A simpler proof for O ( congestion + dilation ) packet routing - - PowerPoint PPT Presentation

A simpler proof for O(congestion + dilation) packet routing

Thomas Rothvoß

Department of Mathematics, MIT

IPCO 2013

Packet Routing

◮ Input:

directed graph G = (V, E)

Packet Routing

◮ Input: Paths Pi in a directed graph G = (V, E)

Packet Routing

◮ Input: Paths Pi in a directed graph G = (V, E) ◮ Goal: Route packets along paths to minimize makespan

Constraint: edge can be crossed by 1 packet per time unit

Packet Routing Time: t = 0

1 2 3 4 5 6

◮ Input: Paths Pi in a directed graph G = (V, E) ◮ Goal: Route packets along paths to minimize makespan

Constraint: edge can be crossed by 1 packet per time unit

Packet Routing Time: t = 0

1 2 3 4 5 6

◮ Input: Paths Pi in a directed graph G = (V, E) ◮ Goal: Route packets along paths to minimize makespan

Constraint: edge can be crossed by 1 packet per time unit

Packet Routing Time: t = 0

1 2 3 4 5 6

◮ Input: Paths Pi in a directed graph G = (V, E) ◮ Goal: Route packets along paths to minimize makespan

Constraint: edge can be crossed by 1 packet per time unit

Packet Routing Time: t = 0

1 2 3 4 5 6

◮ Input: Paths Pi in a directed graph G = (V, E) ◮ Goal: Route packets along paths to minimize makespan

Constraint: edge can be crossed by 1 packet per time unit

Packet Routing Time: t = 1

1 2 3 4 5 6

◮ Input: Paths Pi in a directed graph G = (V, E) ◮ Goal: Route packets along paths to minimize makespan

Constraint: edge can be crossed by 1 packet per time unit

Packet Routing Time: t = 1

1 2 3 4 5 6

◮ Input: Paths Pi in a directed graph G = (V, E) ◮ Goal: Route packets along paths to minimize makespan

Constraint: edge can be crossed by 1 packet per time unit

Packet Routing Time: t = 1

1 2 3 4 5 6

◮ Input: Paths Pi in a directed graph G = (V, E) ◮ Goal: Route packets along paths to minimize makespan

Constraint: edge can be crossed by 1 packet per time unit

Packet Routing Time: t = 1

1 2 3 4 5 6

◮ Input: Paths Pi in a directed graph G = (V, E) ◮ Goal: Route packets along paths to minimize makespan

Constraint: edge can be crossed by 1 packet per time unit

Packet Routing Time: t = 1

1 2 3 4 5 6

◮ Input: Paths Pi in a directed graph G = (V, E) ◮ Goal: Route packets along paths to minimize makespan

Constraint: edge can be crossed by 1 packet per time unit

Packet Routing Time: t = 2

1 2 3 4 5 6

◮ Input: Paths Pi in a directed graph G = (V, E) ◮ Goal: Route packets along paths to minimize makespan

Constraint: edge can be crossed by 1 packet per time unit

Packet Routing Time: t = 2

1 2 3 4 5 6

◮ Input: Paths Pi in a directed graph G = (V, E) ◮ Goal: Route packets along paths to minimize makespan

Constraint: edge can be crossed by 1 packet per time unit

Packet Routing Time: t = 2

1 2 3 4 5 6

◮ Input: Paths Pi in a directed graph G = (V, E) ◮ Goal: Route packets along paths to minimize makespan

Constraint: edge can be crossed by 1 packet per time unit

Packet Routing Time: t = 2

1 2 3 4 5 6

◮ Input: Paths Pi in a directed graph G = (V, E) ◮ Goal: Route packets along paths to minimize makespan

Constraint: edge can be crossed by 1 packet per time unit

Packet Routing Time: t = 2

1 2 3 4 5 6

◮ Input: Paths Pi in a directed graph G = (V, E) ◮ Goal: Route packets along paths to minimize makespan

Constraint: edge can be crossed by 1 packet per time unit

Packet Routing Time: t = 3

1 2 3 4 5 6

◮ Input: Paths Pi in a directed graph G = (V, E) ◮ Goal: Route packets along paths to minimize makespan

Constraint: edge can be crossed by 1 packet per time unit

Packet Routing Time: t = 3

1 2 3 4 5 6

◮ Input: Paths Pi in a directed graph G = (V, E) ◮ Goal: Route packets along paths to minimize makespan

