Minimizing Average Flow-Time

Naveen Garg

IIT Delhi

Joint work with Amit Kumar, Jivi Chadha ,V Muralidhara, S. Anand

Problem Definition Given : A set of M machines

A set of jobs

A matrix of processing times of

job i on machine j.

Each job specifies a release date

pij

r j

r j

Problem Definition Conditions :

r j

r j Pre-emption allowed

Migration not allowed

Problem Definition

r j C j

Flow-time of j,

Goal : Find a schedule which minimizes the average flow-time

F j=C j− r j

Special Cases

Parallel : all machines identical

Related : machines have different speeds

Subset Parallel : parallel except that a job can only go on a subset of machines

Subset Related

pij=p j

pij=p j /s i

pij p j

All of these are NP-hard.

Previous work

• The problem is well studied when the machines are identical.

• For a single machine the Shortest-remaining-processing-time (SRPT) rule is optimum.

• [Leonardi-Raz 97] argued that for parallel machines SRPT is O(min (log n/m, log P)) competitive, where P is max/min processing time. They also show a lower bound of (log P) on competitive ratio

Preemptive, unweighted Flow time

Online OfflineParallel machines

O(log P), (log P) [LR97]

(log1-ε P) [GK07]

Related machines

O(log P) [GK06]

Subset parallel

Unbounded [GK07]

O(log P) (log P/loglogP) [GK07]

Unrelated machines

O(k) [S09]

Fractional flow-time

O(1=²O(1))

pj(t) = remaining processing of job j at time t remaining fraction at time t =

ab

time processing total

(t)pj

Fractional flow-time of j = t≥rj pj(t)/pj

Recall, flow-time of j =

j

j

C

rt

1

Fractional flow-time

O(1=²O(1))

1 5 12

Fractional flow-time = 1*2 + 2/3*3 + 1/3*7

Fractional flow-time can be much smaller than (integral) flow-time

O(1=²O(1))

20

Integer Program

Define 0-1 variables :

x(i,j,t) : 1 iff job j processed on i during [t,t+1] Write constraints and objective in terms of these variables.

Fractional flow-time of j = t ≥ rj (t-rj) x(i,j,t)/pij

LP Relaxation

0t)j,x(i,

ti,allfor1t)j,x(i,

jallfor1p

t)j,x(i,

)r(tp

t)j,x(i,Min

j

ti, ij

jtj,i, ij

One Caveat …

Fractional flow-time

A job can be done simultaneously on many machines : flow-time is almost 0

LP Relaxation

0t)j,x(i,

ti,allfor1t)j,x(i,

jallfor1p

t)j,x(i,

t)j,x(i,)r(tp

t)j,x(i,Min

j

ti, ij

tj,i,j

tj,i, ij

Add a term for processing time

Class of a job

p j : The processing time of j rounded up to nearest power of 2

p j 2 kIf , we say k is the class of job j

Number of different classes = O(log P)

Modified Linear Program

0t)j,x(i,

ti,allfor1t)j,x(i,

jallfor1p

t)j,x(i,

t)j,x(i,)r(tp

t)j,x(i,Min

j

ti, ij

tj,i,j

tj,i, ij

Modified LP

LP value changes by a constant factor only.

But : rearranging jobs of the same class does not change objective value.

From fractional to integral

• The solution to the LP is not feasible for our (integral) problem since it schedules the same job on multiple m/c’s.

• We now show how to get a feasible, non-migratory schedule.

Rounding the LP solution

• Consider jobs of one class, say blue.

• Find the optimum solution to the LP.

Rounding the LP solution (contd.)

• Rearrange blue jobs in the space occupied by the blue jobs so that each job is scheduled on only one m/c.

• If additional space is needed it is created at the end of the schedule

Additional space

Preemptive, unweighted Flow time

Online OfflineParallel machines

O(log P), (log P) (log1-ε P)

Related machines

O(log P)

Subset parallel Unbounded O(log P)(log P/loglogP)

Unrelated machines

O(k)

Assignment as flowFix a class k : arrange the jobs in ascending order of release dates.

r10

iv(i,k,j)

z i , jtx i , j , t

p j

s

Flow = ?

r7r6r5r4r3r2

Unsplittable Flow Problem

s

d1 d2

d3

Unsplittable Flow Problem

s

d1 d2

d3

Flow can be converted to an unsplittable flow such that excess flow on any edge is at most the max demand [Dinitz,Garg, Goemans]

Back to scheduling...Fix a class k : find unsplittable flow

10 2 3 4 5 6 7

iv(i,j,k) p j

s

Gives assignment of jobs to machines

Back to scheduling...J(i,k) : jobs assigned to machine i

10 2 3 4 5 6 7

iv(i,j,k) p j

s

Can we complete J(i,k) on class k slots in I ?

