Morning commute problem with queue-length-dependent bottleneck capacity

Transportation Research Part B 121 (2019) 184–215

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Transportation Research Part B journal homepage: www.elsevier.com/locate/trb

Morning commute problem with queue-length-dependent bottleneck capacity Jin-Yong Chen a, Rui Jiang b,∗, Xin-Gang Li b,∗, Mao-Bin Hu a, Bin Jia b, Zi-You Gao b a

School of Engineering Science, University of Science and Technology of China, Hefei 230026, China Key Laboratory of Transport Industry of Big Data Application Technologies for Comprehensive Transport, Ministry of Transport, Beijing Jiaotong University, Beijing 100044, China

b

a r t i c l e

i n f o

Article history: Received 8 January 2018 Revised 24 January 2019 Accepted 26 January 2019

Keywords: Morning commute Bottleneck model Queue-length-dependent capacity User equilibrium

a b s t r a c t When traffic control is demand-responsive, a bottleneck can have a queue-lengthdependent capacity. Motivated by this fact, we studied the morning commute problem in which the bottleneck capacity increases from s1 to s2 as the queue length exceeds a threshold D1 and decreases back to s1 as the queue length reduces and becomes smaller than another threshold D2 (D2 ≤ D1 ). It has been found that multiple equilibria exist when D2 < D1 . Their stability has been preliminarily studied via the day-to-day dynamics, and we found it is likely that only the equilibrium state with the lowest cost is stable against large disturbances. In the case where D2 = D1 , there is no multiple user equilibrium state, and the capacity does not change more than twice. Moreover, there exists a parameter range in which there is no solution unless the bottleneck capacity can switch back and forth instantaneously. The policy of opening the shoulder lane on a 2-lane highway is discussed as an application example. © 2019 Elsevier Ltd. All rights reserved.

1. Introduction The morning commute problem was first studied by Vickrey (1969) using a bottleneck model. Commuters depart from their residences to go to work every day. Due to the downstream bottleneck, they will experience queue delays. They will also be penalized for an early or late arrival. Commuters therefore adjust their departure times in response to the resulting delays and penalties. The bottleneck model has been widely recognized as a useful tool for modeling the formation and dissipation of queuing at a bottleneck during the morning peak (Small, 2015). Since the Vickrey model can be investigated analytically, it has been generalized to various scenarios, such as pricing (Arnott et al., 1990; Laih, 1994, 2004; Lindsey et al., 2012; Xiao et al., 2013; Wang and Sun, 2014), demand elasticity (Arnott et al., 1993; Yang and Huang, 1997), heterogeneous commuters (Arnott et al., al.,1994; Lindsey, 2004; Van den Berg and Verhoef, 2011; Yao et al., 2012), stochastic capacity and/or demand (Arnott et al., 1999; Xiao et al., 2014a, 2014b), the integration of morning and evening peaks (de Palma and Lindsey, 2002; Zhang et al., 2005; Li et al., 2014), modal split (Tabuchi, 1993; Huang, 2002; Lu et al., 2015), rail transit (Hao et al., 2009), consecutive bottlenecks (Kuwahara, 1990; Lago and Daganzo 2007), and tradable credit schemes (Nie and Yin, 2013; Tian et al., 2013). Most research about the bottleneck model assumes that the bottleneck capacity is a constant over time throughout the day. However, the road capacity may change over time within a day due to various reasons, such as traffic controls. ∗

Corresponding authors. E-mail addresses: [email protected] (R. Jiang), [email protected] (X.-G. Li).

https://doi.org/10.1016/j.trb.2019.01.009 0191-2615/© 2019 Elsevier Ltd. All rights reserved.

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Zhang et al. (2010) investigated how commuters behave in a bottleneck model with a time-varying capacity. In the model developed by Zhang et al., the traffic pattern in the user equilibrium state can still be derived analytically. The applications in dynamic signal timing and in the benefit evaluation of advanced traveler information systems were introduced to demonstrate the usefulness of this analytical model. There are also studies about the situation that occurs when capacity falls. For example, Ren et al. (2016) studied a singlestep-toll equilibrium for the bottleneck model in which the capacity dropped due to lanes being occupied by drivers who stopped and waited for a toll to decrease. Fosgerau and Small (2013) and Hall (2015) studied the situation where the capacity decreases when the queue length exceeds a threshold, which is claimed to be motivated by the capacity drop phenomenon in the traffic flow. Many queue systems in real life usually have extra servers to reduce congestion. For example, when many customers are waiting for service (e.g., at the checkout counters in a supermarket or windows at a bank), the decision-maker often opens more counters or windows to reduce the long waiting line. That is to say, the capacity of the queue system is queue-length dependent. Such systems are queue-dependent servers (Singh, 1973), which have been widely studied in queue theory (Wang and Tai, 20 0 0; Ke et al., 2010). Queue-length-dependent capacity can also be observed in traffic systems. When traffic congestion is serious and the queue is long, the traffic management department may increase the capacity by opening more tollgates, opening additional lanes (e.g., the shoulder of a highway, peak hours lane, or a switchable lane1 ), increasing the green time of traffic signals, decreasing the ramp flow via ramp metering, or closing the ramps on expressways or highways.2 , 3 When the queue decreases, the capacity can be decreased by closing the additional opened tollgates or lanes, re-opening the ramps, or decreasing the green time of traffic signals. Motivated by this fact, Vickrey’s bottleneck model has been generalized to investigate the impact of queue-lengthdependent capacity on travelers’ departure time choices in the morning commute problem. As early as two decades ago, Yang and Huang (1997) studied the situation where the capacity is a continuously differentiable function of the queue length. However, in real traffic, adopting the continuously differentiable function is difficult. In this paper, we study the situation where as the queue length increases to a critical value D1 , the road capacity increases. When the queue length decreases to another critical value D2 (D2 ≤ D1 ), the road capacity is restored. One main goal of this paper is to characterize possible equilibria and compare them with the equilibrium in the standard bottleneck model with a constant capacity. Interestingly, multiple equilibria (and even a continuum of equilibria) can be observed. This is especially intriguing given that in the standard bottleneck model with a constant capacity, a unique deterministic user equilibrium exists under relatively general conditions. Our analysis also revealed that the equilibrium cost can be a locally decreasing function of the number of users. Finally, the policy of opening the shoulder lane on a 2-lane highway is discussed as an application example. The paper is organized as follows. In Section 2, the model is introduced. Sections 3 and 4 study two different cases. Section 5 discusses the practical applications of the model. Finally, a conclusion is given in section 6. 2. Model The notations in Table 1 summarize the parameters and variables used in this paper. 2.1. Vickrey model In the Vickrey model, a fixed number of N commuters go from home to work on a road with a potential bottleneck. Once the bottleneck is activated, a queue will emerge and, thus, commuters incur a queuing delay. Commuters also incur an early/late cost if they arrive at work earlier/later than the desired arrival times. The cost of a commuter is calculated by:

c (t ) = α T (t ) + β max {0, t ∗ − t − T (t )} + γ max {0, t + T (t ) − t ∗ }. Here, T(t) is the travel time (or waiting time, since we ignore the free travel time, as usual) for a departure time t, t∗ is the preferred arrival time, α is the value of the unit travel time, β and γ are the schedule penalties for a unit time of early arrival and late arrival, respectively. As usual, we assumed that γ > α > β > 0. Each individual chooses a departure time to minimize his/her commute cost that combines the queuing delay and early/late cost, and a unique user equilibrium (UE) state can be achieved.

1 See, e.g., https://en.wikipedia.org/wiki/Smart_motorway (accessed on 2018 September 5); https://www.wegenwiki.nl/Plusstrook (in Dutch, accessed on 2018 September 5); https://nl.wikipedia.org/wiki/Spitsstrook (in Dutch, accessed on 2018 September 5); https://www.rijkswaterstaat.nl/wegen/wegbeheer/ spitsstroken/index.aspx (in Dutch, accessed on 2018 September 5). 2 The cost of expanding capacity is clearly an important practical consideration. If it were costless, it would be optimal to maintain capacity at its maximum possible value at all times. For instance, opening the shoulder lane causes a potential risk. Once accidents happen, ambulances and safety personnel may not be able to arrive in time. In Section 5, we present an application example, taking into account the potential risk of opening the shoulder lane. 3 Similar to the setting considered in this paper, incident management entails a trade-off between the cost of the intervention and the benefit of maintaining a greater capacity over the course of the travel period. There are other trade-offs as well (see Ng and Small, 2012).

