Module 2: Transportation planning

Lecture 9 Modal split

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Solution

First, use binary logit model to find the trips when there is only car and bus. Then, again use binary logit model to find the trips when there is only car and train. Finally compare both and see which alternative carry maximum trips.

Cost of travel by car= $c_{car}=0.05\times25+0.2\times22+0.2\times6$ = 6.85

Cost of travel by bus= $c_{bus}=0.05\times35+0.04\times8+0.07\times6+0.2\times8$ = 4.09

Cost of travel by trai= $c_{train}=0.05\times17+0.04\times14+0.07\times5+0.2\times6$ = 2.96

Case 1: Considering introduction of bus, Probability of choosing car, $p_{ij}^{car} = \frac{e^{-6.85}}{{e^{-6.85}+e^{-4.09}}}$ = 0.059

Probability of choosing bus, $p_{ij}^{bus} =\frac{e^{-4.09}}{{e^{-6.85}+e^{-4.09}}}$ = 0.9403

Case 2: Considering introduction of train, Probability of choosing car $p_{ij}^{car} = \frac{e^{-6.85}}{{e^{-6.85}+e^{-2.96}}}$ = 0.02003

Probability of choosing train $p_{ij}^{train} = \frac{e^{-2.96}}{{e^{-6.85}+e^{-2.96}}}$ = 0.979

Trips carried by each mode

Case 1: $T_{ij}^{car}$ = 4200$\times$0.0596 = 250.32 $T_{ij}^{bus}$ = 4200$\times$0.9403 = 3949.546

Case 2: $T_{ij}^{car}$ = 4200$\times$0.02 = 84.00 $T_{ij}^{train}$ = 4200$\times$0.979 = 4115.8

Hence train will attract more trips, if it is introduced.