CE 415-Transportation Engineering II

End Semester Examination

Department of Civil Engineering

Indian Institute of Technology Bombay

Date November 22, 2012
Time 09.30 - 12.30 hours
Marks 100

Instructions

Attempt all questions. Make suitable assumptions, if required, and all the assumptions should be stated clearly. Intermediate steps and appropriate sketches are mandatory. Write page numbers against the questions attempted
  1. [1218] A person is currently using his car to go to office which takes about 15 minutes and costs about Rs. 33 for fuel and parking. One of his friends suggested to use a bus instead which takes about 30 minutes but costs only Rs. 20. However, another friend suggested him to use a metro which takes only 20 minutes but costs Rs. 15. Assuming that the weightages given to travel time and travel cost by the person are 0.7 and 0.5 respectively, (a) What is the probability that the person will stop using car? (b) Given that the person stops using car, what is the probability that he will use metro? (7)

Solution:

The total cost involved in using car = 0.7*15+0.5*33=27 The total cost involved in using bus=0.7*30+0.5*20=31 The total cost involved in using metro=0.7*20+0.5*15=21.5 comparing the two altenatives with the existing one we have For the first strategy: The probability of using car= $ \frac{e^{-27}}{e^{-27}+e^{-31}+e^{-21.5}}$=0.004 The probability of using metro= $ \frac{e^{-21.5}}{e^{-27}+e^{-31}+e^{-21.5}}$=0.995

(a) The probability of not using car=1-0.004=0.996 (b) The probability of using metro given that the person stopped using car== $ \frac{e^{-21.5}}{e^{-31}+e^{-21.5}}$=0.999

[1212] In a traffic stream, 30% of the vehicles travel at a constant speed of 60km/h, 30% at a constant speed of 80km/h, and the remaining vehicles at a constant speed of 100km/h. An observer travelling at a constant speed of 70km/h with the stream over a length of 5km is overtaken by 17 vehicles more than what he has overtaken. The observer met 303 vehicles while traveling against the stream at the same speed and over the same length of highway. What is the mean speed and flow of the traffic stream?  (8)

Solution

From the moving observer method,

(a) $ Q={(x+y)}/{(t_a+t_w)} = {(303+17)}/{(5/70+5/70)} =2240$ veh per hour (b) mean speed = $ \frac{5}{(5/70)-(17/2240)}=78.3$ kmph or mean speed=100/(30/60)+(30/80)+(40/100)=78.43 kmph

[1217] A study area has four zones and it is observed that they generate 70, 89, 120, 110 trips per day. The average income is respectively 1500, 2500, 3500, and 2800 and the population is 3100, 2600, 3600, and 4100. Government proposes two major policy changes in zone 2. First proposal will result in an increase of income by 40 % and the second will increase the population by 50 %. Which proposal will generate more trips?.  (10)

Solution

Zone y trips/day Income $ x_1$ Population $ x_2$ $ x_1y\times 10^3 $ $ x_2y\times 10^3 $ $ x_1^2 \times 10^4 $ $ x_2^2 \times 10^4 $
1 70 1500 3100 105 217 225 961
2 89 2500 2600 222.5 231.4 625 676
3 120 3500 3600 420 432 1225 1296
4 110 2800 4100 308 451 784 1681
$ \sum$ 389 10300 13400 1055.5 1331.4 2859 4614
Mean 97.25 2575 3350        
Relation between $ y$ and $ x_1$ $ y=b_1 x_1+a_1$

$ b_1 = \frac{n \sum x_1 y - \sum x_1 \sum y}{n \sum x_1^2 - [\sum x_1]^2}$= 0.026

$ a_1 = y_{mean} - b  x_{1_{mean}}$ =30.213

$ y = 0.026 x_1+ 30.213$ Similarly relation between $ y$ and $ x_2$ is $ y = 0.0226 x_2+ 21.54$

With increase of 40 % income, Trips/day = 0.026 (1.4 X 2400)+30.213= 118

With 50 % inciarease in population, Trips/day = 0.0226 (1.5 X 2500)+ 21.54= 106

More number of trips are produced in the first case i.e. for increase of 40 % income

Alternate Solution

An alternate solution include considering a power equation consisting of both income and population simultaneously. The final solution is same and marks are given for that also.