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Table 1 Notations in this paper. c D1

The cost of each commuter Threshold at which the bottleneck capacity increases

t∗ t˜

D2

Threshold at which the bottleneck capacity decreases

t1 , t3

kD

kD = D 1 /D 2 ≥ 1

t2 , t4

K

Capacity changes 2K(K ≥ 1) times

t˜1 , t˜2 , t˜3 , t˜4

N s1 s2 ks tq tq

Total number of commuters Normal capacity of the bottleneck Increased capacity of the bottleneck k s =s 2 / s 1 > 1 Departure time of the first commuter Departure time of the last commuter

T(t) T s1 T s2 /Tr

α β (γ ) δ

Preferred arrival time Departure time at which an individual arrives at the destination at time t∗ The time when the queue length increases to D1 for the first time and second time The time when the queue length decreases to D2 for the first time and second time Departure time at which an individual arrives at the destination at time t1 ,t2 , t3 , and t4 Travel time of commuters departing at time t Duration time of low capacity Duration time of high capacity /Opening hours of shoulder lane The value of the unit travel time The schedule penalties for a unit time of early(late) arrival δ = βγ /(β + γ )

2.2. Bottleneck model with queue-length-dependent capacity In our model, the bottleneck capacity is assumed to increase from s1 to s2 as the queue length exceeds a threshold D1 , and decreases back to s1 as the queue length reduces and becomes smaller than another threshold D2 (D2 ≤ D1 ). 3. Results for the case D2 = D1 = D0 3.1. The bottleneck capacity does not change If the bottleneck capacity does not change, the model is reduced to the standard Vickrey model (Scenario 0). In the UE state, the departure rate is:



r (t ) =

α α −β s 1 α α +γ s1

t < t˜ , t > t˜

where t˜ is the departure time of commuters who arrive at preferred time t∗ . γ The earliest departure time is tq = t ∗ − β +γ sN . 1

β N The latest departure time is tq = t ∗ + β + γ s1 . N The cost of each commuter is c = δ s . Here δ = βγ /(β + γ ). 1

The maximum queue length Dmax = αδ N, which should not exceed D0 . As a result, in Scenario 0, the total number of commuters should satisfy the following:

N≤

α D . δ 0

3.2. Capacity changes twice In this case, the departure curve and arrival curve are determined by the order of t1 , t2 , t˜1 , t˜2 , t˜, and t∗ . With the change of parameters, order of the six times could change. However, from the definition, we know t1 < t2 , t˜1 < t˜2 , t˜1 < t1 , t˜2 < t2 , and t˜ < t ∗ . As a result, for the six times, there are 10 possible orders: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

t˜ < t ∗ < t˜1 < t1 < t˜2 < t2 t˜ < t˜1 < t ∗ < t1 < t˜2 < t2 t˜1 < t˜ < t1 < t ∗ < t˜2 < t2 t˜1 < t˜ < t1 < t˜2 < t ∗ < t2 t˜1 < t1 < t˜ < t ∗ < t˜2 < t2 t˜1 < t1 < t˜ < t˜2 < t ∗ < t2 t˜1 < t1 < t˜2 < t˜ < t2 < t ∗ t˜1 < t1 < t˜2 < t2 < t˜ < t ∗ t˜1 < t˜2 < t1 < t2 < t˜ < t ∗ t˜1 < t˜2 < t1 < t˜ < t2 < t ∗

Our analysis shows that Scenario 10 can be further classified into 2 different scenarios: 10(a) and 10(b). We summarize the analysis results here and show the details of the derivation in Appendix A. In the case D1 =D2 , Scenarios 1, 2, 3, 4, 8, 9, and 10(b) do not exist (see Appendix A.11). The four other scenarios are as follows.

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Fig. 1. Typical departure curve and arrival curve (top) and queue length curve (bottom) in Scenario 5.

Fig. 2. Typical departure curve and arrival curve (top) and queue length curve (bottom) in Scenario 6.

3.2.1. Scenario 5, Scenario 6, and Scenario 7 In these 3 scenarios (see Figs. 1–3), the cost of each commuter is:



c=δ N+ • • •

ks − 1 ks



  α−β α+γ + ks D0 /s2 . β γ

In Scenario 5, the total commuter number N should be in the range N > Ns1 . In Scenario 6, the total commuter number N should be in the range Ns2 < N < Ns1 . In Scenario 7, the total commuter number N should be in the range N < Ns2 and



N > Ns3 N > Ns4

Here,



α i f ks < α − β α . i f ks > α − β



  α+γ α−β α+γ (α + γ )2 ks − ks − + D0 , αδ γ β γ     α−β α+γ α−β 1 α+γ Ns2 = + k − + D0 , β ks β s β γ   α+γ α−β 1 Ns3 = +1+ D0 , β ks γ Ns1 =

α−β 1 + β ks

(1)

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Fig. 3. Typical departure curve and arrival curve (top) and queue length curve (bottom) in Scenario 7.

Fig. 4. Typical departure curve and arrival curve (top) and queue length curve (bottom) in Scenario 10(a).

and



 α+γ α−β 1 α−β + D0 . ( ks − 1 ) + β ks β γ

Ns4 =

3.2.2. Scenario 10(a) In this scenario (Fig. 4), the cost of each commuter is

c=α



D0 δ + N− s1 As2

α  D . δ 0

Here,

A=1 +

(k − 1 )2 sα

.

k s α −β − k s

α . In Scenario 10(a), the parameters should satisfy the following: ks < α − β The total commuter number N should be in the range Ns5 < N < Ns3 . Here,

Ns5 =

α D . δ 0

(2)

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α . Fig. 5. Cost function when ks < α − β

α . Fig. 6. Cost function when ks > α − β

3.2.3. Discussion We would like to mention that the existence conditions of Scenarios 0, 5, 6, 7, and 10(a) are mutually exclusive. We summarize the existence conditions of the scenarios as follows. α , 5 scenarios can exist (see Fig. 5). When ks < α − β α , 4 scenarios can exist (see Fig. 6). When ks > α − β

3.3. Absence of situations where capacity changes more than twice We would like to point out that the situations where the capacity changes more than twice are absent when D1 =D2 =D0 . Otherwise, suppose the capacity can change 2K (K > 1) times. If t˜ ≥ t2K−1 , the queue length at time period (t2K − 2 , t2K − 1 ) is always increasing; if t˜ ≤ t2K−1 , the queue length at time period (t2K − 1 , t2K ) is always decreasing. Here, ti (i = 1, 2, ..., 2K) denotes the time when the bottleneck capacity increases from s1 to s2 or decreases from s2 to s1 . α and N < N < N (see Fig. 6), there is no solution unless the capacity As a result, in the parameter range ks > α − s5 s4 β can switch back and forth instantaneously between s1 and s2 . The reason is because when the queue length reaches D0 , the capacity increases. The queue starts to decrease and the capacity decreases. Then, the queue starts to increase again, and the process continues. In the optimal control theory, this is known as a “chattering control” (Zelikin and Borisov, 1994). Such a policy is unrealistic, and it would be reasonable to rule it out. Suppose that the capacity can switch back and forth instantaneously. Then accordingly, the queue length will fluctuate back and forth instantaneously around D0 . The arrival rate fluctuates back and forth instantaneously around a weighted average capacity, which typically does not coincide with the unweighted or simple average.

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Fig. 7. Typical departure curve and arrival curve (top), queue length curve (middle), and waiting time curve (bottom) in Scenario 5.

