Throughput flow

The throughput flow of the fluid queuing model is the strategy as below: Given network C, inflow µ₀, period time θ = 0 ∼ t. Each particle from µ0 is manipulated such that sending the maximal mass of particle throughput C at inflow µ₀ during θ = 0∼ t.

Definition 4. The maximal mass of particle is denoted by M_put(C, µ₀, t). Sometimes we instead inflow function of infinite inflow rate during θ = 0∼ t to afford an upper bound of M_put(C, µ₀, t), denoted by M_put(C,∞, t). When it is clear from the context, we use M_put(t) to denote M_put(C,∞, t).

2.3 Networks of Model

Definition 5 (Parallel­link Network). Given a network C in fluid queuing model, C is Parallel­link networks if e = (s, t) for all e∈ E.

Definition 6 (Series­parallel Network). Series­parallel networks are defined by Induc­

tion: Start from parallel­link networks, each time can do series­linking or parallel­linking operation to another well­defined series­parallel network. All possible results form series­

parallel networks.

Definition 7. Given the series­parallel network C, the maximal number of each path’s nodes among all paths is denoted by D(C).

Definition 8 (Parallel­group­link Network). Given ϵ > 0 and a parallel­link network C.

If there exists [L₁, R₁], ..., [L_N, R_N], ^R_Lⁱ

i ≤ 1 + ϵ and ^L_Rⁱ⁺¹_i ≥ ¹_ϵ, and the delay time of each edge of C is in one of the intervals, then C is called parallel­group­link network.

Connected from the routing game with groups of similar links, refer to [11].

Figure 2.1: The diagram of “(2 + 2)­parallel­link Network”. In the first stage C1, the source is s and the sink is p; In the second stage C₂, the source is p and the sink is t.

Definition 9 ((2 + 2)­parallel­link Network). Given parallel­link networks C₁, C₂ and C1’s inflow µ0. Now, series­link C1 to C2, denoted by C1+ C2. Consider OPT flow and EQU flow on this linking network C₁+C₂, the maximal number of used edges in OPT flow or EQU flow in C1 and C2 part are denoted by m1 and m2 respectively. If m1, m2 ≤ 2, we called C₁+ C₂ is a (2 + 2)­parallel­link network.

Chapter 3

Main

We will show several PoA’s bounds in the fluid queuing model in this chapter. These include some tight upper bound on parallel­link networks or (2+2)­parallel­link networks and some finite upper bound on series­parallel networks.

3.1 Parallel­link Networks

First of all, we need Lemma 1 to evaluate the upper bound of the Price of Anarchy for networks.

Lemma 1. Consider the fluid queuing model with network C, inflow µ₀. For all time θ₀, if

M_put(L(θ₀))≤ k ∗ Q(θ0),

then T_EQU ≤ (k + 1) ∗ TOP T, where M_put is the maximal mass of particle throughput network, Q is the particle existing in network, and T_{OP T}, T_EQU are the arriving time of all particle in OPT flow, EQU flow respectively.

Proof. For the shortest travel time ˆL = L(ˆθ) := l_t(ˆθ)− ˆθ, we have:

T_EQU− TOP T = l_t(ˆθ)− TOP T = ˆL + ˆθ− TOP T (∗)

≤ ˆL,

Figure 3.1: The diagram of “PoA of 2 Example”. Left part is the parallel­link network with edges{e1 = (a, 0), e₂ = (1,¹₃)}; Right part is the inflow function, which is a step function with range={1, a}.

last inequality holds by considering the leaving time and the arriving time of last particle in OPT flow. On the other hand, for the total amount M and the inequality, we have:

M ≥ Q(ˆθ) ≥ 1

last inequality holds since we can copy the infinite inflow during θ = 0 ∼ ¹_kL(ˆθ) for k times as an option of infinite inflow during θ = 0 ∼ L(ˆθ). On the other hand, since we have to send M mass of particle in OPT flow, so:

T_{OP T} ≥ 1

Theorem 1. In the fluid queuing model, the Price of Anarchy of 2 for parallel­link net­

works is tight.

Proof. Gien parallel­link network C, inflow µ₀, denoted e_j = (v_j, τ_j) for all e_j ∈ E. At any moment θ₀, suppose EQU flow use n edges, then we have:

L(θ₀)≤ τn+1.