[1213] The entry and exit width of a rotary intersection are 9m and 11m respectively. The width of approaches at the intersection is 15m. The traffic from the four approaches traversing the intersection is given below. If the traffic composition is 50% car, 40% two-wheelers and 10% trucks and the passenger car units of two-wheelers and trucks are 0.5 and 3 respectively, find the capacity of the rotary using TRL formulae.

Approach Left turn Straight Right turn
North 500 800 300
South 400 350 450
East 250 400 500
West 300 450 500
 (10)

Solution:

Taking into consideration the pcu and the proportion values we have $ 0.5\times1+0.4\times0.5+0.1\times3=1$ Capacity Calculation
Weaving Section 1(W-N) 2(N-E) 3(E-S) 4(S-W)
Entry Width, e1 9 9 9 9
Exit Width, e2 11 11 11 11
Weaving width (e1+e2)/2+3.5 13.5 13.5 13.5 13.5
Weaving Length, l= 4*w 54 54 54 54
Average entry/exit width (e1+e2)/2 10 10 10 10
Non-weaving volume - a 300 500 250 400
Weaving volume - b 950 1100 900 800
Weaving volume - c 850 900 1300 700
Non-weaving volume - d 450 500 300 500
Total volume (a+b+c+d) 2550 3000 2750 2400
Weaving proportion p =(b+c)/(a+b+c+d) 0.706 0.667 0.800 0.625
Capacity of weaving section, Q=280w() 4,025 4,094 3,860 4,167
Capacity of rotary [Min.( Q1, Q2, Q3, Q4)] 3,860

[1215] A line of vehicles are in car following mode and all vehicles are travelling at 18 m/s with distance headway of 20 m. After 1.2 seconds, the lead vehicle suddenly decelerates at a rate of 1.2 $ m/s^2$ until it stops completely. simulate the behaviour of first following vehicle using the GM fifth car following model for the first 2.5 seconds. Tabulate the results. Assume headway exponent 1.2, speed exponent 1.6, sensitivity coefficient 0.8, reaction time 0.6 seconds, and scan interval 0.3 seconds. (10)

Solution:

t a(t) v(t) x(t) a(t) v(t) x(t) dv dx
0 0 18.0 20.00 0.00 18.00 0.00 0.00 20.00
0.3 0 18.0 25.40 0.00 18.00 5.40 0.00 20.00
0.6 0 18.0 30.80 0.00 18.00 10.80 0.00 20.00
0.9 0 18.0 36.20 0.00 18.00 16.20 0.00 20.00
1.2 -1.2 18.0 41.60 0.00 18.00 21.60 0.00 20.00
1.5 -1.2 17.6 46.90 0.00 18.00 27.00 -0.36 19.95
1.8 -1.2 17.3 52.20 0.00 18.00 32.40 -0.72 19.78
2.1 -1.2 16.9 57.30 -0.81 18.00 37.80 -1.08 19.51
2.4 -1.2 16.6 62.30 -1.60 17.76 43.16 -1.20 19.17
2.7 -1.2 16.2 67.30 -2.33 17.28 48.42 -1.08 18.83

[1216] A North-South corridor has three junctions namely A, B, and C. Junction A is on the south end of the corridor and junction C is on the north end. These junctions are coordinated in the north direction. All the junctions are having two phase signals with a cycle of 80 sec. The juctions A, B, and C have green times of 40, 50, and 30 sec respectively in the coordinated direction. The distance between A and B is 600 meters and B and C is 900 meters. The junctions are coordinated considering a speed of 15 m/sec. (a) What will be the resulting band width? (b) While the corridor is operating under the above control conditions, if the vechiles could travel only at a speed of 12 m/sec, what bandwidth will be achieved? (10)

