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 interval of Eq1

01is next above Eq¹1

1(by Proposition 2.3.3 and 2.3.7,). Sequences of equilibrium patterns in Problem II are discovered and classified by the main characteristic values S_α, S_β, ∆ and the additional characteristic values θ, C_α, C_β if necessary.

Lemma 3.3.1. If ∆ > 0 and Sα > 0, Sβ > 0 , the critical intervals in problem II are determined by cr ¹0

1 ↔ ¹1 In this condition, several possible sequences of equilibrium patterns occur and the critical intervals, which are classified by additional characteristic value θ, C_α, C_β, are given below:

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

v) If θ < 0 and C_β = 0, the critical intervals are given below :

vii) If θ = 0, the critical intervals are given below :

- d First, the critical interval of Eq1

01always exists at h cr ¹0

1 ↔ ¹¹

1
, ∞

, and in the next below is the critical interval of Eq1

1 Suppose θ > 0, only three kinds of critical interval assignments will take place and they are determined by C_α. Note that θ = (a₁+ a₅)C_β − (a₄ + a₅)C_α.

⇒ The above result yields the determined intervals for Eq0

11and Eq⁰¹

⇒ The above result yields the determined intervals for Eq0 1 1.

C_α < 0 ⇒ C_β > 0 (If Cβ < 0, it is a contradiction to S_α > 0, S_β > 0.)

⇒ The above result yields the determined intervals for Eq0

11and Eq⁰⁰

Thus, only the critical interval h

0, minn

is determined for Eq0 10.

We enumerate some examples by the pattern in Lemma 3.3.1. All the cases with ∆ > 0 and S_α > 0, S_β > 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.

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, S_α = 10 > 0, S_β = 100 > 0, θ = 110 > 0 and Cα = 0. It satisfies the classification of Lemma 3.3.1-(ii). There is an interval [0, 0.9917] of Eq0

11, an interval [0.9917, 1.1111] of Eq¹1

1, an interval [1.1111, ∞) of Eq1

01.

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, S_α = 100 > 0, S_β = 275 > 0, θ = 385 > 0 and C_α = −5 < 0. It satisfies the classification of Lemma 3.3.1-(iii). There is an interval [0, 1] of Eq0

01, an interval [1.0000, 2.9394] of Eq⁰¹

1, an interval [2.9394, 5.0000]

of Eq1 1

1, an interval [5, ∞) of Eq¹⁰

1.

Example 3.3.4. Given the travel cost on each link as following :

f₁(u) = 10u + 10, f₂(u) = 1u + 50, f₃(u) = 1u + 40, f₄(u) = 10u + 30, f₅(u) = 1u + 0.

We compute the characteristic values that ∆ = 99 > 0, S_α = 320 > 0, S_β = 230 > 0, θ = −110 < 0 and C_β = 20 > 0. It satisfies the classification of Lemma 3.3.1-(iv).

There is an interval [0, 1.8182] of Eq0

10, an interval [1.8182, 2.8099] of Eq¹¹

0, an interval [2.8099, 5.5556] of Eq1

1

1, an interval [5.5556, ∞) of Eq¹⁰

1.

Example 3.3.5. Given the travel cost on each link as following :

f1(u) = 10u + 10, f2(u) = 1u + 50, f3(u) = 1u + 40, f4(u) = 10u + 30, f5(u) = 1u + 20.

We compute the characteristic values that ∆ = 99 > 0, S_α = 100 > 0, S_β = 10 > 0, θ = −110 < 0 and C_β = 0. It satisfies the classification of Lemma 3.3.1-(v). There is an

interval [0, 0.9917] of Eq1 1

0, an interval [0.9917, 1.1111] of Eq¹¹

1, an interval [1.1111, ∞) of Eq1

01.

Example 3.3.6. Given the travel cost on each link as following :

f₁(u) = 10u + 10, f₂(u) = 5u + 15, f₃(u) = 5u + 40, f₄(u) = 10u + 20, f₅(u) = 1u + 0.