For each edge, the travel time minus the delay time of edge is the queuing time. Hence, we can calculate the mass of particle queuing in C at θ0as an lower bound of Q(θ0):

Q(θ₀)≥

∑n i=1

v_i[L(θ₀)− τi],

where Q(θ₀) is the particle existing in C at θ₀. On the other hand, for each edge, the given time of M_put minus the delay time of edge is the total time to pass particle. We have:

M_put(L(θ₀)) =

∑n i=1

v_i[L(θ₀)− τi].

This implies Q(θ0) ≥ Mput(L(θ0)). Follow from Lemma 1, we have P oA ≤ 2. Fur­

thermore, let’s provide an example for the Price of Anarchy of 2. Given 1 > a > 0, consider:

Theorem 2. In the fluid queuing model, if all edges in 1 aggregated network, the Price of Anarchy of 2−_1+ϵ¹ ≈ 1 + ϵ for parallel­group­link networks is tight, where ϵ is given by the network.

Proof. Given parallel­group­link network C whose edges are in 1 aggregated network, inflow µ₀, denoted e_j = (v_j, τ_j) for all e_j ∈ E. At the last leaving time ˆθ, for the shortest travel time ˆL, we have the condition:

Lˆ

τ_j ≤ 1 + ϵ, ∀ej ∈ E.

Similar as Lemma 1, we have:

T_EQU − TOP T = ˆL + ˆθ− TOP T ≤ (1 − 1

1 + ϵ) ˆL + τ₁+ ˆθ− TOP T (∗)

≤ (1 − 1 1 + ϵ) ˆL,

where the last inequality holds by considering the leaving time plus minimal delay time and the arriving time of last particle in OPT flow. Hence, we afford:

2− 1

1 + ϵ ≥ T_EQU

T_{OP T} = PoA.

Finally, similar as Theorem 1, the example for the Price of Anarchy of 2− _1+ϵ¹ is shown as below:

Lemma 2. Consider the fluid queuing model with network C, inflow µ₀, the total capacity of minimal cut face Min­Cut(C), the inequality holds:

M ≤ TOP T ∗ Min­Cut(C),

where M is the total amount and T_{OP T} is the arriving time of all particle in OPT flow.

Proof. In this case, for OPT flow, each particle from s to t has to pass by one of the edges contained in the minimal cut face of C. And, the throughput of the minimal cut face is at most Min­Cut(C) at any moment. As a result, the inequality holds.

Theorem 3. In the fluid queuing model, given network C, inflow µ0. If each of C’s edge is used by OPT flow at some moment, the Price of Anarchy is at most 2|V | − 1.

Proof. At the last leaving time ˆθ, record l_v(ˆθ) for all v ∈ E − {s} and sort them into

Now, for network C, interval (t_i−1, t_i), consider the Min­Cut(C) and an index set of interval S :={k|ek ∈ [ti−1, ti]}, we have:

Conjecture 1. In the fluid queuing model, removing each OPT­unused edge in networks only worsens the Price of Anarchy. Hence, the Price of Anarchy is at most 2|V | − 1 for networks.

Remark 4. In parallel­link networks, this conjecture is true. In general networks, this conjecture could be false, take Pigou’s example for an example. The conjecture is possible to be true in series­parallel networks.

Theorem 4. In the fluid queuing model, the Price of Anarchy is at most D(C) for series­

parallel networks C, where D(C) denotes the maximal number of each path’s node among all s to t paths.

Claim: Given a series­parallel network C, inflow µ₀, any moment θ₀. For the through­

put flow M_put, the shortest travel time L, and the mass of particle existing in network Q.

We have the inequality:

M_put(C,∞, L(C, µ0, θ₀))≤ (D(C) − 1)Q(C, µ0, θ₀), which implies the Theorem from Lemma 1.

Proof. (a)Base Case

In parallel­link network, D(C) = 2. The statement is true by Theorem 1.

(b)Parallel­linking

After parallel­linking operation, series­parallel networks C₁ links to C₂. For C’s inflow µ₀, define the EQU flow’s inflow of C₁ is µ^up₀ , inflow of C₂ is µ^dn₀ , and µ₀ = µ^up₀ + µ^dn₀ ; for any moment θ₀. we have:

L(C₁, µ^up₀ , θ₀) ≥ L(C, µ0, θ₀), L(C₂, µ^dn₀ , θ₀) ≥ L(C, µ0, θ₀),

Q(C, µ₀, θ₀) = Q(C₁, µ^up₀ , θ₀) + Q(C₂, µ^dn₀ , θ₀).