Building the Schedule

Flow increases by at most max processing time = 2k

So all but at most 2 jobs in J(i,k) can be packed into these slotsExtra slots are created at the end to accommodate this spillover

Increase in Flow-timeHow well does

capture the flow-time of j ?

Charge to the processing time of other classes

1

ij

j

ti, p

)r(tt)j,x(i,

Finally...Since there are only log P classes…

Can get OPT + O(log P).processing time flow-time for subset parallel case.

Preemptive, unweighted Flow time

Online OfflineParallel machines

O(log P), (log P) (log1-ε P)

Related machines

O(log P)

Subset parallel Unbounded O(log P)(log P/loglogP)

Unrelated machines

O(k)

Integrality Gap for our LP(identical m/c)

0 mk-1 2mk-1 mk +mk-1

mk

mk-1

m

T

1

Phase 0 1 k-1 k

Integrality Gap for our LP(identical m/c)

• For sufficiently large T, flow time ≥ mT(1+k/2)

0 mk-1 2mk-1 mk +mk-1

T

Phase 0 1 k-1 k

Blue jobs can be scheduled only in this area of volume (mk+mk-1)m/2

At least m/2 blue jobs left At least mk/2 jobs left

Integrality Gap for our LP(identical m/c)

• Optimum fractional solution is roughly mT

0 mk-1 2mk-1 mk +mk-1

mk

mk-1

m

T

1

Phase 0 1 k-1 k

Integrality gap• Optimum flow time is at least mT(1+k/2)• Optimum LP solution has value roughly mT• So integrality gap is (k).• Largest job has size P = mk.• For k = mc, c>1, we get an integrality gap of (log P/loglogP)

Hardness results• We use the reduction from 3-dimensional

matching to makespan minimization on unrelated machines [lenstra,shmoys,tardos] to create a hard instance for subset-parallel.

• Each phase of the integrality gap example would have an instance created by the above reduction.

• To create a hard instance for parallel machines we do a reduction from 3-partition.

Preemptive, unweighted Flow time

Online OfflineParallel machines

O(log P), (log P) (log1-ε P)

Related machines

O(log P)

Subset parallel Unbounded O(log P)(log P/loglogP)

Unrelated machines

O(k)

A bad example

0 T

B x

A x

A+B=TA> T/2

T+L

A x

Flow time is at least AxL > T L/2OPT flow time is O(T2+L) Ω(T) lower bound on any online algorithm

Other Models

• What if we allow the algorithm extra resources ?

• In particular, suppose the algorithm can process (1+ε) units in 1 time-unit. [first proposed by Kalyanasundaram,Pruhs95]

Resource Augmentation Model

Resource Augmentation

• For a single machine, many natural scheduling algorithms are O(1/eO(1))-competitive with respect to any Lp norm [Bansal Pruhs ‘03]

• Parallel machines : randomly assign each job to a machine – O(1/eO(1)) competitive

[Chekuri, Goel, Khanna, Kumar ’04]

• Unrelated Machines : O(1/e2)-competitive, even for weighted case. [Chadha, Garg, Kumar, Muralidhara ‘09]

O(1=²O(1))

Our Algorithm

When a job arrives, we dispatch it to one of the machines.

Each machine just follows the optimal policy : Shortest Remaining Processing Time (SRPT)

What is the dispatch policy ? GREEDY

pj1(t)

The dispatch policy

When a job j arrives, compute for each machine i the increase in flow-time if we dispatch j to i.

j1

j2

jr

jr+1

js

j arrives at time t : pij1(t) ≤ pij2

(t) ≤ …

pijr(t) < pij < pijr+1

(t)

j

Increase in flow-time = pj1

(t) + … + pjr(t) + pij+ pij(s-r)

Our Algorithm

When a job j arrives, compute for each machine i the increase in flow-time if we dispatch j to i.

Dispatch j to the machine for which increase in fractional flow-time is minimum.

