Quiz #2 Solution TTE 4004 3:30 PM 5:30PM, 6/24/09 Summer 2009 · Quiz #2 Solution TTE 4004 3:30 PM...

Quiz #2 Solution TTE 4004

3:30 PM – 5:30PM, 6/24/09 Summer 2009

1

1. At a signalized intersection, assume vehicles arrive at a uniform rate of 450 vehicles per hour

and the saturation flow rate (uniform) for the intersection is 1800 vehicles per hour. The

cycle length is 100 sec with 60 sec (t = 0 – 60) of effective red and 40 sec (t = 60 – 100) of

effective green. Calculate

(1) Delay for a vehicle arriving at the intersection at = 24 sec (4 points)

(2) Total delay in a cycle (4 points)

(3) Average delay per vehicle? (2 points)

Solution:

(1) Queue at t=24 sec

When effective green starts, the time to clear 3 vehicles is

Therefore the delay d = (60 - 24) + 6 = 42 seconds

(2) Assume at time t, the queue is cleared

The total number of accumulated vehicles at t=80 sec

The total delay:

(3) Average delay:

324*3600

450 tqueue

6

3600

1800

3

u

queuetg

802403)60(*3600

1800*

3600

450)60( tttttt

1080*3600

450 tN

sec3002

10*60totald

vehC

d

N

dd total

cycle

totalaverage sec/24

100*3600

450

300


3:30 PM – 5:30PM, 6/24/09 Summer 2009

2

2. Vehicles arrive at a toll plaza in a Poisson distribution with average arrival rate of 160

vehicles per hour. On the average, each vehicle needs to spend 18 sec to pay the toll at each

service station. The service time at each station is exponential distributed. There are two

(N=2) service stations at the toll plaza. Calculate the following:

(1) The percent of time that both service stations are idle. (2 points)

(2) The probability that five vehicles will be in the toll plaza system. (2 points)

(3) The average number of vehicles in the toll plaza system. (2 points)

(4) The average queue length. (2 points)

(5) The average time each vehicle spends in the toll plaza system. (2 points)

Solution:

This is a M/M/2 question

(1) =160 veh/hr, =200 veh/hr, so

(2)

(3)

(4)

(5)

8.0200

160

r

%9.42429.0)8.02!*(1

8.0

!1

8.0

!0

8.0)0(

1210

p

%88.00088.0429.0*2!*2

8.0)5(

)25(

5

p

953.0429.0*)8.02!*(1

8.08.0)(

2

3

XE

153.08.0953.0)()( rXELE q

sec44.2100595625.0160

953.0)()( hr

XETE


3:30 PM – 5:30PM, 6/24/09 Summer 2009

3

3. To conduct a parking study, a license plate survey data from 12:00 PM to 1:00 PM is

obtained as shown below. There are 2 handicapped stalls from Stall No. 1 to 2, and 20

permit-holder-only stalls from Stall No. 3 to 22. Calculate the following four indicators

based on the permit parking:

(1) Occupancy(%) for each 15 minutes (4 points)

(2) Accumulated number for each 15 minutes (4 points)

(3) Turnover during the hour between 12:00 PM and 1:00 PM (3 points)

(4) Average duration of the parking between 12:00 PM and 1:00 PM (4 points)

Solution:

min65.3826

15*)19141618(1

N

t

ationAverageDur

N

i

i


3:30 PM – 5:30PM, 6/24/09 Summer 2009

4

4. Trip generation models usually can be expressed as a linear regress model as:

Y = ao + a1X1 + a2X2 + … + anXn

Where Y = trip generation (dependent variable)

Xi = factors affecting trip generation (independent variable)

There can be several independent variables for the models. Please list

(1) 3 independent variables for a residential development (Production model) (6 points)

(2) 2 independent variables for a shopping area (Attraction model) (4 points)

Solution:

(1) Residential development (any three of the following four factors)

a. Number of household

b. Average household income

c. Average household size

d. Average number of cars per household

(2) Shopping area (any three of the following three factors)

a. Wholesale and retail employment

b. Service employment

c. Average size of the shopping area


3:30 PM – 5:30PM, 6/24/09 Summer 2009

5

5. Given the following trip production and attractiveness data for three zones:

Table 1

Zone Productions Attractiveness

1 1000 2

2 0 5

3 2000 1

Table 2: WIJ:

Model calibration results show that the parameter c = 1.5

Table 3: kIJ:

1 2 3

1 1.1 1.5 0.8

3 1.0 1.4 1.3

Apply the gravity model to calculate

(1) The interchange volumes from zone I to zone J (12 points)

(2) The total attraction trips of Zones 1, 2 and 3 respectively (3 points)

510103

105202

102051

I 321

J

510103

105202

102051

I 321

J

2

I

J


3:30 PM – 5:30PM, 6/24/09 Summer 2009

6

Solution:

For

P1= 1000

J AJ WIJ FIJ KIJ AJFIJKIJ Ratio

QIJ=

P1*Ratio

1 2 5 0.089443 1.1 0.196774 0.64321 643.2103988

2 5 20 0.01118 1.5 0.083853 0.274095 274.0953404

3 1 10 0.031623 0.8 0.025298 0.082694 82.69426085

0.305925 1000

For P3

= 2000

J AJ WIJ FIJ KIJ AJFIJKIJ Ratio

QIJ=

P3*Ratio

1 2 10 0.031623 1 0.063246 0.157767 315.5331795

2 5 10 0.031623 1.4 0.221359 0.552183 1104.366128

3 1 5 0.089443 1.3 0.116276 0.29005 580.1006923

0.400881 2000

Trip Distribution

I \

J--> 1 2 2

1 643 274 83 1000

2 0 0 0 0

3 316 1104 580 2000

959 1378 663

.


3:30 PM – 5:30PM, 6/24/09 Summer 2009

7

6. For mode choice, a calibrated study resulted in the following utility function:

Uk = ak – 0.025X1 – 0.032X2 – 0.015X3 – 0.003x4

Where X1 = access plus egress time, in min

X2 = waiting time, in min

X3 = line-haul time, in min

X4 = out-of-pocket cost, in cents

The trip-distribution forecast for a particular interchange was a target-year volume of Qij =

6000 person-trips per day. During the target trip-makers on this particular interchange will

have a choice between the private automobile (A) and a local bus system (B). The target-

year service attributes of the two competing modes have been estimated to be:

Attribute X1 X2 X3 X4

Automobile 5 0 20 400

Local bus 10 10 40 100

aAuto = 0 and abus = – 0.10, Apply the Multinomial Logit (MNL) model to estimate

(1) The target-year market share of the two modes (8 points)

(2) The resulting fare-box revenue of the bus system (2 points)

Solution:

(1) The utility equation yields

According to the Logit equation,

Therefore the market share of each mode is

(2) The resulting fare-box revenue estimate of the bus system is

625.1400*003.020*015.00*032.05*025.0 AU

57.1100*003.040*015.010*032.010*025.010.0 BU

%63.484863.0208045.019691.0

19691.057.1625.1

625.1

)()(

)(

ee

e

ee

eAp

BUAU

AU

%37.515137.0208045.019691.0

208045.057.1625.1

47.1

)()(

)(

ee

e

ee

eBp

BUAU

BU

daytripsAQIJ /918,26000*4863.0

daytripsBQIJ /082,36000*5137.0

daytripdaytrips /082,3$)/1)($/082,3(


3:30 PM – 5:30PM, 6/24/09 Summer 2009

8

7. Speed data was measured on the Fowler Avenue, Tampa. The data is given as follows:

Speeds (mph): 55, 55, 45, 40, 50, 50, 55, 35, 60

Calculate:

(1) The mean, median, mode, and range (4 points)

(2) The sample standard deviation (4 points)

(3) The Coefficient of Variance (2 points)

Solution:

(1)

