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(1)國立成功大學 應用數學研究所 碩士論文. Braess 運輸問題的均衡解 On Equilibrium Solutions of the Braess Transportation Problem. 研 究 生:呂庭維 指導教授:許瑞麟 中 華 民 國 一 ○ 三 年 七 月.

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(3) 摘要. 在這篇論文裡,我們聚焦在 Dietrich Braess 所提出的四節點五路徑的 路網模型[1]。在依賴車流量的線性旅程函數以及總車流給定的前提之 下,不論平衡達到與否,我們對所有可能的車流分配模式,做一個完整 分類,以直接獲得正確的平衡模式。我們的研究建立在 Marguerit Frank 原先的工作之上[6]。Frank 給出 Braess 悖論存在性在代數上的充分和必 要條件,但我們的分類更廣泛的覆蓋所有的情況,包括悖論不發生的時 候,使得添加新路徑的效用能夠被判斷出來。. 關鍵字:使用者均衡、網路分配、路徑規劃、Braess 悖論。.

(4) ABSTRACT. In this thesis, we focus on the four-node and five-link network model proposed by Dietrich Braess [2]. Given a flow-dependent linear traveling function on each link and the total flow, we can completely classify the flow distribution pattern regardless whether the equilibrium of the network does happen or not. Our study is based on the original work of Marguerit Frank [6] in which algebraic necessary and sufficient conditions for the existence of the Braess Paradox were derived. Our classification extends to the cases when the paradox does not happen so that the effectiveness of the newly added link is able to be judged.. Keywords: user equilibrium, distribution network, route planning, the Braess paradox..

(5) 致謝 本論文得以完成,應歸功於許多人的協助,第一個要感謝的是指導 教授許瑞麟老師,老師以他豐富的經驗與專業,帶領著我學習研究的方 法與態度,並且透過每一次個別指導的機會,修正我思考上的不足與盲 點,另一方面老師也不吝於展現如何把數學縝密的邏輯思維與批判性的 思考,落實在教育、社會以及生活中,讓我在跟隨老師學習的這一年, 在許多方面有所進步、有所改變與有所成長。 另一方面,感謝口試委員王逸琳老師和林仁彥老師的細心審閱,他 們的指導與建議除了讓本論文更臻於完善,也讓我在口試的過程中獲得 許多想法與啟發。 感謝舒宇宸老師在許多方面的解惑與分享,舒老師在研究與教學上 的嚴謹、用心、創意、活力與感染力,亦令我獲益良多。 感謝我的家人,他們擔負起我在外求學時造成的不便,在許多方面 給我很大的支援和照顧,讓我擁有極大的空間做自己想做的事。感謝蔡 哲淵和蔡景州給予的建議和鼓勵,關鍵性的促成我來成大完成碩士的學 業。感謝翁之翊、李奕旻、許智詠、宋政蒲等學長們以及同一屆的許舜 斌、馮鼎軒、吳怡潔、劉建甫在學習上的討論、指點與協助,協助我度 過研究所課程的難關。 感謝成杏合唱團大團、新生小團還有音樂劇青年革命軍的夥伴,一 起練唱、一起排戲還有一起在舞台上的那些時刻,寬慰了我碩二生活的 壓力與苦悶,還讓我找回唱合唱的聲音與自信。感謝舜斌在哲學上的啟 蒙,感謝政蒲在自然保育的帶領,感謝這兩年所遭遇到的一切。 庭維 2014 年 7 月.

(6) CONTENTS. 1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 7. 2. The Braess Problem and the Equilibrium Solutions . . . . . . . . . . . . . . . .. 14. 2.1. Patterns of Equilibrium Condition in the Braess Problem . . . . . . . . . .. 14. 2.2. The Equilibrium Solution and its Constraints in Problem I . . . . . . . . .. 17. 2.3. The Equilibrium Solution and its Constraints in Problem II. . . . . . . . .. 20. 3. The Sequence of Equilibrium Patterns Determined by the Demand . . . . . . .. 33. 3.1. Classify the Equilibrium Patterns by Critical Points . . . . . . . . . . . . .. 33. 3.2. The Sequence of Equilibrium Patterns in Problem I . . . . . . . . . . . . .. 40. 3.3. The Sequence of Equilibrium Patterns in Problem II for ∆ > 0 . . . . . . .. 42. 3.4. The Sequence of Equilibrium Patterns in Problem II for ∆ < 0 . . . . . . .. 55. 3.5. The Sequence of Equilibrium Patterns in Problem II for ∆ = 0 . . . . . . .. 67. 4. Necessary and Sufficient Condition to The Braess Paradox . . . . . . . . . . . .. 71. 5. Discussion and Future Directions . . . . . . . . . . . . . . . . . . . . . . . . . .. 77. Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 78.

(7) 1. INTRODUCTION. In a traffic network, users tend to minimize their individual travel time. A well received assumption in the literature is that: (i) All users have perfect information about the traffic conditions. (ii) And they consistently make the correct decisions regarding route choices.. The simplest model is a two-route choice scenario. See Figure 1.1 below. In this network, there are four nodes {A, B, C, D} with the origin A and the destination D in which there are two routes A → B → D (denoted by “U p”) and A → C → D (denoted by “Down”) for users to choose. We assume that the traffic time on each link is flowdependent; and is a linear function of the traffic flow on the link. For example, in Figure 1.1, if the traffic flow on the link A → B is uAB , the travel time is assumed to be 10uAB +1. route U p. '. A t 10uAB + -1 B t. 50 + uBD. t. -. C &. 50 + uAC. $. 10uCD + 1. t. D. %. route Down. Fig. 1.1: Network example of the Braess Problem. Since there are two routes for choice and due to the perfect information assumption, if the traffic flow on one route causes a shorter traveling time than that on the other route, the traffic flow move from the route having a longer traveling time to the other which is.

(8) 1. Introduction. 8. faster. Then, the faster route slows down gradually by the moving-in traffic. The flow pattern fluctuates between the two alternatives until there is no time difference existing between them, in which case we call the network achieves the equilibrium. When the network achieves the equilibrium, the flow distribution on the two routes can be easily calculated by solving simultaneous linear equation. Take again Figure 1.1 as an example. Suppose the total flow from A to D is 6 and assume that the flow distribution on the U p route is xα ≥ 0 whereas xβ ≥ 0 is the flow on route Down when the equilibrium of the network is achieved. Then, (. xα + xβ. =. 6. (10xα + 1) + (50 + xα ). =. (50 + xβ ) + (10xβ + 1). .. (1.1). The solution of (1.1) is (xα , xβ ) = (3, 3) and the travel time in equilibrium, denoted by T I (3, 3), is (10 × 3 + 1) + (50 + 3) = 84. route U p. '. A t 10uAB + -1 B t. route M iddle. 9 + uBC &. 50 + uAC. 50 + uBD. t. $. -. C. 10uCD + 1. t. D. %. route Down. Fig. 1.2: Network example of the Braess Problem that the shortcut has added in. Now suppose a new link B −→ C with a predefined flow-dependent traveling time 9 + uBC being added into this network, see Figure 1.2. It becomes a three-route scenario in which the users are given a new alternative A → B → C → D (denoted by “M iddle”) in addition to the existed old ones - U p and Down. Again, the flow pattern fluctuates among the three alternatives from the slower ones to the faster, until there no longer exists any time advantage for a particular route. For the example of a total flow 6, the non-negative flow distribution (yα , yβ , yγ ) for the three-route scenario can be similarly set up and solved.

(9) 1. Introduction. as follows    yα + yβ + yγ  . 9. =. 6. (10(yα + yγ ) + 1) + (50 + yα ). =. (10(yα + yγ ) + 1) + (9 + yγ ) + (10(yβ + yγ ) + 1). (50 + yβ ) + (10(yβ + yγ ) + 1). =. (10(yα + yγ ) + 1) + (9 + yγ ) + (10(yβ + yγ ) + 1). (1.2). with the solution of (1.2) (yα , yβ , yγ ) = (2, 2, 2). The equilibrium solution (2, 2, 2) shows that the traffic flow has been effectively shifted to the newly added choice, but the traveling time of the system (1.2) is T II (2, 2, 2) = 93, which is unfortunately longer than that of system (1.1) and contradicts to most people’s common sense that building a new road should somehow ease the congestion of a traffic network. This is the famous Brass Paradox in transportation science, which was published by Dietrich Braess in 1968 [1]. In literature, Daganzo and Sheffi [4] applied the stochastic user equilibrium on networks, for drivers selecting its perceived fastest route. Mahmassani and Herman [7] considered the dynamic user equilibrium, which discussed the network with time-varying flows. Di, He and Liu [5] discussed the Braess Paradox under the bounded rational user equilibrium, which relax the equilibrium assumption and drivers owns acceptable routes in a given tiny time gap. Steinberg and Zangwill [9] gave the Braess Paradox a further work in a general transportation network. Dafermos [3] discussed networks containing two-way streets. Nagurney [8] did many research on Braess Paradox and proved that the paradox disappears when there is a high demand. Apparently, the Brass Paradox may not always happen. It depends on the flowdependent linear traveling time on each link and also on the total flow passing through the network. For example, if the total flow on the networks Figure 1.1 and 1.2 is increased to 10, then the non-negative flow distribution before and after the link B −→ C being added in is (. xα + xβ. =. 10. (10xα + 1) + (50 + xα ). =. (50 + xβ ) + (10xβ + 1). .. (1.3). and    yα + yβ + yγ  . =. 10. (10(yα + yγ ) + 1) + (50 + yα ). =. (10(yα + yγ ) + 1) + (9 + yγ ) + (10(yβ + yγ ) + 1). (50 + yβ ) + (10(yβ + yγ ) + 1). =. (10(yα + yγ ) + 1) + (9 + yγ ) + (10(yβ + yγ ) + 1). ,. (1.4).

(10) 1. Introduction. 10. respectively. The solution of (1.3) is (xα , xβ ) = (5, 5) and the travel time is T I (5, 5) = 106; The solution of (1.4) is (yα , yβ , yγ ) = ( 70 , 70 , − 10 ) and the travel time is T II ( 70 , 70 , − 10 )= 13 13 13 13 13 13 10 7 . Since the system (1.4) admits a negative solution ( 70 , 70 , − 13 ), it is not a valid 102 13 13 13. solution for the total flow 10. The equilibrium can not happen that way. Consider the moment that the network in Figure 1.1 achieves the equilibrium and the new link B −→ C was just connected. The initial flow distribution is (yα , yβ , yγ ) = (5, 5, 0) with travel time on both routes U p and Down is 106. Now suppose the amount 1 and 2 of the total flow moves from route U p and Down, respectively, to the new option B −→ C with 1 , 2 > 0. It can be easily computed that, the travel time on the old routes U p and Down reach to 106 + 102 − 1 and 106 + 101 − 2 , respectively. But the travel time using the M iddle route is 111 + 111 + 112 . The flow distribution on U p and Down can reach equilibrium if 1 = 2 . Let 1 = 2 =  and the above travel time on path U p, Down and M iddle are simplified into 106 + 9, 106 + 9 and 111 + 22, respectively. Since 111 + 22 > 106 + 9 for any possible flow 1 + 2 = 2 on B −→ C, based on the perfect information assumption there will be no user taking this option. The flow pattern (yα , yβ , yγ ) = (5, 5, 0) is the final distribution when the total flow is 10, in which case we shall call a degenerated equilibrium is reached although the classical sense of the equilibrium does not apply. The degenerate equilibrium has the feature that any flow on the route causes a longer travel time than that of the remaining flow traveling on the other two routes and reaching the equilibrium. In addition, even when a non-degenerate equilibrium happens on the three-route scenario, the Brass Paradox may not always happen. That is, the link B −→ C does help to reduce the travel time. See another example below in Figure 1.3. Suppose the total flow 6 is again assumed but the linear travel time on each link is changed. The equilibrium solutions before and after the link B −→ C being added in are (xα , xβ ) = (3, 3) and (yα , yβ , yγ ) = ( 54 , 45 , 22 ), respectively. The travel time of the former is 5 96, which is longer than that of the later, which is 85. It shows that adding the new link does help reduce the travel time for drivers on the network. In this thesis, we focus on the four-node and five-link network in Figure 1.1 and 1.2 and are interested in whether an equilibrium or a degenerated equilibrium is reached for.