Induction hypothesis of Claim:

given C₁, µ^up₀ , θ₀ : M_put(C₁,∞, L(C1, µ^up₀ , θ₀))≤ (D(C1)− 1)Q(C1, µ^up₀ , θ₀), given C₂, µ^dn₀ , θ₀ : M_put(C₂,∞, L(C2, µ^dn₀ , θ₀))≤ (D(C2)− 1)Q(C2, µ^dn₀ , θ₀),

This implies:

M_put(C,∞, L(C, µ0, θ₀))

(∗)

≤ sum of LHS

≤ sum of RHS

≤ max(D(C1)− 1, D(C2)− 1)[Q(C1, µ^up₀ , θ₀) + Q(C₂, µ^dn₀ , θ₀)]

= (D(C)− 1)Q(C, µ0, θ₀),

where D(C) = max(D(C₁), D(C₂)). And, the first inequality holds since the throughput of C is just the sum of throughput of C₁and C₂ with same period time.

(c)Series­linking

After series­linking operation, series­parallel networks C₁ links to C₂. For C₁’s inflow µ₀, define the EQU flow’s outflow of C₁is µ₁, and µ₁is C₂’s inflow; for any moment θ₀, define θ₁ = l_p(θ₀). We have:

L(C, µ₀, θ₀) = l_t(θ₀)− θ0

= l_t(θ₀)− θ1+ l_p(θ₀)− θ0

= L(C₂, µ₁, θ₁) + L(C₁, µ₀, θ₀),

Q(C, µ₀, θ₀)

(∗)

≥ Q(C1, µ₀, θ₀), Q(C, µ₀, θ₀)

(∗∗)

≥ Q(C2, µ₀, θ₀).

Inequality (*) holds by consider the mass of particle existing in C1 at θ = θ0 as a lower bound; Inequality (**) holds by consider the mass of particle existing in C₂at θ = θ₁as a lower bound.

Induction hypothesis of Claim:

given C₁, µ₀, θ₀ : M_put(C₁,∞, L(C1, µ₀, θ₀))≤ (D(C1)− 1)Q(C1, µ₀, θ₀), given C₂, µ₁, θ₁ : M_put(C₂,∞, L(C2, µ₁, θ₁))≤ (D(C2)− 1)Q(C2, µ₁, θ₁),

This implies:

M_put(C,∞, L(C, µ0, θ₀))

(∗)

≤ sum of LHS

≤ sum of RHS

≤ [D(C1)− 1 + D(C2)− 1]Q(C, µ0, θ₀)

= (D(C)− 1)Q(C, µ0, θ₀),

where D(C) = D(C1) + D(C2)− 1. Moreover, the first inequality holds since we can consider C₂sending infinite particle without delay time in θ = 0∼ θ1and C₁sending infinite particle without delay time in θ = θ1 ∼ lt(θ0) as an upper bound of Mput.

3.3 (2 + 2)­parallel­link Network

Remark 5. We remove all edges exceed m₁ and m₂, where m₁ and m₂ are the maximal number of used edges in OPT flow or EQU flow in C₁ and C₂ parts respectively. And, this has no influence on our model. On the other hand, we use e_j = (a_j, σ_j) in C₁ and e_j = (v_j, τ_j) in C₂.

Now, consider the flows on network C₁ and inflow µ₀. Define T_EQU|C1,T_{OP T}|C1 as the arriving time of all particle in EQU flow and OPT flow respectively. Let’s see a new model named by ”Restricted Inflow Model” as below:

Definition 10 (Restricted Inflow Model). In EQU flow, consider any moment θ₀such that µ0(θ0) ≥ a1 + ... + am1 and lt(θ0) = σm1. We imitate the OPT flow, let µ0(θ0)− (a1 + ... + a_m1) mass of particles wait at s until µ₀ ≤ a1+ ... + a_m1 and all previous particles have leaven.

In EQU flow after the operation, the arriving time of each particle is the same as usual.

Since the order of any pair of particles is conserved and each edge passing the same mass of particles as usual. Furthermore, the travel time of each particle is at most σ_m1 in the Restricted inflow model. Finally, each particle in EQU flow would depart earlier or equal to OPT flow, since there is always no more particle wait at s in EQU flow.

Definition 11 (Different Arriving Time). Given parallel­link network C₁, inflow µ₀, each particle p from µ₀. Define d(C₁, µ₀, p) by the different arriving time of EQU flow to OPT flow for particle p. Define d(C₁, µ₀) = sup_pd(C₁, µ₀, p).

