The Equilibrium Solution and its Constraints in Problem II

1. Solving the simultaneous equations:

(

x_α+ x_β = 10

10xα+ 50 = 10xβ+ 110

we find the equilibrium solutions are (x_α, x_β) = (8, 2), and the travel time on route U p and route Down is T_U^I = T_D^I = 130.

ii) If the total flow d = 4 6 6 = cr ¹⁰₀1

01
, it satisfies the pattern Eq⁻¹

0. The equilibrium solution is (x_α, x_β) = (4, 0), and the travel time on route α is

T_U^I = 90 6 110 = TD^I.

In Example 2.2.1, if we exchange the travel cost f1 and f4, and exchange the travel cost f₂ and f₃, then Eq 0

−

1occurs under the total flow d 6 6, where the new critical point cr ⁰⁰

1

1 01
= 6.

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 (y_α, y_β, y_γ) for the three-route satisfies y_α + y_β + y_γ = d, then the flow distribution the travel time needed for U p,Down and M iddle are stated as follows, where T_U^II, T_D^II and T_M^II denotes the travel costs on U p,Down and M iddle, respectively.

T_U^II(y_α, y_β, y_γ) = f₁(y_α+ y_γ) + f₂(y_α).

T_D^II(y_α, y_β, y_γ) = f₃(y_β) + f₄(y_β+ y_γ).

T_M^II(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 routes. Take Eq1

00for example, since only route Up is utilized, the flow distribution is (y_α, y_β, y_γ) = (d, 0, 0) and T_U^II(d, 0, 0) = f₁(d + 0) + f₂(d), T_D^II(d, 0, 0) = f₃(0) + f₄(0 + 0) and T_M^II(d, 0, 0) = f₁(d + 0) + f₅(0) + f₄(0 + 0). However, is it reasonable that the existence

of equilibrium pattern Eq1

00satisfies the inequality (

T_U^II(d, 0, 0) 6 TD^II(d, 0, 0)

T_U^II(d, 0, 0) 6 TM^II(d, 0, 0) 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 Up are reduced to

T_U^II(d − 1− 2, 0 + 1, 0 + 2) = f1(d − 1− 2+ 2) + f2(d − 1− 2)

= (a1+ a2)d + (b1+ b2) − (a1+ a2)1− a22,

and the travel time on routes Down and M iddle are

T_D^II(d − 1− 2, 0 + 1, 0 + 2) = f3(1) + f4(1+ 2)

= (a3+ a4)1+ a42+ (b3+ b4),

T_M^II(d − 1− 2, 0 + 1, 0 + 2) = f1(d − 1− 2+ 2) + f5(2) + f4(1+ 2)

= a1d + (b1+ b5+ b4) − (a1− a4)1+ (a5+ a4)2.

Since T_M^II− T_U^II = −a₂d − (b₂− b₄− b₅) + (a₂+ a₄)₁+ (a₄+ a₅)₂ > −a2d − (b₂− b₄− b₅).

i) If −a₂d − (b₂− b₄− b₅) 6 0,

route M iddle and U p will reach the equilibrium in some non-negative ₁, ₂. ii) If −a₂d − (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 −a₂d−(b₂−b₄−b₅) and d = ^−(b2−b_a^4−b5)

2 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 Eq1 00:

If only U p is utilized, the constraints of y_α, y_β and y_γ would be:



















yα = d

yβ = 0

yγ = 0

T_U^II(yα, yβ, yγ) 6 TM^II(yα, yβ, yγ) T_U^II(yα, yβ, yγ) 6 TD^II(yα, yβ, yγ)

.

The equilibrium solution of Eq1 0

0is (y_α, y_β, y_γ) = (d, 0, 0), and the existence of Eq1 0

0satisfies the inequality constraint:

d 6 minn

We show the interval of pattern Eq1

00 lying on a number line with total flow increasing from zero to infinity, if Eq1 Therefore, d 6 minn

cr ¹0

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.

