Modeling Mode and Route Similarities in Network Equilibrium Problem with Go-Green Modes

Abstract

Environmental sustainability is a common requirement on the development of various real-world systems, especially on road transportation systems. Motorized vehicles generate a large amount of harmful emissions, which have adverse effects to the environment and human health. Environmental sustainability requires more promotions of ‘go-green’ transportation modes such as public transit and bicycle to realize the increasing travel demands while keeping the environmental expenses low. In this paper, we make use of recent advances in discrete choice modeling to develop equivalent mathematical programming formulations for the combined modal split and traffic assignment (CMSTA) problem that explicitly considers mode and route similarities under congested networks. Specifically, a nested logit model is adopted to model the modal split problem by accounting for mode similarity among the available modes, and a cross-nested logit model is used to account for route overlapping in the traffic assignment problem. This new CMSTA model has the potential to enhance the behavioral modeling of travelers’ mode shift between private motorized mode and ‘go-green’ modes as well as their mode-specific route choices, and to assist in quantitatively evaluating the effectiveness of different ‘go-green’ promotion policies.

Appendix

Proof of Proposition 1

The Lagrangian of the minimization problem with respect to (w.r.t.) the equality constraints can be formulated as:

$$ L=Z+{\displaystyle \sum_{ij\in IJ}{\lambda}_{ij}}\left({q}_{ij}-{\displaystyle \sum_{e\in {E}_{ij}}{\displaystyle \sum_{r\in {R}_{ij}}{f}_{er}^{ij}}}\right), $$

(24)

where λ _ij is the dual variable associated with the flow conservation constraint in Eq. (14).

• (1) Given that the Lagrangian L has to be minimized w.r.t. positive route flows, the following conditions have to hold w.r.t. the route-flow variables:

  $$ \frac{\partial L}{\partial {f}_{er}^{ij}}=0\kern0.5em \Rightarrow \kern0.5em {g}_{er}^{ij}+\frac{\mu }{\theta } \ln \frac{f_{er}^{ij}}{{\left({\alpha}_{ij er}\right)}^{\frac{1}{\mu }}}+\frac{1-\mu }{\theta } \ln {\displaystyle \sum_{r\in {R}_{ij}}{f}_{er}^{ij}}-{\lambda}_{ij}=0. $$

  (25)

  Rearranging Eq. (25) gives

  $$ {f}_{er}^{ij}{\left({\displaystyle \sum_{r\in {R}_{ij}}{f}_{er}^{ij}}\right)}^{\frac{1-{\mu}_{ij}}{\mu_{ij}}}={\left({\alpha}_{ij er}\right)}^{\frac{1}{\mu_{ij}}} \exp \left(\frac{\theta_{ij}}{\mu_{ij}}{\lambda}_{ij}-\frac{\theta_{ij}}{\mu_{ij}}{g}_{er}^{ij}\right). $$

  (26)

  Summing all r gives

  $$ {\left({\displaystyle \sum_{r\in {R}_{ij}}{f}_{er}^{ij}}\right)}^{\frac{1}{\mu_{ij}}}= \exp \left(\frac{\theta_{ij}}{\mu_{ij}}{\lambda}_{ij}\right){\displaystyle \sum_{r\in {R}_{ij}}{\left({\alpha}_{ij er}\right)}^{\frac{1}{\mu_{ij}}} \exp \left(-\frac{\theta_{ij}}{\mu_{ij}}{g}_{er}^{ij}\right)}. $$

  (27)
  $$ {\displaystyle \sum_{r\in {R}_{ij}}{f}_{er}^{ij}}= \exp \left({\theta}_{ij}{\lambda}_{ij}\right){\left({\displaystyle \sum_{r\in {R}_{ij}}{\left({\alpha}_{ij er}\right)}^{\frac{1}{\mu_{ij}}} \exp \left(-\frac{\theta_{ij}}{\mu_{ij}}{g}_{er}^{ij}\right)}\right)}^{\mu_{ij}}. $$

  (28)