(11) 1. Introduction. route U p. '. A t. 5uAB + 1 -. B t. route M iddle. 9 + 5uBC &. 50 + 10uAC. 11. 50 + 10uBD. t. $. -. C. 5uCD + 1. t. D. %. route Down. Fig. 1.3: An example that the Braess Paradox does not happen for all given total flow.. different amounts of the total flow and for any possible linear travel functions on the links. Moreover, we will discuss whether the Braess Paradox can happen. We found that there are total 3 patterns that can occur for the network in Figure 1.1 and 7 patterns for Figure 1.2. In Figure 1.1, it could be both routes, U p and Down, are used and an equilibrium is 1 reached. In this case, the notation Eq − is used. In Figure 1.1, it is also possible that 1 1 only the U p route is used in which case we use the notation Eq − where 1 means the 0. total flow d on U p. The constraints of above patterns are arranged in Chapter 2, and we use intuitive α α notations Eq −β and Eq βγ to distinguish these patterns. The location of α, β, γ in the matrix corresponds to routes U p, Down, M iddle. For Figure 1.1, only routes U p and Down can be chosen, so we express it by a 2 × 1 matrix. For Figure 1.2, we have three possible routes U p, Down and M iddle, so we express it by a 3 × 1 matrix. Let   1 if the route is utilized. . α, β, γ =  0 if the route is not utilized. Thus, all possible patterns for Figure 1.1 and 1.2 are included in our notation. For example, 1 Eq 1 means that U p and M iddle are utilized but Down is not for Figure1.2. 0 α1  α2  Another notation, cr βγ1 βγ2 , is designed to represent the value of critical point be2  α1   α2 1 tween Eq βγ1 and Eq βγ2 on the number line with axis relation to total flow d. Base 1. 2. on the solutions set of flow distribution is convex and the linear travel function on each link is increasing and continuous, the constraints of equilibrium patterns forms an interval lying on the number with total flow d in axis. We call each boundary of such intervals is a.

(12) 1. Introduction. 12. critical point. If the various d moving on the number line with initial value in an interval of one equilibrium pattern, we say such pattern occurs under the given d. Suppose d is reduced or increased near the critical point and then pass over it, the original pattern is fail and another pattern replace it. Our study shows that the value of some critical points 1 come from different intervals are equal, for example, the upper bound of Eq − and a pos0 1 1  1  sible lower bound of Eq − are the same. Therefore, we denote the notation cr 0 0 is 0 1 1 1 1 the value of critical point between Eq − and Eq − . Again in Chapter 2, eleven values 0. 1. of critical points are sufficient to ensure that all patterns in Figure 1.1 and 1.2 can be separated. We compare the 11 critical values and find a method to classify the pattern of equilibrium in a suitable order in Chapter 3. If the flow-dependent linear traveling time on each link are given, referring to our classification in Chapter 3, specific equilibrium patterns appear in a sequence on the number line with the magnitude of the total flow from zero to infinity. For example, again in Figure 1.3, the sequence is determined by 1  1  0  1  0  1  0  0  1  1  0  1  cr − − = cr − − = 0 and cr 1 1 = cr 1 1 = cr 1 1 = cr 1 1 = 4. Thus 0. 1. 1. 0. 1. 0. 0. 1. 0. 1. 1. 1. the sequence of the patterns before and after link l5 added in are calculated by the method in Chapter 3 as follows: Eq. 1 − 1. -. d. 0 Fig. 1.4: The sequence of the patterns on a number line before link l5 added in.. Eq. 0. Eq. 1 0. 1 1 1. -. 0. d. 4. Fig. 1.5: The sequence of the patterns on a number line after link l5 added in.. 1 The above number line shows that the pattern Eq − occurs for d ∈ [0, ∞) be0 1 1 1 fore link l5 added in; while pattern Eq and Eq 1 occurs for d ∈ [0, 4] and d ∈ 0. 1.

(13) 1. Introduction. Eq. 13. 1 − 1. -. 0. Eq. 0 1 0. 4. Eq. d. 1 1 1. Fig. 1.6: Combine the above two sequences on a number line.. [4, ∞),respectively, after link l5 added in. Anyone can collate the number line to determine the proper flow distribution and travel time for a given total flow. Compare the travel time before and after link l5 added in, the occurrence of Braess Paradox is determined. Our study is based on the original work of M. Frank [6] in which algebraic necessary and sufficient conditions for the existence of the Braess Paradox were derived. Our classification is somehow more general that covers the cases when the paradox does not happen. If we generalize the flow-dependent linear travel time on each link, see Figure 2.1. A characteristic value ∆ = a1 a4 − a2 a3 , which is proposed to be one of the necessary and sufficient conditions for the existence of the Braess Paradox by Frank, is our main principle to arrange the sequence of the patterns. In Chapter 3, we complete the classification in a suitable order with six characteristic values. It is more exhaustive then the works Frank have done. A Matlab code classifies the proper sequence of the equilibrium patterns and runs the graph of a function from total flow to travel time, if the flow-dependent linear traveling time on each link is given. We show illustrative examples, and it becomes clear which conditions correspond to the paradox and which do not. In Chapter 4, we compile the necessary and sufficient conditions from Frank under which the Braess Paradox exists. If the Braess Paradox exists, the range of Braess Paradox occurrence in demand are determined..

(14) 2. THE BRAESS PROBLEM AND THE EQUILIBRIUM SOLUTIONS. 2.1 Patterns of Equilibrium Condition in the Braess Problem route U p. '. A t. a1 u1 + b1. B t. route M iddle a5 u5 + b5. &. a2 u2 + b2. $. t. t. C. a4 u4 + b4. D. % a3 u3 + b3. route Down. Fig. 2.1: Original Network example of the Braess Problem. In this chapter, we discuss how linear travel costs determine the travel time. Here, we focus on the same four-node and five-link network, see Figure 2.1. Here we denote link AB, link BD, link AC, link CD and link BC as link l1 , l2 , l3 , l4 and l5 , respectively. Our main concern is the influence of the linear travel costs on travel time before and after link l5 is added in, with Problem I and Problem II referring to the cases without and with link l5 , respectively. The traffic time is a linear function of traffic flow, we assume a linear travel cost on each link fi (ui ) = ai ui + bi ,. ai > 0, bi > 0,. i = 1, 2, 3, 4, 5.. The terms bi represents the flow independent cost .It is a constant and the distance of the road individual links is directly proportional to bi . The coefficient of the first degree term ai is a varying time costs, which is determined by the amount of vehicles on this road link. The more vehicles on the road, the more time users have to spend. The variable.

(15) 2. The Braess Problem and the Equilibrium Solutions. 15. ui (vehicles/time) is the flow passing through link li . If there are several routes passing through link li , the flow should be summed. In Problem I, the travel time needed for U p and Down are stated as follows, with TUI and TDI denoting the travel costs of U p and Down, respectively. TUI (xα , xβ ) = f1 (xα ) + f2 (xα ).. (2.1). TDI (xα , xβ ) = f3 (xβ ) + f4 (xβ ).. (2.2). As for in Problem II, the travel time needed for U p,Down and M iddle are stated II as follows, where TUII , TDII and TM denotes the travel costs on U p,Down and M iddle,. respectively. TUII (yα , yβ , yγ ) = f1 (yα + yγ ) + f2 (yα ).. (2.3). TDII (yα , yβ , yγ ) = f3 (yβ ) + f4 (yβ + yγ ).. (2.4). II TM (yα , yβ , yγ ) = f1 (yα + yγ ) + f5 (yγ ) + f4 (yβ + yγ ).. (2.5). When the flow reaches the equilibrium pattern, the travel time for utilized routes are equal but less than any unutilized one. Since there are two possible routes, the equilibrium patterns are classified as 22 −1 = 3 types in Problem I. Similarly, the number of equilibrium patterns can be calculated as 23 − 1 = 7, meaning 7 different equilibrium patterns in α α Problem II. We use an intuitive notation Eq −β and Eq βγ to distinguish these different equilibrium states. The location of α, β, γ in the matrix corresponds to routes U p, Down, M iddle. In Problem I, only route U p and Down can be chosen, so we express it by a 2 × 1 matrix. In Problem II, we have possible routes U p, Down and M iddle, so we express it by a 3 × 1 matrix. Let.   1 α, β, γ =  0. if the route is utilized.. .. if the route is not utilized.. Thus, all possible equilibrium patterns in Problem I and Problem II are included in our notation..

(16) 2. The Braess Problem and the Equilibrium Solutions. i) Equilibrium conditions in Problem I ←−−−−→ Eq. 16. α. ii) Equilibrium conditions in Problem II ←−−−−→ Eq. − β. α γ β. Example 2.1.1. i) Eq. 0. ←− −→ route Down is utilized in Problem I.. ii) Eq. 1. ←− −→ route U p and M iddle are utilized in Problem II.. − 1 1 0. Our designed notations have the advantage of being easy to read; two symbols in the matrix means that we are dealing with Problem I, while three symbols in the matrix means Problem II. Also, the use of notations ‘1’ and ‘0’ make it easy to identify whether or not the route is utilized, with the former representing ‘route utilized’ and the later meaning ‘not taken’. By arranging the symbols of α, β, γ in reflection of U p, Down, M iddle the use of visual spatial arrangements also make it simple to remember the notation’s meaning. In addition, based on the continuity of the linear cost function, any change in demand will result in a subsequent change in the number of routes in use. For further discussion on the effects of change in demand, see section 3.2 and 3.3. α1  α2  Another notation, cr βγ1 βγ2 , is designed to represent the critical point between 1 2  α1   α2   α1  α2  Eq βγ1 and Eq βγ2 . Here, the right square bracket means that cr βγ1 βγ2 is an upper 1 1 2  α1  2  α1  α2  bound of Eq βγ1 and, similarly, the left square bracket identifies cr βγ1 βγ2 as the 1 1 2  α2  1 1 lower bound of Eq βγ2 . A more specific critical point, cr 0 ↔ 1 , is designed because 1 1 2 1 1 it may be an upper or lower bound to the equilibrium patterns Eq 0 or Eq 1 . 1. 1. When we see the relation shown below:. Eq. we understand that Eq.  α2  β2 γ2.  α1 . cr. β1 γ1. becomes Eq. α1 β1 γ1.  α2  β2 γ2. Eq.  α2  β2 , γ 2.  α1  β1 γ1. after passing through cr. α1 β1 γ1.  α2  β2 γ2. . Therefore,. all equilibrium patterns in Problem I and Problem II are connected in section 2.2 and 2.3..