Calculating the critical points, we obtain cr ¹0 0

00condition. The equilibrium solution is (y_α, y_β, y_γ) = (10, 0, 0), and the travel time on U p is T_U^II = 500 <

minT_D^II, T_M^II = min {700, 550}.

Proposition 2.3.2. Equilibrium condition of the pattern Eq0

The equilibrium solution of Eq0

01is (y_α, y_β, y_γ) = (0, d, 0), and the existence of Eq0

01satisfies the inequality constraint:

d 6 minn

We show the interval of pattern Eq0

01 lying on a number line with total flow increasing from zero to infinity, if Eq0

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 Example 2.3.1. And let the total flow d = 10, than Eq0

01occurs.

Proposition 2.3.3. Equilibrium condition of the pattern Eq1 01:

The equilibrium solution of Eq0 01is

(y_α, y_β, y_γ) =

1satisfies the inequality constraint:

d > maxn We show the interval of pattern Eq1

01 lying on a number line with total flow increasing from zero to infinity, if Eq1

Proof. Since M iddle is not utilized in Eq1 0

1, the equilibrium solution and the equilibrium constraints are equivalent to Eq ¹

−

1in Proposition 2.2.3. Therefore, (y_α, y_β, y_γ) = _(a On the other hand, when we simplify the inequality constraint

T_U^II(y_α, y_β, y_γ) = T_D^II(y_α, y_β, y_γ) 6 TM^II(y_α, y_β, y_γ), it follows that

(a₁a₄ − a₂a₃)d > (a2+ a₄)(b₃− b₁− b₅) + (a₁+ a₃)(b₂− b₄− b₅).

This result means y_γ 6 0, and we compare the inequalities to the condition yγ > 0 in Eq1

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 Eq1

01, so we solve the simultaneous equations:

Proposition 2.3.4. Equilibrium condition of the pattern Eq0 10:

The equilibrium solution of Eq0

10is (yα, y_β, y_γ) = (0, 0, d), and the existence of Eq0

10satisfies the inequality constraint:

d 6 minn We show the interval of pattern Eq0

10 lying on a number line with total flow increasing from zero to infinity, if Eq0

Proof.

Therefore, d 6 minn cr ⁰¹

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.

If the total flow d = 10, it satisfies the pattern Eq0

10, and the equilibrium solution is (y_α, y_β, y_γ) = (0, 0, 10), where travel time on M iddle is T_M^II = 610 < minT_U^II, T_D^II = min {700, 800}.

Proposition 2.3.5. Equilibrium condition of the pattern Eq1 10:

The equilibrium solution of Eq1

10is (y_α, y_β, y_γ) = _(a

10satisfies the inequality constraint:

maxn

where we denote cr ¹¹

We show the interval of pattern Eq1

10 lying on a number line with total flow increasing from zero to infinity, if Eq1 The solution of

(

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 Eq1

10, and we solve the simultaneous equations:







y_α+ y_γ = 20

y_β = 0

11y_α+ 10y_γ+ 600 = 10y_α+ 21y_γ+ 400

.

The equilibrium solution is (yα, yβ, yγ) = ⁵₃, 0,⁵⁵₃
, and the travel time on Up and Middle is T_U^II = T_M^II = ²⁴⁰⁵₃ < ²⁶⁵⁰₃ = T_D^II.

Proposition 2.3.6. Equilibrium condition of the pattern Eq0 11:

If Down and M iddle are utilized, but U p is not, the constraints of y_α, y_β and y_γ would

be: 

















yα = 0

yβ+ yγ = d

T_D^II(yα, yβ, yγ) = T_M^II(yα, yβ, yγ) 6 TU^II(yα, yβ, yγ) yβ > 0

yγ > 0

.