  Then, summing all e gives

  $$ {\displaystyle \sum_{e\in {E}_{ij}}{\displaystyle \sum_{r\in {R}_{ij}}{f}_{er}^{ij}}}={q}_{ij}= \exp \left({\theta}_{ij}{\lambda}_{ij}\right){\displaystyle \sum_{e\in {E}_{ij}}{\left({\displaystyle \sum_{r\in {R}_{ij}}{\left({\alpha}_{ij er}\right)}^{\frac{1}{\mu_{ij}}} \exp \left(-\frac{\theta_{ij}}{\mu_{ij}}{g}_{er}^{ij}\right)}\right)}^{\mu_{ij}}}. $$

  (29)

  Dividing Eq. (28) by Eq. (29) gives the marginal probability of choosing nest e:

  $$ \frac{{\displaystyle \sum_{r\in {R}_{ij}}{f}_{er}^{ij}}}{q_{ij}}={P}_e^{ij}=\frac{{\left({\displaystyle \sum_{r\in {R}_{ij}}{\left({\alpha}_{ij er}\right)}^{\frac{1}{\mu_{ij}}} \exp \left(-\frac{\theta_{ij}}{\mu_{ij}}{g}_{er}^{ij}\right)}\right)}^{\mu_{ij}}}{{\displaystyle \sum_{e\in {E}_{ij}}{\left({\displaystyle \sum_{r\in {R}_{ij}}{\left({\alpha}_{ij er}\right)}^{\frac{1}{\mu_{ij}}} \exp \left(-\frac{\theta_{ij}}{\mu_{ij}}{g}_{er}^{ij}\right)}\right)}^{\mu_{ij}}}}. $$

  (30)

  Dividing Eq. (26) by Eq. (27) gives the conditional probability:

  $$ \frac{f_{er}^{ij}}{{\displaystyle \sum_{r\in {R}_{ij}}{f}_{er}^{ij}}}={P}_{r\Big|e}^{ij}=\frac{{\left({\alpha}_{ij er}\right)}^{\frac{1}{\mu_{ij}}} \exp \left(-\frac{\theta_{ij}}{\mu_{ij}}{g}_{er}^{ij}\right)}{{\displaystyle \sum_{r\in {R}_{ij}}{\left({\alpha}_{ij er}\right)}^{\frac{1}{\mu_{ij}}} \exp \left(-\frac{\theta_{ij}}{\mu_{ij}}{g}_{er}^{ij}\right)}}. $$

  (31)

  Hence, these marginal and conditional probabilities correspond to the CNL route choice probability.

• (2) Now we consider the O-D demand variables:

  $$ \frac{\partial L}{\partial {q}_{ij}}=0\kern0.5em \Rightarrow \kern0.5em -\frac{1}{\theta_{ij}} \ln {q}_{ij}-{D}_{ij}^{-1}\left({q}_{ij}\right)+{\lambda}_{ij}=0. $$

  (32)

  From Eq. (29), we have

  $$ {\lambda}_{ij}=\frac{1}{\theta_{ij}} \ln {q}_{ij}-\frac{1}{\theta_{ij}} \ln {\displaystyle \sum_{e\in {E}_{ij}}{\left({\displaystyle \sum_{r\in {R}_{ij}}{\left({\alpha}_{ij er}\right)}^{\frac{1}{\mu_{ij}}} \exp \left(-\frac{\theta_{ij}}{\mu_{ij}}{g}_{er}^{ij}\right)}\right)}^{\mu_{ij}}}. $$

  (33)

  Let w _ij be the CNL expected perceived travel time (EPT) of O-D pair ij:

  $$ {w}^{ij}=-\frac{1}{\theta_{ij}} \ln {\displaystyle \sum_{e\in {E}_{ij}}{\left({\displaystyle \sum_{r\in {R}_{ij}}{\left({\alpha}_{ij er}\right)}^{\frac{1}{\mu_{ij}}} \exp \left(-\frac{\theta_{ij}}{\mu_{ij}}{g}_{er}^{ij}\right)}\right)}^{\mu_{ij}}}. $$

  (34)

  Then, Eq. (32) can be restated as

  $$ -\frac{1}{\theta_{ij}} \ln {q}_{ij}-{D}_{ij}^{-1}\left({q}_{ij}\right)+\frac{1}{\theta_{ij}} \ln {q}_{ij}+{w}^{ij}=0. $$

  (35)

  Finally, the O-D demand is related to the elastic demand function as

  $$ {q}_{ij}={D}_{ij}\left({w}^{ij}\right), $$

  (36)

  where the elastic demand is a function of the EPT received from the route choice following the CNL model. This completes the proof. □

