a stackelberg strategy for routing flow over time

A Stackelberg Strategy for Routing Flow over TimeUmang Bhaskar, Lisa Fleischer

Dartmouth CollegeElliot Anshelevich

Rensselaer Polytechnic Institute

2

Traffic Routing

• Data network• Road network

3

Traffic Routing

Safe zone

• Evacuation planning

4

Traffic Routing

5

Routing Delays

• Delay on edges increases with usage

delay

delay

6

Optimal Routing


delay

delay

• Optimal routing minimizes f(delay)

7

Uncoordinated Routing

• But players choose route independently

delay

delay

8

Routing Games

• Each player picks route to minimize delay

• Delays depend on other players

delay

delay

9

Equilibria

• Equilibrium:- every player minimizes

delay w.r.t. others

• Studied in many contexts

- no player can unilaterally reduce delay

10

Price of Anarchy

• How does equilibrium compare to optimal routing?

• Price of Anarchy(PoA)

=f(delay) of worst equilibriaf(delay) of optimal routing

quality ofequilibria

measures system efficiency

11

Static Flows

s t

xe

• Static: time-invariant• Most routing games assume static flows

xe

xe

12

Flows over Time

s t

• Edges have fixed capacities and delays

• Flow on an edge may vary with time

ce, de

13

Flows over Time

M = 1000 bits

Total time: 12 seconds

d = 2 seconds

c = 100 bps

Bitsper

second

1 2 3 4 5 6 7 8 9 10 11 12

100

Arrival graph:

Time

Quickest flow: flow that gets to destination as soon as possible

A B

14

Temporal Routing Games

M = 1000 bitsd = 2 seconds

c = 100 bps

• Routing Games + Flows over Time = Temporal Routing Games

• Player minimizes time it reaches destination

A B

15

Temporal Routing GamesAssumptions:

• Players are infinitesimal

Model:

• Players are ordered at s• Each player picks a path from s to t• Minimizes the time it arrives at t

s t

M

Time

Flowrateat t

Arrival Graph:M

16


s tc = 2, d = 0 c = 1, d = 0

c = 1, d = 1

vTime

M

12

Flow rate at t

17


Time

s tc = 2, d = 0 c = 1, d = 0

c = 1, d = 1

v

Queue

M

12

Flow rate at t

18


Time

s tc = 2, d = 0 c = 1, d = 0

c = 1, d = 1

v

Queue

• Queues grow when inflow exceeds capacity on an edge• Queues are first in, first out (FIFO)• Queuing models used in modeling traffic [Vickrey ‘69, Yagar ‘71]

M

12

Flow rate at t

19


Time

s tc = 2, d = 0 c = 1, d = 0

c = 1, d = 1

vM

12

Flow rate at t

20


Time

s tc = 2, d = 0 c = 1, d = 0

c = 1, d = 1

vM

12

Flow rate at t

21


Time

s tc = 2, d = 0 c = 1, d = 0

c = 1, d = 1

vM

12

Flow rate at t

22


Time

s tc = 2, d = 0 c = 1, d = 0

c = 1, d = 1

vM

12

MFlow rate

at t

23


Time

s tc = 2, d = 0 c = 1, d = 0

c = 1, d = 1

v

Flow rate at t

Time

s tc = 2, d = 0 c = 1, d = 0

c = 1, d = 1

v

fp = 1

fp = 1

Quickest Flow:

M

M

M

M

12

12

Flow rate at t

24


• For single-source, single sink, equilibrium exists [Koch, Skutella, ’09; Cominetti, Correa, Larre ‘11]

our case

• How does equilibrium in these games compare with optimal?

• What is the PoA?

Flow rate at t

Time

s tc = 2, d = 0 c = 1, d = 0

c = 1, d = 1

vM

M

12

25

Objectives

|Flow| reaching t by time T

Time for M to reach t

Sum of times for M to reach t

Evacuation Price of Anarchy:

Time Price of Anarchy:

Total Delay Price of Anarchy:

Flow rateat t

Time T

M’ = ?

Maximum delay of a player?Flow rate

at t

Time

Flow rateat t

Time ?

Time

Flow rateat t

∫ = ?

M

26

Results



Flow rateat t

Time T

M’ = ?





Flow rateat t

Time

?

Flow rateat t

Time ?