4. Results in the case D2 < D1 4.1. The bottleneck capacity does not change If bottleneck capacity does not change, the total commuter number N should be in the range N < N0 = αδ D1 , and the cost of each commuter is c = δ sN . 1

4.2. The bottleneck capacity changes twice 4.2.1. Realism of the scenarios α +γ α +γ Scenario 1 might occur only when ks > ( α )2 ; Scenario 2, Scenario 3, and Scenario 4 might occur only when ks > α . With assumption α < γ , this implies that ks > 4 and ks > 2, respectively, which seems implausible in practice. Therefore, we only present Scenarios 5, 6, 7, 8, 9, and 10. When a shoulder lane is opened, the capacity increase ratio is even smaller. For example, for 2-lane or multilane highways with one shoulder lane, the increase ratio is 50% or smaller by opening the shoulder lane. 4.2.2. Scenario 5, Scenario 6, and Scenario 7 In the 3 scenarios (see Figs. 7–9), the cost of each commuter is as follows:



c=δ N+ •

ks − 1 ks

  α+γ α−β kD + ks D2 /s2 . β γ

In Scenario 5, the total commuter number N should be in the range:



γ 2 i f kD > α + ks α+αγ 2 . i f kD < α ks

N > N5 N > N6

•



In Scenario 6, the parameters should satisfy:

kD <

 α + γ 2 α

ks .

The total commuter number N should be in the range N < N6 and



N > N5 N > N7

γ i f kD > α + α ks , . α +γ i f kD < α ks

(3)

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Fig. 8. Typical departure curve and arrival curve (top), queue length curve (middle), and waiting time curve (bottom) in Scenario 6.

Fig. 9. Typical departure curve and arrival curve (top), queue length curve (middle), and waiting time curve (bottom) in Scenario 7.

•

In Scenario 7, the parameters should satisfy:

kD <

α+γ k. α s

The total commuter number N should be in the range N < N7 and



N > N8 N > N9

i f kD > i f kD <

α −β α ks α −β . α ks

For N6 , N7 , N8 , and N9 , see Appendix A.

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Fig. 10. Typical departure curve and arrival curve (top), queue length curve (middle), and waiting time curve (bottom) in Scenario 8.

4.2.3. Scenario 8 In this scenario (Fig. 10), the cost of each commuter is:



c=δ N+

 α − β ks − 1 (kD − ks )D2 /s1 . β ks

(4)

In Scenario 8, the parameters should satisfy:

kD <

α−β k. α s

The total commuter number N should be in the range N9 < N < N10 . For N10 , see Appendix A. 4.2.4. Scenario 9 In this scenario (Fig. 11), the cost of each commuter is:



c=δ N−

( kD − 1 ) ( ks − 1 ) α ks − α − β



D2 /s1 .

(5)

In Scenario 9, the parameters should satisfy:

kD <

α−β k. α s

The total commuter number N should be in the range N11 < N < N12 . For N11 , N12 , see Appendix A. 4.2.5. Scenario 10 4.2.5.1. Scenario 10(a). In this scenario (Fig. 12), the cost of each commuter is:



α D1 α + γ D2 c=δ + β s2 γ s1 Here,





+δ

α 1 α 1− X. β α − β ks



    α+γ α α − 2β 1 α − β kD X = N− + kD + D2 / − s1 . β ks γ β α−β ks

In Scenario 10(a), the parameters should satisfy:

ks =

α−β α+γ ,k < k. α − 2β D α s

(6)

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Fig. 11. Typical departure curve and arrival curve (top), queue length curve (middle), and waiting time curve (bottom) in Scenario 9.

Fig. 12. Typical departure curve and arrival curve (top), queue length curve (middle), and waiting time curve (bottom) in Scenario 10(a).

The total commuter number N should be in the range:



˜ N13 < N < N ˜ N < N < N13

if if

1 ks 1 ks

β > αα−2 −β β. < αα−2 −β

˜ , see Appendix A. For N13 , N

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−β ˜ 4.2.5.2. Scenario 10(b). This scenario occurs when ks = αα−2 β . Let t = t1 − t2 , and the cost of each commuter is:



α D1 α + γ D2 c=δ + β s2 γ s1



+δ

α 1 α 1− t. β α − β ks

(7)

In this scenario, the parameters should satisfy:

kD <

α+γ k α s

and

tmin < t < tmax . Here, α α − β ks − α + γ kD D2 , α α ks − α + γ s1 

tmax = tmin =

0

α α −β ks − α−β kD D2 α ks − α−α β s1

β i f kD > α − α ks α −β . i f kD < α ks

The total number of commuters is:

N =N14 =

α−β α+γ D1 + D2 . β γ

−β Note that when ks = αα−2 β , N8 = N11 = N13 =N14 . β α −β ˜ ˜ When t=tmax , t1 = t always holds true. When t=tmin , t1 = t˜2 , if kD > α − α ks ; t2 = t , if kD < α ks . This scenario corresponds to infinite solutions for a given parameter set. When the cost curve is vertical (see Figs. B3 and B5), a continuum of equilibria exists. Fig. 13 shows 3 typical solutions under 2 typical parameter sets.

4.2.6. Discussion 4.2.6.1. Multiple equilibria. The existence of the conditions of Scenarios 0, 5, 6, 7, 8, 9, and 10 are not always mutually exclusive.4 As a result, multiple equilibria can exist. This subsection presents one case that suffices to demonstrate existence of multiple equilibria. Appendix B summarizes the existence conditions of these scenarios in detail. Fig. 14 shows a typical example of the cost vs. commuter numbers. In this case, 7 scenarios can be observed. Specifically, when N13 < N < N0 , Scenario 0 and Scenario 10(a) can exist; when N9 < N < N12 , Scenario 7, Scenario 8, and Scenario 9 can exist; when N12 < N < N10 , Scenario 7 and Scenario 8 can exist. For other values of N, only one scenario exists. We believe that multiple equilibria can exist because the duration times of the high capacity (Ts2 = t2 − t1 ) could be different. One equilibrium solution corresponds to a specific Ts2 that has a matching equilibrium departure pattern. For example, Fig. 15 shows 3 examples of the cost vs. Ts2 at different equilibria. One can see that the longer the high capacity is maintained, the smaller the equilibrium cost is. We can explain intuitively why the multiple equilibria can exist. First, let us consider Scenario 0 and Scenario 10(a) in Fig. 14, which can both be equilibria over a certain range of N. Scenario 0 is the conventional case in which capacity remains constant. The queue never reaches threshold level D1 at which the capacity increases. By contrast, in Scenario 10(a) the queue does reach D1 . Capacity rises from s1 to s2 until the queue drops to D2 . Since Scenario 10(a) has a higher capacity than Scenario 0 over part of the travel period, Scenario 10(a) has a lower equilibrium cost. Once commuters complete their trips after the capacity has increased, the departure rate rises in order to maintain the equilibrium. The higher departure rate causes the queue to grow more quickly than in Scenario 0. Eventually, it exceeds D1 , which triggers the increase in the capacity. If D2
One equilibrium in which the capacity is expected to be low, and indeed remains low (Scenario 0). Another equilibrium in which the capacity is expected to rise, and indeed does rise (Scenario 10(a)).

In contrast, when D2 = D1 , suppose the higher capacity is triggered. It will less than compensate for the accelerated departures early in the travel period. Therefore, the multiple equilibria of Scenario 0 and Scenario 10(a) cannot occur. Other instances of the multiple equilibria (such as those that occur in Scenarios 7, 8 and 9) can be explained similarly, i.e., a longer period of a high capacity more than compensates for the accelerated departures. 4

These scenarios are not always mutually exclusive from the scenarios arising when the capacity changes more than twice, either (see Section 4.3).

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Fig. 13. Three different UE solutions in Scenario 10(b) under 2 different parameter sets. In (a), α =3, β =1, γ = 6, s1 = 10 0 0, s2 = 20 0 0, D1 = 20 0 0, D2 = 10 0 0, β and N = 5500. With these parameter values, kD > α − α ks , so that tmin = 0; In (b), α =4, β =1, γ =8, s1 = 20 0 0, s2 = 30 0 0, D1 = 170 0, D2 = 160 0, and N = 750 0.

β With these parameter values, kD < α − α ks , so that tmin >0.

4.2.6.2. Equilibrium cost is a decreasing function of the commuter number. Furthermore, the equilibrium cost can be a decreasing function of the number of commuters (as in Scenario 10(a) in Fig. 14, as well as Fig. B2(c) in Appendix B). This situation can be understood as follows. From Appendix A.10.1, the duration times of a high capacity and low capacity can be calculated as follows:

Ts2 = t2 − t1 =

D1 − s2



β

α−β

Ts1 = (t1 − tq ) + tq − t2



N − N8 / s2



 α − 2β 1 − , α−β ks

  α − β D1 N − N8 α + γ D2 α − 2β 1 = + ( ks − 1 ) / − + . β s2 s2 α−β ks γ s1

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Fig. 14. A typical example of the cost function. The parameters are α =3.5, β =1, γ =8, s1 = 10 0 0, s2 = 150 0, D1 = 1060, and D2 = 10 0 0.