Constraint: edge can be crossed by 1 packet per time unit

Packet Routing Time: t = 3

1 2 3 4 5 6

◮ Input: Paths Pi in a directed graph G = (V, E) ◮ Goal: Route packets along paths to minimize makespan

Constraint: edge can be crossed by 1 packet per time unit

Packet Routing Time: t = 3

1 2 3 4 5 6

◮ Input: Paths Pi in a directed graph G = (V, E) ◮ Goal: Route packets along paths to minimize makespan

Constraint: edge can be crossed by 1 packet per time unit

Packet Routing Time: t = 3

1 2 3 4 5 6

◮ Input: Paths Pi in a directed graph G = (V, E) ◮ Goal: Route packets along paths to minimize makespan

Constraint: edge can be crossed by 1 packet per time unit

Packet Routing Time: t = 4

1 2 3 4 5 6

◮ Input: Paths Pi in a directed graph G = (V, E) ◮ Goal: Route packets along paths to minimize makespan

Constraint: edge can be crossed by 1 packet per time unit

Packet Routing Time: t = 4

1 2 3 4 5 6

◮ Input: Paths Pi in a directed graph G = (V, E) ◮ Goal: Route packets along paths to minimize makespan

Constraint: edge can be crossed by 1 packet per time unit

Packet Routing Time: t = 4

1 2 3 4 5 6

◮ Input: Paths Pi in a directed graph G = (V, E) ◮ Goal: Route packets along paths to minimize makespan

Constraint: edge can be crossed by 1 packet per time unit

Packet Routing Time: t = 4

1 2 3 4 5 6

◮ Input: Paths Pi in a directed graph G = (V, E) ◮ Goal: Route packets along paths to minimize makespan

Constraint: edge can be crossed by 1 packet per time unit

Packet Routing Time: t = 4

1 2 3 4 5 6

◮ Input: Paths Pi in a directed graph G = (V, E) ◮ Goal: Route packets along paths to minimize makespan

Constraint: edge can be crossed by 1 packet per time unit

Packet Routing Time: t = 5

1 2 3 4 5 6

◮ Input: Paths Pi in a directed graph G = (V, E) ◮ Goal: Route packets along paths to minimize makespan

Constraint: edge can be crossed by 1 packet per time unit

Packet Routing Time: t = 5

1 2 3 4 5 6

◮ Input: Paths Pi in a directed graph G = (V, E) ◮ Goal: Route packets along paths to minimize makespan

Constraint: edge can be crossed by 1 packet per time unit

Packet Routing Time: t = 5

1 2 3 4 5 6

◮ Input: Paths Pi in a directed graph G = (V, E) ◮ Goal: Route packets along paths to minimize makespan

Constraint: edge can be crossed by 1 packet per time unit

Packet Routing Time: t = 5

1 2 3 4 5 6

◮ Input: Paths Pi in a directed graph G = (V, E) ◮ Goal: Route packets along paths to minimize makespan

Constraint: edge can be crossed by 1 packet per time unit

Packet Routing Time: t = 5

1 2 3 4 5 6

◮ Input: Paths Pi in a directed graph G = (V, E) ◮ Goal: Route packets along paths to minimize makespan

Constraint: edge can be crossed by 1 packet per time unit

Packet Routing Time: t = 6

1 2 3 4 5 6

◮ Input: Paths Pi in a directed graph G = (V, E) ◮ Goal: Route packets along paths to minimize makespan

Constraint: edge can be crossed by 1 packet per time unit

Packet Routing (2)

Lower bounds:

Packet Routing (2)

Lower bounds:

◮ congestion = max e∈E {#i : e ∈ Pi}

Packet Routing (2)

Lower bounds:

◮ congestion = max e∈E {#i : e ∈ Pi} ◮ dilation = maxi |Pi|

Packet Routing (2)

Lower bounds:

◮ congestion = max e∈E {#i : e ∈ Pi} ◮ dilation = maxi |Pi|

Theorem (Leighton, Maggs, Rao ’94)

∃ schedule of length O(congestion + dilation) (even with O(1)-size edge buffers).

