a = 0.29, and b = 3.6. Consider the capacity values to be at LOS “E”. During the morning peak hour, 2938 vehicles are using the links between Zones 4 and 5.

A. (15 points) Equilibrium condition. Using FTE Equation 4.13, determine the morning peak flows V(A) and V(B) -- to the nearest 5 vph -- so that user equilibrium occurs.

As in FTE Example 4.19, we can set up the two link performance functions as follows:

[pic] and [pic]

We could use trial and error (as on FTE page 237) to find VB such that tA = tB, or we can use the Solver feature of a spreadsheet:

|Param | | | |

|t0 |18 |31 | |

|Capac "E" |772 |3033 | |

|Capac "C" |579 |2274.75 | |

|a |0.25 |0.29 | |

|b |5.8 |3.6 | |

|V |753.6921 |2184.308 |2938 |

|tL |38.76834 |38.76834 |-1.5E-07 |

tA=tB=38.768 minutes when VB=2184.3 and VA=753.7.

B. (5 points) Equilibrium travel time. Show that the travel times on the two routes are equal.

[pic]=[pic]= 18[1+(0.25*4.615468)] = 38.77

[pic]=[pic] = 31*[1+(0.29*0.864098)] = 38.77

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