Proof of Proposition 2

It is sufficient to prove that objective function Eq. (13) is strictly convex and that the feasible region is convex. The convexity of the feasible region is assured by the linear equality constraints in Eq. (14). The nonnegative constraints in Eq. (15) do not alter this characteristic. By substituting Eq. (14) into the objective function, we have

$$ \begin{array}{c}\hfill \min Z={Z}_1+{Z}_2+{Z}_3+{Z}_4+{Z}_5\hfill \\ {}\hfill ={\displaystyle \sum_{a\in A}{\displaystyle \underset{0}{\overset{v_a}{\int }}{h}_a\left(\omega \right) d\omega}}+{\displaystyle \sum_{ij\in IJ}{\displaystyle \sum_{e\in {E}_{ij}}{\displaystyle \sum_{r\in {R}_{ij}}\frac{\mu_{ij}}{\theta_{ij}}{f}_{er}^{ij}\left( \ln \frac{f_{er}^{ij}}{{\left({\alpha}_{ij er}\right)}^{\frac{1}{\mu_{ij}}}}-1\right)}}}\hfill \\ {}\hfill +{\displaystyle \sum_{ij\in IJ}{\displaystyle \sum_{e\in {E}_{ij}}\frac{1-{\mu}_{ij}}{\theta_{ij}}\left({\displaystyle \sum_{r\in {R}_{ij}}{f}_{er}^{ij}}\right)}\left( \ln \left({\displaystyle \sum_{r\in {R}_{ij}}{f}_{er}^{ij}}\right)-1\right)}-{\displaystyle \sum_{ij\in IJ}\frac{1}{\theta_{ij}}\left({\displaystyle \sum_{e\in {E}_{ij}}{\displaystyle \sum_{r\in {R}_{ij}}{f}_{er}^{ij}}}\right)\left( \ln \left({\displaystyle \sum_{e\in {E}_{ij}}{\displaystyle \sum_{r\in {R}_{ij}}{f}_{er}^{ij}}}\right)-1\right)}\hfill \\ {}\hfill -{\displaystyle \sum_{ij\in IJ}{\displaystyle \underset{0}{\overset{{\displaystyle \sum_{e\in {E}_{ij}}{\displaystyle \sum_{r\in {R}_{ij}}{f}_{er}^{ij}}}}{\int }}{D}_{ij}^{-1}\left(\omega \right) d\omega}}.\hfill \end{array} $$

(37)

The Hessian matrix of Z ₁ + Z ₂ + Z ₃ w.r.t. the route flow variables can be expressed as

$$ \frac{\partial^2\left({Z}_1+{Z}_2+{Z}_3\right)}{\partial {f}_{er}^{ij}\partial {f}_{lk}^{ij}}=\left\{\begin{array}{ll}\frac{d{h}_a}{d{v}_a}{\delta}_{ra}^{ij}+\frac{\mu_{ij}}{\theta_{ij}{f}_{er}^{ij}}+\frac{1-{\mu}_{ij}}{\theta_{ij}{\displaystyle \sum_{r\in {R}_{ij}}{f}_{er}^{ij}}}>0\hfill &; e=l,\kern0.5em r=k\hfill \\ {}0\hfill &; \kern0.5em otherwise\hfill \end{array}\right., $$

(38)

which implies the positive definite matrix from Assumptions 1 and 2. The Hessian matrix of Z ₄ is $-\frac{1}{\theta_{ij}}\frac{\partial^2}{\partial {f}_{er}^{ij}{f}_{lk}^{ij}}{\displaystyle \sum_{j\in IJ}\left({\displaystyle \sum_{e\in {E}_{ij}}{\displaystyle \sum_{r\in {R}_{ij}}{f}_{er}^{ij}}}\right)\left( \ln \left({\displaystyle \sum_{e\in {E}_{ij}}{\displaystyle \sum_{r\in {R}_{ij}}{f}_{er}^{ij}}}\right)-1\right)}$ which is positive semi-definite since all elements of the block matrix w.r.t. O-D pair ij are equal to $-\frac{1}{\theta_{ij}{\displaystyle \sum_{e\in {E}_{ij}}{\displaystyle \sum_{r\in {R}_{ij}}{f}_{er}^{ij}}}}$. Finally, from Assumption 2, the Hessian matrix of Z ₅ is positive semi-definite. Thus, the Hessian matrix of Z is positive definite. The CNL-ED has a unique solution w.r.t. to the route flow variables and hence the O-D demand variables. This completes the proof. □