Packet Routing (2)

Lower bounds:

◮ congestion = max e∈E {#i : e ∈ Pi} ◮ dilation = maxi |Pi|

Theorem (Leighton, Maggs, Rao ’94)

∃ schedule of length O(congestion + dilation) (even with O(1)-size edge buffers). “highly involved”

Packet Routing (2)

Lower bounds:

◮ congestion = max e∈E {#i : e ∈ Pi} ◮ dilation = maxi |Pi|

Theorem (Leighton, Maggs, Rao ’94)

∃ schedule of length O(congestion + dilation) (even with O(1)-size edge buffers). “highly involved” “very difficult to understand”

Packet Routing (2)

Lower bounds:

◮ congestion = max e∈E {#i : e ∈ Pi} ◮ dilation = maxi |Pi|

Theorem (Leighton, Maggs, Rao ’94)

∃ schedule of length O(congestion + dilation) (even with O(1)-size edge buffers). “highly involved” “very difficult to understand” “technical tour de force”

Packet Routing (2)

Lower bounds:

◮ congestion = max e∈E {#i : e ∈ Pi} ◮ dilation = maxi |Pi|

Theorem (Leighton, Maggs, Rao ’94)

∃ schedule of length O(congestion + dilation) (even with O(1)-size edge buffers). “highly involved” “very difficult to understand” “technical tour de force”

◮ Polytime algorithm [Leighton, Maggs, Richa ’99]

Packet Routing (2)

Lower bounds:

◮ congestion = max e∈E {#i : e ∈ Pi} ◮ dilation = maxi |Pi|

Theorem (Leighton, Maggs, Rao ’94)

∃ schedule of length O(congestion + dilation) (even with O(1)-size edge buffers). “highly involved” “very difficult to understand” “technical tour de force”

◮ Polytime algorithm [Leighton, Maggs, Richa ’99] ◮ 39 · (C + D) suffices [Scheideler ’98]

Packet Routing (2)

Lower bounds:

◮ congestion = max e∈E {#i : e ∈ Pi} ◮ dilation = maxi |Pi|

Theorem (Leighton, Maggs, Rao ’94)

∃ schedule of length O(congestion + dilation) (even with O(1)-size edge buffers). “highly involved” “very difficult to understand” “technical tour de force”

◮ Polytime algorithm [Leighton, Maggs, Richa ’99] ◮ 39 · (C + D) suffices [Scheideler ’98] ◮ 24 · (C + D) suffices [Peis, Wiese ’11]

Packet Routing (2)

Lower bounds:

◮ congestion = max e∈E {#i : e ∈ Pi} ◮ dilation = maxi |Pi|

Theorem (Leighton, Maggs, Rao ’94)

∃ schedule of length O(congestion + dilation) (even with O(1)-size edge buffers). “highly involved” “very difficult to understand” “technical tour de force”

◮ Polytime algorithm [Leighton, Maggs, Richa ’99] ◮ 39 · (C + D) suffices [Scheideler ’98] ◮ 24 · (C + D) suffices [Peis, Wiese ’11] ◮ O(1)-apx for finding paths + schedule [Srinivasan, Teo ’00]

Main result

Theorem (R. ’13)

Much simpler proof of O(congestion + dilation)-packet routing (also with O(1)-size edge buffers).

Main result

Theorem (R. ’13)

Much simpler proof of O(congestion + dilation)-packet routing (also with O(1)-size edge buffers).

◮ [Wiese ’12]: Is (1 + o(1)) · (congestion + dilation) possible? ◮ True if congestion ≫ dilation!

Main result

Theorem (R. ’13)

Much simpler proof of O(congestion + dilation)-packet routing (also with O(1)-size edge buffers).

◮ [Wiese ’12]: Is (1 + o(1)) · (congestion + dilation) possible? ◮ True if congestion ≫ dilation!

No!!

Theorem (R. ’13)

∃ instance requiring (1 + ε) · (congestion + dilation) time.

Main result

Theorem (R. ’13)

Much simpler proof of O(congestion + dilation)-packet routing (also with O(1)-size edge buffers).

◮ [Wiese ’12]: Is (1 + o(1)) · (congestion + dilation) possible? ◮ True if congestion ≫ dilation!

No!!

Theorem (R. ’13)

∃ instance requiring (1 + ε) · (congestion + dilation) time. Assumptions:

◮ D := dilation = congestion = |Pi| ∀i

Main result

Theorem (R. ’13)

Much simpler proof of O(congestion + dilation)-packet routing (also with O(1)-size edge buffers).

◮ [Wiese ’12]: Is (1 + o(1)) · (congestion + dilation) possible? ◮ True if congestion ≫ dilation!

No!!

