Precedence Constrained Scheduling

Precedence Constrained Scheduling

Abhiram Ranade

Dept. of CSE

IIT Bombay

InputInput

• Directed Acyclic Graph G, #processors p

A

B

C

D

E

F

G

H

, 3

Vertex = unit time taskedge (u,v) : Time(u) < Time(v)

Output: ScheduleOutput: Schedule

Time 1 2 3 4

Processor 1 A D E G

Processor 2 B F H

Processor 3 C

Schedule Length, to be minimized

ApplicationsApplications

• Project management. Vertex = lay foundation, build walls. Edges: what happens first to what happens later.

• Processors : Number of workmen.

• MS Project, others.

• Our problem: Simplified version.

• Other applications: Parallel computing.

Summary of resultsSummary of results

• Polytime algorithm when p=2. [Fuiji.. 69]

• NP-hard for variable p. [LenKan 78]

• NP-hardness not known for fixed p > 2.

• Polytime algorithm for trees. [Hu 61]

Summary of results - 2Summary of results - 2

• Any greedy algorithm gives 2 - 1/p approximation. i.e. Schedule of length at most (2 - 1/p) times Optimal length.

• [Coffman-Graham 72, Lam-Sethi 77] give 2 - 2/p approximation algorithm.

• [Gangal-Ranade 08] give 2 – 7/(3p+1) approximation for p > 3.

OutlineOutline

• Elementary Lower Bound ideas

• Elementary algorithm and analysis

• Deadline Constraints [GarJoh 76] More complex problem, but generates new ideas. 2 processor optimality, also without deadlines Essentially gives 2 - 2/p approximation

• Ideas behind 2 - 7/(3p+1) approximation algorithm

Elementary Lower BoundsElementary Lower Bounds

• To prove optimality of any algorithm, need to show why it cannot be improved, i.e. lower bound on schedule length.

• OPT H = Length of longest path in G.• OPT [ N / p ], N = #nodes [ x ] = ceiling(x), smallest integer x. Example: H = 3, [N/p] =[8/3] = 3

Generic AlgorithmGeneric Algorithm

1. Assign suitable “priority” to each vertex.2. Of all “ready” vertices pick one with least

priority value, and schedule it at earliest possible time slot.

3. Repeat until done.

ready = no predecessors yet unscheduled.earliest possible = after predecessors.

Priority ExamplePriority Example

• Priority(v) = length of longest path from v.

• Intuition: If many nodes depend upon me, then I should go first.

Proof of 2 approximationProof of 2 approximation

• Full (time-) slot: All processors busy• Number of “full” slots N/p• Number of partial slots H. Why?

Partial slot: Some processor did not get work. All maximally long paths must shrink. This can happen only H times.

• Time N/P + H OPT + OPT = 2 OPT• Improve to 2 - 1/p. Priority?

Deadline Constraints [GarJoh 76]

Deadline Constraints [GarJoh 76]

• Additional Input: Deadline D(v) : time by which v must be processed.

• Need a schedule with p processors in which precedence constraints and deadlines are respected.

Deadline PropagationDeadline Propagation

• v has N(d) descendants with deadline d v must itself finish by d - [N(d)/p].

new deadline: d(v) = min( D(v), mind d -[N(d)/p] )

• In what order to calculate?

• (u,v) edge d(u) < d(v)

• GJ Algo: priority = deadline. Optimal for p=2!

ExampleExample

AB

E

D

C

4

4

4

.

.

.

.

d(A) = 4 - [7/5] = 2d(B) = min(4-[8/5], 2-[1/5]) = 1d(C) = …. = 3d(D) = = 2d(E) = . . . = 0

GJ Deadline PropertiesGJ Deadline Properties

• Deadline < 1 : schedule not possible.• Optimal Schedule length Max deadline - Min deadline + 1. Example: 4 - 0 + 1 = 5• Load bound : [N/p] = [17/5] = 4• Longest path: H = 4• Is this the best lower bound?