We compute the characteristic values that ∆ = 75 > 0, S_α = 275 > 0, S_β = 100 > 0, θ = −385 < 0 and C_β = −5 < 0. It satisfies the classification of Lemma 3.3.1-(vi).

There is an interval [0, 1.0000] of Eq1

00, an interval [1.0000, 2.9394] of Eq¹1

0, an interval [2.9394, 5] of Eq1

11, an interval [5, ∞) of Eq¹⁰

1.

Example 3.3.7. Given the travel cost on each link as following :

f₁(u) = 10u + 10, f₂(u) = 1u + 20, f₃(u) = 1u + 20, f₄(u) = 10u + 10, f₅(u) = 1u + 5.

We compute the characteristic values that ∆ = 99 > 0, S_α = 55 > 0, S_β = 55 > 0 and θ = 0. It satisfies the classification of Lemma 3.3.1-(vii). There is an interval [0, 0.4545]

of Eq0

10, an interval [0.4545, 1.1111] of Eq¹1

1, an interval [1.1111, ∞) of Eq¹0 1.

We will investigate the Braess Paradox which only occurs under the conditions of ∆ > 0 and S_α > 0, S_β > 0 in Chapter 4.

Lemma 3.3.2. If ∆ > 0 and S_α > 0, S_β 6 0 , the critical intervals in problem II are determined by cr ¹⁰

0

1 0

1
. In this condition, one possible sequence of equilibrium patterns occurs and the critical intervals are given below:

- d

1
> 0, there is no doubt that the critical interval [cr ¹⁰

1 ↔ ¹¹

0. Hence, the critical interval of Eq1

00is [0, cr ¹0 0

1 01
].

ii) If cr ¹⁰

1 ↔ ¹¹

1
6 0, the critical interval of Eq¹⁰₀is [0, cr ¹⁰

0

1 0 1
].

Given that the linear travel costs f_i(u_i) = a_iu_i+ b_i satisfies ∆ > 0 and S_α > 0, S_β 6 0, we consider the sequence of equilibrium patterns under two separate conditions of

cr ¹⁰

1 ↔ ¹¹

1
> 0 and cr ¹⁰

1 ↔ ¹¹

1
6 0. We list an example for each below (see Example 3.3.8).

Example 3.3.8. Given the travel cost on each link as following :

f₁(u) = 10u + 20, f₂(u) = 1u + 20, f₃(u) = 1u + 50, f₄(u) = 10u + 20, f₅(u) = 1u + 10.

We compute the characteristic values that ∆ = 99 > 0, S_α = 190 > 0 and S_β = −80 < 0.

It satisfies the classification of Lemma 3.3.2. There is an interval [0, 2.7273] of Eq1 00, an interval [2.7273, ∞) of Eq1

01.

Lemma 3.3.3. If ∆ > 0 and S_α 6 0, Sβ > 0, the critical intervals in problem II are determined by cr ⁰⁰

1

1 0

1
and cr ¹⁰

1 ↔ ¹¹

1
. In this condition, one possible sequence of equilibrium patterns occurs and the critical intervals are given below:

- d

0 ^{cr 0}0

1

1 0 1

Eq0 01

 Eq1

01



Proof. Similarly to the proof in Lemma 3.3.2.

Given that the linear travel costs f_i(u_i) = a_iu_i+ b_i satisfies ∆ > 0 and S_α 6 0, Sβ > 0, we consider the sequence of equilibrium patterns under two separate conditions of

cr ¹0

1 ↔ ¹1

1
> 0 and cr ¹0

1 ↔ ¹1

1
6 0 . We list an example for each below (see Example 3.3.9).

Example 3.3.9. Given the travel cost on each link as following :

f₁(u) = 10u + 20, f₂(u) = 1u + 50, f₃(u) = 1u + 20, f₄(u) = 10u + 20, f₅(u) = 1u + 5.

We compute the characteristic values that ∆ = 99 > 0, S_α = −25 < 0 and S_β = 245 > 0.

It satisfies the classification of Lemma 3.3.3. There is an interval [0, 2.7273] of Eq0 01, an interval [2.7273, ∞) of Eq1

01.