Solution:

Image 1216ts

[1211] In a traffic study, the observed densities were 150, 120, 50, 70 and 20 veh/km and the corresponding speeds were 10, 25, 45, 40 and 32km/h. Find the jam density according to Greenberg's logarithmic traffic stream model. (Hint: Linearize the expression)

Solution:

$ u=u_0\times\ln(\frac{k_j}{k})$ Expanding the above equation we have $ u=u_0\times \ln{k_j}-u_0\times \ln{k}$ Now the linear relationship is between $ u$ and $ \ln k$ if $ y$=$ a$+$ b$$ \times$$ x$ then y=$ u$, x=$ ln (k)$, $ a$=$ u_0$$ \times$$ ln({k_j})$ and $ b$=-$ u_0$

x=(ln k) y=(u) x-xmean y-ymean (x-xmean)(y-ymean) $ (x-xmean)^2$
5.011 10 0.820 -20 -16.723 0.672
4.787 25 0.597 -5.4 -3.222 0.356
3.912 45 -0.279 14.6 -4.071 0.078
4.248 40 0.058 9.6 0.553 0.003
2.996 32 -1.195 1.6 -1.912 1.428
4.191 34     -25.375 2.537
From the calibration of linear speed-density model between $ u$ and $ ln (k)$ we can find out $ k_{j}$ as follows
  1. b=$ u_0$=10
  2. a=$ u_0$$ \times$$ ln({k_j})$=72.311
  3. $ k_{j}=1381.246$

[1219] Illustrate with neat sketches: (i) A diamond interchange showing the movement of all the flows. (ii) Road markings on a two lane bi-directional horizontal curve when the sight distance is less than the length of the curve. (iii) The concept of flow prediction in a transportation system when the supply is improved. (15)

Solution

[1214] The traffic flow and phase plan for a four-legged intersection is as shown in Figure. The E-W flow is 800, W-E flow is 740, N-S flow is 450, and S-N flow is 490 vehicles per hour. Assume for all the phases the yellow time is 3 seconds, the lost time is 2 seconds, saturation headway is 1.8 seconds, and degree of saturation is 0.9. Assume left turn adjustment factor 1.10 and right turn adjustment factor 1.30. Assume the left turn and right turn traffic proportion is 10% and 20% respectively. Assuming no pedestrian traffic, compute the cycle time and green time for each phase. Compute also the stopped delay for the traffic from north.

Image 1452
  (20)

Solution:

$\displaystyle W-E (LT) lane flow$ $\displaystyle =$ $\displaystyle total vol * left turn proportion * correction factor$ (1)
  $\displaystyle =$ $\displaystyle 740*0.1*1.10 = 81.40$ (2)
$\displaystyle N-S (LT+TH) lane flow$ $\displaystyle =$ $\displaystyle 450*0.1*1.10 + 450*0.7/2 = 207.00$ (3)

Direction wise flows LT TH RT LT+TH RT+TH
N-S       207.00 274.50
W-E 81.4 518.00 148.00    
S-N       225.40 298.90
E-W 88 560.00 208.00    
Phases 1 2 3 4  
critical flows Vci 518.00 298.90 560.00 274.50  
  1. Total critical flows $ \sum{Vci}= 1,651.4$
  2. Saturation Flow S= 2,000.00
  3. Cycle Time C= 96.90
  4. Total effective Green Time Tg=C-NL= 88.90
Effective green gi 27.89 16.09 30.15 14.78
Actual green Gi 26.89 15.09 29.15 13.78
Rounded green 27.00 16.00 30.00 14.00
Final cycle C 99.00
Effective green 28.00 17.00 31.00 15.00
Approach (1)WE (2)SN (3)EW (4)NS
Delay       41.31
-November 27, 2012-
Prof. Tom V. Mathew 2012-11-27