Mechanism Design for Real-Time Scheduling Carl Bussema III 18 April 2005 Based on "Mechanism Design for Real-Time Scheduling" by Ryan Porter

Teaser Definition of problem Previous results Designing a mechanism Restricted cases Conclusions and questions

Clueless: Problem Definition I Online real-time scheduling (uniprocessor) Jobs released at various times, no knowledge of upcoming jobs Jobs released at various times, no knowledge of upcoming jobs Goal: complete highest value of jobs before their deadlines Goal: complete highest value of jobs before their deadlines Zero value if job not complete by deadline Compare results to optimal performance if all jobs known from start Compare results to optimal performance if all jobs known from start Formally define a job to have a release time r, a length ℓ, a value v, and a deadline d Formally define a job to have a release time r, a length ℓ, a value v, and a deadline d

Clueless: Problem Definition II Mechanism design Each jobs released to a different self- interested agent Each jobs released to a different self- interested agent Agent gets value v if its job finishes before d Agent gets value v if its job finishes before d Agent submits job with some type (r′, ℓ′, v′, d′) Agent submits job with some type (r′, ℓ′, v′, d′) Scale values as needed so no job has v < ℓ Server decides schedule from what it's told Server decides schedule from what it's told Server charges agents some price p Server charges agents some price p Goal: Design competitive mechanism with dominant strategy Goal: Design competitive mechanism with dominant strategy

A Different World: Motivation Problem traditionally considered in non- strategic setting Jobs released directly to server; no chance to lie; no payments required Jobs released directly to server; no chance to lie; no payments required Strategic setting makes more sense with modern environments like grid computing RT setting allows model to be used for mission-critical applications

The Others: Previous results Most previous work focused on non- strategic settings Express competitive ratios based on k =  max /  min, where  is v/ℓ for any job For k = 1, a 4-competitive algorithm exists For k = 1, a 4-competitive algorithm exists For k ≥ 1, a (1+√k) 2 - competitive algorithm exists For k ≥ 1, a (1+√k) 2 - competitive algorithm exists Both bounds also shown tight for deterministic algorithms Both bounds also shown tight for deterministic algorithms

Step by Step: Designing a Mechanism Agents being self-motivated want to profit Encourage participation: Non-participants get 0 utility Non-participants get 0 utility Participants get at least 0 utility Participants get at least 0 utility System must be designed to handle this System must be designed to handle this Competitive performance easier if truthfulness ensured Design system such that lying can't increase utility and may decrease it Design system such that lying can't increase utility and may decrease it

CSI * : An example (I) *CSI = Conflicting Schedule Illustrated Consider jobs in the table (ℓ = v  k = 1) Deciding to preempt: Use some metric to assign priority Use some metric to assign priority This example: find expected loss based on “what could be done by new job’s deadline” This example: find expected loss based on “what could be done by new job’s deadline” Don't preempt if expected loss under desired ratio Don't preempt if expected loss under desired ratio This example, 2 waits for 1 and is preempted by 3 This example, 2 waits for 1 and is preempted by 3 Job riririri didididi ℓiℓiℓiℓi vivivivi

CSI: Example revisited (II) If instead, job 2 lies about its deadline, an algorithm might preempt job 1 for it, so 2 would finish before 3 arrives Job riririri didididi lililili vivivivi

The Practice: Porter's Mechanism At every time t, run the job i with highest priority = v i ′ + √k (e i (t))  min e i (t) is elapsed processing time for job i by t e i (t) is elapsed processing time for job i by tNotice: Priority based on reported value and elapsed processing time (k and  min are independent of any single job) Priority based on reported value and elapsed processing time (k and  min are independent of any single job) Priority of a job only increases when active Priority of a job only increases when active Requires a priori knowledge of k and  min Requires a priori knowledge of k and  min

The Price is Right: Payment Scheme Agents whose jobs are completed by deadlines must pay server the minimum value that would have completed their job, given everything else all agents reported This is similar to a second-price auction This is similar to a second-price auction These agents then have utility = v – p These agents then have utility = v – p Agents aware of this charging scheme and will abide by it (could demand v up front and use refund) Agents aware of this charging scheme and will abide by it (could demand v up front and use refund) Other agents pay nothing (therefore have zero utility) Thus we satisfy Individual Rationality: agents will voluntarily participate because it can only help

To Tell the Truth: Preventing Lies—Overview Agents could lie about 4 things: Release time Release time Prohibit: claiming r′ < r Length Length Prohibit: claiming ℓ´ < ℓ Value Value Deadline Deadline Necessary to show under- or over-stating any of these is not beneficial

To Tell the Truth: Preventing Lies to Improve Because of 2 nd -price auction, lying about value can't help Claiming an early release time or a shorter length are prohibited Claiming a later deadline could possibly help We avoid this by "holding" jobs on the server until reported deadline, only then return to agent We avoid this by "holding" jobs on the server until reported deadline, only then return to agent Thus agents will not get their jobs by their real deadlines, and thus get no utility Thus agents will not get their jobs by their real deadlines, and thus get no utility More interesting is the problem of lying to worsen