Lemma 3.3.4. If ∆ > 0 and S_α 6 0, Sβ 6 0 , the critical intervals in problem II are determined by cr ⁰0

1

1

01
, cr ¹⁰

0

1

01
. In this condition, the critical intervals are identical as Problem I:

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

-0 ^{cr 00}

1

1 0 1

cr ⁰0

1

0 1 1

Eq0 01

 Eq1

01



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

-0 ^{cr 10}

0

¹ 0 1

cr ¹0

0

¹ 1 0

Eq1 00

 Eq1

01



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

-0

Eq1 01



Proof. Since ∆ > 0 and S_α 6 0, Sβ 6 0, by Lemma 3.1.1 and 3.1.2, cr ¹⁰₁ ↔ ¹1

1
6 0 and cr ¹⁰

1 ↔ ¹¹

1
6 minn cr ⁰¹

1

1

11
, cr ¹¹

0

1 11

o

are true, so the equilibrium solution yγ < 0 for all d > 0. It means that the path γ is not utilized which happens to be the same situation as proposed in Problem I. Therefore, we return to Problem I, where only the patterns Eq1

01, Eq1

00and Eq⁰0

1may occur, and they are determined by maxn cr ⁰0

1

1

01
, cr ¹0 0

1 01

o . Given that the linear travel costs f_i(u_i) = a_iu_i+ b_i satisfies ∆ > 0 and S_α 6 0, Sβ 6 0, we consider the sequence of equilibrium patterns under three separate conditions of C_β > C_α, C_β < C_α and C_β = C_α. We list an example for each below (see Example 3.3.10 - 3.3.12).

Example 3.3.10. Given the travel cost on each link as following :

f₁(u) = 10u + 20, f₂(u) = 1u + 50, f₃(u) = 1u + 20, f₄(u) = 10u + 10, f₅(u) = 1u + 45.

We compute the characteristic values that ∆ = 99 > 0, S_α = −455 < 0, S_β = −95 < 0 and C_β − C_α = (−5) − (−45) = 40 > 0. It satisfies the classification of Lemma 3.3.4-(i).

There is an interval [0, 3.6364] of Eq0 0

1, an interval [3.6364, ∞) of Eq¹⁰

1.

Example 3.3.11. Given the travel cost on each link as following :

f₁(u) = 10u + 10, f₂(u) = 1u + 20, f₃(u) = 1u + 50, f₄(u) = 10u + 20, f₅(u) = 1u + 45.

We compute the characteristic values that ∆ = 99 > 0, Sα = −95 < 0, Sβ = −455 < 0 and C_β− C_α = (−45) − (−5) = −40 < 0. It satisfies the classification of Lemma 3.3.4-(ii).

There is an interval [0, 3.6364] of Eq1

00, an interval [3.6364, ∞) og Eq¹⁰

1.

Example 3.3.12. Given the travel cost on each link as following :

f1(u) = 10u + 10, f2(u) = 1u + 20, f3(u) = 1u + 20, f4(u) = 10u + 10, f5(u) = 1u + 40.

We compute the characteristic values that ∆ = 99 > 0, S_α = −330 < 0, S_β = −330 < 0 and C_β − C_α = (−30) − (−30) = 0. It satisfies the classification of Lemma 3.3.4-(iii).

There is an interval [0, ∞) og Eq1 0 1.

Proposition 3.3.5. If ∆ > 0 and, at the same time, either S_α 6 0 or Sβ 6 0 is satisfied it implies that T^I − T^II ≡ 0, ∀d.

Proof. Path γ is not used in this condition, T^I− T^II ≡ 0, ∀d.

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 3.3.1 - 3.3.4. With the increasing of the amounts of total flow from zero to infinity, the order of the patterns is shown in a sequence. Figure 3.5 lists all the possible sequence in Problem II for ∆ > 0.

Eq1

Fig. 3.5: Relation Between Equilibrium Conditions in Problem II for ∆ > 0.

在文檔中 Braess運輸問題的均衡解 (頁 42-55)