The equilibrium solution of Eq0

11is (y_α, y_β, y_γ) =

0,^(a1+a_a^{5)d−(b3−b1−b5)}

3+a1+a5 ,^a3d+(b_a ^3−b1−b5)

3+a1+a5

 , and the existence of Eq0

11satisfies the inequality constraint:

maxn cr ⁰0

1

0

11
, cr ⁰1 0

0 11

o

< d 6 cr ⁰¹₁1 11
, where we denote cr ⁰¹

1

1

11
= ^(a3+a1+a_a^{5)(b2−b4−b5)−a5(b3−b1−b5)}

3(a4+a5)+a4(a1+a5) . Note that the inequality d 6 cr ⁰¹₁1

11
shows that y_α 6 0, we can compare the inequality to the condition yα > 0 in Eq1

11later.

We show the interval of pattern Eq0

11 lying on a number line with total flow increasing from zero to infinity, if Eq0

11exists.

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 Example 2.3.4. Let the total flow d = 20, then Eq0

11will occur.

- d

Proposition 2.3.7. Equilibrium condition of the pattern Eq1 11:

The equilibrium solution of Eq1 11is

11satisfies the above mentioned inequality constraints. Therefore, in brief, we have proven that:

i) To fit the constraints y_α > 0, y_β > 0 , the total flow d is imposed by

- d

We show the interval of pattern Eq1

11 lying on a number line with total flow increasing from zero to infinity, if Eq1

The solution of



⇒















d < cr ¹0

1 ↔ ¹1 1

if a₁a₄− a₂a₃ > 0 d > cr ¹⁰

1 ↔ ¹¹

1

if a₁a₄− a₂a₃ < 0 d > ^(a1_(a^{+a3)(b2−b4−b5)+(a2+a4)(b3−b1−b5)}

1+a3)(a2+a4)+a5(a1+a2+a3+a4) > 0 if a₁a₄− a₂a₃ = 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 Eq1

11, and we solve the simultaneous equations:







yα+ yβ+ yγ = 40

11yα+ 10yγ+ 600 = 10yα+ 10yβ+ 21yγ+ 400 11y_β+ 10y_γ+ 700 = 10y_α+ 10y_β+ 21y_γ+ 400

.

The equilibrium solution is (y_α, y_β, y_γ) = ²⁷⁴⁰₁₄₃,¹⁴⁴⁰₁₄₃,¹⁵⁴⁰₁₄₃
, and the travel time on Up,Down and M iddle is T_U^II = T_D^II = T_M^II = ¹¹⁹⁴⁰₁₃ .

Recall the notation of critical points which we denote in section 2.2 and 2.3 as follows:

cr ⁻¹

0

1

− 1

= (b₃+ b₄) − (b₁+ b₂) a₁+ a₂ . cr ⁻⁰

1

1

− 1

= (b₁+ b₂) − (b₃+ b₄) a₃+ a₄ . cr ¹0

0

1 01

= (b₃+ b₄) − (b₁+ b₂) a₁+ a₂ . cr ⁰0

1

1 01

= (b₁+ b₂) − (b₃+ b₄) a3+ a4

. cr ¹0

0

1 10

= −(b₂− b₄ − b₅) a2

. cr ⁰0

1

0 11

= −(b₃− b₁ − b₅)

a₃ .

cr ⁰¹

0

1 10

= b₂− b₄− b₅ a₄+ a₅ . cr ⁰¹

0

0 11

= b3− b1− b5

a₁+ a₅ . cr ¹¹

0

1 11

= (a2+ a4+ a5)(b3− b1− b5) − a5(b2− b4− b5) a₂(a₁+ a₅) + a₁(a₄+ a₅) . cr ⁰¹

1

1 11

= (a3+ a1+ a5)(b2− b4− b5) − a5(b3− b1− b5) a₃(a₄+ a₅) + a₄(a₁+ a₅) . cr ¹⁰

1 ↔ ¹¹

1

= (a₁+ a₃)(b₂− b₄ − b₅) + (a₂+ a₄)(b₃− b₁− b₅) a₁a₄− a₂a₃ .

