Scheduling with AND/OR Precedence Constraints

Scheduling with AND/OR Precedence Constraints Seminar Mathematische Optimierung - SS 2007 23th April 2007

Synthesis synthesis: transfer from the behavioral domain (e. g. system specifications, algorithms) to the structural domain (e. g. processor, controllers) methods: allocation assignment scheduling compute start times for operations minimize costs today: (by me) next week: Solving Project Scheduling Problems by Minimum Cut Computations (by Christian Rinder)

Example

Table of Contents 1 Definitions and Graph Representation 2 Realizations 3 Implicit Constraints 4 Minimal Representation 5 Computing Earliest Job Start Times

AND/OR Precedence Constraints Definition V : set of jobs W: set of waiting conditions w = (X, j) where job j V cannot be started until at least one of the jobs X (V \{j}) is done represented by a directed graph D = (V, A): V = V W: set of nodes every waiting condition w = (X, j) W is represented by arcs (i, w) A for every i X and an additional arc (w, j) A

AND/OR Precedence Constraints V = {j 1, j 2, j 3, j 4 } W = {w 1 =({j 1 }, j 2 ), w 2 =({j 2, j 3 }, j 1 ), w 3 =({j 1, j 4 }, j 3 )}

Realizations Definitions realization: partial order R = (V, R ) on V where for each (X, j) W there exists an i X with i R j linear realization: realization which is a total order W is feasible if and only if there exists a (linear) realization for W. Note: Every extension R = (V, R ) of a realization R (i. e. i R j, then i R j) is also a realization.

Realizations V = {j 1, j 2, j 3, j 4 } W = {w 1 =({j 1 }, j 2 ), w 2 =({j 2, j 3 }, j 1 ), w 3 =({j 1, j 4 }, j 3 )} Example j 4 R j 3 R j 1 R j 2 is a linear realization.

Idea for Checking Feasibility Is a set W of AND/OR precedence constraints feasible? Construct a linear realization L in a greedy way: Does there exist a job j V which is not a waiting job of a waiting condition in W? insert j at the end of L Is a waiting condition (X, j) satisfied? delete (X, j) from W Iterate these steps until no other job can be planned (e. g. all jobs are planned)

Algorithm for Checking Feasibility Input: set V of jobs and set W of waiting conditions Output: list L of jobs from V Q := L := for jobs j V do a(j) := {(X, j) W} if a(j) = 0 then add j to Q end end

Algorithm for Checking Feasibility (continued) while Q do remove a job i from Q insert i at the end of L for waiting conditions (X, j) W with i X do decrease a(j) by 1 if a(j) = 0 then add j to Q end remove (X, j) from W end end

Algorithm for Checking Feasibility W = {w 1, w 2, w 3 } Q = {} L = () a(j 1 ) = a(j 2 ) = a(j 3 ) = a(j 4 ) =

Algorithm for Checking Feasibility W = {w 1, w 2, w 3 } Q = {j 4 } L = () a(j 1 ) = 1 a(j 2 ) = 1 a(j 3 ) = 1 a(j 4 ) = 0

Algorithm for Checking Feasibility W = {w 1, w 2 } Q = {j 3 } L = (j 4 ) a(j 1 ) = 1 a(j 2 ) = 1 a(j 3 ) = 0 a(j 4 ) = 0

Algorithm for Checking Feasibility W = {w 1 } Q = {j 1 } L = (j 4, j 3 ) a(j 1 ) = 0 a(j 2 ) = 1 a(j 3 ) = 0 a(j 4 ) = 0

Algorithm for Checking Feasibility W = {} Q = {j 2 } L = (j 4, j 3, j 1 ) a(j 1 ) = 0 a(j 2 ) = 0 a(j 3 ) = 0 a(j 4 ) = 0

Algorithm for Checking Feasibility W = {} Q = {} L = (j 4, j 3, j 1, j 2 ) a(j 1 ) = 0 a(j 2 ) = 0 a(j 3 ) = 0 a(j 4 ) = 0

Algorithm for Checking Feasibility (Proof) Theorem A set of AND/OR precedence constraints is feasible if and only if the list L obtained from the algorithm contains all jobs of V.

Definition of Implicit AND/OR Precedence Constraints Goal: Derive new constraints from a given set W Definitions Let U V and j V \U. Then the waiting condition (U, j) is implied by W if and only if for every realization R = (V, R ) of W there exists an i U with i R j. Let Y V. Then W Y = {(X Y, j) (X, j) W, j Y } is the set of induced waiting conditions. But: How can we compute such constraints?