Proof of Proposition 3

The Lagrangian of the equivalent minimization problem w.r.t. the equality constraints can be formulated as:

$$ L=Z+{\displaystyle \sum_{ij\in IJ}{\displaystyle \sum_{u\in {U}_{ij}}{\displaystyle \sum_{m\in {M}_{ij u}}{\phi}_{um}^{ij}\left({q}_{um}^{ij}-{\displaystyle \sum_{e\in {E}_{ij u m}}{\displaystyle \sum_{r\in {R}_{ij u m e}}{f}_{um er}^{ij}}}\right)+{\displaystyle \sum_{ij\in IJ}{\displaystyle \sum_{u\in {U}_{ij}}{\lambda}_{ij}\left({q}_{ij}-{\displaystyle \sum_{u\in {U}_{ij}}{\displaystyle \sum_{m\in {M}_{ij u}}{q}_{um}^{ij}}}\right)}}}}}, $$

(39)

where ϕ _um ^ij and λ _ij are respectively the dual variables associated with the conservation constraints in Eqs. (20) and (21).

• (1) Given that the Lagrangian L has to be minimized w.r.t. positive route flows, the following conditions have to hold w.r.t. the route-flow variables:

  $$ \frac{\partial L}{\partial {f}_{um er}^{ij}}=0\kern0.5em \Rightarrow \kern0.5em {g}_{um er}^{ij}+\frac{\mu_{ij um}}{\theta_{ij um}} \ln \frac{f_{um er}^{ij}}{{\left({\alpha}_{ij um e r}\right)}^{\frac{1}{\mu_{ij um}}}}+\frac{1-{\mu}_{ij um}}{\theta_{ij um}} \ln {\displaystyle \sum_{r\in {R}_{ij um e}}{f}_{um er}^{ij}}-{\phi}_{um}^{ij}=0. $$

  (40)

  Rearranging Eq. (40) gives

  $$ {f}_{um er}^{ij}{\left({\displaystyle \sum_{r\in {R}_{ij um e}}{f}_{um er}^{ij}}\right)}^{\frac{1-{\mu}_{ij um}}{\mu_{ij um}}}={\left({\alpha}_{ij um e r}\right)}^{\frac{1}{\mu_{ij um}}} \exp \left(\frac{\theta_{ij um}}{\mu_{ij um}}{\phi}_{um}^{ij}-\frac{\theta_{ij um}}{\mu_{ij um}}{g}_{um er}^{ij}\right). $$

  (41)

  Summing all r gives

  $$ {\left({\displaystyle \sum_{r\in {R}_{ij um e}}{f}_{um er}^{ij}}\right)}^{\frac{1}{\mu_{ij um}}}= \exp \left(\frac{\theta_{ij um}}{\mu_{ij um}}{\phi}_{um}^{ij}\right){\displaystyle \sum_{r\in {R}_{ij um e}}{\left({\alpha}_{ij um e r}\right)}^{\frac{1}{\mu_{ij um}}} \exp \left(-\frac{\theta_{ij um}}{\mu_{ij um}}{g}_{um er}^{ij}\right)}. $$

  (42)
  $$ {\displaystyle \sum_{r\in {R}_{ij um e}}{f}_{um er}^{ij}}= \exp \left({\theta}_{ij um}{\phi}_{um}^{ij}\right){\left({\displaystyle \sum_{r\in {R}_{ij um e}}{\left({\alpha}_{ij um e r}\right)}^{\frac{1}{\mu_{ij um}}} \exp \left(-\frac{\theta_{ij um}}{\mu_{ij um}}{g}_{um er}^{ij}\right)}\right)}^{\mu_{ij um}}. $$

  (43)