Fig. 15. Cost vs. duration time of the high capacity. The critical values are N9 =4304, N12 = 4474, N10 = 4540.

β α −2β 1 When 1 < kD < α − α ks and ks > α −β , we can derive ∂ Ts2 /∂ N > 0, ∂ Ts1 /∂ N < 0, and ∂ Ts2 /∂ N < −∂ Ts1 /∂ N, i.e., the decreasing rate of Ts1 exceeds the increasing rate of Ts2 . Thus, the total duration time of the peak hours (tq − tq ) decreases with the increase of the commuter number. As the equilibrium cost c(t) = δ (tq − tq ), the equilibrium cost would decrease with the increase of the commuter number. The intuitive explanation is similar to that in Section 4.2.6.1. With the increase of the commuter number, the higher capacity is triggered earlier and remains longer. The increase of the duration time of a high capacity more than compensates for the increase of the commuter number. Therefore, the cost decreases with N.

4.3. Capacity changes 2K (K > 1) times When K = 2, the departure curve and arrival curve are determined by the order of t1 , t2 , t3 , t4 , t˜1 , t˜2 , t˜3 , t˜4 , t˜, and t∗ . Here, t3 is the time that the queue length increases to D1 for the second time, and t4 is the time that the queue length decreases to D2 for the second time. t˜3 and t˜4 are the departure times corresponding to arrival times t3 and t4 , respectively. The number of possible orders of the 10 times equals 62, i.e., there are 62 possible scenarios. To fully discuss the scenarios is very complex.5 As an example, Fig. 16(a) shows an equilibrium solution at a given parameter set: α =4, β =1, γ =8, s1 = 10 0 0, s2 = 150 0, D1 = 1067, D2 = 10 0 0, and N = 5090. In the solution, the capacity changes 4 times. Note that for this parameter set, there is more than one solution. At the very least, the solution that capacity changes twice also exists (see Fig. 16(b)). With the increase of K, the situation becomes more and more complex. When K→∞, the number of possible scenarios also becomes infinite. 5

Our preliminary study indicates that some of the 62 scenarios could not exist. Further investigations are needed to examine how many scenarios exist.

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Fig. 16. A solution where the capacity (a) changes 4 times and (b) changes twice.

4.4. A preliminary study on solution stability This subsection preliminarily studies stability of the solutions via day-to-day dynamics (see Xiao et al., 2016; Guo et al., 2018). We can write a simple day-to-day dynamics equation as follows.6

r n+1 (tmin ) = r n (tmin ) + η (cn (tmax ) − cn (tmin ))r n (tmax ) r n+1 (tmax ) = r n (tmax ) − η (cn (tmax ) − cn (tmin ))r n (tmax ) Here, tmin is the time interval with the minimum travel cost and tmax is the time interval with the maximum travel cost. For the time interval with the travel cost in between, the departure rate does not change. η is a constant parameter, and it is set to 0.01. The iteration is terminated if n > 80 0,0 0 0 or the difference between the maximum and minimum travel cost is smaller than ε = 0.0 0 0 01. In our study, the parameters were set to α = 1.0, β = 0.2, γ = 2.0, s1 = 40 0 0, s2 = 720 0, D1 = 1400, and N = 6500. The time horizon was [0,3], and it was discretized into 1080 time slices. The preferred arrival time of all commuters was t∗ = 2. According to the analytical results, the selected parameters are in the parameter range C3 (see Appendix B), and Scenarios 0, 7, 8, and 9 might exist. Fig. 17 shows the evolution of the average cost of the commuters. The initial departure rates at each time interval were calculated from the analytical results. Theoretically, the departure rates should not change with time. However, the discretization brings small disturbances into the system. Scenarios 0 and 7 are stable against the small disturbances, which recover after a transient time (see the time series of the average cost before the 10 0 0 0 0th day in Fig. 17(a) and (b)). In contrast, Scenarios 8 and 9 are unstable. They evolve into Scenario 7, which has the lowest cost (Fig. 17(c) and (d)). Next, we introduced a large disturbance into Scenarios 0 and 7 at the 10 0 0 0 0th day. The large disturbance was implemented as follows: Commuters departing in the earliest 30 time intervals change their departure time to t∗ . We see that Scenario 7 recovers after a transient time, but Scenario 0 also becomes unstable and transitions into Scenario 7. To further study how likely is the transition from Scenario 0 to Scenario 7, we investigated the effect of the disturbance with different amplitudes. Here, a disturbance with amplitude w is defined as follows: Commuters departing in the earliest w time intervals change their departure time to t∗ . The typical results are shown in Fig. 18. We see that when w is small, Scenario 0 is stable. However, when w is large, Scenario 0 is unstable and transitions into Scenario 7. The simulation indicates that the threshold value of w is about 23. Our preliminary study indicates that it is likely that only the solution with the lowest cost is stable against large disturbances. However, further studies are needed to investigate this issue, including a stability analysis based on the day-to-day dynamics, an experimental study based on a virtual experiment (e.g., Ye et al., 2018; Iida et al., 1992; Rapoport et al., 2014), and empirical data validation.

6 In this case, the adjustment process is very slow. We tested the rule that rescheduling occurs from 5% of the departure times with the highest costs to the 5% of the departure times with the lowest costs. However, the adjustment process does not converge. The convergence of the day-to-day dynamics is an interesting topic and warrants a thorough investigation in future work.

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Fig. 17. The evolution of the average travel cost. The system is initially in (a) Scenario 0, (b) Scenario 7, (c) Scenario 8, and (d) Scenario 9. The horizontal dashed lines indicate the average cost in different scenarios.

Fig. 18. The evolution of the average travel cost in Scenario 0 with different disturbance amplitudes. (a) w = 10; (b) w = 22, (c) w = 23, and (d) w = 30.

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5. Practical application: opening the shoulder of a highway In engineering applications, s1 , N, α , β , and γ are usually given. Sometimes s2 is also given, and sometimes the optimal value of s2 needs to be determined. One needs to calculate the optimal value of D1 , D2, and sometimes s2 . This section discusses an application example in which the shoulder lane is opened on a 2-lane highway when the queue length exceeds D0 and closed when the queue length decreases to D0 .7 As a result, only one parameter, D0 , needs to be optimized. We α . Under the circumstances, a solution always exists, which could be from Scenarios 0, 5, 6, 7, consider the situation ks < α − β

and 10(a) (Section 3.2.3). Opening the shoulder lane of a highway causes potential risks. Once accidents happen, the ambulances and emergency personnel may not be able to arrive in time to provide adequate assistance. We denote the potential risk as cr . To quantify the risk, we assume it is a quadratic function of the time Tr that the shoulder lane is opened.8 Thus, cr = aTr + bTr2 (a > 0, b > 0), where a and b are coefficients. Now the total system cost ctotal = cr + ct , where ct = cN is the sum of total delay and total schedule penalties of all commuters. We derive the dependence of ctotal on D0 analytically. Based on a straightforward derivation, we can derive that

⎧

 ⎨ N − α−β β k1s + α+γ γ D0 /s2

 Tr = β /(α −β )  ⎩ 1 − α−2β N − α−β β + α+γ γ D0 /s2 ks

α −β

D0 < N m

N m

< D0 < αδ N

,

and

ct =δ N 2 /s1 − δ N (ks − 1 )Tr . Here,

m=

α−β 1 α+γ +1+ . β ks γ

Therefore,

ctotal = aTr + bTr2 − δ N (ks − 1 )Tr + δ N 2 /s1 = bTr2 + [a − δ N (ks − 1 )]Tr + δ N 2 /s1 . - When a > δ N(ks − 1), we have ∂ ctotal /∂ Tr = 2bTr + [a − δ N (ks − 1 )] > 0. Therefore, the shoulder lane should not be open. The optimal value D0 = ∞. - When a < δ N (ks − 1 ) − 2bN/s2 , the minimum value of ctotal will be achieved at Tr = N/s2 . In this case, the shoulder lane should be always open. The optimal value D0 = 0. )−a - When δ N (ks − 1 ) − 2bN/s2 < a < δ N (ks − 1 ), the minimum value of ctotal will be achieved at Tr = δ N (ks2−1 . Accordingly, b the optimal value

D0 =

⎧ 1− δN(ks −1)−a 2bN/s ⎪ ⎨ α−β 1 + α2+γ N β

ks

α −β ⎪ ⎩ 1− β



γ



β δ N (ks −1 )−a − αα−2 −β 2bN/s2 α −β α +γ β + γ

1 ks

N

a < δ N ( ks − 1 ) −

2bN/s2 m

a > δ N ( ks − 1 ) −

2bN/s2 m

.