(17) 2. The Braess Problem and the Equilibrium Solutions. 17. 2.2 The Equilibrium Solution and its Constraints in Problem I We have classified three different equilibrium patterns in Problem I and seven equilibrium patterns in Problem II. In this section, we list both the inequality and equality constraints for each equilibrium state. Let us assume a linear travel cost on each link fi (ui ) = ai ui + bi ,. ai > 0, bi > 0,. i = 1, 2, 3, 4, 5. and let the total flow be d. Under specified constraints, we can find its corresponding equilibrium patterns , as well as the combination of inequalities - as such, we can also obtain the critical points - for which the equilibrium patterns is based upon. The notation α1  α2  cr βγ1 βγ2 was introduced in section 2.1, and the values of these critical points are given 1. 2. in Proposition 2.2.1 - 2.3.7, which will be further explained in section 2.2 and 2.3. Proposition 2.2.1. Equilibrium condition of the pattern Eq. 1 − : 0. If only route U p is utilized, the constraints of xα and xβ would be:   xα = d  . xβ = 0   I I TU (xα , xβ ) 6 TD (xα , xβ ). The equilibrium solution of Eq. 1 1 − is (xα , xβ ) = (d, 0), and the existence of Eq − satisfies 0. 0. the inequality constraint: d6. (b3 + b4 ) − (b1 + b2 ) , a1 + a2. if (b3 + b4 ) − (b1 + b2 ) > 0.. 1 which is an upper bound of d if Eq − exists. 0 1 We show the interval of pattern Eq − lying on a number line with total flow increasing 0 1 from zero to infinity, if Eq − exists. We denote cr. 1  1  − − 0 1. =. (b3 +b4 )−(b1 +b2 ) a1 +a2. 0. Eq. 1 − 0. -. 0. cr. 1 − 0. . 1 − 1. . d.

(18) 2. The Braess Problem and the Equilibrium Solutions. 18. Proof. TUI (xα , xβ ) = f1 (d) + f2 (d) = (a1 + a2 )d + (b1 + b2 ) TDI (xα , xβ ) = f3 (0) + f4 (0) = (b3 + b4 ) ∵ TUI (xα , xβ ) 6 TDI (xα , xβ ) ⇒ (a1 + a2 )d + (b1 + b2 ) 6 (b3 + b4 ) (b3 + b4 ) − (b1 + b2 ) , if (b3 + b4 ) − (b1 + b2 ) > 0. ⇒d6 a1 + a2 Proposition 2.2.2. Equilibrium condition of the pattern Eq. 0 − : 1. If only Down is utilized, the constraints of xα and xβ would be:   xα = 0  . xβ = d   I I TD (xα , xβ ) 6. The equilibrium solution of Eq. TU (xα , xβ ). 0 0 − is (xα , xβ ) = (0, d), and the existence of Eq − satisfies 1. 1. the inequality constraint: d6. (b1 + b2 ) − (b3 + b4 ) , a3 + a4. if (b1 + b2 ) − (b3 + b4 ) > 0.. 0 which is an upper bound of d if Eq − exists. 1 0 We show the interval of pattern Eq − lying on a number line with total flow increasing 1 0 from zero to infinity, if Eq − exists. We denote cr. 0  1  (b +b )−(b3 +b4 ) − − = 1 2 a3 +a4 1 1. 1. Eq. 0 − 1. -. 0. cr. 0  1  − − 1 1. Proof. TDI (xα , xβ ) = f3 (d) + f4 (d) = (a3 + a4 )d + (b3 + b4 ) TUI (xα , xβ ) = f1 (0) + f2 (0) = (b1 + b2 ) ∵ TDI (xα , xβ ) 6 TUI (xα , xβ ) ⇒ (a3 + a4 )d + (b3 + b4 ) 6 (b1 + b2 ) (b1 + b2 ) − (b3 + b4 ) , if (b1 + b2 ) − (b3 + b4 ) > 0. ⇒d6 a3 + a4. d.

(19) 2. The Braess Problem and the Equilibrium Solutions. Proposition 2.2.3. Equilibrium condition of the pattern Eq. 19. 1 − : 1. If both U p and Down are utilized, the constraints of xα and xβ would be:   xα + xβ = d     T I (x , x ) = T I (x , x ) α β β U D α .  xα > 0     x > 0 β.   1 1 +b2 )+(b3 +b4 ) (a1 +a2 )d−(b3 +b4 )+(b1 +b2 ) The equilibrium solution of Eq − is (xα , xβ ) = (a3 +a4a)d−(b , , a1 +a2 +a3 +a4 1 +a2 +a3 +a4 1 1 and the existence of Eq − satisfies the inequality constraint: 1 o n 0  1  1  1  d > max cr − − , cr − − . 1. 0. We show the interval of pattern Eq 1 from zero to infinity, if Eq − exists.. 1 − 1. 1. 1. lying on a number line with total flow increasing. 1. Eq. 1 − 1. -. 0. cr. 1  1  − − 0 1. or cr. d. 0  1  − − 1 1. Proof. TUI (xα , xβ ) = f1 (xα ) + f2 (xα ) = (a1 + a2 )xα + (b1 + b2 ) (xα , xβ ) = f3 (xβ ) + f4 (xβ ) = (a3 + a4 )xβ + (b3 + b4 ) TDI( ∵. xα + xβ. = d. I (xα , xβ ) = TD. TUI (xα , xβ ). ⇒ (xα , xβ ) = ( and ⇒d> ⇒d>. . (a3 +a4 )d−(b1 +b2 )+(b3 +b4 ) (a1 +a2 )d−(b3 +b4 )+(b1 +b2 ) , a1 +a2 +a3 +a4 a1 +a2 +a3 +a4. xα. > 0. xβ. > 0. 0  1  (b1 +b2 )−(b3 +b4 ) − − and = cr a3 +a4 1 1 n o 1  1  0  1  max cr − − , cr − − . 0 1 1 1. d>. (b3 +b4 )−(b1 +b2 ) a1 +a2. . = cr. 1  1  − − 0 1. Example 2.2.1. Given the travel time on each link as following: f1 (u) = 8u + 10, f2 (u) = 2u + 40, f3 (u) = 4u + 80, f4 (u) = 6u + 30. Through calculation, we obtain the critical points cr. 1  1  0 0 = 0 1. 6 ,cr. 0  1  0 0 = 1 1. −6..

(20) 2. The Braess Problem and the Equilibrium Solutions. n 1  1  i) If the total flow d = 10 > 6 = max cr 0 0 , cr 0 1 1 − Eq . Solving the simultaneous equations: 1 (. 0  1  0 0 1 1. xα + xβ. =. 10. 10xα + 50. =. 10xβ + 110. 20. o , it satisfies the pattern. we find the equilibrium solutions are (xα , xβ ) = (8, 2), and the travel time on route U p and route Down is TUI = TDI = 130. ii) If the total flow d = 4 6 6 = cr. 1  1  0 0 , 0 1. it satisfies the pattern Eq. 1 − . The 0. equilibrium solution is (xα , xβ ) = (4, 0), and the travel time on route α is TUI = 90 6 110 = TDI . In Example 2.2.1, if we exchange the travel cost f1 and f4 , and exchange the travel 0 cost f2 and f3 , then Eq − occurs under the total flow d 6 6, where the new critical point 1 0  1  0 0 cr = 6. 1. 1. 2.3 The Equilibrium Solution and its Constraints in Problem II In Problem II, assume that the total flow is d and the non-negative flow distribution (yα , yβ , yγ ) for the three-route satisfies yα + yβ + yγ = d, then the flow distribution the II travel time needed for U p,Down and M iddle are stated as follows, where TUII , TDII and TM. denotes the travel costs on U p,Down and M iddle, respectively. TUII (yα , yβ , yγ ) = f1 (yα + yγ ) + f2 (yα ). TDII (yα , yβ , yγ ) = f3 (yβ ) + f4 (yβ + yγ ). II TM (yα , yβ , yγ ) = f1 (yα + yγ ) + f5 (yγ ) + f4 (yβ + yγ ).. When the degenerate equilibrium occurs, some routes may not utilized and we need to compare the travel time on utilized routes with the expect travel time on unutilized 1 routes. Take Eq 0 for example, since only route U p is utilized, the flow distribution is 0. (yα , yβ , yγ ) = (d, 0, 0) and TUII (d, 0, 0) = f1 (d + 0) + f2 (d), TDII (d, 0, 0) = f3 (0) + f4 (0 + 0) II and TM (d, 0, 0) = f1 (d + 0) + f5 (0) + f4 (0 + 0). However, is it reasonable that the existence.

(21) 2. The Braess Problem and the Equilibrium Solutions. ( 1 of equilibrium pattern Eq 0 satisfies the inequality 0. TUII (d, 0, 0). II 6 TD (d, 0, 0). TUII (d, 0, 0). II 6 TM (d, 0, 0). 21. and the. critical points are determined by solving the inequalities? Suppose the amount 1 and 2 of the total flow moves from route U p to Down and M iddle, respectively, with 1 , 2 > 0. The travel time on routes U p are reduced to TUII (d − 1 − 2 , 0 + 1 , 0 + 2 ). = f1 (d − 1 − 2 + 2 ) + f2 (d − 1 − 2 ) =. (a1 + a2 )d + (b1 + b2 ) − (a1 + a2 )1 − a2 2 ,. and the travel time on routes Down and M iddle are II TD (d − 1 − 2 , 0 + 1 , 0 + 2 ). = f3 (1 ) + f4 (1 + 2 ) =. II (d − 1 − 2 , 0 + 1 , 0 + 2 ) TM. (a3 + a4 )1 + a4 2 + (b3 + b4 ),. = f1 (d − 1 − 2 + 2 ) + f5 (2 ) + f4 (1 + 2 ) = a1 d + (b1 + b5 + b4 ) − (a1 − a4 )1 + (a5 + a4 )2 .. II − TUII = −a2 d − (b2 − b4 − b5 ) + (a2 + a4 )1 + (a4 + a5 )2 > −a2 d − (b2 − b4 − b5 ). Since TM. i) If −a2 d − (b2 − b4 − b5 ) 6 0, route M iddle and U p will reach the equilibrium in some non-negative 1 , 2 . ii) If −a2 d − (b2 − b4 − b5 ) > 0, the travel time on route M iddle always longer than that on U p even if 1 , 2 −→ 0. Therefore, whether the route M iddle will be utilized or not depends on −a2 d−(b2 −b4 −b5 ) and d =. −(b2 −b4 −b5 ) a2. is one critical value of total flow d to determine the above conditions.. All the classification and all the constraints are under such methods, see Proposition 2.3.1 - 2.3.7. Proposition 2.3.1. Equilibrium condition of the pattern Eq. 1 0 : 0. If only U p is utilized, the constraints of yα , yβ and yγ would be:   yα = d      yβ = 0  . yγ = 0   II II  TU (yα , yβ , yγ ) 6 TM (yα , yβ , yγ )     II II TU (yα , yβ , yγ ) 6 TD (yα , yβ , yγ ).