Algorithm for Detecting Implicit Constraints Theorem For given U V let L be the output of the algorithm with input V \U and W V \U. The set of waiting conditions of the form (U, j) which are implied by W is precisely {(U, j) : j V \(L U)}. Example V = {j 1, j 2, j 3, j 4 }, W = {({j 1 }, j 2 ), ({j 2, j 3 }, j 1 ), ({j 1, j 4 }, j 3 )} Choose U = {j 3 } W V \U = {({j 1 }, j 2 ), ({j 2 }, j 1 )} Algorithm: L = (j 4 ) implied waiting conditions (U, j): ({j 3 }, j 1 ), ({j 3 }, j 2 )

Minimal Representation of AND/OR Precedence Constraints Definitions W minimal: 1 no waiting condition (X, j) W is implied by W\{(X, j)}, and 2 for each waiting condition (X, j) W the set X is minimal with respect to inclusion, i. e. for all i X the waiting condition (X \{i}, j) is not implied by W. W, W equivalent: their sets of (linear) realizations coincide W minimal reduction of W: W, W equivalent and W minimal

Examples for Minimal Representation Examples V = {j 1, j 2, j 3, j 4 }, W = {({j 1 }, j 2 ), ({j 2, j 3 }, j 1 ), ({j 1, j 4 }, j 3 )} W not minimal since ({j 3 }, j 1 ) is implied by W W = {({j 1 }, j 2 ), ({j 3 }, j 1 ), ({j 4 }, j 3 )} minimal reduction of W

Unique Minimal Reduction Theorem Each feasible set of waiting conditions has a unique minimal reduction.

Definitions Definition d iw Z: time lag for job i V and waiting condition w = (X, j) W S = (S 1,..., S n ) Z V : vector of start times for all jobs in V S (Z { }) V : partial schedule, i. e. S i = means that job i V is not planned Job processing times p i can be modeled by setting d iw := p i for all w = (X, j) W with i X. d iw < 0 can be interpreted as maximal time lags.

First Model Find (optimal) S such that S j min i X (S i + d iw ) is satisfied for all waiting conditions w = (X, j) W.

Graph Representation and Example Definition AND-nodes: jobs in V OR-nodes: jobs in W

Introducing Dummy Jobs We want to simplify our model: interpret waiting conditions w W as dummy jobs add dummy job s to V and additional waiting conditions ({s}, j) to W for every j V reformulate previous constraints to S w min i X (S i + d iw ) and S j S w for every waiting condition w = (X, j) W and S s 0 since s is our starting point. It follows: S j 0 for all AND-nodes j V

Final Model for (ES) Find a componentwise minimal schedule S Z V such that S s 0 S j S w max (S w + d wj ), (w,j) A min (S j + d jw ), (j,w) A j V w W Without loss of generality: d wj = 0 (otherwise set d iw to d iw + d wj for all i in(w))

More about Schedules and (ES) S = (,..., ) is feasible for (ES) S, S feasible S = min(s, S ) feasible There exists a (unique) componentwise minimal partial schedule S. In (ES) we can replace by = without changing the optimal solution.

Computing Earliest Job Start Times Consider now only positive arc weights, i. e. d iw > 0. How can we compute the earliest job start times? Start with S s = 0, S w = 0 for all w out(s) and S w = for all other OR-nodes w Iteration: Choose not yet planned OR-node w 0 = (X, j) with minimal start time S w0 Plan w 0 with start time S w0 Plan AND-node j if all waiting conditions are now satisfied Update S w for all not yet planned OR-nodes w

Computing Earliest Job Start Times 1 Input: digraph D = (V = V W, A) with positive arc weights on the arcs in V W A Output: feasible (partial) schedule S (Z { }) V Heap := for AND-nodes j V do a(j) := in(j) S s := 0 // AND-node s is planned at time 0 for OR-nodes w W do if w out(s) then insert w in Heap with key S w := 0 else insert w in Heap with key S w := end end

Computing Earliest Job Start Times 2 3 while Heap do remove next OR-node w 0 = (X, j) from Heap // OR-node is planned reduce a(j) by 1 if a(j) = 0 then S j := max w in(j) S w // AND-node is planned for OR-nodes w out(j) do S w := min{s w, S j + d jw } decrease key of w in Heap to S w end end delete node w 0 and all incident arcs from D end

Computing Earliest Job Start Times Example Heap = {S s1 = 0, S s2 = 0, S s3 = 0, S s4 = 0, S w1 =, S w2 =, S w3 = } Planned AND-nodes = {S s = 0} Planned OR-nodes = {} a(j 1 ) = 2 a(j 2 ) = 2 a(j 3 ) = 2 a(j 4 ) = 1

