Mechanism design for aggregating energy consumption and quality of service in speed scaling scheduling.

Mechanism design for

aggregating energy consumption and quality of service in speed

scaling scheduling.

Christoph Dürr

Óscar Carlos Vásquez Pérez Łukasz Jeż


Saint Étienne — nov 2016

Get these slides at goo.gl/6RKuo3

Informal Setting

• There are n players, each has a job to submit on a server

• The player want their job to complete early

• The server wants to consume few energy

• How to charge the players so they adapt a good behavior?

🔌 ⌛

bill

€bill

€bill

€

2

Execution model

• Every job has a workload wj

• The machine can vary its speed


[speed scaling model introduced in 2005 by Irani, Pruhs]

• Power consumption is

∫speed(t)^αdt for some physical constant 2≤α≤3

• Machine has to decide on

• speed

• schedule

• charging the players (bill bj)

area

=wj

speed

time

0 C

j

completion time

3

2 games studied

Deadline Game

• player=job

• has private penalty factor pj

• he announces deadline dj — job cannot be scheduled after

• machine produces minimum energy schedule respecting deadlines [easy]

• his goal is to minimize pjdj+bj

Penalty Game

• player=job

• he announces a penalty factor pj

• his job will complete at time Cj

• machine produces schedule minimizing energy plus weighted total completion times [hard?]

• his goal is to minimize pjCj+bj

4

[D.Jeż,Vásquez,DAM’2015] [D.Jeż,Vásquez,WINE’2013]

Deadline game

• Energy optimal schedule can be easily computed: Sort

deadlines. Find highest density prefix: dj with Σwi/dj maximal

over i with di≤dj. Schedule those jobs, and repeat.

• How to charge the consumed energy?

• We want pure Nash equilibria to exist

• We want the total charges to cover the power consumption

bill

€bill

€bill

€

🔌 ⌛

speed

time

0 d

j

density

5

Deadline game 


Proportional cost sharing

• charge every player the exact amount of energy used to

schedule the job

• does not guarantee pure Nash equilibria [not a surprise]

• example: 2 players, p1=p2=w1=w2=1

• best response diagram: no fix point 1

2

speed

d

2

0 d

¹

6

Deadline game 


Marginal cost sharing

• E[S] := minimum energy needed to schedule player set S

• Charge player j the difference E[N]-E[N\{j}] — N=all players

• Game is an exact potential game, it is truthful

• Convergence can take forever (continuous strategy space)

• Total charge is at most α･E[N]

1 2

4 3

speed

d

3

0 d

1

d

2

d

4

1 4 3

speed

d

3

0 d

1

d

4

E[N]

E[N\{2}]

7

Penalty game

• Player announces penalty pj

• Machine computes some schedule

• his job will complete at Cj

• he will be charged bj 
 := the marginal cost

• he wants to minimize pjCj+bj

• machine should minimize social cost := energy + total weighted completion time

Seems to be a hard scheduling problem

• Say job j has rank πj


and length lj 


in the schedule

• Energy consumption is 


• total weighted completion time is

• goal: find l,π that minimize E[l,w]

+F[π,l,p]

E[`, w] :=

Xn

j=1

w_j^↵`¹_j ^↵

F[⇡,`, p] :=

Xn

j=1

pj

C_j

z }| {X

i:⇡_i⇡_j

`i π1 π2

length

Penalty game 


Hard scheduling problem

• For fixed job order π, computing the optimal lengths l is easy

• Minimize E[l,w]+F[π,l,p] reduces to a classical scheduling problem:

• machine can run only at unit speed

• job j has processing time p_j and penalty weight w_j 


(notice the inversion)

• goal is to minimize

• Open: is it NP-hard ?

• PTAS has is known

• dominance prop. are known

[Megow,Verschae,ICALP’2013]

Xn

j=1

w_jC_j^{1 1/↵}

0 C

j

time

completion time

pj

√ C

^j

[Bansal,D,Nguyễn,Vásquez,Journal of Scheduling 2016]

The reduction

• For fixed order π, say π(i)=i

• Second order analysis shows l minimizing E[l,w]+F[π,l,p] is

• plugging it into E[l,w]+F[π,l,p]

leads after transformation to








the total weighted completion time of a schedule in inverse π order

`_j = w_j · ↵ 1 Pn

k=j p_k

!1/↵ 2

3 4

speed

0 time

↵(↵ 1)

^{1 1/↵}

X

n j=1

w

_j

0

@

X

n k=j

p

_k

1 A

1 1/↵

1

[Megow,Verschae,ICALP’2013]