(22) 2. The Braess Problem and the Equilibrium Solutions. The equilibrium solution of Eq. 1 1 0 is (yα , yβ , yγ ) = (d, 0, 0), and the existence of Eq 0 satisfies 0. 0. the inequality constraint: n 1  1  d 6 min cr 0 0 , cr 0. 22. 1. 1  1  0 1 0 0. o. n , if min cr. o. 1  1  1  1  0 0 , cr 0 1 0 1 0 0. > 0,. 1  1  1  1  1 +b2 ) − . − and cr 0 0 = −(b3 +ba41 )−(b =cr +a2 0 1 1 0 1 We show the interval of pattern Eq 0 lying on a number line with total flow increasing 0 1 from zero to infinity, if Eq 0 exists.. where we denote cr. 1  1  −(b2 −b4 −b5 ) 0 1 = a2 0 0. 0. Eq. 1 0 0. -. 0. cr. 1  1  0 0. 0 1. or cr. d. 1  1  0 0. 1 0. Proof. TUII (yα , yβ , yγ ) = f1 (d) + f2 (d) = (a1 + a2 )d + (b1 + b2 ) TDII (yα , yβ , yγ ) = f3 (0) + f4 (0) = (b3 + b4 ) II TM (yα , yβ , yγ ) = f1 (d) + f5 (0) + f4 (0) = a1 d + (b1 + b5 + b4 ) II (yα , yβ , yγ ) ∵ TUII (yα , yβ , yγ ) 6 TM. ⇒ (a1 + a2 )d + (b1 + b2 ) 6 a1 d + (b1 + b5 + b4 ) 1  1  4 −b5 ) 0 1 =cr ⇒ d 6 −(b2 −b a2 ∵. TUII (yα , yβ , yγ ). 6. 0 0 II TD (yα , yβ , yγ ). ⇒ (a1 + a2 )d + (b1 + b2 ) 6 (b3 + b4 ) 1  1  1 +b2 ) =cr 0 0 ⇒ d 6 (b3 +ba41)−(b +a2 0 1 n o n 1  1  1  1  Therefore, d 6 min cr 0 0 , cr 0 1 , if min cr 0. 1. 0. 0. 1  1  1  1  0 0 , cr 0 1 0 1 0 0. o. > 0.. Example 2.3.1. Given the travel time on each link as following: f1 (u) = 10u + 150, f2 (u) = 5u + 200, f3 (u) = 5u + 450, f4 (u) = 10u + 250, f5 (u) = u + 50. 1  1  1  1  Calculating the critical points, we obtain cr 0 0 = 40 , cr 0 1 = 20. 3 0 1 0 0 o n 1 1  1  1  1  If the total flow d = 10 < min cr 0 0 , cr 0 1 , it satisfies Eq 0 condition. The 0. 1. 0. 0. 0. equilibrium solution is (yα , yβ , yγ ) = (10, 0, 0), and the travel time on U p is TUII = 500 < . II min TDII , TM = min {700, 550}..

(23) 2. The Braess Problem and the Equilibrium Solutions. Proposition 2.3.2. Equilibrium condition of the pattern Eq. 23. 0 0 : 1. If only Down is utilized, the constraints of yα , yβ and yγ would be:   yα = 0      yβ = d  . yγ = 0   II II   TD (yα , yβ , yγ ) 6 TM (yα , yβ , yγ )    II II TD (yα , yβ , yγ ) 6 TU (yα , yβ , yγ ). The equilibrium solution of Eq. 0 0 0 is (yα , yβ , yγ ) = (0, d, 0), and the existence of Eq 0 satisfies 1. 1. the inequality constraint: n 0  1  d 6 min cr 0 0 , cr 1. 1. 0  0  0 1 1 1. o. n , if min cr. o. 0  1  0  0  0 0 , cr 0 1 1 1 1 1. > 0,. 0  1  0  1  3 +b4 ) − − . =cr and cr 0 0 = −(b1 +ba23 )−(b +a4 1 1 1 1 0 We show the interval of pattern Eq 0 lying on a number line with total flow increasing 1 0 from zero to infinity, if Eq 0 exists.. where we denote cr. 0  0  −(b3 −b1 −b5 ) 0 1 = a3 1 1. 1. Eq. 0 0 1. d. -. 0. cr. 0  1  0 0 1 1. or cr. 0  0  1 0 1 1. Proof. Similarly to the proof of Proposition2.3.1. If we exchange the travel cost f1 and f4 , and exchange the travel cost f2 and f3 from 0 Example 2.3.1. And let the total flow d = 10, than Eq 0 occurs. 1. Proposition 2.3.3. Equilibrium condition of the pattern Eq. 1 0 : 1. If U p and route Down are utilized, but M iddle is not, the constraints of yα , yβ and yγ would be:.               . yα + yβ. =. d. yγ. =. 0. TUII (yα , yβ , yγ ). =. II II TD (yα , yβ , yγ ) 6 TM (yα , yβ , yγ ). yα. >. 0. yβ. The equilibrium solution of Eq. > 0. 0 0 1. is. ..

(24) 2. The Braess Problem and the Equilibrium Solutions. (yα , yβ , yγ ) = And the existence of Eq. . 24. . (a3 +a4 )d−(b1 +b2 )+(b3 +b4 ) (a1 +a2 )d−(b3 +b4 )+(b1 +b2 ) , ,0 a1 +a2 +a3 +a4 a1 +a2 +a3 +a4. .. 1 0 satisfies the inequality constraint: 1. n d > max cr We show the interval of pattern Eq 1 from zero to infinity, if Eq 0 exists.. 1 0 1. 1  1  0  1  0 0 , cr 0 0 0 1 1 1. o. .. lying on a number line with total flow increasing. 1. Eq. 1 0 1. -. 0. cr. 1  1  0 0. 0 1. or cr. d. 0  1  0 1. 0 1. 1 Proof. Since M iddle is not utilized in Eq 0 , the equilibrium solution and the equilibrium 1 1 constraints are equivalent to Eq − in Proposition 2.2.3. Therefore, 1   1 +b2 )+(b3 +b4 ) (a1 +a2 )d−(b3 +b4 )+(b1 +b2 ) , , 0 (yα , yβ , yγ ) = (a3 +a4a)d−(b a1 +a2 +a3 +a4 1 +a2 +a3 +a4 n o 1  1  0  1  and d > max cr 0 0 , cr 0 0 . 0. 1. 1. 1. On the other hand, when we simplify the inequality constraint II TUII (yα , yβ , yγ ) = TDII (yα , yβ , yγ ) 6 TM (yα , yβ , yγ ),. it follows that (a1 a4 − a2 a3 )d > (a2 + a4 )(b3 − b1 − b5 ) + (a1 + a3 )(b2 − b4 − b5 ). This result means yγ 6 0, and we compare the inequalities to the condition yγ > 0 in 1 Eq 1 later. To fit the constraints yγ 6 0 , the total flow d is imposed by 1.  1 1  d > cr 0 ↔ 1 if   1 1    1 1 d 6 cr 0 ↔ 1 if 1 1       d 6 (a1 +a3 )(b2 −b4 −b5 )+(a2 +a4 )(b3 −b1 −b5 ) if (a1 +a3 )(a2 +a4 )+a5 (a1 +a2 +a3 +a4 ) where we denote cr. 1 0 1. ↔. 1 1 1. =. a1 a4 − a2 a3 > 0 a1 a4 − a2 a3 < 0 , a1 a4 − a2 a3 = 0. (a1 +a3 )(b2 −b4 −b5 )+(a2 +a4 )(b3 −b1 −b5 ) . a1 a4 −a2 a3.

(25) 2. The Braess Problem and the Equilibrium Solutions. 25. Example 2.3.2. Given the travel time on each link as following: f1 (u) = 10u + 100, f2 (u) = u + 500, f3 (u) = u + 400, f4 (u) = 10u + 300, f5 (u) = u. If the total flow d = 60, it satisfies the pattern Eq. 1 0 , so we solve the simultaneous 1. equations:     . yα + yβ. =. 60. yγ. =. 0. 11yα + 600. =. 11yβ + 700. .. 380 280 , ,0 11 11.  , and the travel time on U p and Down. The equilibrium solution is (yα , yβ , yγ ) = II is TUII = TDII = 980 < 1000 = TM .. Proposition 2.3.4. Equilibrium condition of the pattern Eq. 0 1 : 0. If only M iddle is utilized, the constraints of yα , yβ and yγ would be:  yα = 0      yβ = 0   . yγ = d   II  TM (yα , yβ , yγ ) 6 TUII (yα , yβ , yγ )     II II TM (yα , yβ , yγ ) 6 TD (yα , yβ , yγ ). The equilibrium solution of Eq. 0 1 0. is (yα , yβ , yγ ) = (0, 0, d), and the existence of Eq. 0 1 0. satisfies. the inequality constraint: n d 6 min cr. 0  1  0  0  1 1 , cr 1 1 0 0 0 1. o. n , if min cr. o. 0  1  0  0  1 1 , cr 1 1 0 0 0 1. > 0,. 0  0  1 −b5 and cr 1 1 = b3a−b . 1 +a5 0 1 0 We show the interval of pattern Eq 1 lying on a number line with total flow increasing 0 0 from zero to infinity, if Eq 1 exists.. where we denote cr. 0  1  b −b −b 1 1 = 2 4 5 a4 +a5 0 0. 0. Eq. 0 1 0. -. 0. cr. 0  1  1 1 0 0. or cr. 0  0  1 1 0 1. d.