Theorem (R. ’13)

∃ instance requiring (1 + ε) · (congestion + dilation) time. Assumptions:

◮ D := dilation = congestion = |Pi| ∀i ◮ O(1) packets can cross an edge per time unit

Preliminaries

Lemma (Lov´ asz Local Lemma)

Let A1, . . . , Am be events such that (1) Pr[Ai] ≤ p (2) each Ai depends on ≤ d other events (3) 4 · p · d ≤ 1 Then Pr m

i=1 ¯

Ai

• > 0.

◮ Constructive via [Moser, Tardos ’10]

Preliminaries

Lemma (Lov´ asz Local Lemma)

Let A1, . . . , Am be events such that (1) Pr[Ai] ≤ p (2) each Ai depends on ≤ d other events (3) 4 · p · d ≤ 1 Then Pr m

i=1 ¯

Ai

• > 0.

◮ Constructive via [Moser, Tardos ’10]

Lemma (Chernov-Hoeffding)

Let Z1, . . . , Zk ∈ [0, δ] be independently RV, sum Z := k

i=1 Zi.

Then Pr[Z > (1 + ε) E[Z]] ≤ exp

• − ε2

3 · E[Z] δ

• .

A probabilistic schedule

A probabilistic schedule

A probabilistic schedule

b b b b b b b b b b

A probabilistic schedule

b b b b b b b b b b

level 0 D

A probabilistic schedule

b b b b b b b b b b

level 0 level 1 D1 := √ D

A probabilistic schedule

b b b b b b b b b b

level 0 level 1 level 2 D2 := √D1

A probabilistic schedule

b b b b b b b b b b

level 0 level 1 level 2 D2 := √D1 Waiting rule:

◮ At source: wait α0 ∼ [D] ◮ When entering kth level ℓ interval: Wait αℓ,k ∼ [D1/4 ℓ

]

◮ When leaving kth level ℓ interval: Wait D1/4 ℓ

− αℓ,k

A probabilistic schedule

b b b b b b b b b b

level 0 level 1 level 2 D2 := √D1 time O(D) Level 0 waiting time Level 1 waiting time Level 2 waiting time move

A probabilistic schedule

b b b b b b b b b b

level 0 level 1 level 2 D2 := √D1 D1/4

1

∼ [D] ∼ [D1/4

1

] time O(D)

A probabilistic schedule

b b b b b b b b b b

level 0 level 1 level 2 D2 := √D1 Observations:

◮ total waiting time: D + ℓ≥1 D Dℓ · D1/4 ℓ

= O(D)

A probabilistic schedule

e

b b b b b b b b b b

level 0 level 1 level 2 D2 := √D1 e

b b

Observations:

◮ total waiting time: D + ℓ≥1 D Dℓ · D1/4 ℓ

= O(D)

◮ time that i crosses e depends only on waiting times of

intervals containing e

A probabilistic schedule

e

b b b b b b b b b b

level 0 level 1 level 2 D2 := √D1 e

b b

Observations:

◮ total waiting time: D + ℓ≥1 D Dℓ · D1/4 ℓ

= O(D)

◮ time that i crosses e depends only on waiting times of

intervals containing e

◮ Pr[packet i crosses e at time t] ≤ 1 D & E[load(e, t)] ≤ 1

A probabilistic schedule

b b b b b b b b b b

level 0 level 1 level 2 D2 := √D1

Claim

There exist waiting times s.t. load(e, t) ≤ O(1) ∀e, t

A probabilistic schedule

b b b b b b b b b b

level 0 level 1 level 2 D2 := √D1

Claim

There exist waiting times s.t. load(e, t) ≤ O(1) ∀e, t Idea:

◮ Fix waiting times on level ℓ = 0, 1, 2, . . . iteratively.

A probabilistic schedule

b b b b b b b b b b

level 0 level 1 level 2 D2 := √D1

Claim

There exist waiting times s.t. load(e, t) ≤ O(1) ∀e, t Idea:

◮ Fix waiting times on level ℓ = 0, 1, 2, . . . iteratively. ◮ Show maxe,t{E[load(e, t)]} increases ≤ D − 1

32

ℓ

in step ℓ.

A probabilistic schedule

b b b b b b b b b b

level 0 level 1 level 2 D2 := √D1

Claim

There exist waiting times s.t. load(e, t) ≤ O(1) ∀e, t Idea:

◮ Fix waiting times on level ℓ = 0, 1, 2, . . . iteratively. ◮ Show maxe,t{E[load(e, t)]} increases ≤ D − 1

32

ℓ

in step ℓ.