In the example where the shoulder lane is opened on a 2-lane highway, ks = 1.5. We set other parameters to N = 2500,

α =2, β =1, γ =4, a = 400, b = 300, s1 = 1000, and s2 = 1500. Therefore, we can calculate the minimum value ctotal = 4700, and the corresponding optimal value D0 = 462, Tr = 1 (Fig. 19). 6. Conclusion Queue-length-dependent capacity can be observed in traffic systems under demand-response control. Motivated by this fact, Vickrey’s bottleneck model has been generalized to investigate the impact of queue-length-dependent capacity on travelers’ departure time choices in the morning commute problem. In this paper, we studied the situation where the bottleneck capacity increases from s1 to s2 as the queue length exceeds a threshold D1 , and decreases back to s1 as the queue length reduces and becomes smaller than another threshold D2 . The main findings are as follows. 1. When D2 < D1 , the bottleneck capacity might not change, change twice, or change more than twice. We analyzed the 10 scenarios where the bottleneck capacity changes twice in detail. Our analysis shows that multiple user equilibrium states can be observed at a given set of parameters. We performed a preliminary study on the stability of the solutions via the 7 The case D2 < D1 has multiple solutions. We have preliminarily studied the stability of some solutions, but we are still far from understanding the true stability of the solutions. Therefore, we considered D2 = D1 = D0 in the application. 8 To accurately quantify the risk is important and difficult, and is beyond the scope of this paper. Considering that driving fatigue increases with time, it is reasonable to assume the risk is a quadratic function of the time Tr .

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Fig. 19. The dependence of ctotal on D0 .

day-to-day dynamics. We found that while some solutions might be stable against very small disturbances, all solutions that have a larger cost become unstable and transition into the solution that has the lowest cost when the disturbance is large. 2. When D2 = D1 , the situation becomes simpler. The multiple user equilibrium state does not exist. The bottleneck capacity does not change more than twice. Moreover, there exists a parameter range in which there is no solution unless the bottleneck capacity can switch back and forth instantaneously. To summarize, the main methodological contributions of this work are that (i) multiple equilibria and even a continuum of equilibria were observed; (ii) the equilibrium cost can be a locally decreasing function of the number of users; and (iii) no solution exists in a certain parameter range. As an application example, the policy of opening the shoulder lane on a 2-lane highway was discussed in which the shoulder lane was opened when the queue length exceeded D0 and closed when queue length decreased to D0 . Taking into account the potential risk of opening the shoulder lane, the optimal value of D0 was derived. More practice of opening shoulder lanes during peak-period times can be found in e.g., US FHWA (2017). In our future work, we will investigate the issue of the multiple user equilibrium solutions. Apart from the stability analysis based on the day-to-day dynamics, we will carry out a virtual experiment to study this issue. Another interesting topic is to study the situation under uncertain demand.

Acknowledgments The authors are grateful to the anonymous reviewers for their constructive and insightful comments for improving this paper. This work is supported by the National Key R&D Program of China(No. 2017YFC0803300), National Natural Science Foundation of China (Grant Nos. 71621001, 71631002, 71371175, and 71771021), and Beijing Natural Science Foundation (Grant No. 9172013).

Appendix A. Analysis of Scenarios 1–10 In this Appendix, we present a detailed analysis of Scenarios 1–10. Before we analyze each scenario, let us recall the Theorem below, see also Zhang et al. (2010). Theorem 0. Consider a bottleneck model in which the capacity changes with time. Assume that a commuter who departs from home at time t arrives at a workplace at time t , and the bottleneck capacity is s at time t . The departure rate at time t under the UE state is:



r (t ) =

α α −β s α α +γ s

t < t˜ . t > t˜

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Table A1 The arrival rate, bottleneck capacity, and queue length in different time intervals in Scenario 1. Interval Departure rate Bottleneck capacity Queue length

tq < t < t˜ α s r1 = α − β 1 s1 Increase

t˜1 < t < t1 α s r3 = α + γ 2 s1 Increase or decrease? See text

t˜ < t < t˜1 α s r2 = α + γ 1 s1 Decrease

t1 < t < t˜2 α s r4 = α + γ 2 s2 Decrease

t˜2 < t < t2 α s r5 = α + γ 1 s2 Decrease

t2 < t < tq α s r6 = α + γ 1 s1 Decrease

Proof. Assume that the waiting time of the commuter who departs from home at time t is T(t), then his/her total travel cost is:





α T (t ) + β t ∗ − t  c (t ) =

α T (t ) + γ t  − t ∗

t < t˜ t > t˜

.

Here, t =t + T(t). At equilibrium, all commuters’ costs are equal, so dc (t )/dt = 0. Thus, we have:

dT (t ) = dt



β α −β γ − α+ γ

t < t˜ . t > t˜

β Therefore, if t < t˜, we have T (t ) = α − β (t − tq ), and

α β t− t . α−β α−β q

t  =t + T (t ) =  t tq

(A1)

The number of departing commuters from tq to t equals

t tq

r (t )dt. The number of arriving commuters from tq to t equals

s(t )dt. The two numbers should equal each other. Thus,



t

tq

r (t )dt =



t

tq

s(t )dt.

Taking the derivative to t, and taking into account Eq. (A1), we obtain

r (t ) =

α

α−β



s t =

α s. α−β

Similarly, we can derive that the departure rate of commuters who are late for work as:

r (t )=

α s.  α+γ

Now we can analyze the 10 scenarios. A.1. Scenario 1 t˜ < t ∗ < t˜1 < t1 < t˜2 < t2 In this scenario, the arrival rate between tq and t1 is s1 . Thus, from Theorem 0, the departure rate is the following:



r (t ) =

α α −β s 1 α α +γ s1

t < t˜ . t˜ < t < t˜1

The arrival rate between t1 and t2 is s2 , therefore

r (t ) =

α s t˜ < t < t˜2 . α+γ 2 1

The arrival rate between t2 and tq is s1 , therefore

r (t ) =

α s t > t˜2 . α+γ 1

Table A1 compares the arrival rate and bottleneck capacity in different time intervals. It is clear that the queue length increases when tq < t < t˜, since the arrival rate exceeds the bottleneck capacity. Similarly, the queue length decreases when t˜ < t < t˜1 , t1 < t < t˜2 , t˜2 < t < t2 , t2 < t < tq , since the arrival rate is smaller than the bottleneck capacity. α s > s and decrease if α s < s . As shown below, the conWhen t˜1 < t < t1 , the queue length would increase if α + 1 1 γ 2 α +γ 2 α s > s is always met in Scenario 1, see (A14). Thus, the queue length increases when t˜ < t < t . dition α + 1 1 1 γ 2 Now we can write out the following two equations:

α

α−β





s1 t˜ − tq +

α

α+γ





s1 t˜1 − t˜ +

α

α+γ





s2 t˜2 − t˜1 +

α

α+γ





s1 tq − t˜2 = N

(A2)

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s1 (t1 − tq ) + s2 (t2 − t1 ) + s1 tq − t2 = N.

(A3)

Eq. (A2) means the cumulative departure number of commuters equals N, and Eq. (A3) means the cumulative arrival number of commuters also equals N. Since the queue length equals D1 at t1 , we obtain:







α α α s t˜ − tq + s t˜ − t˜ + s t − t˜ − s1 (t1 − tq )=D1 . α−β 1 α+γ 1 1 α+γ 2 1 1

(A4)

Similarly, since the queue length equals D2 at t2 , we obtain:

N−





 α s t  − t2 − N − s1 tq − t2 =D2 . α+γ 1 q

(A5)

In the UE state, the cost of commuters that depart at tq , t˜, t˜1 , t˜2 , tq equals each other, thus we obtain the following equations:

and



β (t ∗ − tq ) = γ tq − t ∗ ,

(A6)





α t ∗ − t˜ = γ tq − t ∗ ,

(A7)





α t1 − t˜1 + γ (t1 − t ∗ ) = γ tq − t ∗ ,

(A8)





α t2 − t˜2 + γ (t2 − t ∗ ) = γ tq − t ∗ .

