Parallel Scheduling of Complex DAGs under Uncertainty

1. Parallel Scheduling of Complex DAGs under Uncertainty Grzegorz Malewicz

2. Terminology • Graph – G(V,E) • DAG – Directed Acyclic Graph • Chain – A path • Antichain – subset of V, A, such that there is no path between two nodes of A.

3. Terminology - cont • Width of a DAG – the largest cardinality of an antichain in the DAG • A set of chains is said to cover a DAG if every node in the DAG is a node in at least one of the chains. • Dilworth’s theorem states that the width of a graph equals to the minimum number of covering chains. • Source – A node without arcs from it. • Sink – A node without arcs pointing to it.

4. The model • Set of workers – {1,2,…,n} denoted by [n]. • Set of tasks – {1,2,…,t} denoted by [t]. • The tasks has mutual dependencies which is represented by a DAG where every node is a task and every directed edge u→v represents task v depends upon completion of task u. • Task t is eligible if all the tasks it’s depends on was successfully executed. • We’ll denote by E(X) the set of all eligible tasks when all the tasks in set X had executed.

5. The model - cont • A set of tasks satisfies precedence constraintsif, for every task in the set, all parents of the task are also in the set. • The execution is consist of synchronous steps. • In each of those steps, a central daemon assign a task from the group E(X) È {∅} to every worker. • A worker i, completes task j successfully in a certain step with probability of Pi,j. • The execution completes when every task was executed successfully by at least one worker.

6. Regimen definition • Formally, the set theory representation of a regimen is: Σ : 2|V| → ([n] → (VÈ{∅})) • For any subset Z of tasks that satisfies precedence constraints, the value Σ(Z) is a function from [n] to the set E(Z) È {∅}

7. ROPAS • Recomputation and Overbooking allowed Probabilistic dAg Scheduling Problem • Given such DAG, G, where for every task j there exist at least one i such that Pi,j > 0, find a regimen ∑* which minimize the expected completion time.

8. Lemma 2.1 • Denote by BX, The minimum expected completion across all regimen which start with tasks X already executed. • Lemma 2.1: For any set X of tasks that satisfies precedence constraints, BX is finite. • Proof: We define a regimen that has finite expectation; thus minimum expectation must be finite, too. • We take a topological sort t1, ..., tmof the subdag of G induced by tasks other than X. • The probability that every worker fails to execute task tjis: • The expected value to execution:

9. Optimized algorithm for restricted ROPAS • We show an dynamic programming algorithms which solves the ROPAS problem in polynomial time under the following restrictions: • The number of workers is at most constant. • The DAG width is at most constant. • Later, we show that if we drop any of those restrictions, the problem will become NP complete.

10. Definitions: X - Set of tasks that satisfies precedence constraints and does not contain all sinks of G. D0, D1, ..., Dk - All the distinct subsets of E(X) where D0=∅. ai - The probability that Di is exactly the set of tasks executed by workers in the assignment Σ(X). Xi = X È Di. TXi - The expected time to completion for regimen Σ starting with tasks Xi already executed. Then the following relation holds: Theorem 3.1

11. Theorem 3.1 - Proof • T - The time to completion for the regimen that starts with tasks X already executed. • T = TL+TR, where TL is the time until at least one task of E(X) has been executed, and TR is the rest of the time. • Exp[T] = Exp[TL] + Exp[TR] • TL~G(P), where P = 1-a0 = a1+a2+…+ak. • Exp[TL] = 1/P = 1/(a1+a2+…+ak)

12. Calculating Exp[TR] – The remaining time to execution Let Di be the event: “Di is exactly the set of tasks first executed by the workers”.

13. Theorem 3.1 - conclusions

14. Admissible evolution of execution graph • For every DAG G we can construct a graph G’(V’,E’) called “admissible evolution of execution” which has single source and single sink. • Every node in G’ represents a set of nodes (or tasks) from G. • This special graph is also a DAG. • Every execution flow of every ROPAS regimen on G can be represented by a path from the source to the sink of this graph.