Maximum delay of a player

Ω(log n) [KS ‘09]

Ω(n) [MLS ‘10]

≥ e/(e-1) [KS ‘09]

Flow rateat t

Time

∫ = ?

M

27

Results









≥ e/(e-1) [KS ‘09]

Ω(n) [MLS ‘10]

Our Results

= e/(e-1) †

= 2 e/(e-1) †

† can be enforced by reducing capacities

Flow rateat t

Time T

M’ = ?

Flow rateat t

Time

?

Flow rateat t

Time ?

Flow rateat t

Time

∫ = ?

M

28


- compute quickest flow (can compute in poly time [FF ’62])- reduce edge capacities to that used by quickest flow

Can enforce bound of e / (e-1) on Time PoAby reducing edge capacities

• Main Result:

• Procedure:

s tc = 2, d = 0 c = 1.5

c = 1, d = 1

vc = 1, d = 0

wfp = 1

fp = 1 , d = 0

?

M

29


- reduce edge capacities to that used by quickest flow

s tc = 2, d = 0 c = 1

c = 1, d = 1

vc = 1, d = 0

wfp = 1

fp = 1

• Main Result:

• Procedure:- compute quickest flow (can compute in poly time [FF ’62])

, d = 0


?

M

30

Temporal Routing Games• Main Result:

- reduce edge capacities to that used by quickest flow

• Procedure:

If quickest flow saturates graph, thenTime PoA ≤ e / (e-1)

• Main Theorem:

- compute quickest flow (can compute in poly time [FF ’62])


?

M

31


If quickest flow saturates graph, thenTime PoA ≤ e / (e-1)


Theorem 1:

If quickest flow saturates graph, thenTotal Delay PoA ≤ 2 e / (e-1)

Can enforce bound of 2 e / (e-1) on Total Delay PoAby reducing edge capacities

Theorem 2:∫ = ?

M

?

32

Talk Outline

For this talk:

• Structure of equilibria

• Sketch weaker version of Theorem 1.

• Structure of Quickest flow

33

Quickest Flows

OPT:Flow rate

at t

Times t

c = 2, d = 0 c = 1, d = 0

c = 1, d = 1

v

• Quickest flow is static flow repeated over time [FF ‘62]

12

M

34

Quickest Flows

OPT:

Times t

c = 2, d = 0 c = 1, d = 0

c = 1, d = 1

v


fp = 1

Flow rate at t

12

M

35

Quickest Flows

OPT:

Times t

c = 2, d = 0 c = 1, d = 0

c = 1, d = 1

v


fp = 1

fp = 1

Flow rate at t

12

M

36

Phases in Equilibria

Time

s tc = 2, d = 0 c = 1, d = 0

c = 1, d = 1

v

QueueFlow rate

at t

12

M

37


Time

s tc = 2, d = 0 c = 1, d = 0

c = 1, d = 1

v

Flow rate at t

12

M

38


Time

s tc = 2, d = 0 c = 1, d = 0

c = 1, d = 1

v

Flow rate at t

12

M

39

• Equilibrium proceeds in “phases” [KS ‘09]

• Phase characterized by

• edges being used • edges with queue

• Within phase, flow rate at t is fixed


40

1-Phase Equilibrium

For one-phase equilibrium, Time PoA ≤ 2

For this talk:

• Phases in equilibria

• Sketch of proof

• Quickest flow

41

A Bound on EQ

EQ

Arrival Graphs:

Equilibrium

M M

s tc = 2, d = 0 c = 1, d = 0

c = 1, d = 1

vTime

ce = 1

Flow rate at t

42

A Bound on EQArrival Graphs:

Equilibrium

Quickest Flow

M M

Flow rate at t

s tc = 2, d = 0 c = 1, d = 0

c = 1, d = 1

v

fp = 1

fp = 1

s tc = 2, d = 0 c = 1, d = 0

c = 1, d = 1

vTime

Time

co = 2

OPT

EQ

ce = 1

Flow rate at t

43

A Bound on EQ

Time

Arrival Graphs:

Equilibrium

Quickest Flow

M M

Time

s tc = 2, d = 0 c = 1, d = 0

c = 1, d = 1

v

fp = 1

fp = 1

s tc = 2, d = 0 c = 1, d = 0

c = 1, d = 1

v

Claim 1: ce EQ ≤ co OPT

Flow rate at t

Flow rate at t

EQ ≤ OPT ce

co

co = 2

OPT

EQ

ce = 1

44

A Second Bound on EQ

Time

Arrival Graphs:

Equilibrium

s tc = 2, d = 0 c = 1, d = 0

c = 1, d = 1

v

Flow rate at t

EQ

ce = 1

45


Time

Arrival Graphs:

Equilibrium

s tc = 2, d = 0 c = 1, d = 0

c = 1, d = 1

v

Flow rate at t

EQ

ce = 1

46


Time

Arrival Graphs:

Equilibrium

s tc = 2, d = 0 c = 1, d = 0

c = 1, d = 1

vτ(θ) increasing

τ(θ) : minimum time taken to reach t at time θ

Flow rate at t

EQ

ce = 1

47


Time

Arrival Graphs:

Equilibrium

s tc = 2, d = 0 c = 1, d = 0

c = 1, d = 1

v

Flow rate at t

τ(θ) increasing


Claim 2: τ(EQ) ≥ EQco

co - ce

EQ

ce = 1

48


Time

Arrival Graphs:

Equilibrium

s tc = 2, d = 0 c = 1, d = 0

c = 1, d = 1

v

• But graph has path p of length ≤ OPT not being used (ce < co)• When τ(θ) ≥ OPT, this path p should be used (phase ends)

Flow rate at t

τ(θ) increasing



co - ce

EQ

ce = 1

49


Time

Arrival Graphs:

Equilibrium

s tc = 2, d = 0 c = 1, d = 0

c = 1, d = 1

v

Claim 3: τ(EQ) ≤ OPTco

co - ce

Flow rate at t

Claim 2: θ ≥ EQClaim 2: τ(EQ) ≥ EQco

co - ce

τ(θ) increasing


• But graph has path p of length ≤ OPT not being used (ce < co)• When τ(θ) ≥ OPT, this path p should be used (phase ends)

EQ

ce = 1

50


Time

Arrival Graphs:

Equilibrium

s tc = 2, d = 0 c = 1, d = 0

c = 1, d = 1

v

Claim 4: EQ ≤ OPTco - ce

co

Flow rate at t

τ(θ) increasing



co - ce Claim 3: τ(EQ) ≤ OPT+

EQ

ce = 1

51

Proof Sketch for 1-Phase

Claim 4: EQ ≤ OPTco - ce

co

EQ ≤ 2 OPT

+

(upper bounds equalwhen ce = co / 2)

Claim 1: EQ ≤ OPT ce

co

For one-phase equilibrium, Time PoA ≤ 2

52

1-Phase Examples



Flow rateat t

Time T

M’ = ?





Flow rateat t

Time

?

Flow rateat t

Time ?



Ω(n) [MLS ‘10]

≥ e/(e-1) [KS ‘09]

Flow rateat t

Time

∫ = ?

All 1-phase examples!

53

Summary

In paper: If quickest flow saturates graph, for arbitrary phases,

• Time PoA ≤ e/(e-1)

• Total Delay PoA ≤ 2 e/(e-1)

Thus, network administrator can enforce these bounds,by reducing edge capacities.

54

Open Questions1. Can PoA exceed e/(e-1) if capacity is not restricted?

3. What if players have imperfect information?

4. ...

2. PoA for multiple sources

Thanks for listening!

55

Optimal Routing


delay

delay

56

Optimal Routing

• Optimal routing minimizes f(delay)• Delay on edges increases with usage

delay

delay

57

Uncoordinated Routing

• But players choose route independently

delay

delay

58

Routing Games

• Each player picks route to minimize delay• Delays depend on other players

delay

delay

59

Equilibrium

• Equilibrium: every player minimizes delay w.r.t. others- no player can unilaterally reduce delay

• Equilibria in routing games studied in many contexts

60

Equilibrium

• How does equilibrium compare to optimal routing?

• Price of Anarchy(PoA)

=f(delay) of worst equilibriaf(delay) of optimal routing

quality ofequilibria measures system efficiency

61

A 1-Phase Lower Bound on PoA

1-Phase example with PoA of e/(e-1) ~ 1.6 [KS ‘09]

s t

Flow rate at t

Time

a stackelberg strategy for routing flow over time

Documents

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temporal routing games