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CE 415-Transportation Engineering-II

Mid Semester Examination

Department of Civil Engineering

Indian Institute of Technology Bombay

Date September 12, 2012
Time 8:30-10.30 hours
Marks 50

Instructions

Attempt all questions. Answers should be brief and to the point. Draw neat sketches. Make suitable assumptions, if required, and all the assumptions should be stated clearly. Intermediate steps should be show along with appropriate tabular forms to get full marks. Write page numbers against the questions attempted
  1. [1609] The free flow travel time on each link of the following network is given below. The demand from 1 to 5, 1 to 6, and 2 to 6 is 100 trips each, and the BPR's travel time parameters, $ \alpha$, $ \beta$, and k are 0.1, 2, and 120 respectively. If the capacity of the link having highest flow is doubled, what will be the change in Total System Travel Time? (10)
    From To Free flow travel time
    1 2 10
    1 3 12
    2 4 20
    2 3 8
    2 5 30
    3 5 25
    4 5 10
    4 6 8
    5 6 9
    Figure 1: Road Network
    Image 1609

Solution


OD Pair Paths Free flow tt
  1-2-4-6 38
  1-2-5-6
  1-3-5-6
  1-2-3-5-6
  1-2-4-5-6
  Min
  2-3-5-6 42
  2-5-6-
  2-4-6-
  2-4-5-6
  Min
  1-2-5- 40
  1-3-5-
  1-2-3-5
  1-2-4-5
  Min


From To t0 alpha beta K x1-6 x2-6 x3-6 x t x*t
1 2 10 0.1 2 120 100     100 10.69 1,069.44
1 3 12 0.1 2 120     100 100 12.83 1,283.33
2 3 8 0.1 2 120       0 8.00 0.00
2 4 20 0.1 2 120 100 100   200 25.56 5,111.11
2 5 30 0.1 2 120       0 30.00 0.00
3 5 25 0.1 2 120     100 100 26.74 2,673.61
4 5 10 0.1 2 120       0 10.00 0.00
4 6 8 0.1 2 120 100 100   200 10.22 2,044.44
5 6 9 0.1 2 120       0 9.00 0.00
  TSTT 12181.9

If the capacity of the link 2-4 is doubled, change in TSTT=12181.94-11348.61=833.33veh-min=13.89veh-hrs.
If the capacity of the link 4-6 is doubled, change in TSTT=12181.94-11848.61=333.33veh-min=5.5veh-hrs.
If the capacities of both 2-4 and 4-6 are doubled, change in TSTT=12181.94-11015.28=1166.66veh-min=19.44veh-hrs [2019] (i) Suppose you were asked to do travel demand modelling for Greater Mumbai, illustrate the study area, internal zones, external zones, cordon line, and the screen lines (ii)Define the goal of transportation system analysis and state the assumptions. (iii)State the trip distribution equation of the gravity model and derive the expression for production constant. (iv) State Wardrop's first principles and show its mathematical representation. (12)

Solution:

[1008] The trip productions from zones 1, 2 and 3 are 110, 122 and 140 respectively and the trip attractions to these zones are 120,134 and 118 respectively. Assume the peoples choice to destination is influenced inversely by the square of the distance.The distance matrix is given below.
\begin{displaymath}\left[ \begin{array}{ccc} 1.0&1.2&1.8 \\ 1.2&1.0&1.5 \\ 1.8&1.5&1.0 \\ \end{array}\right]\end{displaymath}      

Compute the trip matrix using doubly constrained gravity model. Provide one complete iteration.  (7)

Solution

Solution
$ i$ $ j$ $ B_j$ $ D_J$ $ f(c_{ij})$ $ {B_j D_j f(c_{ij})}$ $ {\Sigma{B_j D_j f(c_{ij})}}$ $ A_i=\frac{1}{{\Sigma{B_j D_j f(c_{ij})}}}$
  1 1.0 120 1.0 120.00    
1 2 1.0 134 0.694 93.056 249.475 0.004
  3 1.0 118 0.309 36.420    
  1 1.0 120 0.694 83.333    
2 2 1.0 134 1.0 134 269.778 0.004
  3 1.0 118 0.444 52.444    
  1 1.0 120 0.309 37.037    
3 2 1.0 134 1.000 59.556 214.593 0.005
  3 1.0 118 1.00 118    

The second step is to find $ B_j$. This can be found out as $ B_j = 1/{\Sigma{A_i O_i f(c_{ij})}}$, where $ A_i$ is obtained from the previous step.

