Module 2: Transportation planning

Lecture 9 Modal split

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Solution

Cost of travel by car (Equation)= $c_{car}=0.03\times20+18\times0.1+4\times0.1$ = 2.8

Cost of travel by bus (Equation)= $c_{bus}=0.03\times30+0.04\times5+0.06\times3+0.1\times6$ = 1.88

Cost of travel by train (Equation)= $c_{train}=0.03\times12+0.04\times10+0.06\times2+0.1\times4$ = 1.28

Probability of choosing mode car (Equation) $p_{ij}^{car} =\frac{e^{-2.8}}{{e^{-2.8}+e^{-1.88}+e^{-1.28}}}$ = 0.1237

Probability of choosing mode bus (Equation $p_{ij}^{bus} =\frac{e^{-1.88}}{{e^{-2.8}+e^{-1.88}+e^{-1.28}}}$ = 0.3105

Probability of choosing mode train (Equation)= $p_{ij}^{train} =\frac{e^{-1.28}}{{e^{-2.8}+e^{-1.88}+e^{-1.28}}}$ = 0.5657

Proportion of trips by car,$T_{ij}^{car}$ = 5000$\times$0.1237 = 618.5

Proportion of trips by bus, $T_{ij}^{bus}$ = 5000$\times$0.3105 = 1552.5

Similarly, proportion of trips by train, $T_{ij}^{train}$ = 5000$\times$0.5657 = 2828.5 We can put all this in the form of a table as shown below 1:

Table 1: Multinomial logit model problem: solution

	$t_{ij}^v$	$t_{ij}^{walk}$	$t_{ij}^t$	$F_{ij}$	$\phi_{ij}$	$C$	$e^C$	$p_{ij}$	$T_{ij}$
coeff	0.03	0.04	0.06	0.1	0.1	-	-	-	-
car	20	-	-	18	4	2.8	0.06	0.1237	618.5
bus	30	5	3	6	-	1.88	0.15	0.3105	1552.5
train	12	10	2	4	-	1.28	0.28	0.5657	2828.5

 $$

Fare collected from the mode bus = $T_{ij}^{bus}\times F_{ij}$ = 1552.5$\times$6 = 9315

 $$

Fare collected from mode train = $T_{ij}^{train}\times F_{ij}$ = 2828.5$\times$4 = 11314