Precedence Constrained
Scheduling
Abhiram Ranade
Dept. of CSE
IIT Bombay
Input
• Directed Acyclic Graph G, #processors p
A
B
E
G
F
H
, 3
C
D
Vertex = unit time task
edge (u,v) : Time(u) < Time(v)
Output: Schedule
Time
Processor 1
Processor 2
Processor 3
1 2 3 4
A D E G
B
F H
C
Schedule Length,
to be minimized
Applications
• 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 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 - 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.
• [Svensson 10] Better than 2-ε unlikely.
Outline
• 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 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
Input
• Directed Acyclic Graph G, #processors p
A
B
E
G
F
H
, 3
C
D
Vertex = unit time task
edge (u,v) : Time(u) < Time(v)
Generic Algorithm
1. Pick any “ready” vertex.
2. Schedule it at earliest possible time.
3. Repeat until done.
ready = no predecessors yet unscheduled.
earliest possible = after predecessors.
Proof 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.
Deadline Constraints
[GarJoh 76]
• Additional Input: D(v) : time by which v
must be processed.
• Need a schedule with p processors in
which precedence constraints and
deadlines are respected.
Deadline 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!
Example
4
4
B
A
.
.
D
C
.
E
.
4
d(A) = 4 - [7/5] = 2
d(B) = min(4-[8/5], 2-[1/5]) = 1
d(C) = …. = 3
d(D) =
=2
d(E) = . . . = 0
GJ Deadline Properties
• Deadline < 1 : schedule not possible.
• Optimal Schedule length 
•
•
•
•
Max deadline - Min deadline + 1.
Opt for example  4 - 0 + 1 = 5
Load bound : [N/p] = [17/5] = 4
Longest path: H
=
4
Is this the best lower bound?
Partial Slot Bound
• Max Deadline – Min Deadline + 1 >= H
Time 1 2 . . . t . . . . .
P1
uv
P2
u is ancestor of v, so d(u) < d(v)
Load Bound
• Max deadline – Min deadline + 1 >= [N/p]
• Add a universal parent z
• d(z) <= Max deadline – [Number of
descendants with deadline d/p]
•
= Max – [N/p]
• Min deadline <= Max – [N/p]
Scheduling without 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 Optimality
• v : earliest scheduled vertex not meeting
deadline
• w : latest scheduled vertex before v
scheduled alone. Always exists?
Time: 1 2 3
Proc 1:
Proc 2:
t’
w
-
t
v
Nodes in region must be
Descendants of w.
Region has 2(t-t’)-1nodes with
Deadline  t-1.
• d(w)  t’, d(v)  t-1
d(w)  t-1 - (2(t-t’)-1)/2 = t’-1 Contradiction
Remarks
• 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]
• 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]
Example
4
4
B
A
.
.
D
C
.
E
.
4
d(A) = 4 - [7/5] = 2
d(B) = min(4-[8/5], 2-[1/5]) = 1
d(C) = …. = 3
d(D) =
=2
d(E) = . . . = 0
d(E)  4 – 2 – [12/5]
= -1
Max - min + 1 = 6. Optimal!
Algorithm
• 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 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 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
Time> 1 2 3 ….
1
2
3
.
.
p
uv
w
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
1
2
3
.
.
p
1 2 3 ….increasing time.
uv
w
x
y
-
Deadline must increase in first row after a partial slot:
D(u) < D(v) … why was not v scheduled earlier?
1-slot rule
1
2
3
.
.
p
1 2 3 ….increasing time.
u
-
v
Let M denote the number of nodes scheduled after u.
Then D(u)  D(v) - [M/p]
Intuition: 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.
• 1-slot rule gives optimality.
Intuition: 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.
Extreme case 1:
•
•
•
•
•
Time> <--L--> <---m-->
P1 : a v v v v v v v v v w
P2 : b v v v v v v v v v
P3 :
vvvvv
P4 :
vvvvv
•
•
•
•
A,b must be ancestors of all to right.
One of them say a, must be ancestor of mp/2
d(a)  d(w) - L - mp/2.
Drop = #2slots + full slots/2
Extreme Case 2
•
•
•
•
•
Time:1
P1 : v
P2 : v
P3 :
P4 :
2 3 4 5....
v v v v v
v v v v v
v v
v
v v
v
• Use ancestor rearrangement to argue
large drop.
Actual Analysis
• Keep track of how many 1-slots, 2-slots,
full slots .. encountered.
• Relate numbers to deadline drop,
schedule length.
• Solve for schedule length/deadline drop.
Remarks
• Analysis is complicated, but not much
more than 2-2/p analysis of Lam-Sethi.
• Algorithm is simpler than CoffmanGraham.
• Technique will not work beyond 2 - 3/p.
Even getting there is hard.