  Then, summing all e gives

  $$ {\displaystyle \sum_{e\in {E}_{ij um}}{\displaystyle \sum_{r\in {R}_{ij um e}}{f}_{um er}^{ij}}}={q}_{um}^{ij}= \exp \left({\theta}_{ij um}{\phi}_{um}^{ij}\right){\displaystyle \sum_{e\in {E}_{ij um}}{\left({\displaystyle \sum_{r\in {R}_{ij um e}}{\left({\alpha}_{ij um e r}\right)}^{\frac{1}{\mu_{ij um}}} \exp \left(-\frac{\theta_{ij um}}{\mu_{ij um}}{g}_{um er}^{ij}\right)}\right)}^{\mu_{ij um}}}. $$

  (44)

  Dividing Eq. (43) by Eq. (44) gives the marginal probability of choosing nest e:

  $$ \frac{{\displaystyle \sum_{r\in {R}_{ij um e}}{f}_{um e r}^{ij}}}{q_{um}^{ij}}={P}_{um e}^{ij}=\frac{{\left({\displaystyle \sum_{r\in {R}_{ij um e}}{\left({\alpha}_{ij um e r}\right)}^{\frac{1}{\mu_{ij um}}} \exp \left(-\frac{\theta_{ij um}}{\mu_{ij um}}{g}_{um e r}^{ij}\right)}\right)}^{\mu_{ij um}}}{{\displaystyle \sum_{e\in {E}_{ij um}}{\left({\displaystyle \sum_{r\in {R}_{ij um e}}{\left({\alpha}_{ij um e r}\right)}^{\frac{1}{\mu_{ij um}}} \exp \left(-\frac{\theta_{ij um}}{\mu_{ij um}}{g}_{um e r}^{ij}\right)}\right)}^{\mu_{ij um}}}}, $$

  (45)

  and dividing Eq. (41) by Eq. (42) gives the conditional probability:

  $$ \frac{f_{um er}^{ij}}{{\displaystyle \sum_{r\in {R}_{ij um e}}{f}_{um er}^{ij}}}={P}_{um\left(r\Big|e\right)}^{ij}=\frac{{\left({\alpha}_{ij um e r}\right)}^{\frac{1}{\mu_{ij um}}} \exp \left(-\frac{\theta_{ij um}}{\mu_{ij um}}{g}_{um er}^{ij}\right)}{{\displaystyle \sum_{r\in {R}_{ij um e}}{\left({\alpha}_{ij um e r}\right)}^{\frac{1}{\mu_{ij um}}} \exp \left(-\frac{\theta_{ij um}}{\mu_{ij um}}{g}_{um er}^{ij}\right)}}, $$

  (46)

  which corresponds to the CNL choice probability model in Eqs. (9) and (10).

• (2) Now we consider the mode choice:

  $$ \frac{\partial L}{\partial {q}_{um}^{ij}}=0\kern0.5em \Rightarrow \kern0.5em \left(\frac{\varphi_{ij u}}{\gamma_{ij}}-\frac{1}{\theta_{ij u m}}\right) \ln {q}_{um}^{ij}+\frac{1-{\varphi}_{ij u}}{\gamma_{ij}} \ln \left({\displaystyle \sum_{m\in {M}_{ij u}}{q}_{um}^{ij}}\right)-{\varPsi}_{ij u m}+{\phi}_{um}^{ij}-{\lambda}_{ij}=0. $$

  (47)

  From Eq. (44), ϕ _um ^ij can be defined as

  $$ {\phi}_{um}^{ij}=\frac{1}{\theta_{ij um}} \ln {q}_{um}^{ij}-\frac{1}{\theta_{ij um}} \ln {\displaystyle \sum_{e\in {E}_{ij um}}{\left({\displaystyle \sum_{r\in {R}_{ij um e}}{\left({\alpha}_{ij um e r}\right)}^{\frac{1}{\mu_{ij um}}} \exp \left(-\frac{\theta_{ij um}}{\mu_{ij um}}{g}_{um er}^{ij}\right)}\right)}^{\mu_{ij um}}}. $$

  (48)

  Note that the second term on the right hand side is the CNL EPT w _um ^ij. Substituting Eq. (48) into (47) gives

  $$ \frac{\varphi_{ij u}}{\gamma_{ij}} \ln {q}_{um}^{ij}+\frac{1-{\varphi}_{ij u}}{\gamma_{ij}} \ln \left({\displaystyle \sum_{m\in {M}_{ij u}}{q}_{um}^{ij}}\right)-{\varPsi}_{ij u m}+{w}_{um}^{ij}-{\lambda}_{ij}=0 $$