(26) 2. The Braess Problem and the Equilibrium Solutions. 26. Proof. TUII (yα , yβ , yγ ) = f1 (d) + f2 (0) = a1 d + (b1 + b2 ) TDII (yα , yβ , yγ ) = f3 (0) + f4 (d) = a4 d + (b3 + b4 ) II TM (yα , yβ , yγ ) = f1 (d) + f5 (d) + f4 (d) = (a1 + a5 + a4 )d + (b1 + b5 + b4 ) II ∵ TM (yα , yβ , yγ ) 6 TUII (yα , yβ , yγ ). ⇒ (a1 + a5 + a4 )d + (b1 + b5 + b4 ) 6 a1 d + (b1 + b2 ) (b2 − b4 − b5 ) 0  1  = cr 1 1 , if (b2 − b4 − b5 ) > 0. ⇒d6 0 0 a4 + a5 II II ∵ TM (yα , yβ , yγ ) 6 TD (yα , yβ , yγ ) ⇒ (a1 + a5 + a4 )d + (b1 + b5 + b4 ) 6 a4 d + (b3 + b4 ) (b3 − b1 − b5 ) 0  0  = cr 1 1 , if (b3 − b1 − b5 ) > 0. ⇒d6 0 1 a1 + a5 n o n 0  1  0  0  0  1  Therefore, d 6 min cr 1 1 , cr 1 1 , if min cr 1 1 , cr 0. 0. 0. 1. 0. 0. 0  0  1 1 0 1. o. > 0.. Example 2.3.3. Given the travel time on each link as following: f1 (u) = 10u + 100, f2 (u) = u + 500, f3 (u) = u + 400, f4 (u) = 10u + 300, f5 (u) = u. 0 1 , and the equilibrium solution is 0 . II (yα , yβ , yγ ) = (0, 0, 10), where travel time on M iddle is TM = 610 < min TUII , TDII = If the total flow d = 10, it satisfies the pattern Eq. min {700, 800}. Proposition 2.3.5. Equilibrium condition of the pattern Eq. 1 1 : 0. If U p and M iddle are utilized, but Down is not, the constraints of yα , yβ and yγ would be:.        . yα + yγ. = d. yβ. =. TUII (yα , yβ , yγ ).       . 0. II II = TM (yα , yβ , yγ ) 6 TD (yα , yβ , yγ ). yα. > 0. yγ. > 0. ..   1 a2 d+(b2 −b4 −b5 ) 2 −b4 −b5 ) , 0, , The equilibrium solution of Eq 1 is (yα , yβ , yγ ) = (a4 +aa5 )d−(b a2 +a4 +a5 2 +a4 +a5 0 1 and the existence of Eq 1 satisfies the inequality constraint: 0. n. max cr. 1  1  0  1  0 1 , cr 1 1 0 0 0 0. o. < d 6 cr. 1  1  1 1 , 0 1.

(27) 2. The Braess Problem and the Equilibrium Solutions 1  1  (a2 +a4 +a5 )(b3 −b1 −b5 )−a5 (b2 −b4 −b5 ) 1 1 = . a2 (a1 +a5 )+a1 (a4 +a5 ) 1 0   1 1 Note that the inequality d 6 cr 1 1 reflects that yβ 6 0 1 1 inequality to the condition yβ > 0 in Eq 1 later.  1 1 We show the interval of pattern Eq 1 lying on a number 0 1 1 from zero to infinity, if Eq exists. 0. 27. where we denote cr. Eq. 0, and we can compare the. line with total flow increasing. 1 1 0. -. 0. cr. 1  1  0 0. 1 0. or cr. 0  1  1 0. cr. 1 0. d. 1  1  1 0. 1 1. Proof. TUII (yα , yβ , yγ ) = f1 (yα + yγ ) + f2 (yα ) = a1 d + a2 yα + (b1 + b2 ) TDII (yα , yβ , yγ ) = f3 (0) + f4 (yγ ) = a4 yγ + (b3 + b4 ) II TM (yα , yβ , yγ ) =(f1 (yα + yγ ) + f5 (yγ ) + f4 (yγ ) = a1 d + (a5 + a4 )yγ + (b1 + b5 + b4 ). yα + yγ. = d. is The solution of II TUII (yα , yβ , yγ ) = TM (yα , yβ , yγ )   a2 d+(b2 −b4 −b5 ) 2 −b4 −b5 ) (yα , yβ , yγ ) = (a4 +aa5 )d−(b , 0, . a2 +a4 +a5 2 +a4 +a5 ( ∵. yα. > 0. yγ. > 0. ,. 0  1  = cr 1 1 and d > 0 0 n o 1  1  0  1  ⇒ d > max cr 0 1 , cr 1 1. ⇒d>. (b2 −b4 −b5 ) a4 +a5. 0. ∵. II T( U (yα , yβ , yγ ). ⇒. =. 0 0 0 II TM (yα , yβ , yγ ). a1 d + a2 yα + (b1 + b2 ) 6 a4 yγ + (b3 + b4 ). ⇒d6. 6. −(b2 −b4 −b5 ) a2. = cr. 1  1  0 1 . 0 0. 6 TDII (yα , yβ , yγ ). a1 d + (a5 + a4 )yγ + (b1 + b5 + b4 ). ,. a1 d + (a5 + a4 )yγ + (b1 + b5 + b4 ). (a2 +a4 +a5 )(b3 −b1 −b5 )−a5 (b2 −b4 −b5 ) a2 (a1 +a5 )+a1 (a4 +a5 ). = cr. 1  1  1 1 . 0 1. Therefore, n max cr if. 1  1  0  1  0 1 , cr 1 1 0 0 0 0. o. < d 6 cr n o 1  1  0  1  max cr 0 1 , cr 1 1 < cr 0. 0. 0. 0. 1  1  1 1 , 0 1 1  1  1 1 . 1 0.

(28) 2. The Braess Problem and the Equilibrium Solutions. 28. Example 2.3.4. Given the travel time on each link as following: f1 (u) = 10u + 100, f2 (u) = u + 500, f3 (u) = u + 400, f4 (u) = 10u + 300, f5 (u) = u. If the total flow d = 20, it satisfies the pattern Eq. 1 1 0. , and we solve the simultaneous. equations:     . yα + yγ. =. 20. yβ. =. 0. 11yα + 10yγ + 600. =. 10yα + 21yγ + 400. The equilibrium solution is (yα , yβ , yγ ) = II is TUII = TM =. 2405 3. <. 2650 3. 5 , 0, 55 3 3. ..  , and the travel time on U p and M iddle. = TDII .. Proposition 2.3.6. Equilibrium condition of the pattern Eq. 0 1 : 1. If Down and M iddle are utilized, but U p is not, the constraints of yα , yβ and yγ would be:.               . yα. =. 0. yβ + yγ. =. d. II TD (yα , yβ , yγ ). =. II TM (yα , yβ , yγ ) 6 TUII (yα , yβ , yγ ). yβ. >. 0. yγ. >. 0. ..   0 (a1 +a5 )d−(b3 −b1 −b5 ) a3 d+(b3 −b1 −b5 ) , a3 +a1 +a5 , The equilibrium solution of Eq 1 is (yα , yβ , yγ ) = 0, a3 +a1 +a5 1 0 and the existence of Eq 1 satisfies the inequality constraint: 1 n o 0  0  0  0  0  1  max cr 0 1 , cr 1 1 < d 6 cr 1 1 , 1. 1. 0. 1. 1. 1. 0  1  5 )(b2 −b4 −b5 )−a5 (b3 −b1 −b5 ) . where we denote cr 1 1 = (a3 +a1 +a a3 (a4 +a5 )+a4 (a1 +a5 ) 1 1   0 1 Note that the inequality d 6 cr 1 1 shows that yα 6 0, we can compare the inequality 1 1 1 to the condition yα > 0 in Eq 1 later. 1 0 We show the interval of pattern Eq 1 lying on a number line with total flow increasing 1 0 from zero to infinity, if Eq 1 exists. 1. Proof. Similarly to the proof of Proposition 2.3.5. If we exchange the travel cost f1 and f4 , and exchange the travel cost f2 and f3 from 0 Example 2.3.4. Let the total flow d = 20, then Eq 1 will occur. 1.

(29) 2. The Braess Problem and the Equilibrium Solutions. Eq. 0 1 1. -. 0. cr. 0  0  0 1 1 1. or cr. 29. 0  0  1 1 0 1. cr. d. 0  1  1 1 1 1. Proposition 2.3.7. Equilibrium condition of the pattern Eq. 1 1 : 1. If all three routes U p, Down and M iddle are utilized, the constraints of yα , yβ and yγ would be:.   yα + yβ + yγ      TUII (yα , yβ , yγ )     T II (y , y , y ) α β γ D  yα      yβ     yγ. The equilibrium solution of Eq   yα =      yβ =       yγ = and the existence of Eq. 1 1 1. = d II = TM (yα , yβ , yγ ) II = TM (yα , yβ , yγ ). .. > 0 > 0 > 0. is. [a3 (a4 +a5 )+a4 (a1 +a5 )]d−(a1 +a3 +a5 )(b2 −b4 −b5 )+a5 (b3 −b1 −b5 ) (a1 +a3 )(a2 +a4 )+a5 (a1 +a2 +a3 +a4 ) [a2 (a1 +a5 )+a1 (a4 +a5 )]d−(a2 +a4 +a5 )(b3 −b1 −b5 )+a5 (b2 −b4 −b5 ) (a1 +a3 )(a2 +a4 )+a5 (a1 +a2 +a3 +a4 ). ,. [−(a1 a4 −a2 a3 )]d+(a1 +a3 )(b2 −b4 −b5 )+(a2 +a4 )(b3 −b1 −b5 ) (a1 +a3 )(a2 +a4 )+a5 (a1 +a2 +a3 +a4 ). 1 1 1. satisfies the above mentioned inequality constraints. Therefore,. in brief, we have proven that: i) To fit the constraints yα > 0, yβ > 0 , the total flow d is imposed by n d > max cr. o. 1  1  0  1  1 1 , cr 1 1 1 0 1 1. .. ii) To fit the constraints yγ > 0 , the total flow d is imposed by  1 1  0 ↔ 1 d < cr if   1 1    1 1 d > cr 0 ↔ 1 if 1 1       3 )(b2 −b4 −b5 )+(a2 +a4 )(b3 −b1 −b5 ) d > (a1(a+a > 0 if 1 +a3 )(a2 +a4 )+a5 (a1 +a2 +a3 +a4 ). a1 a4 − a2 a3 > 0 a1 a4 − a2 a3 < 0 . a1 a4 − a2 a3 = 0.