2

The reduction

• For fixed order π, say π(i)=i

• Second order analysis shows l minimizing E[l,w]+F[π,l,p] is

• plugging it into E[l,w]+F[π,l,p]

leads after transformation to








the total weighted completion time of a schedule in inverse π order

`_j = w_j · ↵ 1 Pn

k=j p_k

!1/↵

1

3

4

↵(↵ 1)

^{1 1/↵}

X

n j=1

w

_j

0

@

X

n k=j

p

_k

1 A

1 1/↵

time

completion time

^1-1/α

constant

completion time

0

Penalty game 


Mechanism design

• Mechanism fixes some order π on the jobs

• optimal order is hard to compute

• if order is independent from job strategies then game has good properties

• Machine executes optimal schedule for this order π

• Machine charges marginal cost

Properties

• Game is truthful (players announce real penalty p_j)

• Total charge is at most α+1 times consumed energy

Future directions

• Machine could organize an auction on the rank π

Scheduling problem

• Given n jobs with processing times pj and priority weights wj

• Find order π that minimizes ΣwjCjβ for some β>0


Easy variant β=1

• It is optimal order jobs in decreasing Smith-ratio wj/pj

• Proof is by exchange argument:

• exchange improved objective if and only if

i j

i j

w_i(t + p_i) + w_j(t + p_i + p_j)  w_j(t + p_j) + w_i(t + p_i + p_j) w_jp_i  w_ip_j

w_j/p_j  w_i/p_i

t

Scheduling problem

For general β

• Whether the exchange of the adjacent jobs j,i improves the objective value depends on

• wj,pj,wi,pi

• … and t


Question

• [local rule] Under which condition on wj,pj,wi,pi do we know that j followed immediately by i is not optimal (for all t)

• [global rule] Under which condition do we know that j scheduled somewhere

before i is not optimal

i j

i j

t

this implication is trivial

what do we know on the other direction? hard

easy

Results

4 Nikhil Bansal et al.

0 < < 1 > 1 = 2

our contributionsbefore

pi pj

wi

wj

β=2

(c)

(b) (d) (e)

j ≺g i

j ≺^l i

i ≺g j i ≺^l j

pi pj

wi

wj

(c)

j ≺g i (b)

i ≺g j

pi pj

wi

wj

j ≺g i j ≺l i

0<β<1

(c) (b)

j ≺^l i

i ≺l j

i ≺^l j i ≺g j

pi pj

wi

wj

i ≺^l j

(a) (b) (c)

j ≺g i

i ≺g j j ≺^l i

pi pj

wi

wj

β>1

(c)

(b)

i ≺^l j j ≺g i j ≺l i

j ≺^l i

i ≺^l j i ≺g j

pi pj

wi

wj

(c)

j ≺g i (b)

i ≺g j

our contributionsbefore

pi pj

wi

wj

β=2

(c)

(b) (d) (e)

j ≺g i

j ≺^l i

i ≺g j i ≺^l j

pi pj

wi

wj

(c)

j ≺g i (b)

i ≺g j

pi pj

wi

wj

j ≺g i j ≺l i

0<β<1

(c) (b)

j ≺^l i

i ≺l j

i ≺^l j i ≺g j

pi pj

wi

wj

i ≺^l j

(a) (b) (c)

j ≺g i

i ≺g j j ≺^l i

pi pj

wi

wj

β>1

(c)

(b)

i ≺^l j j ≺g i j ≺l i

j ≺^l i

i ≺^l j i ≺g j

pi pj

wi

wj

(c)

j ≺g i (b)

i ≺g j

our contributionsbefore

pi pj

wi

wj

β=2

(c)

(b) (d) (e)

j ≺g i

j ≺^l i

i ≺g j i ≺^l j

pi pj

wi

wj

(c)

j ≺g i (b)

i ≺g j

pi pj

wi

wj

j ≺g i j ≺l i

0<β<1

(c) (b)

j ≺^l i

i ≺l j

i ≺^l j i ≺g j

pi pj

wi

wj

i ≺^l j

(a) (b) (c)

j ≺g i

i ≺g j j ≺^l i

pi pj

wi

wj

β>1

(c)

(b)

i ≺^l j j ≺g i j ≺l i

j ≺^l i

i ≺^l j i ≺g j

pi pj

wi

wj

(c)

j ≺g i (b)

i ≺g j

our contributionsbefore

pi pj

wi

wj

β=2

(c)

(b) (d) (e)

j ≺g i

j ≺^l i

i ≺g j i ≺^l j

pi pj

wi

wj

j ≺g i j ≺l i

0<β<1

(c) (b)

j ≺^l i

i ≺l j

i ≺^l j i ≺g j

pi pj

wi

wj

β>1

(c)