Follow from definition, we have:

d(C1, µ0, p)≤ σm1. TEQU|C1 − TOP T|C1

(∗)

≤ d(C1, µ0)≤ σm1,

last inequality holds by considering the last particle as a special case of particle p.

Definition 12 (EO flow). Given (2 + 2)­parallel­link network C1 + C₂, inflow µ₀. For EQU flow in network C₁ using inflow µ₀, we have the outflow of C₁ as the inflow of C₂,

denoted by µ₁. Now, consider the OPT flow of network C₂using inflow µ₁, this is the EO flow in (2 + 2)­parallel­link network C₁+ C₂. And, define the arriving time of all particle in EO flow by the arriving time of all particle in OPT flow of network C₂using inflow µ₁.

Define n₂ is the maximal number of used edges in EO flow or EQU flow in C₂part.

We apply the above lemma on networks C₂, inflow µ₁, we have:

T_EQU − TEO ≤ τn2 ≤ τm2,

where And, m₂ is the maxiaml number in OPT flow or EQU flow in C₂ part. Since we remove all edges exceed m₂, n₂ ≤ m2.

Lemma 3. In the fluid queuing model, given (2 + 2)­parallel­link network C₁ series­

linking to C2 and inflow µ0, denoted ej = (aj, σj) in C1 and ej = (vj, τj) in C2. For the OPT flow and EQU flow, we have:

T_EQU − TOP T ≤ σm1 + τ_m2.

Proof. Follow from above, for the EO flow in this (2 + 2)­parallel­link network, we have:

T_EQU− TEO ≤ τm2, T_EQU|C1 − TOP T|C1 ≤ σm1.

In EO flow, consider a strategy as below: For all particle arriving source of C₂, let them wait at source for σm1 − d(C1, µ0, p) time and leave. In this case, EO flow has the same inflow of C₂as OPT flow, but delay for exactly σ_m1 time. That is:

OPT’s inflow of C₂ : µ₂(θ), EO’s inflow of C₂ : µ₂(θ + σ_m1),

where µ₂(θ) is the inflow of C₂in OPT flow. This relationship implies T_EO−TOP T = σ_m1. However, EO flow may choose some better solutions, implies

T_EO− TOP T ≤ σm1,

T_EQU − TOP T = T_EQU− TEO+ T_EO− TOP T ≤ σm1 + τ_m2.

Remark 6. Inthe proof of Lemma 3, for EO flow, we let the particle wait at the intermedi­

ate point p. Despite this is not allowed in our model, it only worsens T_EQUand makes the PoA bigger. The upper bound of PoA is still true if we allowed waiting at the intermediate point p. So, we allow it.

Lemma 4. Given (2 + 2)­parallel­link network C₁+ C₂, T_{OP T} = T_EOor T_{OP T} ≥ σ2+ τ₂, where T_{OP T}, T_EOis the arriving time of all particle in OPT flow and EO flow respectively and (a₁, σ₁), (a₂, σ₂) denotes the edges of C₁, (v₁, τ₁), (v₂, τ₂) denotes the edges of C₂.

Proof. Given (2 + 2)­parallel­link network C₁ + C₂, inflow µ₀, the maximal number of used edges in OPT flow or EQU flow in C₁ part and C₂ part are m₁ and m₂ respectively.

Now, if m₁ = 1, then T_{OP T} = T_EO. Else, consider m₁ = 2. Let’s divide it into two cases:

(a)m₁ = 2 and a₁ ≥ v1

During θ = 0 ∼ σ2, OPT flow and EO flow have the same inflow and outflow at C₂. Now, if m₂ = 1, after θ = σ + 2, OPT flow and EO flow still have the same outflow at C₂, implying T_{OP T} = T_EO. Else, m₂ = 2, both OPT and EO active (v₂, τ₂) at θ = σ₁.

• If EO flow shutdown (v₂, τ₂)f irst, then T_EO ≤ TOP T, implies T_EQ = T_{OP T}.

• If OPT flow shutdown (v₂, τ₂)f irst but before θ = σ₂+ τ₂, then OPT flow and EO flow have the same outflow at C₂ later, which will pass the same mass of particle.

We let EO flow shutdown (v₂, τ₂) at the same time to afford T_EQ = T_{OP T}.

• If OPT flow shutdown (v₂, τ₂)f irst and after θ = σ₂+ τ₂, then T_{OP T} ≥ σ2+ τ₂.