(A9)

However, Eq. (A9) is not independent, and it can be derived from Eqs. (A2) to (A8). Therefore, in this scenario, we have 7 unknowns tq , t˜,t˜1 , t1 , t˜2 , t2 , tq , and 7 Eqs. (A2)–(A8). Thus, the 7 unknowns can be solved. The solution is:

  δ α + γ ks − 1 tq = t − N− (kD − ks )D2 /s1 , β γ ks   δ α + γ ks − 1 ∗  tq = t + N− (kD − ks )D2 /s1 , γ γ ks ∗

and

t1 = tq −

α + γ D1 , γ s2

t2 = tq −

α + γ D2 , γ s1

t˜1 = tq −

(α + γ )2 D1 , αγ s2

t˜2 = tq −

(α + γ )2 D2 , αγ s1

  δ α + γ ks − 1 t˜ = t − N− (kD − ks )D2 /s1 . α γ ks ∗

In the solution, tq < t˜ is always satisfied since α > β ; t˜ < t ∗ requires that

N>

α + γ ks − 1 ( k D − k s )D 2 . γ ks

(A10)

t ∗ < t˜1 requires that



 (α + γ )2 kD α + γ ks − 1 N> + ( kD − ks ) D2 . αδ ks γ ks

Note that the condition in Eq. (A10) is always met when Eq. (A11) is satisfied. t˜1 < t1 is always satisfied sinceα > 0, γ > 0;

(A11)

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t1 < t˜2 requires that

kD >

α+γ k. α s

(A12)

t˜2 < t2 is always satisfied sinceα > 0, γ > 0; t2 < tq is always satisfied sinceα > 0, γ > 0. The queue length at t˜ can be calculated

  δ α + γ ks − 1 Dt˜ = N− (kD − ks )D2 , α γ ks

which should not exceed D1 . Thus, we can derive:



N<

 α + γ ks − 1 α kD + ( kD − ks ) D2 . δ γ ks

(A13)

Comparing Eqs. (A11) and (A13), we obtain:

(α + γ )2 kD α < kD , αδ ks δ which yields:

 α + γ 2

ks >

α

.

(A14)

To summarize, in Scenario 1, the parameters should satisfy:

kD >

 α + γ 2 α+γ ks and ks > . α α

The total commuter number N should be in the range N2 < N < N1 . Here,



N1 =

 N2 =

 α + γ ks − 1 α kD + ( kD − ks ) D2 , δ γ ks

 (α + γ )2 kD α + γ ks − 1 + ( kD − ks ) D2 , αδ ks γ ks

(A15)

(A16)

A.2. Scenario 2 t˜ < t˜1 < t ∗ < t1 < t˜2 < t2 In this scenario, we can write out 7 equations exactly the same as those in (A2)–(A8). Therefore, we obtain the same solutions of tq , t˜, t˜1 , t1 , t˜2 , t2 , and tq . In the solution, t˜ < t˜1 requires that



 α + γ kD α + γ ks − 1 N> + ( kD − ks ) D2 δ ks γ ks

(A17)

t˜1 < t ∗ requires that



N<

 (α + γ )2 kD α + γ ks − 1 + ( kD − ks ) D2 . αδ ks γ ks

(A18)

From t∗ < t1 , we can derive the same constraint as Eq. (A17). Other constraints can be derived as shown in Scenario 1, which are



N<

 α + γ ks − 1 α kD + ( kD − ks ) D2 δ γ ks

and

KD >

α+γ α+γ k and ks > . α s α α +γ

α +γ 2

Note that when ks > ( α )2 [ks < ( α ) ], the RHS of Eq. (A18) is smaller (larger) than the RHS of Eq. (A19). To summarize, in Scenario 2, the parameters should satisfy

kD >

α+γ α+γ k and ks > . α s α

(A19)

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The total commuter number N should be in the range N > N3 and



γ 2 i f ks > α + , α+α γ 2 i f ks < α .

N < N2 N < N1

Here,



 α + γ kD α + γ ks − 1 N3 = + ( kD − ks ) D2 , δ ks γ ks

(A20)

A.3. Scenario 3 t˜1 < t˜ < t1 < t ∗ < t˜2 < t2 In Scenario 3, we can write out the following 7 independent equations:



α

α−β



s1 t˜1 − tq +

α

α−β





s2 t˜ − t˜1 +



α

α+γ





s2 t˜2 − t˜ +

α

α+γ





s1 (t1 − tq ) + s2 (t2 − t1 ) + s1 tq − t2 = N,



α

α−β N−

and



s1 t˜1 − tq +



α

α+γ

α

α−β







s2 t˜ − t˜1 +



α

α+γ



s1 tq − t2 − N − s1 tq − t2



s1 tq − t˜2 =N,

(A21) (A22)





s2 t1 − t˜ − s1 (t1 − tq )=D1 ,

=D2 ,

(A23) (A24)



β (t ∗ − tq ) = γ tq − t ∗ ,

(A25)





α t1 − t˜1 + β (t ∗ − t1 ) = γ tq − t ∗ ,

(A26)





α t ∗ − t˜ = γ tq − t ∗ .

(A27)

The solution is:

  δ α + γ ks − 1 N− (kD − ks )D2 /s1 , β γ ks   δ α + γ ks − 1 tq = t ∗ + N− (kD − ks )D2 /s1 , γ γ ks   α+γ t1 = tq + N − ( k D − k s + 1 )D 2 / s 1 , γ tq = t ∗ −

α + γ D2 , γ s1   α −β α +γ t˜1 = tq + N− (kD − ks + 1 )D2 /s1 , α γ t2 = tq −

t˜2 = tq − and

t˜ = t ∗ −

(α + γ )2 D2 , αγ s1

  δ α + γ ks − 1 N− (kD − ks )D2 /s1 . α γ ks

By substituting the solutions into tq < t˜1 < t˜ < t1 < t ∗ < t˜2 < t2 < tq , and considering the queue length constraint, we derive that the parameters should satisfy:

kD >

α+γ α+γ k and ks > . α s α

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The total commuter number N should be in the range N < N3 and:



γ 2 i f kD > α + ks , α γ 2 . i f kD < α + ks α

N > N5 N > N4

Here,



 (α + γ )2 α + γ ks − 1 + ( kD − ks ) D2 , αδ γ ks   α kD α + γ ks − 1 N5 = + ( kD − ks ) D2 . δ ks γ ks N4 =

(A28)

(A29)

A.4. Scenario 4 t˜1 < t˜ < t1 < t˜2 < t ∗ < t2 In this scenario, we can write out the same 7 equations as those found in Eqs. (A21)–(A27), and obtain the same solutions of tq , t˜, t˜1 , t1 , t˜2 , t2 , and tq . In Scenario 4, the parameters should satisfy:

 α + γ 2 α+γ α+γ ks < kD < ks and ks > . α α α

The total commuter number N should be in the range N5 < N < N4 . A.5. Scenario 5 t˜1 < t1 < t˜ < t ∗ < t˜2 < t2 In this scenario, we can write out the following 7 independent equations:



α

α−β



s1 t˜1 − tq +

α

α−β





s2 t˜ − t˜1 +





α

α+γ





s2 t˜2 − t˜ +

s1 (t1 − tq ) + s2 (t2 − t1 ) + s1 tq − t2 = N,





α α s t˜ − t + s t − t˜ − s1 (t1 − tq )=D1 , α−β 1 1 q α−β 2 1 1



 α N− s1 tq − t2 − N − s1 tq − t2 =D2 , α+γ

β (t ∗ − tq ) = γ tq − t ∗ ,

and

α

α+γ





s1 tq − t˜2 =N,

(A30) (A31) (A32) (A33) (A34)





α t1 − t˜1 + β (t ∗ − t1 ) = γ tq − t ∗ ,

(A35)





α t ∗ − t˜ = γ tq − t ∗ .