◮ Eventually load(e, t) ≤ 1 + ℓ≥0 D − 1

32

ℓ

≤ O(1).

Proof

◮ Pick level-0 waiting times α ∼ [D]n. ◮ Y (e, t) := E[load(e, t) | α] = ave. load on e at t dep. on α

Lemma

Pr[Y (e, t) ≤ 1 + D− 1

32 ∀e, t] > 0

Proof

◮ Pick level-0 waiting times α ∼ [D]n. ◮ Y (e, t) := E[load(e, t) | α] = ave. load on e at t dep. on α

Lemma

Pr[Y (e, t) ≤ 1 + D− 1

32 ∀e, t] > 0

◮ E[Y (e, t)] ≤ 1

Proof

◮ Pick level-0 waiting times α ∼ [D]n. ◮ Y (e, t) := E[load(e, t) | α] = ave. load on e at t dep. on α

Lemma

Pr[Y (e, t) ≤ 1 + D− 1

32 ∀e, t] > 0

◮ E[Y (e, t)] ≤ 1 ◮ Y (e, t) = n i=1 Pr[i crosses e at t | α]

• ∈[0,

1 ( √ D)1/4 ]

Proof

◮ Pick level-0 waiting times α ∼ [D]n. ◮ Y (e, t) := E[load(e, t) | α] = ave. load on e at t dep. on α

Lemma

Pr[Y (e, t) ≤ 1 + D− 1

32 ∀e, t] > 0

◮ E[Y (e, t)] ≤ 1 ◮ Y (e, t) = n i=1 Pr[i crosses e at t | α]

• ∈[0,

1 ( √ D)1/4 ]

◮ Pr[Y (e, t) > 1 + D− 1

32 ] ≤ e−Ω(D1/16)

Proof

◮ Pick level-0 waiting times α ∼ [D]n. ◮ Y (e, t) := E[load(e, t) | α] = ave. load on e at t dep. on α

Lemma

Pr[Y (e, t) ≤ 1 + D− 1

32 ∀e, t] > 0

◮ E[Y (e, t)] ≤ 1 ◮ Y (e, t) = n i=1 Pr[i crosses e at t | α]

• ∈[0,

1 ( √ D)1/4 ]

◮ Pr[Y (e, t) > 1 + D− 1

32 ] ≤ e−Ω(D1/16)

• Rand. var.

packets Y (e, t) Y (e′, t′) i ≤ D ≤ O(D2)

Proof

◮ Pick level-0 waiting times α ∼ [D]n. ◮ Y (e, t) := E[load(e, t) | α] = ave. load on e at t dep. on α

Lemma

Pr[Y (e, t) ≤ 1 + D− 1

32 ∀e, t] > 0

◮ E[Y (e, t)] ≤ 1 ◮ Y (e, t) = n i=1 Pr[i crosses e at t | α]

• ∈[0,

1 ( √ D)1/4 ]

◮ Pr[Y (e, t) > 1 + D− 1

32 ] ≤ e−Ω(D1/16)

◮ Dependence degree ≤ O(D3)

• Rand. var.

packets Y (e, t) Y (e′, t′) i ≤ D ≤ O(D2)

Proof

◮ Suppose waiting times on level 0, . . . , ℓ − 1 already fixed. ◮ Pick level-ℓ waiting times α ∼ [∆1/4]n× D

∆ .

∆ := Dℓ

◮ Y (e, t) := E[load(e, t) | α] = ave. load on e at t dep. on α

Lemma

Pr[Y (e, t) ≤ E[Y (e, t)] + ∆− 1

32 ∀e, t] > 0

◮ E[Y (e, t)] ≤ 1+o(1) ◮ Y (e, t) = n i=1 Pr[i crosses e at t | α]

• ∈[0,

1 ( √ ∆)1/4 ]

◮ Pr[Y (e, t) > E[Y (e, t)] + ∆− 1

32 ] ≤ e−Ω(∆1/16)

◮ If nonzero,

Pr[i crosses e at t] ≥

ℓ′≥ℓ 1 D1/4

ℓ′

≥

1 ∆2 ◮ Possible positions & time frame ≤ O(∆) ◮ Dependence degree ≤ O(∆4)

• Rand. var.

packets Y (e, t) Y (e′, t′) i ≤ O(∆2) ≤ O(∆2)

Proof

◮ Suppose waiting times on level 0, . . . , ℓ − 1 already fixed. ◮ Pick level-ℓ waiting times α ∼ [∆1/4]n× D

∆ .