(b)m₁ = 2 and a₁ ≤ v1

If m₂ = 1, consider the EO flow and OPT flow. Maybe EO flow and OPT flow have the total same outflow, then T_EO = T_{OP T}; Else, EO flow and OPT flow must have the same outflow during θ = 0∼ σ2+ τ₁, implies T_{OP T} ≥ σ2 + τ₁. Both of the cases are enough to prove PoA≤ 2.

Else, m₂ = 2, which implies a₁+ a₂ ≥ v2 ≥ a1 since v₂is used. Now, if OPT flow

use (v₂, τ₂), then T_{OP T} ≥ σ2+ τ₂. Else, EQU flow use (v₂, τ₂), we can compute: On the other hand, in EQU flow, let’s find a lower bound of the total mass of particle arriving at the source of C₁+ C₂ before activating (v₂, τ₂) as an lower bound of M :

Theorem 5. In the fluid queuing model, the Price of Anarchy of 2 for (2 + 2)­parallel­link networks is tight.

Chapter 4

Extension

We will show a self­defined tax scheme and improve the PoA’s bound with it in the fluid queuing model. Although it seems helpless for the networks with constant inflow, it is helpful for the parallel­link networks with some extreme inflow cases.

Definition 13 (Delay­time Tax Scheme). In the fluid queuing model, we increase the delay time of edges by imposing tax in the given networks. The tax is not considered as the cost of society’s welfare. That is, the tax scheme will only change the behavior of particles, and the computation of delay time and travel time still uses the original setting. This is called the delay­time tax scheme, refer to [11].

Remark 7. Under the definition of the Delay­time Tax Scheme, OPT flow would not change anything after taxing. This is because the computation of delay time and travel time still uses the original setting. As a result, we only discuss EQU flow and the changes on the arriving time of all particle in EQU flow, T_EQU.

Theorem 6. In the fluid queuing model with constant inflow, the Price of Anarchy is at least ⁴₃ for parallel­link networks after taxing.

Proof. Let’s show an example of the Price of Anarchy of ⁴₃, refer to [10]. Consider the parallel­link network with two edges, the total amount M , and the constant inflow function

Figure 4.1: The diagram of “PoA of ⁴₃ Example” This is the parallel­link network with edges {e1 = (0.5, 0), e2 = (1, 1)}, the total amount M = 1, and the constant inflow function µ₀ = 1.

µ0 as below:

E ={e1 = (0.5, 0), e₂ = (1, 1)}, M = 1,

µ₀(θ) = 1.

In this example, we have:

T_{OP T} = 1.5, T_EQU = 2, PoA = 4 3.

Now, for this example, we apply the Delay­time tax scheme into the below 3 cases, and discuss the influence on T_EQU:

(a)Taxing on both edges

This can reduce to ”Taxing on single edge” case, since the behavior of particles only be influenced by the difference of two edges’ delay time, but not the exactly value of each edge’s delay time.

(b)Taxing on down edge e₂ = (1, 1)

In EQU flow, all particle still choose up edge as usual. Nothing is changed. We have:

T_EQU = 2 and PoA = ⁴₃ after taxing.

(c)Taxing on up edge e₁ = (0.5, 0)

Denote the delay time of e₁ by x after taxing. Consider EQU flow:

• if x > 1, then all particle choose down edge. We have: T_EQU = 2 and PoA = ⁴₃ after taxing.

• if x = 1, then passing two edges spends the same time. That is, there are infinitely many Nash equilibrium, and we consider the worst­case, ”the last particle go down edge”. We have: TEQU = 2 and PoA = ⁴₃ after taxing.

• if x < 1, then the behavior of particle is as below: During θ = 0 ∼ 1 − x, all particles go up edge and queue length is ¹₂(1− x) finally. From now on, passing two edges spends the same time. That is, there are infinitely many Nash equilibrium, and we consider the worst­case. We have: T_EQU = 2 and PoA = ⁴₃ after taxing.

As a result, the Price of Anarchy for this example is ⁴₃ after taxing.

Theorem 7. In the fluid queuing model with constant inflow, the Price of Anarchy is at least _e−1^e after taxing on single edge.