(A36)

The solution is:

    δ ks − 1 α − β α+γ tq = t − N+ k + ks D2 /s2 , β ks β D γ     δ ks − 1 α − β α+γ tq = t ∗ + N+ kD + ks D2 /s2 , γ ks β γ ∗

t1 = tq +

α − β D1 , β s2

t2 = tq −

α + γ D2 , γ s1

t˜1 = tq +

(α − β )2 D1 , αβ s2

t˜2 = tq −

( α + γ )2 D 2 , αγ s1

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and

    δ ks − 1 α − β α+γ N+ kD + ks D2 /s2 . α ks β γ

t˜ = t ∗ −

By substituting the solutions into tq < t˜1 < t˜ < t1 < t ∗ < t˜2 < t2 < tq , and considering the queue length constraint, we can derive that the total commuter number N should be in the range:



γ 2 i f kD > α + ks α+αγ 2 . i f kD < α ks

N > N5 N > N6

Here,



  ks − 1 α − β α+γ (α + γ )2 N6 = ks − k + ks D2 . αδ ks β D γ

(A37)

A.6. Scenario 6 t˜1 < t1 < t˜ < t˜2 < t ∗ < t2 In this scenario, the 7 equations and the solutions of tq , t˜, t˜1 , t1 , t˜2 , t2 , and tq are the same as those in Scenario 5. In Scenario 6, the parameters should satisfy:

kD <

 α + γ 2 α

ks .

The total commuter number N should be in the range N < N6 and



γ i f kD > α + α ks , . α +γ i f kD < α ks

N > N5 N > N7

Here,



   α−β α+γ α − β kD α+γ N7 = + k − k + D2 . β ks β s β D γ

(A38)

A.7. Scenario 7 t˜1 < t1 < t˜2 < t˜ < t2 < t ∗ In this scenario, we can write out the following 7 independent equations:









α α α α s t˜ − t + s t˜ − t˜ + s t˜ − t˜2 + s t  − t˜ =N, α−β 1 1 q α−β 2 2 1 α−β 1 α+γ 1 q

s1 (t1 − tq ) + s2 (t2 − t1 ) + s1 tq − t2 = N,

α

α−β



s1 t˜1 − tq +

α

α−β







s2 t1 − t˜1 − s1 (t1 − tq )=D1 ,



 α s t  − t2 − N − s1 tq − t2 =D2 , α+γ 1 q

β (t ∗ − tq ) = γ tq − t ∗ , N−

and

(A39) (A40) (A41) (A42) (A43)





α t1 − t˜1 + β (t ∗ − t1 ) = γ tq − t ∗ ,

(A44)





α t2 − t˜2 + β (t ∗ − t2 ) = γ tq − t ∗ .

(A45)

The solution is:

  δ α − β ks − 1 α tq = t − N+ (kD − ks )D2 + (ks − 1 )D2 /s2 , β β ks δ   δ α − β ks − 1 α tq = t ∗ + N+ k − k D + k − 1 D ( D ( ) 2 /s2 , s) 2 γ β ks δ s ∗

t1 = tq +

α − β D1 , β s2

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207



  α+γ α − β kD α − β t2 = tq + N − − k + D2 /s2 , β ks β D γ (α − β )2 D1 , αβ s2     α−β α+γ α − β kD α − β t˜2 = tq + N− − kD + D2 /s2 , α β ks β γ t˜1 = tq +

and

t˜ = t ∗ −

  δ α − β ks − 1 α N+ (kD − ks )D2 + (ks − 1 )D2 /s2 . α β ks δ

In Scenario 7, the parameters should satisfy:

kD <

α+γ k. α s

The total commuter number N should be in the range N < N7 and:



N > N8 N > N9

Here,

β i f kD > α − α ks α −β . i f kD < α ks



 α+γ α − β kD N8 = + kD + D2 , β ks γ   α α − β ks − 1 N9 = − ( kD − ks ) D2 . δ β ks

(A46)

(A47)

A.8. Scenario 8 t˜1 < t1 < t˜2 < t2 < t˜ < t ∗ In this scenario, we can write out the following 7 independent equations:









α α α α s t˜ − t + s t˜ − t˜ + s t˜ − t˜2 + s t  − t˜ =N, α−β 1 1 q α−β 2 2 1 α−β 1 α+γ 1 q

s1 (t1 − tq ) + s2 (t2 − t1 ) + s1 tq − t2 = N,

and

(A48) (A49)





α α s t˜ − t + s t − t˜ − s1 (t1 − tq )=D1 , α−β 1 1 q α−β 2 1 1





 α α N− s t  − t˜ − s t˜ − t2 − N − s1 tq − t2 =D2 , α+γ 1 q α−β 1

β (t ∗ − tq ) = γ tq − t ∗ ,

(A52)





α t1 − t˜1 + β (t ∗ − t1 ) = γ tq − t ∗ ,

(A53)





α t2 − t˜2 + β (t ∗ − t2 ) = γ tq − t ∗ .

(A54)

The solution is:

  δ α − β ks − 1 N+ (kD − ks )D2 /s1 , β β ks   δ α − β ks − 1 tq = t ∗ + N+ (kD − ks )D2 /s1 , γ β ks tq = t ∗ −

t1 = tq +

α − β D1 , β s2

(A50) (A51)

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t2 = tq +

α − β D2 , β s1

t˜1 = tq +

(α − β )2 D1 , αβ s2

t˜2 = tq +

(α − β )2 D2 , αβ s1

and

  δ α − β ks − 1 N+ (kD − ks )D2 /s1 . α β ks

t˜ = t ∗ −

In Scenario 8, the parameters should satisfy

kD <

α−β k. α s

The total commuter number N should be in the range N9 < N < N10 . Here,



N10 =

 α − β ks − 1 α kD − ( kD − ks ) D2 . δ β ks

(A55)

A.9. Scenario 9 t˜1 < t˜2 < t1 < t2 < t˜ < t ∗ In this scenario, we can write out the following 7 independent equations:









α α α α s t˜ − t + s t˜ − t˜ + s t˜ − t˜2 + s t  − t˜ =N, α−β 1 1 q α−β 2 2 1 α−β 1 α+γ 1 q

s1 (t1 − tq ) + s2 (t2 − t1 ) + s1 tq − t2 = N, α

α−β





s1 t˜1 − tq +

α





s2 t˜2 − t˜1 +

α





s1 t1 − t˜2 − s1 (t1 − tq )=D1 ,

α−β α−β





 α α N− s t  − t˜ − s t˜ − t2 − N − s1 tq − t2 =D2 , α+γ 1 q α−β 1

β (t ∗ − tq ) = γ tq − t ∗ ,

and

(A56) (A57) (A58) (A59) (A60)





α t1 − t˜1 + β (t ∗ − t1 ) = γ tq − t ∗ ,

(A61)





α t2 − t˜2 + β (t ∗ − t2 ) = γ tq − t ∗ .

(A62)

The solution is:

  δ ( kD − 1 ) ( ks − 1 ) N− D 2 /s1 , α β ks − α − β   δ ( kD − 1 ) ( ks − 1 ) tq = t ∗ + N− D 2 /s1 , α γ ks − α − β tq = t ∗ −

t1 = tq +

D1 − D2 α − β D2 − , β s1 s2 − α−α β s1

t2 = tq +

α − β D2 , β s1

t˜1 = tq +

(α − β )2 D2 α − β D1 − D2 − , αβ s1 α s2 − α−α β s1

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t˜2 = tq + and

t˜ = t ∗ −

209

(α − β )2 D2 , αβ s1

  δ ( kD − 1 ) ( ks − 1 ) N− D2 /s1 . α α ks − α − β

In Scenario 9, the parameters should satisfy:

kD <

α−β k. α s

The total commuter number N should be in the range N11 < N < N12 . Here,



 α ( kD − 1 ) ( ks − 1 ) + D2 , α δ ks − α − β   α ( kD − 1 ) ( ks − 1 ) N12 = kD + D2 . α δ ks − α − β N11 =

(A63)

(A64)

A.10. Scenario 10 t˜1 < t˜2 < t1 < t˜ < t2 < t ∗ In this scenario, we can write out the following 7 equations:





s1 (t1 − tq ) + s2 (t2 − t1 ) + s1 tq − t2 = N

(A65)







α α α s t˜ − t + s t˜ − t˜ + s t − t˜ − s1 (t1 − tq )=D1 α−β 1 1 q α−β 2 2 1 α−β 1 1 2