$ j$ $ i$ $ A_i$ $ O_i$ $ f(c_{ij})$ $ {A_i O_i f(c_{ij})}$ $ \Sigma{A_i O_i f(c_{ij})}$ $ B_j = 1/{\Sigma{A_i O_i f(c_{ij})}}$
  1 0.004 110 1.000 0.441    
1 2 0.004 122 0.694 0.314 0.956 1.046
  3 0.005 140 0.309 0.201    
  1 0.004 110 0.694 0.306    
2 2 0.004 122 1.000 0.452 1.048 0.954
  3 0.005 140 0.444 0.290    
  1 0.004 110 0.309 0.136    
3 2 0.004 122 0.444 0.201 0.989 1.011
  3 0.005 140 1.000 0.652    
The function $ f(c_{ij})$ can be written in the matrix form as:

$\displaystyle \left[ \begin{array}{ccc} 1.0&0.69&0.31 \\ 0.69&1.0&0.44 \\ 0.31&0.44&1.0 \\ \end{array} \right]$ (1)

Then $ T_{ij}$ can be computed using the formula

$\displaystyle T_{ij} = A_i O_i B_j D_jf(c_{ij})$ (2)

$ O_i$ is the actual productions from the zone and $ O_i^1$ is the computed ones. Similar is the case with attractions also.

  1 2 3 $ A_i$ $ O_i$ $ O_i^1$
1 55.327 39.137 16.229 0.004 110 110.694
2 39.406 57.802 23.969 0.004 122 121.177
3 25.266 37.061 77.802 0.005 140 140.129
$ B_j$ 1.046 0.954 1.011      
$ D_j$ 120 134 118      
$ D_j^1$ 120 134 118      

$ O_i$ is the actual productions from the zone and $ O_i^1$ is the computed ones. Similar is the case with attractions also.

Therefore error can be computed as ; $ Error = \Sigma{\vert O_i - O_i^1\vert} +\Sigma{\vert D_j - D_j^1\vert}$ $ Error = \vert 110-110.694\vert+\vert 122-121.177\vert+\vert 140-140.129\vert+\vert 120-120\vert+\vert 134-134\vert+\vert 118-118\vert=1.646$ [2101] Build the minimum path tree from origin node 1 to all the other nodes for the following case: (10)

Link Travel time
1-2 2
1-3 2
2-4 3
2-5 2
3-5 4
3-6 3
4-8 4
4-9 2
5-7 2
5-8 3
6-7 4
7-10 5
8-10 4
8-11 3
9-11 2
10-12 3
11-12 3

Solution:

O-D Shortest Path
1-2 1-2
1-3 1-3
1-4 1-2-4
1-5 1-2-5
1-6 1-3-6
1-7 1-2-5-7
1-8 1-2-5-8
1-9 1-2-4-9
1-10 1-2-5-7-10 (or) 1-2-5-8-10
1-11 1-2-4-9-11
1-12 1-2-4-9-11-12
[1713] Calculate the system travel time and link flows by system optimum assignment using Frank Wolfe algorithm for a network with two nodes having two paths as links. The travel time functions are $ 14+4 x_1$ and $ 20+2 x_2$. The total flow on the two links is limited to 14. Assume a precison of 0.01 for convergence.
Image 1713
(7)

Solution:

Given $ x_1+x_2 = 14$
Free flow time $ t_1 = 14, t_2 = 20$
% latex2html id marker 1049 $ \therefore$ By AON, $ x_1 = 14, x_2 = 0$
First iteration
Step 1:
$ t_1 = 14+4*14 = 70$
$ t_2 = 20+2*0 = 20$
% latex2html id marker 1057 $ \therefore y_1 = 0, y_2 = 14$
Step 2:

$\displaystyle \bar{x_1}$ $\displaystyle =$ $\displaystyle x_1+\alpha_n(y_1-x_1)$  
  $\displaystyle =$ $\displaystyle 14+\alpha(0-14)$  
  $\displaystyle =$ $\displaystyle 14(1-\alpha)$  
$\displaystyle \bar{x_2}$ $\displaystyle =$ $\displaystyle x_2+\alpha_n(y_2-x_2)$  
  $\displaystyle =$ $\displaystyle 0+\alpha(14-0)$  
  $\displaystyle =$ $\displaystyle 14\alpha$  
$\displaystyle By SO assignment Z$ $\displaystyle =$ $\displaystyle \sum X_at_a$  
  $\displaystyle =$ $\displaystyle x_1(14+4x_1)+x_2(20+2x_2)$  
  $\displaystyle =$ $\displaystyle 14x_1+4X^2_1+20x_2+2x^2_2$  
  $\displaystyle =$ $\displaystyle 14*14(1-\alpha)+4*14^2(1-\alpha)^2+20*14\alpha+2*14^2*\alpha^2$  
  $\displaystyle =$ $\displaystyle 1176\alpha^2-1484\alpha+980$  
$\displaystyle \frac{dz}{d\alpha}$ $\displaystyle =$ $\displaystyle 2*1176\alpha-1484 = 0$  
% latex2html id marker 1116 $\displaystyle \therefore \alpha$ $\displaystyle =$ $\displaystyle 0.631$  

Step 3:
% latex2html id marker 1122 $ \therefore \bar{x_1} = 14(1-\alpha) = 14(1-0.631) = 5.17$
$ \bar{x_2} = 14\alpha = 8.83$
Second Iteration
Step 1:
$ t_1 = 14+4*5.17 = 34.68$
$ t_2 = 20+2*8.83 = 37.66$
Step 2:

$\displaystyle \bar{x_1}$ $\displaystyle =$ $\displaystyle x_1+\alpha(y_1-x_1)$  
  $\displaystyle =$ $\displaystyle 5.17+\alpha(14-5.17)$  
  $\displaystyle =$ $\displaystyle 5.17+8.83\alpha$  
$\displaystyle \bar{x_2}$ $\displaystyle =$ $\displaystyle x_2+\alpha(y_2-x_2)$  
  $\displaystyle =$ $\displaystyle 8.83+\alpha(0-8.83)$  
  $\displaystyle =$ $\displaystyle 8.83(1-\alpha)$  
$\displaystyle Z$ $\displaystyle =$ $\displaystyle 14x_1+4x^2_1+20x_2+20x^2_2$  
  $\displaystyle =$ $\displaystyle 14(5.17+8.83\alpha)+4(5.17+8.83\alpha)^2+20*8.83(1-\alpha)+2*8.83^2*(1-\alpha)^2$  
  $\displaystyle =$ $\displaystyle 511.84+0.42\alpha+467.82\alpha^2$  
$\displaystyle \frac{dz}{d\alpha}$ $\displaystyle =$ $\displaystyle 0.42+2*467.82\alpha =0$  
% latex2html id marker 1179 $\displaystyle \therefore \alpha$ $\displaystyle =$ $\displaystyle -0.000448$  

Step 3:
% latex2html id marker 1185 $ \therefore \bar{x_1} = 5.17+8.83(-0.000448) = 5.17 $
$ \bar{x_2} = 8.83(1+0.000448) = 8.834$
% latex2html id marker 1189 $ \therefore$ Link flows: $ x_1=5.17, x_2= 8.83$
% latex2html id marker 1193 $ \therefore$ System travel time = $ x_1*t_1+x_2*t_2$
$ = 5.17*34.68+8.83*37.66$
$ = 511.83$

 E O F 

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Prof. Tom V. Mathew 2012-09-24