Proof. Let’s show an example of the Price of Anarchy of_e−1^e , refer to [10]. Consider the series­parallel­link network with 2m edges, the total amount M , and the constant inflow function µ₀ as below:

Now, for this example, we apply the Delay­time tax scheme on single edge into the below 2 cases, and discuss the influence on T_EQU:

(a)Taxing on up edge e_i = (uⁱ, αµ^m(_u ¹

OP T −_u¹_i)):

In EQU flow, all particles go down edge firstly and become part of queuing until t = α.

At this moment, inflow stop! For non­taxing­up­edges, they are about to be activated;

For the only taxing­up­edge e_i, it is not yet to be activated. So, T_EQU > αµ^m and

Figure 4.2: The diagram of “PoA of_e−1^e Example” This is the series­parallel­link network with edges {ei = (u_i, 0), eⁱ = (uⁱ, αµ^m(_u ¹

OP T − _u¹_i))}i=1∼m, the total amount M = αµ^m, and the constant inflow function µ₀ = µ^m, where m∈ N, α > 0, µⁱ =∑i

k=1µ_k.

lim_m→∞PoA > _e−1^e .

(b)Taxing on down edge eⁱ = (u_i, 0):

Denote the delay time of eⁱ by x after taxing.

• In EQU flow, if x > αµ^m(_u ¹

OP T −_u¹_i) or x = αµ^m(_u ¹

OP T −_u¹_i), similar as Theorem 6, then all particles coming v_i+1go up road ei+1. This makes TEQU the same and lim_m→∞PoA = _e−1^e .

• In EQU flow, if x < αµ^m(_u ¹

OP T −_u¹_i), then all particles go down edge firstly. And, the up road e_i+1 is activated earlier than t = α, implies l_vi+1(θ) = τ_i+1. Despite the up orad e_i+1 is activated earlier, but the subsystem from v_i+1 to v₁ would not finish earlier. This is because the inflow from v_i+1coming e_i+1and eⁱ is the same as usual. So, T_EQU = αµ^m and lim_m→∞PoA = _e−1^e .

As a result, the Price of Anarchy for this example is _e−1^e after taxing on single edge.

Figure 4.3: The diagram of “PoA of 2 Example”. Left part is the parallel­link network with 2 edges{e1 = (a, σ), e₂ = (b, τ )}; Right part is the inflow function, which is a step function with range={1, a}.

Theorem 8. In the fluid queuing model with dynamic inflow, the Price of Anarchy can be reduced by the delay­time tax scheme in some cases.

Proof. Let’s show the parallel­link network with 2 edges as example, whose Price of An­

archy is 1 + ϵ after taxing with given ϵ. This example is similar as Chapter 3­Therorem 1.

Given a, σ, b, τ such that a + b≥ 1, σ < τ, consider:

Now, we apply the Delay­time tax scheme on this example. Given ϵ > 0, tax on up edge e₁ = (a, σ) such that e₁ = (a, τ−ϵ). In EQU flow, the behavior of particles are as follow:

All particles go up­road and queue length achieves aϵ, 0 ∼ ϵ a 1− a, Particles go up road at speed a; down road at speed 1− a, ϵ a

1− a ∼ (τ − σ) a 1− a, Inflow stop! Consume the queuing, (τ − σ) a

1− a ∼ (τ − σ) a

1− a + τ + ϵ.

As a result, T_EQU = (τ − σ)_1−a^a + τ + ϵ and PoA = 1 + ⁽¹_τ−aσ^−a)ϵ → 1 as ϵ → 0⁺.

Chapter 5

Conclusions and Future Works

We find upper bounds of the PoA of networks with dynamic inflow. We prove the PoA of 2 of parallel­link networks and (2 + 2)­parallel­link networks, the PoA of D(C) of series­

parallel networks, the PoA of 2|V | − 1 of general networks with assumption. We reduce the upper bound of PoA of parallel­link networks or series­parallel networks from infinite to 2 and D(C) respectively. The bounds we proved are different from networks with constant inflow in the fluid queuing model. On the other hand, similar to the work of tax scheme [11], we design the Delay­time tax scheme to improve the system’s inefficiency, which may help a lot on networks with dynamic inflow. Our main work is to use the total amount of inflow as an upper bound of the maximal throughput of networks to afford the lower bound of the cost of optimal flows (or said optimal time) in the fluid queuing model.

This technique helps us finding PoA’s tight bound of parallel­link networks and a simple example of series­parallel networks, even a loose bound of series­parallel networks. This

This technique helps us finding PoA’s tight bound of parallel­link networks and a simple example of series­parallel networks, even a loose bound of series­parallel networks. This

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