 α N− s1 tq − t2 − N − s1 tq − t2 =D2 α+γ

β (t ∗ − tq ) = α t1 − t˜1 + β (t ∗ − t1 )

(A68)





α t1 − t˜1 + β (t ∗ − t1 ) = α t2 − t˜2 + β (t ∗ − t2 )

(A69)





α t2 − t˜2 + β (t ∗ − t2 ) = α t ∗ − t˜

(A70)





α t ∗ − t˜ = γ tq − t ∗

(A71)

A.10.1. Scenario 10(a) −β When ks = αα−2 β , these 7 equations are independent. The solution is:

  δ α D1 α + γ D2 αδ  α 1 + − 2 1− X, β β s2 γ s1 α − β ks β   δ α D1 α + γ D2 αδ  α 1 tq = t ∗ + + + 1− X, γ β s2 γ s1 βγ α − β ks  1 α−β D α tq = t ∗ −

t1 = tq + t2 = tq +

t˜1 = tq + t˜2 = tq +

β

1

s2

+

β

1−

ks

X,

α D1 α  α 1 + 1− X, β s2 β α − β ks   1 (α − β )2 D1 α − β + 1− X, αβ s2 β ks α−β D α−β α 1 β

1

s2

+

β

1−

α − β ks

X,

(A66) (A67)

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and

  δ α D1 α + γ D2 δ α 1 + − 1− X. α β s2 γ s1 β α − β ks

t˜ = t ∗ − Here,





X = N−

    α+γ α α − 2β 1 α − β kD + kD + D2 / − s1 . β ks γ β α−β ks

In Scenario 10(a), the parameters should satisfy:

kD <

α+γ k. α s

The total commuter number N should be in the range:



˜ N13 < N < N ˜ < N < N13 N

Here,

 N13 =



if if

1 ks 1 ks

β > αα−2 −β α −2β . < α −β

α+γ α − β kD + kD + β ks γ

 +

   α α − β α − 2β 1 ks − α +γ kD − D2 , α β α−β ks ks − α + γ

(A72)

β i f kD > α − α ks α −β . i f kD < α ks

˜ = N8 N N11

A.10.2. Scenario 10(b) −β ˜ When ks = αα−2 β , only 6 equations are independent. Let t = t1 − t2 , together with (A66)–(A71), we can solve:

  δ α D1 α + γ D2 αδ  α 1 + − 2 1− t, β β s2 γ s1 α − β ks β   δ α D1 α + γ D2 αδ  α 1 tq = t ∗ + + + 1− t, γ β s2 γ s1 βγ α − β ks  1 α − β D1 α  t1 = tq + + 1− t, β s2 β ks α D1 α  α 1 t2 = tq + + 1− t, β s2 β α − β ks tq = t ∗ −

t˜1 = tq +

  1 (α − β )2 D1 α − β + 1− t, αβ s2 β ks

t˜2 = tq +

α − β D1 α − β  α 1 + 1− t, β s2 β α − β ks

and

  δ α D1 α + γ D2 δ α 1 + − 1− t. α β s2 γ s1 β α − β ks

t˜ = t ∗ −

Substituting the solution into Eq. (A65), we obtain:

N =N14 =

α−β α+γ D1 + D2 . β γ

In the solution, tq < t˜1 requires that

t > −

α − β ks D1 . α ks − 1 s2

(A73)

t˜1 < t˜2 requires that

t <

α − β D1 α s1

(A74)

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t˜2 < t1 requires that

t > 0

(A75)

t1 < t˜ requires that α α − β ks − α + γ kD D2 α α ks − α + γ s1

t <

(A76)

t˜ < t2 requires that



D α  α−β α 2 t > ks − kD α−β α α−β s1

ks −

(A77)

t2 < t∗ requires that



  α  α+γ α D2 ks − t < ks − k . α−β α α + γ D s1

(A78)

t∗ < tq requires that



  α  β α+γ D2 ks − t > − ks − kD . α−β α γ s1

(A79)

One can easily prove that Eq. (A73) is always met provided Eq. (A75) is satisfied, Eq. (A79) is always met provided Eq. (A77) is satisfied. From Eqs. (A75) to (A76), we can derive

ks >

α k , α+γ D

(A80)

One can also easily prove that Eq. (A74) is always met provided Eqs. (A75) and (A76) are satisfied. α k , From Eq. (A77), if ks > α − β D α α − β k s − α −β k D D 2 . α α ks − α − β s1

t >

(A81)

Otherwise, Eq. (A77) is always met provided Eqs. (A75) and (A76) are satisfied. From Eq. (A76), we obtain



ks −

   α−β α D2 t < ks − kD . α+γ α α+γ s1 α

(A82)

Since the LHS of Eq. (A78) is smaller than the LHS of Eq. (A82), and the RHS of Eq. (A78) is larger than the RHS of Eqs. (A82), (A78) is always met provided Eq. (A82) is satisfied. To summarize, in Scenario 10(b), the parameters should satisfy:

kD <

α+γ k, α s

and

tmin < t < tmax . A.11. Case D2 =D1 α +γ

Note that kD ≥ α ks would never be satisfied when kD =1. Thus Scenarios 1–4 would not exist. If we substitute kD =1 into Eqs. (A47), (A55), (A63), and (A64), then we have N9 =N10 and N11 =N12 . Therefore, neither Scenario 8 nor Scenario 9 would exist. −β α −β α −β D0 When kD =1 and ks = αα−2 β , we can easily prove that kD < α ks always holds, which yields tmin =tmax = α s . There1

fore, Scenario 10(b) would not exist. Appendix B. Multiple equilibria

This appendix shows details of the different parameter ranges in which multiple equilibria are exhibited. α +γ Parameter range A: kD > α ks . See Fig. B1 for a typical example. α +γ β Parameter range B1: α − α ks < kD < α ks ,

1 ks

β 9 > αα−2 . See Fig. B2 for typical examples. −β

212

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Fig. B1. A typical example of the cost function in the parameter range A. The parameters are α =2, β =1, γ =4, s1 = 10 0 0, s2 = 150 0, D1 = 40 0 0, and D2 = 80 0.

Fig. B2. Typical examples of the cost function in parameter range B1. The parameters are β =1, γ =4, s1 = 10 0 0, s2 = 150 0, D1 = 30 0 0, and D2 = 10 0 0; (a) α =2, (b) α =3, and (c) α =3.5. Note the slope of the cost function is positive, zero, and negative in (a), (b), and (c), respectively.

α +γ β α −2β 10 1 Parameter range B2: α − α ks < kD < α ks , ks = α −β . See Fig. B3 for a typical example. α +γ β α −2β 11 1 Parameter range B3: α − α ks < kD < α ks , ks < α −β . See Fig. B4 for a typical example. α −β α −2β 1 Parameter range C1: kD < α ks , k > α −β . See Fig. 14 for a typical example. s

9 10 11

−β When α < 2β , this inequality is always met. When α > 2β , this inequality is ks < αα−2 β.

−β When α < 2β , this equation is never met. When α > 2β , this equation is ks = αα−2 β.

−β When α < 2β , this inequality is never met. When α > 2β , this inequality is ks > αα−2 β.

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213

Fig. B3. A typical example of the cost function in parameter range B2. The parameters are α =4, β =1, γ =8, s1 = 10 0 0, s2 = 150 0, D1 = 30 0 0, and D2 = 10 0 0.

Fig. B4. A typical example of the cost function in parameter range B3. The parameters are α =5, β =1, γ =8, s1 = 10 0 0, s2 = 150 0, D1 = 30 0 0, and D2 = 10 0 0.

Fig. B5. A typical example of the cost function in parameter range C2. The parameters are α =4, β =1, γ =8, s1 = 10 0 0, s2 = 150 0, D1 = 170 0, and D2 = 160 0.

214

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Fig. B6. A typical example of the cost function in parameter range C3. The parameters are α =5, β =1, γ =8, s1 = 10 0 0, s2 = 150 0, D1 = 110 0, and D2 = 10 0 0.

β Parameter range C2: kD < α − α ks , β Parameter range C3: kD < α − α ks ,

β 1 = αα−2 . See Fig. B5 for a typical example. ks −β α −2β 1 < α −β . See Fig. B6 for a typical example. ks

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