Law & Order: Proof of Truthfulness – r i (I) Ground rule: can't declare r′ < r Assume for some reported (r i ′, ℓ i ′, v i ′, d i ′) job i finishes before d, but not under (r i, ℓ i ′, v i ′, d i ′) ℓ′ ≥ ℓ by ground rules, d′ < d by rationality ℓ′ ≥ ℓ by ground rules, d′ < d by rationality We will show this is impossible

Law & Order: Proof of Truthfulness – r i (II) Consider case with truthful r: Let t p be the first time i is preempted; t a be time I abandoned (not enough time left) Let t p be the first time i is preempted; t a be time I abandoned (not enough time left) Claim: all jobs that execute during (t p,t a ] either arrive during that same time and have higher priority at their release than i or are I If a job x was released before t p, clearly i has priority over x, because at some point before t p, i ran instead of x If a job x was released before t p, clearly i has priority over x, because at some point before t p, i ran instead of x If a job z is released during the interval, then by definition it has lower priority at its release If a job z is released during the interval, then by definition it has lower priority at its release Inactive jobs do not increase in priority, so the claim holds Inactive jobs do not increase in priority, so the claim holds

Law & Order: Proof of Truthfulness – r i (III) Now consider case where agent declares r i ′ > r i S must change during (t p, t a ] vs. when r i declared S must change during (t p, t a ] vs. when r i declared i can't run longer outside this range when r i ′ > r i The change could be any job for another, not necessarily i Call t c first time change occurs, y c job running then under original S. How can any job have higher priority at t c than y c under S′? Call t c first time change occurs, y c job running then under original S. How can any job have higher priority at t c than y c under S′? Before t p the job ran longer under S′ (had to replace i)—won’t increase priority enough to outweigh a job that beats I at t p Before t p the job ran longer under S′ (had to replace i)—won’t increase priority enough to outweigh a job that beats I at t p During (t p,t c ) the job ran longer under S′ (again has to replace or be i). Contradicts “first time change occurs.” During (t p,t c ) the job ran longer under S′ (again has to replace or be i). Contradicts “first time change occurs.” Since no job can have higher priority than y c at t c under S′ than under S, no change can occur, and a later release time can’t help Since no job can have higher priority than y c at t c under S′ than under S, no change can occur, and a later release time can’t help

Law & Order: Proof of Truthfulness – ℓ i & d i Recall: ℓ′ d i unhelpful Only effect these have is on defining set of "available" jobs Incomplete: those with elapsed time < ℓ′ Incomplete: those with elapsed time < ℓ′ Abandoned: elapsed time + time to d′ ≤ ℓ′ Abandoned: elapsed time + time to d′ ≤ ℓ′ If job completes with ℓ′ and d′, the 1 st condition must become false before 2 nd becomes true If job completes with ℓ′ and d′, the 1 st condition must become false before 2 nd becomes true Declaring earlier release and later deadline will only make "incomplete" false earlier and "abandoned" false later, so it will not help to lie about these.

NUMB3RS: Competitive Ratio Proofsketch (recall k=  max /  min,  =v/ℓ): List only jobs completed by schedule and divide time into intervals (t f open,t f close ]: List only jobs completed by schedule and divide time into intervals (t f open,t f close ]: t f close = time when job f finished; t f open = t f-1 close (0 for t 1 ) Prove no interval length ≥ (1+1/√k)v f (induction) Prove no interval length ≥ (1+1/√k)v f (induction) Prove max OPT can get per interval for jobs we abandoned is (1+√k)v f (we wouldn't have abandoned them) Prove max OPT can get per interval for jobs we abandoned is (1+√k)v f (we wouldn't have abandoned them) Give OPT per interval: Give OPT per interval: k times interval length (almost entire interval on max. dense job) + bonus for doing some job we abandoned (short, dense) + v f (it might finish this later) Add results: (t f close - t f open )k + v f + (1+ √k)v f ≤ ((1+ √k)² +1)v f Add results: (t f close - t f open )k + v f + (1+ √k)v f ≤ ((1+ √k)² +1)v f v f is what we achieve each interval; so ratio is O((1+ √k)² +1) v f is what we achieve each interval; so ratio is O((1+ √k)² +1)

The Outer Limits: Special case If k=1 and agents can't lie about length, payments will no longer be required to keep agents honest Proof mirrors above, only simpler Same as above, achieves 5-competitive ratio, but has advantage of not needing payments

Happy Days: Conclusion Porter also proves his bound is a lower bound for any deterministic online mechanism with non-negative payments RT grid computing increasingly popular Need for competitive scheduling algorithms Need for competitive scheduling algorithms Insights here show strategic setting not significantly worse than non-strategic

Unsolved Mysteries Extend results to multiprocessor environment? Apply strategic concepts to bidding for CPU time in resource augmentation environment (online algorithm gets faster CPU(s) than OPT)? Find better bound with randomized mechanisms? Would knowing exact range of densities instead of just  min and k lead to stronger bound? Your questions?

The A-Team: Reference [1] Porter, Ryan. Mechanism Design for Real-Time Scheduling. In Proc. of the ACM Conference on Electronic Commerce (EC'04), 2004