The Sequence of Equilibrium Patterns in Problem II for ∆ = 0

The last part of chapter 3, we discuss the equilibrium patterns under the assumption of

∆ = 0. It is a special boundary condition because the equilibrium solution y_γ in Eq1 11is a fixed value (see Proposition 2.3.7). Therefore, the existence of Eq1

11is separated into two situation as follows:

Lemma 3.5.1. If ∆ = 0 and S_α + S_β 6 0, according to the equilibrium solutions of

11is fail. Under the assumption of ∆ = 0, S_α

S_β = a₄ a₂ = a₃

a₁, then S_α 6 0 and Sβ 6 0. The critical intervals in Problem II are determined by cr ⁰0

1

1

01
and cr ¹0 0

1

01
, which happens to be the same situation as proposed in Lemma 3.3.4 and the sequences of equilibrium patterns are classified as follows:

i) If Cβ > Cα. In this condition, the critical intervals are given below :

ii) If C_β < C_α. In this condition, the critical intervals are given below :

iii) If C_β = C_α. In this condition, the critical intervals are given below :

Lemma 3.5.2. If ∆ = 0 and S_α + S_β > 0, according to the equilibrium solutions of Eq1

1

1as follows:

 transportation problem becomes a new linear travel cost as follows:



and there remain total demand (d − yγ) to be separated into path U p and Down. There-fore, we repeat the process of Problem I (see Lemma 3.2.1) and the equilibrium pattern Eq1

11exists. A new relation between cr ¹¹

0

It means that we are not interfere by the fixed value yγ because the pattern Eq1 11faces

1. The critical intervals in Problem II are determined by cr ¹¹

0

first , which happens to be the

same situation as proposed in Lemma 3.4.1 and the sequences of equilibrium patterns are classified as follows:

i) If θ > 0 and Cα> 0, the critical intervals are given below :

ii) If θ > 0 and C_α= 0. In this condition, the critical intervals are given below :

- d

iv) If θ < 0 and C_β > 0. In this condition, the critical intervals are given below :

- d

vi) If θ < 0 and C_β < 0. In this condition, the critical intervals are given below :

- d

vii) If C_β = C_α > 0. In this condition, the critical intervals are given below :

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 3.5.1 - 3.5.2. With the increasing of the amounts of total flow from zero to infinity, the order of the patterns is shown in a sequence. Figure 3.7 lists all the possible sequence in Problem II for ∆ = 0.

Eq¹

Fig. 3.7: Relation Between Equilibrium Conditions in Problem II for ∆ = 0

PARADOX

In this chapter, we introduce the necessary and sufficient condition for the Braess Paradox, which are proposed by M. Frank ??. Note that the Braess Paradox is just a part of cases appearing in our classification. However, observing the examples in Chapter 3, it is a obviously that the Braess Paradox only occurs in the examples of Lemma 4.3.1. As such we are interested in whether or not ∆ > 0, S_α > 0, S_β > 0 always leads to the Braess Paradox, and the conclusions are as follows:

Lemma 4.0.3. The equilibrium solutions in Problem II must utilize path γ (y_γ>0) if the Braess Paradox occurs. The Braess Paradox interval only belongs to Eq0

1

0, Eq¹¹

0, Eq0

11or Eq¹¹

1.

Proof. Suppose that y_γ = 0, then the equilibrium solution (y_α, y_β, 0) belongs to Eq1 00, Eq0

0

1and Eq¹⁰

1. These solutions are equle to the solutions in Eq⁻¹

0, Eq⁻⁰

1and Eq⁻¹

1;

thus the travel time between Problem II and I is T^II − T^I = 0.

Lemma 4.0.4. The equilibrium solutions in Problem I must be Eq1

−

1 if the Braess Para-dox occurs.

Proof. If the Braess Paradox occurs for a given flow d, the travel time T^II > T^I. Suppose that the equilibrium condition Eq1

−

0holds, then the travel time should be T^I = f₃(d) + f₄(d) < f₁(0) + f₂(0).

However, the equilibrium solution in Problem II is (y_α, y_β, y_γ) where y_γ > 0. By Lemma 4.0.3, only Eq0

10, Eq¹1

0, Eq⁰1

1and Eq¹1

1need to be considered.

i) If y_β > 0 (see Eq0 1

1and Eq¹¹

1), path β is also utilized in Problem II and travel time is

T^II = f₃(y_β) + f₄(y_β+ y_γ) 6 f3(d) + f₄(d) = T^I. ii) If yβ = 0 and yα > 0(see Eq1

10), path Up is utilized but path Down is not. The travel time in Problem II is

T^II = f1(yα+ yγ) + f2(yα) < f3(0) + f4(yγ) 6 f³(d) + f4(d) = T^I.

iii) If y_α = y_β = 0(see Eq0

10), both path Up and path Down are not utilized. Since y_γ = d, the travel time in Problem II is

T^II = f₁(d) + f₅(d) + f₄(d) < f₃(0) + f₄(d) 6 f3(d) + f₄(d) = T^I.