15. Admissible evolution of execution graph – Construction • The construction of this graph is inductive: • We begin with a set V’={∅}. • For any node XÎV’ that doesn’t contain all sinks of G calculate the set of eligible tasks E(X) in G. • Take all non-empty subsets D Í E(X) and add to V’ the node XÈD if it is not there already. • Add an arc (X, XÈD) to V’. • Since G is finite, the inductive process clearly defines a unique directed graph G’.

17. Admissible evolution of execution graph – Properties • G’ is a DAG – it cannot has any cycle because every arc is in the from (X, XÈD) and X has strictly more elements than XÈD. • Nodes of G’ are exactly the sets of tasks that satisfy precedence constraints • => By induction – if X satisfy precedence constraints so does XÈD. • <= Pick any subset Y of tasks that satisfies precedence constraints. Let t1, ..., t|Y |be a topological sort of the sub-DAG of G induced by Y. Clearly, tjbelongs to the set of tasks eligible when tasks t1, ..., tj−1 have been executed, for any j. So if {t1, ..., tj−1} is a node of G’, so is {t1, ..., tj}. Since ∅ is a node of G’, Y must also be.

18. Admissible evolution of execution graph Properties – cont’d • G’ has a single source - We add a node Y to G’only if there is an arc leading to Y from some other node. So Y cannot be a source. However,∅ is a source because no arc leads to a set with the same number or fewer elements. • G’ has a single sink – For any XÎV’, E(X)¹∅ except X=V.

19. The algorithm OPT • When processing each XÎV’ define two values: • A number SX - The expected time to completion for regimen ΣOPTstarting with tasks X already executed. • An assignment ΣOPT(X). • Pick a topological sort Y1, ..., Ym of G’ and process it in the reverse order. • Start by setting Sym to 0 and ΣOPT(Ym) to ∅. • For any 1£h<m consider all possible (|E(Yh)|+1)n assignment to n workers: • For any assignment, calculate the probability ai that Di is exactly the set of tasks executed by workers in the assignment. • Find the assignment which minimizes the expression: • Set SYh to the value of this expression and ΣOPT(Yh) to this assignment.

20. Theorem 3.3 • The regimen ΣOPT‘s expected time to completion when starting with tasks X already executed is the minimal expected time to completion that any regimen can achieve (denoted by BX).

21. Theorem 3.3 - Proof • The following invariant holds during the reverse order processing of V’: • For all i such that h ≤ i ≤ m, SYi = BYi and SYi is equal to the expected time to completion of ΣOPT that starts with tasks Yi already executed. • The invariant is clearly true for i=m because G’ has exactly one sink. • By induction, assume the invariant holds for 2≤i≤m (i.e.SYi = BYi). According to theorem 3.1: And it’s exactly equals to SYi

22. Theorem 3.4 • The OPT algorithm works at polynomial time when the DAG’s width and the workers number are at most constants. • Proof: Let the width of the graph be w and number of workers n. • By Dilworth’s theorem, the DAG can be covered by w chains, each chain can have a length of at most t so there are at most (t+1)w distinct subsets of tasks that satisfy precedence constraints. • In each step, the algorithm consider assignment of at most t+1 tasks to each worker. So there are at most (t+1)n calculations in each step. • We only need to evaluate probabilities aifor at most 2nof the sets Di, because the workers can get assigned to at most n different tasks, • The computations step of OPT is at most C1(t+1)w (t+1)n 2n+C2, thus polynomial.

23. Part II NP hardness

24. X3C - Exact 3-Set cover • Instance: F = {S1, …, Sm} of subsets of a finite set U, and |Si| = 3, |U| = 3n for some n. • Objective: Find n sets in F that are disjoint and have U as their union. • The problem is a sub-problem of the exact set cover and thus NP hard.

25. MP – Multiplicative partition • Instance: Number n, integer sizes s1, ..., sn where si≥ 2. • Question: Is there a set P such that: • The MP problem is NP hard. • We show a polynomial time reduction from X3C.

26. MP is NP hard • Given an instance of X3C, we associate consecutive prime numbers with each element of the universe (U). • Each 3-subset is converted into a size that is the product of the associated primes. • We add two more sizes: • C which is p1a1-2 * p2a2-2 * …pkak-2 where ai is the number of occurrences of pi and pi occur more than once. • D which is pi1 * pi2 * pi3 * … * pin where pij appear once.