(b)

i ≺^l j j ≺g i j ≺l i

j ≺^l i

i ≺^l j i ≺g j

pi pj

wi

wj

(c)

j ≺g i (b)

i ≺g j

pi pj

wi

wj

i ≺^l j

(a) (b) (c)

j ≺g i

i ≺g j j ≺^l i

Thm1 Thm1 Thm1

Thm1

Thm2

Thm3 Thm2

Thm3

our contributionsbefore

pi pj

wi

wj

β=2

(c)

(b) (d) (e)

j ≺g i

j ≺^l i

i ≺g j i ≺^l j

pi pj

wi

wj

j ≺g i j ≺l i

0<β<1

(c) (b)

j ≺^l i

i ≺l j

i ≺^l j i ≺g j

pi pj

wi

wj

β>1

(c)

(b)

i ≺^l j j ≺g i j ≺l i

j ≺^l i

i ≺^l j i ≺g j

pi pj

wi

wj

(c)

j ≺g i (b)

i ≺g j

pi pj

wi

wj

i ≺^l j

(a) (b) (c)

j ≺g i

i ≺g j j ≺^l i

Thm1 Thm1 Thm1

Thm1

Thm2

Thm3 Thm2

Thm3

(a) w_j = w_i(p_j/p_i)² (b) w_j = w_i^{(pi+pj) pi}

(p_i+p_j) p_j

(c) w_j = w_ip_j/p_i (d) w_j = w_ip_j/2p_i (e) w_j = 2w_ip_j/p_i

Fig. 1: Illustration of our contribution (bottom row) compared to previous results (top row), using a similar representation as in [9]. Every point in the diagram represents a job j with respect to some fixed job i. The space is divided into regions where i _{`</sub> j holds or j <sub>`} i or none. These regions contain subregions where we know that the stronger condition g holds. The boundaries are defined by functions which are named from (a) to (e). The last 2 diagrams also indicate the related theorems.

1. f(x) 0 for x 0.

2. f is convex and non-decreasing, i.e. f⁰, f⁰⁰ 0.

3. f⁰ is log-concave (i.e. log(f⁰) is concave), which im- plies that f⁰⁰/f⁰ is non-increasing. Intuitively this means that f does not increase much faster than e^x.

4. For every b > 0, the function g_b(x) = f(b + e^x) f(b) is log-convex in x. Intuitively this means that f(b + e^x) f(b) increases faster than e^cx for some c > 0. Formally this means

yf⁰(b +y)/ f(b + y) f(b) (1)

is increasing in y.

The proof is based on standard functional analysis and is omitted.

Lemma 3 For a < b, the fraction f⁰(b) f⁰(a)

f(b) f(a)

– is decreasing in a and decreasing in b for any > 1 – and is increasing in a and increasing in b for any

0 < < 1.

Proof First we consider the case > 1. We can write f⁰(b) f⁰(a) = Rb

a f⁰⁰(x)dx andf(b) f(a) = Rb

a f⁰(x)dx.

Note that f⁰⁰ and f⁰ are non-negative. By > 1 and Lemma 2f⁰ is log-concave, which means thatf⁰⁰(x)/f⁰(x) is non-increasing in x. This implies

Z b

a

f⁰⁰(b)

f⁰(b) f⁰(x)dx  Z b

a

f⁰⁰(x)

f⁰(x) f⁰(x)dx

 Z b

a

f⁰⁰(a)

f⁰(a) f⁰(x)dx

Hence Rb

a f⁰⁰(b)dx Rb

a f⁰(b)dx  R b

a f⁰⁰(x)dx Rb

a f⁰(x)dx  R b

a f⁰⁰(a)dx R b

a f⁰(a)dx .

For positive values u1, u2, u3, v1, v2, v3 with u1/v1  u₂/v₂  u₃/v₃ we have

u₁ + u₂

v1 + v2  u₂

v2  u₂ + u₃ v2 + v3

.

was known our r esults

[Höhn,Jacobs,2012]
 [Sen,Dileepan,Ruparel,1990] [Mondal,Sen,2000]

[Bansal,D,Nguyễn,Vásquez,
 Journal of Scheduling 2016]

Summary

• Speed scaling model: non linear cost

• Find compromise quality of service versus energy

consumption

• Use marginal cost sharing

• Two games studied

• Optimize social cost is interesting scheduling problem

• Some dominance properties are known

Thank you for executing my

job

You are welcome. 


Especially since you paid for it.