Module 2: Transportation planning

Lecture 9 Modal split

1

2

3

4

5

6

7

8

9

10

11

12

13

14

Solution

The base case is given below.

Cost of travel by car (Equation)= $c_{car}=0.03\times20+18\times0.06+4\times0.1$ = 2.08

Cost of travel by bus (Equation)= $c_{bus}=0.03\times30+0.04\times5+0.06\times3+0.1\times9$ = 2.18

Probability of choosing mode car (Equation)= $p_{ij}^{car} =\frac{e^{-2.08}}{{e^{-2.08}+{e^{-2.18}}}}$ = 0.52

Probability of choosing mode bus (Equation)= $p_{ij}^{bus} =\frac{e^{-2.18}}{{e^{-2.08}+{e^{-2.18}}}}$ = 0.475

Proportion of trips by car = $T_{ij}^{car}$ = 5000$\times$0.52 = 2600

Proportion of trips by bus = $T_{ij}^{bus}$ = 5000$\times$0.475 = 2400

Fare collected from bus = $T_{ij}^{bus}\times F_{ij}$ = 2400$\times$9 = 21600

When the fare of bus gets reduced to 6,

Cost function for bus= $c_{bus}=0.03\times30+0.04\times5+0.06\times3+0.1\times6$ = 1.88

Probability of choosing mode bus (Equation)= $p_{ij}^{bus} =\frac{e^{-1.88}}{{e^{-2.08}+{e^{-1.88}}}}$ = 0.55

Proportion of trips by bus = $T_{ij}^{bus}$ = 5000$\times$0.55 = 2750

Fare collected from the bus $T_{ij}^{bus}\times F_{ij}$= 2750$\times$6 = 16500

The results are tabulated in table