GJ Longest Path BoundGJ Longest Path Bound

• (u,v) is edge d(u) < d(v)

• Path u to v of length H d(u) < d(v) - H

• H < Max Deadline – Min Deadline

GJ Load BoundGJ Load Bound

• Add a universal parent z

• d(z) Max deadline – [Number of descendants with Max deadline/p]

• = Max – [N/p]

• Min deadline Max – [N/p]

Scheduling without external deadlines

Scheduling without external deadlines

• Set d(terminal vertices) = k, some number.

• Propagate deadlines. m = least deadline.

• Schedule from time m using deadlines.

Theorem: Algorithm is optimal for p=2.

2 Processor Optimality2 Processor Optimality

• v : earliest scheduled vertex not meeting deadline

• w : latest scheduled vertex before v scheduled alone. Always exists?

Nodes in region must be Descendants of w.

Region has 2(t-t’)-1nodes with

Deadline t-1.

d(w) t-1 - (2(t-t’)-1)/2 = t’-1 Contradiction

Time: 1 2 3 t’ t

Proc 1: w v

Proc 2: -

• d(w) t’, d(v) t-1

RemarksRemarks

• Why does this not work for p > 2?

• Algorithm gives 2 - 2/p approximation for even p. More complex proof.

Improvements to GJ [GR 09]Improvements to GJ [GR 09]

• Node v has N(d,L) descendants at distance at least L+1 having deadline at least d

• Then d(v) mind,L d - L - [N(d,L)/p]

ExampleExample

AB

E

D

C

4

4

4

.

.

.

.

d(A) = 4 - [7/5] = 2d(B) = min(4-[8/5], 2-[1/5]) = 1d(C) = …. = 3d(D) = = 2d(E) = . . . = 0

d(E) 4 – 2 – [12/5] = -1Max - min + 1 = 6. Optimal!

Algorithm [GR 08]Algorithm [GR 08]

• Set d(terminal vertices) = 0

• Propagate deadlines. New rule.

• For each v in non-decreasing deadline order: (Rearrange ancestors of v if possible). Schedule v in earliest possible slot, and

smallest numbered processor.

Rearrange ancestors of vRearrange ancestors of v

• Suppose t = last slot with ancestors of v.

• Suppose vertices in slots t-1,t have same deadline.

• Suppose v has < p ancestors in t-1,t.

• Then move ancestors of v to slot t-1, move other vertices to slot t.

• If slot t is not full, v can be scheduled in t.

Analysis OutlineAnalysis Outline

• Key part of proof: If algorithm constructs a long schedule, then deadline must drop a lot moving from last column to first.

• Max deadline - min deadline + 1 optimal schedule length.

• Optimal schedule must also be long, so good approximation factor.

How deadline varies in the schedule

How deadline varies in the schedule

123..p

Time> 1 2 3 ….

u vw

Deadline can only increase in first row: d(u) d(v)

Deadline can only increase in any column: d(u) d(w)

Partial slot rule 1Partial slot rule 1

123..p

1 2 3 ….increasing time.

u vwxy--

Deadline must increase in first row after a partial slot:

d(u) < d(v) … why was not v scheduled earlier?

Partial slot rule 2Partial slot rule 2

123..p

1 2 3 ….increasing time.

u1 vu2

…uk

--

Let M denote the number of nodes scheduled after

u1..uk. Then d(u1) d(v) - [[M/k]/p]

Analysis DetailsAnalysis Details

• Count number of 1-slots, 2-slots, full slots present in schedule.

• Relate counts to deadline drop, using rules discussed.

• Find patterns of slot occupancies.

• Special rules for specific patterns.

• Combine estimates and take best.

Intuition: Easy schedulesIntuition: Easy schedules

• Suppose all slots are partial: drop per slot.• Thus total deadline drop = length of

schedule.• Optimal!• Suppose all slots are either 1 slots or full

slots.• Partial slot rule 2 gives optimality.

Intuition: Difficult SchedulesIntuition: Difficult Schedules

• Schedules with mixture of 2-slots and full slots.

• Extreme case 1: 2-slots at the beginning, full slots at the end.

• Extreme case 2: 2 slots and full slots alternate.

• Omitted. See paper.

Concluding RemarksConcluding Remarks

• Analysis is complicated, but not much more than 2-2/p analysis of Lam-Sethi.

• Algorithm is simpler than Coffman-Graham.

• Technique will not work beyond 2 - 3/p. Even getting there is hard.