  (49)
  $$ {q}_{um}^{ij}{\left({\displaystyle \sum_{m\in {M}_{ij u}}{q}_{um}^{ij}}\right)}^{\frac{1-{\varphi}_{ij u}}{\varphi_{ij u}}}= \exp \left(\frac{\gamma_{ij}}{\varphi_{ij u}}\left({\lambda}_{ij}+{\varPsi}_{ij u m}-{w}_{um}^{ij}\right)\right). $$

  (50)

  Using the same principle in the route choice level, we have the marginal probability

  $$ \frac{{\displaystyle \sum_{m\in {M}_{ij u}}{q}_{um}^{ij}}}{q_{ij}}={P}_u^{ij}=\frac{{\left({\displaystyle \sum_{m\in {M}_{ij u}} \exp \left(\frac{\gamma_{ij}}{\varphi_{ij u}}\left({\varPsi}_{ij u m}-{w}_{um}^{ij}\right)\right)}\right)}^{\varphi_{ij u}}}{{\displaystyle \sum_{u\in {U}_{ij}}{\left({\displaystyle \sum_{m\in {M}_{ij u}} \exp \left(\frac{\gamma_{ij}}{\varphi_{ij u}}\left({\varPsi}_{ij u m}-{w}_{um}^{ij}\right)\right)}\right)}^{\varphi_{ij u}}}}, $$

  (51)

  and the conditional probability

  $$ \frac{q_{um}^{ij}}{{\displaystyle \sum_{m\in {M}_{ij u}}{q}_{um}^{ij}}}={P}_{m\Big|u}^{ij}=\frac{ \exp \left(\frac{\gamma_{ij}}{\varphi_{ij u}}\left({\varPsi}_{ij u m}-{w}_{um}^{ij}\right)\right)}{{\displaystyle \sum_{m\in {M}_{ij u}} \exp \left(\frac{\gamma_{ij}}{\varphi_{ij u}}\left({\varPsi}_{ij u m}-{w}_{um}^{ij}\right)\right)}}, $$

  (52)

  which corresponds to the NL model in Eqs. (4) and (5). This completes the proof. □

Proof of Proposition 4

It is sufficient to prove that objective function in Eq. (19) is strictly convex in the vicinity of route flows and that the feasible region is convex.

The Hessian matrix of Z ₁ + Z ₂ + Z ₃ w.r.t. the route flow variables can be expressed as

$$ \frac{\partial^2\left({Z}_1+{Z}_2+{Z}_3\right)}{\partial {f}_{umer}^{ij}\partial {f}_{umlk}^{ij}}=\left\{\begin{array}{ll}\frac{d{h}_a}{d{v}_a}{\delta}_{umra}^{ij}+\frac{\mu_{ij um}}{\theta_{ij um}{f}_{umer}^{ij}}+\frac{1-{\mu}_{ij um}}{\theta_{ij um}{\displaystyle \sum_{r\in {R}_{ij um e}}{f}_{umer}^{ij}}}>0\hfill &; \kern0.5em e=l,\kern0.5em r=k\hfill \\ {}0\hfill &; \kern0.5em otherwise\hfill \end{array}\right., $$

(53)

which implies the positive definite matrix from Assumptions 2 and 3. The Hessian matrix of the Z ₄ + Z ₅ + Z ₆ w.r.t. the modal demand variables can be expressed as

$$ \frac{\partial^2\left({Z}_4+{Z}_5+{Z}_6\right)}{\partial {q}_{um}^{ij}\partial {q}_{tn}^{ij}}=\left\{\begin{array}{ll}\left(\frac{\varphi_{ij u}}{\gamma_{ij}}-\frac{1}{\theta_{ij u m}}\right)/{q}_{um}^{ij}+\left(\frac{1-{\varphi}_{ij u}}{\gamma_{ij}}\right)/{\displaystyle \sum_{m\in {M}_{ij u}}{q}_{um}^{ij}}>0\hfill &; \kern0.5em u=t,\kern0.5em m=n\hfill \\ {}0\hfill &; \kern0.5em otherwise\hfill \end{array}\right., $$

(54)

which also implies the positive definite matrix. Therefore, the NL-CNL model has a unique solution. This completes the proof. □