∆ := Dℓ

◮ Y (e, t) := E[load(e, t) | α] = ave. load on e at t dep. on α

Lemma

Pr[Y (e, t) ≤ E[Y (e, t)] + ∆− 1

32 ∀e, t] > 0

◮ E[Y (e, t)] ≤ 1+o(1) ◮ Y (e, t) = n i=1 Pr[i crosses e at t | α]

• ∈[0,

1 ( √ ∆)1/4 ]

◮ Pr[Y (e, t) > E[Y (e, t)] + ∆− 1

32 ] ≤ e−Ω(∆1/16)

◮ If nonzero,

Pr[i crosses e at t] ≥

ℓ′≥ℓ 1 D1/4

ℓ′

≥

1 ∆2 ◮ Possible positions & time frame ≤ O(∆) ◮ Dependence degree ≤ O(∆4)

• Rand. var.

packets Y (e, t) Y (e′, t′) i ≤ O(∆2) ≤ O(∆2)

O(1)-size edge buffers

O(1)-size edge buffers

Theorem

∃ O(congestion + dilation)-time, O(1)-load schedule where packets wait {0, 1} time units per node.

O(1)-size edge buffers

Theorem

∃ O(congestion + dilation)-time, O(1)-load schedule where packets wait {0, 1} time units per node.

b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b

level 0 level 1 level 2

◮ Assign path edges to intervals s.t. level ℓ interval gets D1/4 ℓ

edges

O(1)-size edge buffers

Theorem

∃ O(congestion + dilation)-time, O(1)-load schedule where packets wait {0, 1} time units per node.

b b b b b b b b b b b b b b b b b b b

2 2 2 2 2 2 2 2 2

b b b b b b b b b b b b b b b b b b b

level 0 level 1 level 2

◮ Assign path edges to intervals s.t. level ℓ interval gets D1/4 ℓ

edges

O(1)-size edge buffers

Theorem

∃ O(congestion + dilation)-time, O(1)-load schedule where packets wait {0, 1} time units per node.

b b b b b b b b b b b b b b b b b b b

2 2 2 2 2 2 2 2 2 1 1 1 1 1

b b b b b b b b b b b b b b b b b b b

level 0 level 1 level 2

◮ Assign path edges to intervals s.t. level ℓ interval gets D1/4 ℓ

edges

O(1)-size edge buffers

Theorem

∃ O(congestion + dilation)-time, O(1)-load schedule where packets wait {0, 1} time units per node.

b b b b b b b b b b b b b b b b b b b

2 2 2 2 2 2 2 2 2 1 1 1 1 1

b b b b b b b b b b b b b b b b b b b

level 0 level 1 level 2

◮ Assign path edges to intervals s.t. level ℓ interval gets D1/4 ℓ

edges

O(1)-size edge buffers

Theorem

∃ O(congestion + dilation)-time, O(1)-load schedule where packets wait {0, 1} time units per node.

b b b b b b b b b b b b b b b b b b b

2 2 2 2 2 2 2 2 2 1 1 1 1 1

b b b b b b b b b b b b b b b b b b b

level 0 level 1 level 2

◮ Assign path edges to intervals s.t. level ℓ interval gets D1/4 ℓ

edges

◮ Wait on first α ∼ [D1/4 ℓ

] assigned edges

A lower bound construction

source sink n packets e1 e2 en

A lower bound construction

source sink n packets e1 e2 en

◮ Choose P1, . . . , Pn : go through e1, . . . , en in random order

A lower bound construction

source sink n packets e1 e2 en

◮ Choose P1, . . . , Pn : go through e1, . . . , en in random order

A lower bound construction

source sink n packets e1 e2 en

◮ Choose P1, . . . , Pn : go through e1, . . . , en in random order

A lower bound construction

source sink n packets e1 e2 en

◮ Choose P1, . . . , Pn : go through e1, . . . , en in random order

A lower bound construction

source sink n packets e1 e2 en

◮ Choose P1, . . . , Pn : go through e1, . . . , en in random order

A lower bound construction

source sink n packets e1 e2 en

◮ Choose P1, . . . , Pn : go through e1, . . . , en in random order

A lower bound construction

source sink n packets e1 e2 en

◮ Choose P1, . . . , Pn : go through e1, . . . , en in random order

A lower bound construction

source sink n packets e1 e2 en

◮ Choose P1, . . . , Pn : go through e1, . . . , en in random order

A lower bound construction

source sink n packets e1 e2 en

◮ Choose P1, . . . , Pn : go through e1, . . . , en in random order

A lower bound construction

source sink n packets e1 e2 en

◮ Choose P1, . . . , Pn : go through e1, . . . , en in random order

A lower bound construction

source sink n packets e1 e2 en

◮ Choose P1, . . . , Pn : go through e1, . . . , en in random order

A lower bound construction

source sink n packets e1 e2 en

◮ Choose P1, . . . , Pn : go through e1, . . . , en in random order

A lower bound construction

source sink n packets e1 e2 en

◮ Choose P1, . . . , Pn : go through e1, . . . , en in random order

Observations:

◮ Congestion n

A lower bound construction

source sink n packets e1 e2 en

◮ Choose P1, . . . , Pn : go through e1, . . . , en in random order

Observations:

◮ Congestion n ◮ Dilation 2n + 3

A lower bound construction

source sink n packets e1 e2 en

◮ Choose P1, . . . , Pn : go through e1, . . . , en in random order

Observations:

◮ Congestion n ◮ Dilation 2n + 3 ◮ makespan ≥ 3n

A lower bound construction

source sink n packets e1 e2 en

◮ Choose P1, . . . , Pn : go through e1, . . . , en in random order

Observations:

◮ Congestion n ◮ Dilation 2n + 3 ◮ makespan ≥ 3n ◮ Suppose makespan ≤ (3 + ε)n

A lower bound construction

source sink n packets e1 e2 en

◮ Choose P1, . . . , Pn : go through e1, . . . , en in random order

Observations:

◮ Congestion n ◮ Dilation 2n + 3 ◮ makespan ≥ 3n ◮ Suppose makespan ≤ (3 + ε)n ◮ Goal: # schedules · Pr[fixed schedule collision free] ≪ 1

A lower bound construction

source sink n packets e1 e2 en

◮ Choose P1, . . . , Pn : go through e1, . . . , en in random order

Observations:

◮ Congestion n ◮ Dilation 2n + 3 ◮ makespan ≥ 3n ◮ Suppose makespan ≤ (3 + ε)n ◮ Goal: # schedules · Pr[fixed schedule collision free] ≪ 1 ◮ A routing strategy for a single packet is of the form

(park, park, go, wait, go, go, wait, wait, go, go, park)

A lower bound construction

source sink n packets e1 e2 en

◮ Choose P1, . . . , Pn : go through e1, . . . , en in random order

Observations:

◮ Congestion n ◮ Dilation 2n + 3 ◮ makespan ≥ 3n ◮ Suppose makespan ≤ (3 + ε)n ◮ Goal: # schedules · Pr[fixed schedule collision free] ≪ 1 ◮ A routing strategy for a single packet is of the form

(park, park, go, wait, go, go, wait, wait, go, go, park) time horizon (3 + ε)n

A lower bound construction

source sink n packets e1 e2 en

◮ Choose P1, . . . , Pn : go through e1, . . . , en in random order

Observations:

◮ Congestion n ◮ Dilation 2n + 3 ◮ makespan ≥ 3n ◮ Suppose makespan ≤ (3 + ε)n ◮ Goal: # schedules · Pr[fixed schedule collision free] ≪ 1 ◮ A routing strategy for a single packet is of the form

(park, park, go, wait, go, go, wait, wait, go, go, park) time horizon (3 + ε)n #go ≈ 2n

A lower bound construction

source sink n packets e1 e2 en

◮ Choose P1, . . . , Pn : go through e1, . . . , en in random order

Observations:

◮ Congestion n ◮ Dilation 2n + 3 ◮ makespan ≥ 3n ◮ Suppose makespan ≤ (3 + ε)n ◮ Goal: # schedules · Pr[fixed schedule collision free] ≪ 1 ◮ A routing strategy for a single packet is of the form

(park, park, go, wait, go, go, wait, wait, go, go, park) time horizon (3 + ε)n #go ≈ 2n E[# park] ≥ n

A lower bound construction

source sink n packets e1 e2 en

◮ Choose P1, . . . , Pn : go through e1, . . . , en in random order

Observations:

◮ Congestion n ◮ Dilation 2n + 3 ◮ makespan ≥ 3n ◮ Suppose makespan ≤ (3 + ε)n ◮ Goal: # schedules · Pr[fixed schedule collision free] ≪ 1 ◮ A routing strategy for a single packet is of the form