(30) 2. The Braess Problem and the Equilibrium Solutions. Eq 0. cr. 1  1  1 1 0 1. or cr. 1 1 1. 0  1  1 1 1 1. cr. 1 0 1. Eq 0. cr. 1  1  0 0. 1 0. or cr. 0  1  1 0. We show the interval of pattern Eq 1 from zero to infinity, if Eq 1 exists.. 1 0. or cr. 1 1 1. 30. ↔. -. d. -. d. 1 1 1. 1 1 1. 1  1  1 0. 1 1. lying on a number line with total flow increasing. 1. Proof. TUII (yα , yβ , yγ ) = f1 (yα + yγ ) + f2 (yα ) = a1 (yα + yγ ) + a2 yα + (b1 + b2 ) TDII (yα , yβ , yγ ) = f3 (yβ ) + f4 (yβ + yγ ) = a3 yβ + a4 (yβ + yγ ) + (b3 + b4 ) II TM (yα , yβ , yγ ) = f1 (yα + yγ ) + f5 (yγ ) + f4 (yγ ). = a1 (yα + yγ ) + a5 yγ + a4 (yβ + yγ ) + (b1 + b5 + b4 )    yα + yβ + yγ = d II The solution of (yα , yβ , yγ ) is TUII (yα , yβ , yγ ) = TM   II II TD (yα , yβ , yγ ) = TM (yα , yβ , yγ )  [a3 (a4 +a5 )+a4 (a1 +a5 )]d−(a1 +a3 +a5 )(b2 −b4 −b5 )+a5 (b3 −b1 −b5 )    yα = (a1 +a3 )(a2 +a4 )+a5 (a1 +a2 +a3 +a4 )    (a4 +a5 )]d−(a2 +a4 +a5 )(b3 −b1 −b5 )+a5 (b2 −b4 −b5 ) , yβ = [a2 (a1 +a5 )+a1(a 1 +a3 )(a2 +a4 )+a5 (a1 +a2 +a3 +a4 )       y = [−(a1 a4 −a2 a3 )]d+(a1 +a3 )(b2 −b4 −b5 )+(a2 +a4 )(b3 −b1 −b5 ) γ (a1 +a3 )(a2 +a4 )+a5 (a1 +a2 +a3 +a4 ) ( ∵. yα. > 0. yβ. > 0. ⇒d> ⇒d>. ,. (a1 +a3 +a5 )(b2 −b4 −b5 )−a5 (b3 −b1 −b5 ) 5 )+a4 (a1 +a5 ) n a3 (a4 +a o 0  1  1  1  max cr 1 1 , cr 1 1 1 1 0 1. and d >. (a2 +a4 +a5 )(b3 −b1 −b5 )−a5 (b2 −b4 −b5 ) . a2 (a1 +a5 )+a1 (a4 +a5 ). ∵ yα > 0, ⇒ [−(a1 a4 − a2 a3 )]d + (a1 + a3 )(b2 − b4 − b5 ) + (a2 + a4 )(b3 − b1 − b5 ) > 0. 1 1 5 )+(a2 +a4 )(b3 −b1 −b5 ) Denote cr 0 ↔ 1 = (a1 +a3 )(b2 −b4 −b , a1 a4 −a2 a3 1. 1.

(31) 2. The Braess Problem and the Equilibrium Solutions.  1 1  0 ↔ 1 d < cr if   1 1    1 1 ⇒ d > cr 0 ↔ 1 if 1 1       d > (a1 +a3 )(b2 −b4 −b5 )+(a2 +a4 )(b3 −b1 −b5 ) > 0 if (a1 +a3 )(a2 +a4 )+a5 (a1 +a2 +a3 +a4 ). 31. a1 a4 − a2 a3 > 0 a1 a4 − a2 a3 < 0 . a1 a4 − a2 a3 = 0. Example 2.3.5. Given the travel time on each link as following: f1 (u) = 10u + 100, f2 (u) = u + 500, f3 (u) = u + 400, f4 (u) = 10u + 300, f5 (u) = u. If the total flow d = 40, it satisfies the pattern Eq. 1 1 1. , and we solve the simultaneous. equations:     . yα + yβ + yγ. =. 40. 11yα + 10yγ + 600. =. 10yα + 10yβ + 21yγ + 400. 11yβ + 10yγ + 700. =. 10yα + 10yβ + 21yγ + 400. The equilibrium solution is (yα , yβ , yγ ) = II = and M iddle is TUII = TDII = TM. 2740 1440 1540 , , 143 143 143. ..  , and the travel time on U p,Down. 11940 . 13. Recall the notation of critical points which we denote in section 2.2 and 2.3 as follows:.

(32) 2. The Braess Problem and the Equilibrium Solutions. (b3 + b4 ) − (b1 + b2 ) . a1 + a2 (b1 + b2 ) − (b3 + b4 ) . a3 + a4 (b3 + b4 ) − (b1 + b2 ) . a1 + a2 (b1 + b2 ) − (b3 + b4 ) . a3 + a4 −(b2 − b4 − b5 ) . a2 −(b3 − b1 − b5 ) . a3 b2 − b4 − b5 . a4 + a5 b3 − b1 − b5 . a1 + a5 (a2 + a4 + a5 )(b3 − b1 − b5 ) − a5 (b2 − b4 − b5 ) . a2 (a1 + a5 ) + a1 (a4 + a5 ). cr. 1  1  − − 0 1. =. cr. 0  1  − − 1 1. =. cr. 1  1  0 0 0 1. =. cr. 0  1  0 0 1 1. =. cr. 1  1  0 1 0 0. =. cr. 0  0  0 1 1 1. =. cr. 0  1  1 1 0 0. =. cr. 0  0  1 1 0 1. =. cr. 1  1  1 1 0 1. =. cr. 0  1  1 1 1 1. =. (a3 + a1 + a5 )(b2 − b4 − b5 ) − a5 (b3 − b1 − b5 ) . a3 (a4 + a5 ) + a4 (a1 + a5 ). cr. 1 0 1. =. (a1 + a3 )(b2 − b4 − b5 ) + (a2 + a4 )(b3 − b1 − b5 ) . a1 a4 − a2 a3. ↔. 1 1 1. 32.

(33) 3. THE SEQUENCE OF EQUILIBRIUM PATTERNS DETERMINED BY THE DEMAND. 3.1 Classify the Equilibrium Patterns by Critical Points In chapter 2, we obtained a total of three equilibrium patterns in Problem I and seven equilibrium patterns in Problem II. When we fix the given linear travel cost, fi (ui ) = ai ui + bi , the relation between equilibrium patterns are determined by the total demand d. As such, when we are given the total demand d, we can find the equilibrium pattern, which can then be used to obtain the equilibrium solutions and the travel time can be found. The 1  1  0  1  1  1  equilibrium patterns are dependent on the critical points cr − − , cr − − , cr 0 0 , 0 0 1 1 1  11  1  1  0  1  0  0  0  1  0  0  1  1  0  1  1 cr 0 1 , cr 0 0 , cr 0 1 , cr 1 1 , cr 1 1 , cr 1 1 , cr 1 1 , cr 0 ↔ 1 . 0. 0. 1. 1. 1. 1. 0. 0. 0. 1. 0. 1. 1. 1. 1. 1. However, how should we manipulate the critical values in order to determine the sequence of equilibrium pattern interchange? First we denote our characteristic values as follows: Cα = (b3 − b1 − b5 ), Cβ = (b2 − b4 − b5 ), ∆. = a1 a4 − a2 a3 .. Sα = a4 Cα + a3 Cβ , Sβ = a2 Cα + a1 Cβ , θ. = (a1 + a5 )Cβ − (a4 + a5 )Cα.

(34) 3. The Sequence of Equilibrium Patterns Determined by the Demand. 34. Then, the value of critical points can be expressed briefly. Cα − Cβ . a1 + a2 Cβ − Cα . a3 + a4 Cα − Cβ . a1 + a2 Cβ − Cα . a3 + a4 −Cβ . a2 −Cα . a3 Cβ . a4 + a5 Cα . a1 + a5 (a2 + a4 + a5 )Cα − a5 Cβ . a2 (a1 + a5 ) + a1 (a4 + a5 ). cr. 1  1  − − 0 1. =. cr. 0  1  − − 1 1. =. cr. 1  1  0 0 0 1. =. cr. 0  1  0 0 1 1. =. cr. 1  1  0 1 0 0. =. cr. 0  0  0 1 1 1. =. cr. 0  1  1 1 0 0. =. cr. 0  0  1 1 0 1. =. cr. 1  1  1 1 0 1. =. cr. 0  1  1 1 1 1. =. (a3 + a1 + a5 )Cβ − a5 Cα . a3 (a4 + a5 ) + a4 (a1 + a5 ). cr. 1 0 1. =. (a1 + a3 )Cβ + (a2 + a4 )Cα . a1 a4 − a2 a3. ↔. 1 1 1. These characteristic values are the common factors from the difference of critical points, see Lemma 3.1.1 - 3.1.3. The existence conditions of the equilibrium patterns are discussed in Chapter 2. The propositions in Chapter 2 show that we need to compare the value of 0 0 critical points. Take Eq − for example, the existence of Eq − satisfies 1. d6 If Cβ < Cα , then d 6. Cβ −Cα a3 +a4. 1. (b1 + b2 ) − (b3 + b4 ) Cβ − Cα = . a3 + a4 a3 + a4 0 < 0, thus the pattern Eq − does not exist for any d > 0. 1. Therefore, Cβ − Cα is chosen to judge the existence of equilibrium patterns in Problem I, see section 3.2. In problem II, the method to judge the existence of equilibrium patterns is more complex. We subtract all the pairs of critical values and factorize the difference of each pair. The characteristic values ∆,Sα ,Sβ and θ are chosen by us to classify the.

(35) 3. The Sequence of Equilibrium Patterns Determined by the Demand. 35. equilibrium patterns in a suitable order, see Lemma 3.1.1 - 3.1.4. ∆ involves the or1 1 0  1  0  1  der of Eq 1 and Eq 0 ; Sα helps us to determine the order of cr 1 1 ,cr 0 0 , and 1 1 1 1 1 1 0  0  1  1  1  1  1  1  cr 0 1 ; Sβ helps us to determine the order of cr 1 1 ,cr 0 0 , and cr 0 1 ; θ is 1 1 0 1 0 1 0 0 0  1  0  1  0  0  1  1  involved in the order of cr 1 1 ,cr 1 1 ,cr 1 1 , and cr 1 1 . 1. 1. 0. 0. 0. 1. 0. 1. The following flowchart shows our classification which covers all the possible order of patterns in Problem II, see Figure 3.1.. Lemma 3.3.1. see Figure 3.2. 60. Lemma 3.3.2. (done). Sα 6 0, Sβ > 0. Lemma 3.3.3. (done). Lemma 3.3.4. see Figure 3.3. >0. Lemma 3.4.1. see Figure 3.2. 60 Sα > 0, Sβ. Lemma 3.4.2. (done). Sα 6 0, S β >0. Lemma 3.4.3. (done). Lemma 3.4.4. see Figure 3.3. >0. Lemma 3.5.2. see Figure 3.2. Sα + Sβ 6 0. Lemma 3.5.1. see Figure 3.3. Sα. , Sβ >0. S Sα > 0, β. >0. section 3.3. ∆. >. 0. Sα 6 0, S β 60. 0, S β Sα >. . ∆<0. section 3.4. Sα 6 0, S β. 60. ∆ = 0 Sα + S β section 3.5. Fig. 3.1: The Flowchart of the Classification in Problem II. Follow the flowchart, if the constraints (∆, Sα , Sβ ) are enough to remove the nonexis-.

(36) 3. The Sequence of Equilibrium Patterns Determined by the Demand. 36. tence patterns and specific equilibrium patterns appear in a sequence on the number line with the magnitude of the total flow from zero to infinity, we write down the sequence. If the constraints are not enough to determine the sequence, we continue our classification by θ, Cα and Cβ . The method for further classification in Lemma 3.3.1, 3.4.1 and 3.5.2 are the same and the classification will be done, see Figure 3.2. On the other hand, The method for further classification in Lemma 3.3.4, 3.4.4 and 3.5.1 are the same and the classification will be done, see Figure 3.3. Therefore, any one can base on our classification in Chapter 3 to determine the right pattern if the linear travel function on each link and the total flow are given. Certainly, the flow distribution and average travel time are taken. Cα. .. >0. Cα = 0. <0. (done). case ii). (done). case iii). (done). case iv). (done). case v). (done). case vi). (done). θ>. 0. Cα. case i). Cβ. .. Lemma 3.3.1. >0. Cβ = 0. θ<0 Cβ. <0. θ= 0 case vii). (done). Fig. 3.2: The Flowchart of the Classification in Lemma 3.3.1 or 3.4.1 or 3.5.2.