27. S1 S2 S3 S5 MP ≥ X3C - example S4 =11 =5 =2 =7 =13 =3

28. MP ≥ X3C - example • S1=2*3*13=78 • S2=2*5*11=110 • S3=5*7*11=385 • S4=3*5*11=165 • S5=2*7*11=154 • C=21*30*51*70*112=1210 • D=13 • 2*3*13 * 5*7*11 * 21*30*51*70*112 = 2*5*11 * 3*5*11 * 2*7*11 * 13

29. Partition with Minimum Sum of Products (PMSP) • Instance: Number n, rational numbers: 0 < r1, ..., rn < 1. • Objective: Find a set P that minimizes: • Claim: PMSP is NP hard: • Proof: Pick any instance of MP with integer s1, …, sn. • Look at the instance of PMSP where ri=1/si2.

30. PMSP is NP hard • The function ex+ed-x in the domain {0≤x≤d} is minimized exactly when x=d/2 and it’s value is 2ed/2. • s1, …, sn can be partitioned to two sets of equal multiplication iff: • Thus if we would know the answer to PMSP in polynomial time – we’ll know the answer to MP in polynomial time.

31. Theorem 4.3 • The ROPAS Problem restricted to the DAG with two independent tasks (where the number of workers may grow) is NP-hard. • Proof: Polynomial time reduction from PMSP: • Consider a DAG with only two independent tasks: right task and left task. • Each worker executes a task successfully with probability of pi. • Assume we split the tasks to N workers where l is the probability that at least one of the workers assigned to the left task will succeed and r is same probability regarding the right task.

32. Theorem 4.3 – cont. • By theorem 3.1, the expected time to completion: • the regimen initially assigns every worker to a task, and when one task remains to be executed, the regimen assigns all workers to the task. So T{left}=T{right}=T and we get: • Since L and R partition {1, ..., n}, then 1-(1-l)(1-r)is just the probability that at least one of the n workers succeeds and is independent of how workers are assigned to left and right task. • Thus the expectation is minimized if, and only if, lris maximized across partitions L and R of {1, ..., n}.

33. Theorem 4.3 – cont. • Since L and R is a partition – The value in the 3rd term is independent of R and L. • So lr is minimized iff is minimized.

34. FIRMSSU Problem • Fixed Ratio Many Subsets with Small Union • Instance: Number k, nonempty subsets S1, ..., S3kof the set {1, ..., 3k} whose union is {1, ..., 3k}. • Question: Are there 2k of these subsets whose union has cardinality at most 2k?

35. ROPAS ≥FIRMSSU • Tasks A1, B1 can only be executed by the first worker. • Tasks A2, B2,C2 can only be executed by the second worker.

36. Round 2/3n2 (24)

37. Round 2/3n2+2/3n (28)

38. Round n2+2/3n (40)

39. Round n2+n (42)

40. Suppose the instance of FIRMSSU is negative… • Pick any regimen. For the reason that the instance is negative, the union of any 2k subsets has cardinality at least 2k + 1. • But the cardinality of any group is n, so at least 2/3n2+n sources of A1 must be executed in order to have 2/3n sinks of B2 eligible. • Therefore, at most 2/3n−1 sinks of B2 can get executed by the end of round 2/3n2 + 2/3n. • By the end of this round, no sinks of C2 can be executed, because each depends on execution of 2/3n2+2/3n other tasks. • Hence, by the end of that round, there are at least 1/3n+1+1/3n2 unexecuted tasks that can only be executed by the second worker. • As a result the regimen takes at least n2 + n + 1 rounds to complete computation.

41. ROPAS is inaproximable • When there are no restriction on both the DAG width and the number of workers - ROPAS is inaproximable with a factor better than 5/4, unless P=NP. • Similar proof like the former. • Assume there are 2/3n2 workers of type 1 and 2/3n2 workers of type 2. • The next DAG can be scheduled in 4 rounds iff an instance of FIRMSSU is positive.

43. Round 1

44. Round 3

45. Round 4