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  Module 2: Transportation planning
Lecture 8 Trip distribution
  

Solution

The sum of the attractions in the horizon year, i.e. $\Sigma{O_i}$ = 98+106+122 = 326. The sum of the productions in the horizon year, i.e. $\Sigma{D_j}$ = 102+118+106 = 326. They both are found to be equal. Therefore we can proceed. The first step is to fix $b_j=1$, and find balancing factor $a_i$. $a_i = O_i/o_i$, then find $T_{ij}=a_i\times t_{ij}$

So $a_1 = 98/78 = 1.26$

$a_2 = 106/92 = 1.15$

$a_3 = 122/82 = 1.49$ Further $T_{11} = t_{11} \times a_1 = 20 \times 1.26= 25.2$. Similarly $T_{12} = t_{12} \times a_2 = 36 \times 1.15 = 41.4$. etc. Multiplying $a_1$ with the first row of the matrix, $a_2$ with the second row and so on, matrix obtained is as shown below.

  1 2 3 $o_i$
1 25.2 37.8 35.28 98
2 41.4 36.8 27.6 106
3 32.78 50.66 38.74 122
$d_j^1$ 99.38 125.26 101.62  
$D_j$ 102 118 106  
Also $d_j^1=25.2+41.4+32.78=99.38$

In the second step, find $b_j$ = $D_j$/$d_j^1$ and $T_{ij}=t_{ij} \times b_j$. For example $b_1 = 102/99.38 = 1.03$, $b_2 = 118/125.26 = 0.94$ etc., $T_{11} = t_11 \times b_1 = 25.2 \times 1.03 = 25.96 $ etc. Also $O_i^1 = 25.96+35.53+36.69 = 98.18$. The matrix is as shown below:

  1 2 3 $o_i$ $O_i$
1 25.96 35.53 36.69 98.18 98
2 42.64 34.59 28.70 105.93 106
3 33.76 47.62 40.29 121.67 122
$b_j$ 1.03 0.94 1.04    
$D_j$ 102 118 106    
  1 2 3 $O_i^1$ $O_i$
1 25.96 35.53 36.69 98.18 98
2 42.64 34.59 28.70 105.93 106
3 33.76 47.62 40.29 121.67 122
$d_j$ 102.36 117.74 105.68 325.78  
$D_j$ 102 118 106 326  
Therefore error can be computed as ; $Error = \Sigma{\vert O_i - O_i^1\vert} +\Sigma{\vert D_j - d_j\vert}$

Error = $\vert 98.18-98\vert+\vert 105.93-106\vert+\vert 121.67-122\vert+\vert 102.36-102\vert+\vert 117.74-118\vert+\vert 105.68-106\vert = 1.32$