Computing Earliest Job Start Times Example Heap = {S s2 = 0, S s3 = 0, S s4 = 0, S w1 =, S w2 =, S w3 = } Planned AND-nodes = {S s = 0} Planned OR-nodes = {S s1 = 0} a(j 1 ) = 1 a(j 2 ) = 2 a(j 3 ) = 2 a(j 4 ) = 1

Computing Earliest Job Start Times Example Heap = {S s3 = 0, S s4 = 0, S w1 =, S w2 =, S w3 = } Planned AND-nodes = {S s = 0} Planned OR-nodes = {S s1 = 0, S s2 = 0} a(j 1 ) = 1 a(j 2 ) = 1 a(j 3 ) = 2 a(j 4 ) = 1

Computing Earliest Job Start Times Example Heap = {S s4 = 0, S w1 =, S w2 =, S w3 = } Planned AND-nodes = {S s = 0} Planned OR-nodes = {S s1 = 0, S s2 = 0, S s3 = 0} a(j 1 ) = 1 a(j 2 ) = 1 a(j 3 ) = 1 a(j 4 ) = 1

Computing Earliest Job Start Times Example Heap = {S w1 =, S w2 =, S w3 = 3} Planned AND-nodes = {S s = 0, S j4 = 0} Planned OR-nodes = {S s1 = 0, S s2 = 0, S s3 = 0, S s4 = 0} a(j 1 ) = 1 a(j 2 ) = 1 a(j 3 ) = 1 a(j 4 ) = 0

Computing Earliest Job Start Times Example Heap = {S w1 =, S w2 = 4} Planned AND-nodes = {S s = 0, S j4 = 0, S j3 = 3} Planned OR-nodes = {S s1 = 0, S s2 = 0, S s3 = 0, S s4 = 0, S w3 = 3} a(j 1 ) = 1 a(j 2 ) = 1 a(j 3 ) = 0 a(j 4 ) = 0

Computing Earliest Job Start Times Example Heap = {S w1 = 6} Planned AND-nodes = {S s = 0, S j4 = 0, S j3 = 3, S j1 = 4} Planned OR-nodes = {S s1 = 0, S s2 = 0, S s3 = 0, S s4 = 0, S w3 = 3, S w2 = 4} a(j 1 ) = 0 a(j 2 ) = 1 a(j 3 ) = 0 a(j 4 ) = 0

Computing Earliest Job Start Times Example Heap = {} Planned AND-nodes = {S s = 0, S j4 = 0, S j3 = 3, S j1 = 4, S j2 = 6} Planned OR-nodes = {S s1 = 0, S s2 = 0, S s3 = 0, S s4 = 0, S w3 = 3, S w2 = 4, S w1 = 6} a(j 1 ) = 0 a(j 2 ) = 0 a(j 3 ) = 0 a(j 4 ) = 0

Computing Earliest Job Start Times Example Start times: AND-nodes: (4, 6, 3, 0) OR-nodes: (6, 4, 3) Possible realization: j 4 w 3 j 3 w 2 j 1 w 1 j 2

Computing Earliest Job Start Times Theorem For a given set of AND/OR precedence constraints represented by a digraph D = (V W, A) with nonnegative arc weights and without cycles of length 0, the algorithm computes an optimal partial schedule S. In particular, the instance is infeasible if and only if S w = for some OR-node w.

Computing Earliest Job Start Times Theorem For a given set of AND/OR precedence constraints represented by a digraph D = (V W, A) with nonnegative arc weights and without cycles of length 0, the algorithm computes an optimal partial schedule S. In particular, the instance is infeasible if and only if S w = for some OR-node w. Lemma The algorithm can be implemented to run in O( W log W + A + V ) time.

Generalization to Nonnegative Arc Weights We also want to compute earliest job start times for d jw 0 instead only d jw > 0. Previous algorithm gives us: Theorem For a given set of AND/OR precedence constraints represented by a digraph D = (V W, A) with nonnegative arc weights and without cycles of length 0, the algorithm computes an optimal partial schedule S. In particular, the instance is infeasible if and only if S w = for some OR-node w.

Generalization to Nonnegative Arc Weights We also want to compute earliest job start times for d jw 0 instead only d jw > 0. There exists an algorithm with: Theorem For a given set of AND/OR precedence constraints represented by a digraph D = (V W, A) with nonnegative arc weights, the algorithm computes an optimal partial schedule S. In particular, the instance is infeasible if and only if S w = for some OR-node w.

Arbitrary Arc Weights Consider d iw with M < d iw < M, M 0 We cannot use feasibility results from previous sections since they require d iw = p i > 0, i. e. j V can be started if and only if all waiting conditions (X, j) W are satisfied. Definition A set W of waiting conditions is feasible if and only if there exists a feasible schedule S for (ES).

Thank you for listening! Questions?