The discussion above is a contradiction to T^II > T^I. Suppose the equilibrium condition Eq0

−

1holds, the proof is similar. In conclusion, both paths Up and Down are utilized in Problem I.

Lemma 4.0.5. The constraints of the Braess Paradox in Eq1 11are

If flow d leads the Braess Paradox, maxn

When the inequality holds, by Lemma 4.0.1, either the conditions of ∆ > 0, S_α> 0, S_β > 0

Lemma 4.0.6. The constraints of the Braess Paradox in Eq1 1

If flow d leads the Braess Paradox, maxn

When the inequality holds, by Lemma 4.3.1 and Lemma 4.3.3 in Chapter 4, S_β > 0 and

θ < 0 must also be true. Consider should also be added to the constraints.

In conclusion, ∆ > 0, S_α > 0, S_β > 0 must hold, and the lower bound of the Braess

Lemma 4.0.7. The constraints of the Braess Paradox in Eq0 11are Proof. The proof is similar to Lemma 4.0.4.

Lemma 4.0.8. The constraints of the Braess Paradox in Eq0 10are

If flow d leads the Braess Paradox, possi-ble the lower bound value. Due to the existence of Eq0

1

Note that the numerator of Bp0 1

Therefore, consider all of the above mentioned Lemma 5.0.1 - 5.0.6, we present the theory below:

iii) The Braess Paradox of Eq0 1

1exists if and only if

∆ > 0, S_α > 0, S_β > 0, θ > 0.

iv) The Braess Paradox of Eq0 1

0exists if and only if

∆ > ∆0, C_α > 0, C_β > 0, where ∆₀ = max (a₃+ a₄)θ

C_α ,−(a₂+ a₁)θ C_β



> 0.

Proof. Only iv) needs to be proved. By Lemma 5.0.6, If the Braess Paradox of Eq0

10exists, Cα > 0, Cβ > 0. Since the constraints of the Braess Paradox in Eq0

10are maxn

cr ¹0 0

1

01
, cr ⁰0 1

1

01
, Bp⁰1 0

o

< d < minn cr ⁰1

0

1

10
, cr ⁰1 0

0 11

o , we have





 cr ⁰¹

0

0

11
− Bp⁰¹

0

 = _(a ^{∆Cα−(a3+a4)θ}

1+a5)[(a1+a2)(a1+a5)+(a4+a3)(a4+a5)+∆] > 0 cr ⁰1

0

1

10
− Bp⁰1 0

 = _(a ^{∆Cβ+(a2+a1)θ}

1+a5)[(a1+a2)(a1+a5)+(a4+a3)(a4+a5)+∆] > 0 ,

or 





∆C_α− (a₃+ a₄)θ > 0

∆C_β + (a₂+ a₁)θ > 0 .

Therefore, the constraintss ∆ > ∆0 = maxn

(a3+a4)θ

Cα ,^−(a2_C^+a1)θ

β

o

> 0 should be added in.

In this thesis, we investigate all possible equilibrium patterns on a simple four-node and five-link network. If the flow-dependent linear traveling time on each link are given, specific equilibrium patterns appear in a sequence on the number line with the magnitude of the total flow from zero to infinity. We completely classify the flow distribution pattern of equilibrium in a suitable order and present the order on a number line of total flow. If the total flow on the network is given, the corresponding pattern is then determined, and thus subsequently the flow distribution and the travel time are obtained and formulated.

If ∆ > 0,S_α > 0 and S_β > 0, the Braess Paradox exists and the interval of existence is determined. If ∆ > 0 but S_α ≤ 0 or S_β ≤ 0, route M iddle is not utilized for any given total flow, and the travel times are identical whether the new link is added, where the new link is ineffective. On the other hand, in hope of reducing the average travel time, the new link is suggested to be added under the condition of ∆ 6 0, which represents the case where l₁ and l₄ are less sensitive to the increasing of flows than l₂ and l₃ are.

A more complex network will make us spend more work on classification. However, once the classification is done, the flow distribution is easily obtained from the simultaneous equations since equilibrium patterns are immediate from the classification. Moreover, the flow distribution directly gives the average travel time of the network and, certainly, the existence of the Braess Paradox.

BIBLIOGRAPHY

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[5] X. Di, X. He, X. Guo, and H. X. Liu, Braess paradox under the boundedly rational user equilibria, Transportation Research Part B: Methodological 67 (2014), 86–108.

[6] M. Frank, The braess paradox, Mathematical Programming 20 (1981), no. 1, 283–302.

[7] H. Mahmassani and R. Herman, Dynamic user equilibrium departure time and route choice on idealized traffic arterials, Transportation Science 18 (1984), no. 4, 362–384.

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在文檔中 Braess運輸問題的均衡解 (頁 67-0)