Analyzing our algorithm

Primal LPDual LP

LP opt. valueAlgorithm’s value

Construct a dualsolution

Show that the dual solution value and algorithm’s flow-time are close to each other.

Dual LP

0t)j,x(i,

ti,allfor1t)j,x(i,

jallfor1p

t)j,x(i,

t)j,x(i,)r(tp

t)j,x(i,Min

j

ti, ij

tj,i,j

tj,i, ij

αj

βit

Dual LP

0,

rtt,j,i,allfor1p

rt

p

Max

itj

jij

jit

ij

j

ti,it

jj

βα

βα

βα

pj1(t)

αj = pj1

(t) + … + pjr(t) + pij + pij(s-r)

Setting the Dual Values

When a job j arrives, set αj to the increase in flow-time when j is dispatched greedily.

j1

j2

jr

jr+1

js

j arrives at time t : pij1(t) ≤ pij2

(t) ≤ …

pijr(t) < pij < pijr+1

(t)

Thus j αj is equal to the total flow-time.

βit = s

Setting the Dual Values

j1

j2

jr

jr+1

js

pj1(t)

Set βit to be the number of jobs waiting at time t for machine i.

Thus i,t βit is equal to the total flow-time.

Need to verify

Dual Feasibility

j1

j2

jl

jl+1

js

pj1(t)

)( ls ji'ji'jjj pp(t)p...(t)pl1

Fix a machine i’, a job j and time t’.

1p

tt'p ji'

t'i'ji'

j

βα

Suppose pi’jl(t) < pi’j < pi’jl+1

(t)

Dual Feasibility

j1

j2

jl

jl+1

js

pj1(t)

ls ji'ji'jjj pp(t)p...(t)pl1

11p

(t)p...(t)p

p ti'ji'

jj

ji'

j l1

βslsα

1

What happens when t’ = t ?

Dual Feasibility

j1

j2

jl

jl+1

js

t'i'ji'ji'

ji'

jj

ji'

jjjj

ji'

j

1p

t)(t'1)k-(s 1

pt)(t'

1p

(t)p...(t)pt)(t'

1p

(t)p...(t)p(t)p...(t)p

p

lk

lk1-k1

β

ls

lsα

δ

What happens when t’ = t + δ? Suppose at time t’ job jk is being processed

Case 1: k ≤ l

Dual Feasibility

j2

jr

jr+1

js

t'i'ji'ji'

ji'

jjjj

ji'

ji'ji'jj

ji'

j

1p

t)(t'1)k-(s 1

pt)(t'

1p

(t)p...(t)p(t)p...(t)p

p

pp(t)p...(t)p

p

1-k1lr1

l1

β

ks

lsα

1

Case 2: k > l ¢

Dual Feasibility

So, αj, βit are dual feasible

But i,t βit and j αj both equal the total flow time and hence the dual objective value is

1p

t)(t'p ji'

t'i'ji'

j

βα

Hence, for any machine i’, time t’ and job j

0,

, ti

tij

j

Incorporating machine speed-up

So the values αj, βit/(1+ε) are dual feasible for an instance with processing times larger by a factor (1+ε)

1pt)(t'

p ji'

t'i'

ji'

j

111

βα

For any machine i’, time t’ and job j

Equivalently, schedule given instance on machines ofspeed (1+ε) to determine αj, βit. The values αj, βit/(1+ε) are dual feasible.

Dual Objective Value

The dual value is less than the optimum fractional flow time.

jj

ti

ti

jj

11,

,

Since i,t βit = j αj the value of the dual is

Hence, the flow time of our solution, j αj, is at most (1+1/ε) times the optimum fractional flow time.

Extensions

• Can extend this analysis to the Lp-norm of the flow time to get a similar result.

• Analysis also extends to the case of minimizing sum of flow time and energy on unrelated machines.

Open Problems

Single Machine :

Constant factor approximation algorithm for weighted flow-time.

loglog n approx [Bansal Pruhs ’10]

2+ε quasi polynomial time algorithm [Chekuri Khanna Zhu ‘01]

Open Problems

Parallel machines :

Constant factor approximation algorithm if we allow migration of a job from one machine to another.

The (log1-ε P) hardness is for non-migratory schedules

Open Problems

Unrelated Machines :

poly-log approximation algorithm

(LP integrality gap ?)

O(k) approximation [Sitters 08] is known, where k is the number of different processing times.

Thank You