(park, park, go, wait, go, go, wait, wait, go, go, park) time horizon (3 + ε)n #go ≈ 2n E[# park] ≥ n E[#wait] ≤ εn

A lower bound construction

source sink n packets e1 e2 en

◮ Choose P1, . . . , Pn : go through e1, . . . , en in random order

Observations:

◮ Congestion n ◮ Dilation 2n + 3 ◮ makespan ≥ 3n ◮ Suppose makespan ≤ (3 + ε)n ◮ Goal: # schedules · Pr[fixed schedule collision free] ≪ 1 ◮ total # routing strategies ≤ (2n ·

4n

εn

• )n ≤ 2o(n2) for ε → 0

Lemma

Fix a schedule. Pr[schedule feasible] ≤ ( 1

2)Θ(n2)

source sink e1 e2 en

Lemma

Fix a schedule. Pr[schedule feasible] ≤ ( 1

2)Θ(n2)

source sink e1 e2 en

n 4 packets

Lemma

Fix a schedule. Pr[schedule feasible] ≤ ( 1

2)Θ(n2)

source sink e1 e2 en

Lemma

Fix a schedule. Pr[schedule feasible] ≤ ( 1

2)Θ(n2)

source sink e1 e2 en

Lemma

Fix a schedule. Pr[schedule feasible] ≤ ( 1

2)Θ(n2)

source sink e1 e2 en

Lemma

Fix a schedule. Pr[schedule feasible] ≤ ( 1

2)Θ(n2)

source sink e1 e2 en

Lemma

Fix a schedule. Pr[schedule feasible] ≤ ( 1

2)Θ(n2)

source sink e1 e2 en

Lemma

Fix a schedule. Pr[schedule feasible] ≤ ( 1

2)Θ(n2)

source sink e1 e2 en

Lemma

Fix a schedule. Pr[schedule feasible] ≤ ( 1

2)Θ(n2)

source sink e1 e2 en

Lemma

Fix a schedule. Pr[schedule feasible] ≤ ( 1

2)Θ(n2)

source sink e1 e2 en

Lemma

Fix a schedule. Pr[schedule feasible] ≤ ( 1

2)Θ(n2)

source sink e1 e2 en

Lemma

Fix a schedule. Pr[schedule feasible] ≤ ( 1

2)Θ(n2)

source sink e1 e2 en

Lemma

Fix a schedule. Pr[schedule feasible] ≤ ( 1

2)Θ(n2)

source sink e1 e2 en

◮ Pr[no collision] ≤ ( 1 8)n/8

Lemma

Fix a schedule. Pr[schedule feasible] ≤ ( 1

2)Θ(n2)

source sink e1 e2 en

◮ Pr[no collision] ≤ ( 1 8)n/8 ◮ ∃ n 16 time steps in which n 4 packets cross a random edge ej.

Lemma

Fix a schedule. Pr[schedule feasible] ≤ ( 1

2)Θ(n2)

source sink e1 e2 en

◮ Pr[no collision] ≤ ( 1 8)n/8 ◮ ∃ n 16 time steps in which n 4 packets cross a random edge ej. ◮ Problem: Steps not independent!

Lemma

Fix a schedule. Pr[schedule feasible] ≤ ( 1

2)Θ(n2)

source sink e1 e2 en

◮ Pr[no collision] ≤ ( 1 8)n/8 ◮ ∃ n 16 time steps in which n 4 packets cross a random edge ej. ◮ Problem: Steps not independent! But more careful

analysis works.

The end

Open question

Can acyclic job shop with preemption be done in O(congestion + dilation)?

◮ O((C + D) · log log(C + D)) suffices [Feige, Scheideler ’02]

The end

Open question

Can acyclic job shop with preemption be done in O(congestion + dilation)?

◮ O((C + D) · log log(C + D)) suffices [Feige, Scheideler ’02]

Thanks for your attention

Acyclic job shop with preemption

Given:

◮ Directed (simple) paths Pi ◮ Processing times pi,e ∀i ∈ [n] ∀e ∈ Pi

Constraints:

◮ Packet i takes time pi,e to cross e ∈ Pi ◮ At most one packet can actively move on an edge per time

unit

◮ Preemption: Packet can “stop” in the middle of an edge

(and another packet can be processed) Parameters:

◮ Congestion C := maxe∈E{ i:e∈Pi pi,e} ◮ Dilation D := maxi{ e∈Pi pi,e} ◮ L := max{C, D}

Question: Is O(L) possible? (Known: O(L · log log L))