Mean=(55+55+45+40+50+50+55+35+60)/9=49.4

Median=middle(35, 40, 45, 50, 50, 55, 55, 55, 60)=50

Mode=55

Range=60-35=25

(2)

(3)

08.81

)(1

2

n

XxSD

n

ii

163.0X

SDCV


3:30 PM – 5:30PM, 6/24/09 Summer 2009

9

8. Crash history was investigated for 100 intersections in the State of Florida. The total crash

frequency is found to be 350 crashes per year. Assuming that crashes at these intersections

are Poisson-distributed, calculate

(1) The probability of having 0 crashes at an intersection per year (3 points)

(2) The probability of having less than 3 crashes at an intersection per year (3 points)

(3) The probability of having more than 3 crashes at an intersection per year (4 points)

Solution:

(1)

(2)

(3)

030.0!0

5.30 5.3

0

ep

321.0185.0106.0030.0

!2

5.3

!1

5.3

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2

5.3

1

5.3

0

eeepppxP

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5.3

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5.3

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))3(210(1)3(1)3(

5.3

3

5.3

2

5.3

1

5.3

0

eeee

ppppxPxP

5.31*100/350 t


3:30 PM – 5:30PM, 6/24/09 Summer 2009

10

9. The probability that a vehicle will turn right at an intersection is known to be 0.25.

Assuming independence, calculate the probabilities of the following events:

(1) The ninth vehicle is not turning right. (2 points)

(2) Exactly four out of ten vehicles will turn right. (2 points)

(3) At least three out of ten vehicles will turn right. (2 points)

(4) The first right-turning vehicle will be the fifth vehicle. (2 points)

(5) The eighth vehicle will be the third to turn right. (2 points)

Solution:

(1) The movement of each vehicle is a Bernoulli trail with p=0.25 and q=0.75

The probability that the ninth vehicle, and any other vehicle, will not turn right is q =

0.75

(2) According to the binomial distribution, the probability that four out of ten vehicles will

turn right is

(3) The complement of event A: at least three out of ten is event B: Zero, or one or two out

of ten and

(4) This question may be answered by using either the geometric distribution or the negative

binomial (Pascal) distribution with k=1. Thus

(5) The negative binomial (Pascal) distribution provides the answer to the question of the

probability that the kth (i.e., third) left turner is the eighth vehicle, x, is the sequence

146.075.025.0!6!*4

!10

)!(!

! 64

xnxqpxnx

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pFor x=8 and k=3


3:30 PM – 5:30PM, 6/24/09 Summer 2009

11

10. The speeds of vehicles on the I-75 follow the normal distribution with mean speed of 68 mph

and standard deviation of 9.6 mph. Calculate the following

(1) Calculate of the portion of speeds above 80 mph (3 points)

(2) Calculate the minimum speed that places a speed in the top 5% (2 points)

Solution:

(1) The speeds are distributed as a normal distribution as N(,), the variable

Check the Standard Normal Cumulative Distribution table and found that the probability

that Z is less than or equal to 1.25 is 0.8944. Therefore, the probability that z is greater

than 1.25 is given as 1 – 0.8944 = 0.1056; that is 10.56% of speeds are expected to be

above the 80 mph.

(2) F(z) = 0.95 lies between z = 1.64 and z = 1.65. The corresponding values of speed are

(68 + 1.64*9.6 = 83.7 mph) and (68 + 1.64*9.6 = 83.8 mph), respectively. Consequently,

a speed above 83.8 mph ensures that the speed is in the top 5%.

25.16.9

6880

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3:30 PM – 5:30PM, 6/24/09 Summer 2009

12

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3:30 PM – 5:30PM, 6/24/09 Summer 2009

13


3:30 PM – 5:30PM, 6/24/09 Summer 2009

14

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3:30 PM – 5:30PM, 6/24/09 Summer 2009

15

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3:30 PM – 5:30PM, 6/24/09 Summer 2009

16

Quiz #2 Solution TTE 4004 3:30 PM 5:30PM, 6/24/09 Summer 2009 · Quiz #2 Solution TTE 4004 3:30 PM...

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