(37) 3. The Sequence of Equilibrium Patterns Determined by the Demand. Cβ. Lemma 3.3.4. −C. α. case i). (done). case ii). (done). case iii). (done). 37. >0. Cβ − Cα < 0. Cβ. −C. α=. 0. Fig. 3.3: The Flowchart of the Classification in Lemma 3.3.4 or 3.4.4 or 3.5.1. Lemma 3.1.1. The following identities holds: cr. 1 0 1. ↔. 1 1 1. − cr. 0  1  0 0 1 1. =. cr. 1 0 1. ↔. 1 1 1. − cr. 0  1  1 1 1 1. =. cr. 1 0 1. ↔. 1. − cr. 0  0  0 1 1 1. =. cr. 0  1  1 1 1 1. − cr. 0  1  0 0 1 1. =. cr. 0  1  1 1 1 1. − cr. 0  0  0 1 1 1. =. cr. 0  1  0 0 1 1. − cr. 0  0  0 1 1 1. =. 1 1. a1 + a2 + a3 + a4 Sα × . (3.1) a3 + a4 ∆ (a1 + a3 )(a2 + a4 ) + a5 (a1 + a2 + a3 + a4 ) Sα × .(3.2) (a3 (a4 + a5 ) + a4 (a1 + a5 )) ∆ a3 + a1 Sα × . (3.3) a3 ∆ a1 + a3 × Sα . (3.4) (a3 + a4 )(a3 (a4 + a5 ) + a4 (a1 + a5 )) a1 + a3 + a5 × Sα . (3.5) a3 (a3 (a4 + a5 ) + a4 (a1 + a5 )) 1 × Sα . (3.6) a3 (a3 + a4 ). Proof. i) cr. 1 0 1. ↔. 1 1 1. − cr. 0  1  0 0 1 1. = = = =. C −C (a1 +a3 )Cβ +(a2 +a4 )Cα − aβ3 +a4α a1 a4 −a2 a3 a4 (a1 +a2 +a3 +a4 )Cα +a3 (a1 +a2 +a3 +a4 )Cβ (a1 a4 −a2 a3 )(a3 +a4 ) (a1 +a2 +a3 +a4 )(a4 Cα +a3 Cβ ) (a1 a4 −a2 a3 )(a3 +a4 ) (a1 +a2 +a3 +a4 ) × S∆α . (a3 +a4 ). ii) cr. 1 0 1. ↔. 1 1 1. − cr. 0  1  1 1 1 1. = = =. (a1 +a3 )Cβ +(a2 +a4 )Cα (a +a1 +a5 )Cβ −a5 Cα − a3 3(a4 +a a1 a4 −a2 a3 5 )+a4 (a1 +a5 ) ((a1 +a3 )(a2 +a4 )+a5 (a1 +a2 +a3 +a4 ))(a4 Cα +a3 Cβ ) (a1 a4 −a2 a3 )(a3 (a4 +a5 )+a4 (a1 +a5 )) (a1 +a3 )(a2 +a4 )+a5 (a1 +a2 +a3 +a4 ) × S∆α . (a3 (a4 +a5 )+a4 (a1 +a5 )).

(38) 3. The Sequence of Equilibrium Patterns Determined by the Demand. 38. iii) cr. 1 0 1. ↔. 1 1 1. − cr. 0  0  0 1 1 1. = = =. (a1 +a3 )Cβ +(a2 +a4 )Cα a1 a4 −a2 a3 (a1 +a3 )(a4 Cα +a3 Cβ ) (a1 a4 −a2 a3 )a3 (a3 +a1 ) × S∆α . a3. −. −Cα a3. iv) cr. 0  1  1 1 1 1. − cr. 0  1  0 0 1 1. C −C (a3 +a1 +a5 )Cβ −a5 Cα − aβ3 +a4α a3 (a4 +a5 )+a4 (a1 +a5 ) (a1 +a3 )(a4 Cα +a3 Cβ ) (a3 (a4 +a5 )+a4 (a1 +a5 ))(a3 +a4 ) (a1 +a3 ) × (a3 (a4 +a5 )+a4 (a1 +a5 ))(a3 +a4 ). = = =. Sα .. v) cr. 0  1  1 1 1 1. − cr. 0  0  0 1 1 1. (a3 +a1 +a5 )Cβ −a5 Cα α − −C a3 (a4 +a5 )+a4 (a1 +a5 ) a3 (a3 +a1 +a5 )(a4 Cα +a3 Cβ ) (a3 (a4 +a5 )+a4 (a1 +a5 ))a3 (a3 +a1 +a5 ) × Sα . (a3 (a4 +a5 )+a4 (a1 +a5 ))a3. = = =. vi) cr. 0  1  0 0 1 1. − cr. 0  0  0 1 1 1. Cβ −Cα α − −C a3 +a4 a3 (a4 Cα +a3 Cβ ) (a3 +a4 )a3. = =. 1 (a3 +a4 )a3. =. × Sα .. Lemma 3.1.2. The following identities holds: cr. 1 0 1. ↔. 1 1 1. − cr. 1  1  0 0 0 1. =. cr. 1 0 1. ↔. 1 1 1. − cr. 1  1  1 1 0 1. =. cr. 1 0 1. ↔. 1. − cr. 1  1  0 1 0 0. =. cr. 1  1  1 1 0 1. − cr. 1  1  0 0 0 1. =. cr. 1  1  1 1 0 1. − cr. 1  1  0 1 0 0. =. cr. 1  1  0 0 0 1. − cr. 1  1  0 1 0 0. =. 1 1. a1 + a2 + a3 + a4 Sβ × . (3.7) a2 + a1 ∆ (a1 + a3 )(a2 + a4 ) + a5 (a1 + a2 + a3 + a4 ) Sβ × . (3.8) (a2 (a1 + a5 ) + a1 (a4 + a5 )) ∆ a2 + a4 S β × . (3.9) a2 ∆ a4 + a2 × Sβ . (3.10) (a2 + a1 )(a2 (a1 + a5 ) + a1 (a4 + a5 )) a4 + a2 + a5 × Sβ . (3.11) a2 (a2 (a1 + a5 ) + a1 (a4 + a5 )) 1 × Sβ . (3.12) a2 (a2 + a1 ).

(39) 3. The Sequence of Equilibrium Patterns Determined by the Demand. 39. Proof. Similarly to the proof of Lemma 3.1.1. By the difference in the value of critical points, in relation to Sβ and ∆, the identities holds. Lemma 3.1.3. The following identities holds: a5 × θ. (a4 + a5 )(a4 (a1 + a5 ) + a3 (a4 + a5 )) 1 0  1  0  0  cr 1 1 − cr 1 1 = × θ. 0 0 0 1 (a4 + a5 )(a1 + a5 ) a5 0  0  1  1  cr 1 1 − cr 1 1 = × θ. 0 1 0 1 (a1 + a5 )(a4 (a1 + a5 ) + a3 (a4 + a5 )). cr. 0  1  1 1 1 1. − cr. 0  1  1 1 0 0. =. Proof. i) cr. 0  1  1 1 1 1. − cr. 0  1  1 1 0 0. = =. Cβ (a3 +a1 +a5 )Cβ −a5 Cα − a4 +a a3 (a4 +a5 )+a4 (a1 +a5 ) 5 a5 ((a1 +a5 )Cβ −(a4 +a5 )Cα ) (a3 (a4 +a5 )+a4 (a1 +a5 ))(a4 +a5 ). =. a5 (a3 (a4 +a5 )+a4 (a1 +a5 ))(a4 +a5 ). =. Cβ α − a1C+a a4 +a5 5 (a1 +a5 )Cβ −(a4 +a5 )Cα (a4 +a5 )(a1 +a5 ). × θ.. ii) cr. 0  1  1 1 0 0. − cr. 0  0  1 1 0 1. = =. 1 (a4 +a5 )(a1 +a5 ). =. (a +a4 +a5 )Cα −a5 Cβ Cα − a2 2(a1 +a a1 +a5 5 )+a1 (a4 +a5 ) a5 ((a1 +a5 )Cβ −(a4 +a5 )Cα ) (a1 +a5 )(a2 (a1 +a5 )+a1 (a4 +a5 )). × θ.. iii) cr. 0  0  1 1 0 1. − cr. 1  1  1 1 0 1. = =. a5 (a1 +a5 )(a2 (a1 +a5 )+a1 (a4 +a5 )). × θ.. Lemma 3.1.4. i) cr. 0  1  1 1 1 1. > cr. 0  1  1 1 0 0. > cr. 0  0  1 1 0 1. > cr. 1  1  1 1 0 1. if and only if θ > 0.. ii) cr. 0  1  1 1 1 1. < cr. 0  1  1 1 0 0. < cr. 0  0  1 1 1 0. < cr. 1  1  1 1 0 1. if and only if θ < 0.. iii) cr. 0  1  1 1 1 1. = cr. 0  1  1 1 0 0. = cr. 0  0  1 1 0 1. = cr. 1  1  1 1 0 1. if and only if θ = 0.. (3.13) (3.14) (3.15).

(40) 3. The Sequence of Equilibrium Patterns Determined by the Demand. 40. Proof. By the identities in Lemma 3.1.4, the proportionality holds.. Through the comparison of the differences in critical points in relation to the six different characteristic values (Sα , Sβ , ∆, θ, Cα , Cβ ), we discover that the resulting expression for that of Lemma 3.1.1. and Lemma 3.1.2. show a remarkable similarity of form. In 1 1 addition, we find that ∆ exerts a direct influence on the direction of Eq 0 and Eq 1 (see 1. 1. (3.1),(3.2),(3.3),(3.7),(3.8),(3.9)). By using these two observations, we can utilize Sα , Sβ and ∆ as our main means to determine the sequence of equilibrium patterns, with θ, Cα , Cβ as an additional reference.. 3.2 The Sequence of Equilibrium Patterns in Problem I In this section, we assume that fixed costs and variable costs for users traveling on the road network are given, and we investigate which equilibrium pattern will arise. Here we discuss how the fixed costs and variable costs influence the composition of 1  1  0  1  equilibrium patterns. In Problem I, only cr 0 0 and cr 0 0 need to be compared. We 0. 1. 1. 1. arrange three kinds of critical intervals (see Lemma 3.2.1.).. Lemma 3.2.1. The critical intervals in problem I are determined by cr. 0  1  − − and 1 1. i) If Cβ > Cα . In this condition, the critical intervals are given below : Eq. 0. Eq. − 1. 1 − 1. -. 0. cr. d. 0  1  − − 1 1. ii) If Cβ < Cα . In this condition, the critical intervals are given below : Eq. 1. Eq. − 0. 1 − 1. -. 0. cr. 1  1  − − 1 0. d. cr. 1  1  − − . 0 1.