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) = a_iu_i+ b_i, 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 equilibrium patterns are dependent on the critical points cr ⁻¹

0

¹

−

1
, cr ⁻⁰

1

¹

−

1
, cr ¹⁰

0

1 01
, cr ¹0

0

1

10
, cr ⁰0 1

1

01
, cr ⁰0 1

0

11
, cr ⁰1 0

1

10
, cr ⁰1 0

0

11
, cr ¹1 0

1

11
, cr ⁰1 1

1

11
, cr ¹0

1 ↔ ¹1 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:

C_α = (b₃− b₁− b₅), C_β = (b₂− b₄− b₅),

∆ = a₁a₄− a₂a₃. S_α = a₄C_α+ a₃C_β, S_β = a₂C_α+ a₁C_β,

θ = (a₁+ a₅)C_β − (a₄+ a₅)C_α

Then, the value of critical points can be expressed briefly.

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 critical points. Take Eq 0

−

1for example, the existence of Eq⁻⁰

1satisfies

Therefore, C_β− C_α 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 ∆,S_α,S_β and θ are chosen by us to classify the

equilibrium patterns in a suitable order, see Lemma 3.1.1 - 3.1.4. ∆ involves the involved in the order of cr ⁰¹

1

The following flowchart shows our classification which covers all the possible order of patterns in Problem II, see Figure 3.1.

.

see Figure 3.2

(done)

(done)

see Figure 3.3

see Figure 3.2

(done)

(done)

see Figure 3.3

see Figure 3.2

see Figure 3.3

∆

Fig. 3.1: The Flowchart of the Classification in Problem II

Follow the flowchart, if the constraints (∆, S_α, S_β) are enough to remove the

nonexis-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, C_α and C_β. 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.

Lemma 3.3.1

.

.

case vii)

case i)

case ii)

case iii)

case iv)

case v)

case vi)

(done)

(done)

(done)

(done)

(done)

(done)

(done)

θ>0

θ < 0

θ= 0

C^α> 0 Cα= 0

C_α

< 0

Cβ> 0 Cβ= 0

C_β

< 0

Fig. 3.2: The Flowchart of the Classification in Lemma 3.3.1 or 3.4.1 or 3.5.2

Lemma 3.3.4

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 ¹⁰

iii)

Lemma 3.1.2. The following identities holds:

cr ¹0

Proof. Similarly to the proof of Lemma 3.1.1. By the difference in the value of critical points, in relation to S_β and ∆, the identities holds.

Lemma 3.1.3. The following identities holds:

cr ⁰¹

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 dif-ferent 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 addition, we find that ∆ exerts a direct influence on the direction of Eq1

01and Eq¹¹

1(see (3.1),(3.2),(3.3),(3.7),(3.8),(3.9)). By using these two observations, we can utilize S_α, S_β 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 equilibrium patterns. In Problem I, only cr ¹⁰

0

1 0

1
and cr ⁰⁰

1

1 0

1
need to be compared. We 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

¹

−

1
and cr ⁻¹

0

¹

− 1
.

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

- d

0 cr ⁻⁰

1

1

− 1

Eq0

− 1

 Eq1

− 1



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

- d

0 cr ⁻¹

0

¹

− 1

Eq1

− 0

 Eq1

− 1



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

Proof. By the equilibrium conditions in Proposition 2.2.1 - ??.

The existence of Eq1

−

0satisfies the inequality constraint:

d 6 (b₃+ b₄) − (b₁+ b₂)

1satisfies the inequality constraint:

d 6 (b₁+ b₂) − (b₃+ b₄)

1satisfies the inequality constraint:

d > max

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.

Eq1

Fig. 3.4: Relation Between Equilibrium Conditions in Problem I

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 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

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

在文檔中 Braess運輸問題的均衡解 (頁 20-0)