(41) 3. The Sequence of Equilibrium Patterns Determined by the Demand. 41. iii) If Cβ = Cα . In this condition, the critical intervals are given below : Eq. 1 − 1. -. 0 =cr. 0  1  − − 1 1. =cr. d. 1  1  − − 0 1. Proof. By the equilibrium conditions in Proposition 2.2.1 - ??. 1 The existence of Eq − satisfies the inequality constraint: 0. (b3 + b4 ) − (b1 + b2 ) 1  1  = cr − − , if (b3 + b4 ) − (b1 + b2 ) > 0. 0 1 a1 + a2 0 The existence of Eq − satisfies the inequality constraint: d6. 1. (b1 + b2 ) − (b3 + b4 ) 0  1  = cr − − , if (b1 + b2 ) − (b3 + b4 ) > 0. 1 1 a3 + a4 1 The existence of Eq − satisfies the inequality constraint: d6. 1. n d > max cr. o. 1  1  0  1  − − , cr − − 0 1 1 1. .. Discuss the positive or negative conditions to the numerator terms of the critical points, the critical intervals are determined as the Lemma shows. Once we fix the linear travel time function on each link, then the relation between equilibrium patterns is determined by total flow d. The relation between the patterns in detail is discussed in Lemma 3.2.1. With the increasing of the amounts of total flow from zero to infinity, the order of the patterns is shown in a sequence. Figure 3.4 lists all the possible sequence in Problem I. Eq. 1 − 0. cr. 1 − 0.  1  − 1. Eq Eq. 0 − 1. cr. 0  − 1. 1 − 1. 1 − 1. Fig. 3.4: Relation Between Equilibrium Conditions in Problem I.

(42) 3. The Sequence of Equilibrium Patterns Determined by the Demand. 42. 3.3 The Sequence of Equilibrium Patterns in Problem II for ∆ > 0 We now consider the situation where ∆ > 0, it exerts a direct influence that the critical 1 1 interval of Eq 0 is next above Eq 1 (by Proposition 2.3.3 and 2.3.7,). Sequences of 1. 1. equilibrium patterns in Problem II are discovered and classified by the main characteristic values Sα , Sβ , ∆ and the additional characteristic values θ, Cα , Cβ if necessary. Lemma 3.3.1. If ∆ > 0 and Sα > 0, Sβ > 0 , the critical intervals in problem II are n o n o 1 1 0  1  1  1  0  1  0  0  determined by cr 0 ↔ 1 , max cr 1 1 , cr 1 1 and min cr 1 1 , cr 1 1 . 1. 1. 1. 1. 0. 0. 1. 0. 0. 1. In this condition, several possible sequences of equilibrium patterns occur and the critical intervals, which are classified by additional characteristic value θ, Cα , Cβ , are given below: i) If θ > 0 and Cα > 0, the critical intervals are given below : Eq. 0. Eq. 1 0. 0. cr. 0. Eq. 1 1. 0  0  1 1 0 1. cr. 1. Eq. 1 1. 0  1  1 1 1 1. cr. 1 0 1. ↔. 1 0 1. -. d. -. d. -. d. -. d. 1 1 1. ii) If θ > 0 and Cα = 0, the critical intervals are given below : Eq 0 =cr. =cr. 0  0  1 1 0 1. 0. Eq. 1 1. 0  0  0 1 1 1. cr. 1. Eq. 1 1. 0  1  1 1 1 1. cr. 1 0 1. ↔. 1 0 1. 1 1 1. iii) If θ > 0 and Cα < 0, the critical intervals are given below : Eq. 0. Eq. 0 1. 0. cr. 0. Eq. 1 1. 0  0  0 1 1 1. cr. 1. Eq. 1 1. 0  1  1 1 1 1. cr. 1 0 1. ↔. 1 0 1. 1 1 1. iv) If θ < 0 and Cβ > 0, the critical intervals are given below : Eq 0. 0. Eq. 1 0. 0  1  1 0 0. cr 1. 1. Eq. 1 0. 1  1  1 0 1. cr 1. 1. Eq. 1 1. 1. cr 0 1. ↔. 1 1 1. 1 0 1.

(43) 3. The Sequence of Equilibrium Patterns Determined by the Demand. 43. v) If θ < 0 and Cβ = 0, the critical intervals are given below : Eq 0 =cr. =cr. 0  1  1 1 0 0. 1. Eq. 1 0. 1  1  1 0 0 0. 1. 1  1  1 1 1 0. cr. Eq. 1 1. cr. 1 0 1. ↔. 1 0 1. -. d. -. d. -. d. 1 1 1. iv) If θ < 0 and Cβ < 0, the critical intervals are given below : Eq 0. 1. Eq. 0 0. cr. 1. Eq. 1 0. 1  1  1 0 0 0. 1. 1  1  1 1 1 0. cr. Eq. 1 1. cr. 1 0 1. ↔. 1 0 1. 1 1 1. vii) If θ = 0, the critical intervals are given below : Eq. 0. Eq. 1 0. 0. cr. =cr. 0  1  1 1 1 1. 1. Eq. 1 1. 1  1  1 1 0 1. cr. 1 0 1. ↔. 1 0 1. 1 1 1. Proof. Since ∆ > 0 and Sα > 0, Sβ > 0, by Lemma 3.1.1 and 3.1.2, cr. 1 0 1. ↔. 1 1 1. 1 0 1. > 0 and cr. ↔. 1 1 1. n > max cr. o. 0  1  1  1  1 1 , cr 1 1 1 1 0 1. .. h 1 1  1 First, the critical interval of Eq 0 always exists at cr 0 ↔ 1 , ∞ , and in the next 1 1 1 o n 1 h  1 i 0  1  1  1  1 below is the critical interval of Eq 1 at max cr 0 0 , cr 0 0 , cr 0 ↔ 1 . 1. 1. 1. 0. 1. 1. 1. Suppose θ > 0, only three kinds of critical interval assignments will take place and they are determined by Cα . Note that θ = (a1 + a5 )Cβ − (a4 + a5 )Cα . Cα > 0 ⇒ C β > 0 0  1  0  0  1  1  0  0  ⇒ cr 1 1 > 0, cr 1 1 > 0, cr 0 1 < 0, cr 0 1 < 0 0 0 n 1 1 0n 0 0 o 1 o   0  0  0  0  0  0  0  1  0  0  ⇒ max cr 10 11 , cr 01 11 = cr 10 11 , min cr 10 10 , cr 10 11 = cr 010 011 0 0 ⇒ The above result yields the determined intervals for Eq 1 and Eq 1 . 1. Cα = 0 ⇒ Cβ > 0 0  1  0  0  1  1  0  0  ⇒ cr 1 1 > 0, cr 1 1 = 0, cr 0 1 < 0, cr 0 1 = 0 0n 0 0 o 1 1o 1 n0 0 0  0  0  0  0  1  0  0  ⇒ max cr 10 11 , cr 01 11 = 0, min cr 10 10 , cr 10 11 = 0 0 ⇒ The above result yields the determined intervals for Eq 1 . 1. 0.

(44) 3. The Sequence of Equilibrium Patterns Determined by the Demand. 44. Cα < 0 ⇒ Cβ > 0 (If Cβ < 0, it is a contradiction to Sα > 0, Sβ > 0.) 0  1  0  0  1  1  0  0  ⇒ cr 1 1 > 0, cr 1 1 < 0, cr 0 1 6 0, cr 0 1 > 0 0n 0 0 o 1 0 0 n 1 1 o             ⇒ max cr 010 011 , cr 001 011 = cr 001 011 , min cr 001 101 , cr 001 011 = cr 001 011 0 0 ⇒ The above result yields the determined intervals for Eq 1 and Eq 0 . 1. 1. Suppose θ < 0, the proof is similar to θ > 0.. Suppose θ = 0, both Cα and Cβ are equal and positive. Then 0  0  1  1  0  0  = cr 1 1 > 0, cr 0 1 < 0, cr 0 1 < 0. 0 1 0 0 1 1 h n oi     0 0 1 0 0 Thus, only the critical interval 0, min cr 1 1 , cr 1 1 is determined for Eq 1 .. cr. 0  1  1 1 1 1. = cr. 1  1  1 1 0 1. = cr. 0  1  1 1 0 0. 0. 0. 0. 1. 0. We enumerate some examples by the pattern in Lemma 3.3.1. All the cases with ∆ > 0 and Sα > 0, Sβ > 0 are classified into seven categories.(see Example 3.3.1 - 3.3.7). Example 3.3.1. Given the travel cost on each link as following : f1 (u) = 10u + 30, f2 (u) = 1u + 40, f3 (u) = 1u + 50, f4 (u) = 10u + 10, f5 (u) = 1u + 0. We compute the characteristic values that ∆ = 99 > 0, Sα = 230 > 0, Sβ = 230 > 0, θ = 110 > 0 and Cα = 20 > 0. It satisfies the classification of Lemma 3.3.1-(i). 0 0 There is an interval [0, 1.8182] of Eq 1 , an interval [1.8182, 2.8099] of Eq 1 , an interval 0 1 1 1 [2.8099, 5.5556] of Eq 1 , an interval [5.5556, ∞) of Eq 0 . 1. 1.

(45) 3. The Sequence of Equilibrium Patterns Determined by the Demand. 45. The dotted line in the left of the figure shows the travel time (T I ) depending on total flow d in Problem I. The solid line in the left of the figure shows the travel time (T II ) depending on total flow d in Problem II. The right of the figure shows the time difference T II − T I and we have painted a line about time = 0 on it. If the right of the figure shows any positive part than the Braess Paradox exists. The explanations of the figures in Example 3.3.1 - 3.3.12 and Example 3.4.1 - 3.4.12 are the same. Example 3.3.2. Given the travel cost on each link as following : f1 (u) = 10u + 30, f2 (u) = 1u + 40, f3 (u) = 1u + 50, f4 (u) = 10u + 10, f5 (u) = 1u + 20. We compute the characteristic values that ∆ = 99 > 0, Sα = 10 > 0, Sβ = 100 > 0, θ = 110 > 0 and Cα = 0. It satisfies the classification of Lemma 3.3.1-(ii). There is an 0 1 interval [0, 0.9917] of Eq 1 , an interval [0.9917, 1.1111] of Eq 1 , an interval [1.1111, ∞) 1 1 1 of Eq 0 . 1. Example 3.3.3. Given the travel cost on each link as following : f1 (u) = 10u + 20, f2 (u) = 5u + 40, f3 (u) = 5u + 15, f4 (u) = 10u + 10, f5 (u) = 1u + 0. We compute the characteristic values that ∆ = 75 > 0, Sα = 100 > 0, Sβ = 275 > 0, θ = 385 > 0 and Cα = −5 < 0. It satisfies the classification of Lemma 3.3.1-(iii). There is 0 0 an interval [0, 1] of Eq 0 , an interval [1.0000, 2.9394] of Eq 1 , an interval [2.9394, 5.0000] 1. 1.

數據

Fig. 1.1: Network example of the Braess Problem
Fig. 1.2: Network example of the Braess Problem that the shortcut has added in
Fig. 1.3: An example that the Braess Paradox does not happen for all given total flow.
Fig. 1.4: The sequence of the patterns on a number line before link l 5 added in.
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