Robust and reactive scheduling - LAAS · An initial set of activities and an estimation on problem parameters are available but subject to unexpected changes!Reactive scheduling C

Robust and reactive scheduling

Christian Artigues

LAAS-CNRS, Universite de Toulouse, France

TU Munchen- Technology-Based Services and Operations ManagementJuly 20, 2011

C. Artigues TU Munchen- Technology-Based Services and Operations Management, July 20, 2011 1 / 67

1 Introduction: scheduling under uncertainty

2 Reactive scheduling: the activity insertion problem

3 Robust scheduling: project scheduling with interval dataMinimum float computationThe Robust RCPSP


Resource-constrained schedulingRCPSP, Resource-Constrained Project Scheduling Problem

Example: 3 resource units available

i dur . res. usage

1 2 12 2 23 2 14 1 25 2 1

Schedule


DefinitionsRCPSP, Resource-Constrained Project Scheduling Problem

V = {0, 1, . . . , n, n + 1} = set of activities (project) with 0 dummmystart activity and n + 1 dummy end activity

R set of resource types

ak availability of resource type k ∈ R

rik demand of activity i for resource type k ∈ R

E ⊂ V × V : precedence constraints

pi duration of activity i

Minimize the makespan Cmax

Cmax ≥ Si + pi i ∈ V

Sj ≥ Si + pi (i , j) ∈ E∑i∈A(t)

rik ≤ ak k ∈ R with A(t) = {i ∈ V |Si ≤ t ≤ Si + pi − 1}


Uncertainties

Various sources of uncertaintiesUncertain durationsUncertain resource demandsUnexpected activitiesResource breakdownsOthers (due dates, costs,...)

Different uncertainty levels and modelsThe set of activities and/or resource availabilities is completely unkownin advance→ On-line schedulingProbability distributions on problem parameters are available→ Stochastic schedulingNo probability distributions on problem parameters are availablebut a set of scenarios is available→ Robust schedulingAn initial set of activities and an estimation on problemparameters are available but subject to unexpected changes→ Reactive scheduling


Robust Scheduling

Application to scheduling of robust discrete optimizationS denote a set of scenariosXs denote the discrete set of feasible solutions under scenario s ∈ Sf (x , s) denote the objective function value for solution x ∈ Xs underscenario s.x∗s denote an optimal solution for scenario s,The robust discrete optimization problem consists in finding a solutionx feasible for all scenarios which either minimizes the worst-caseperformance under the set of scenarios, i.e.

minx∈∩s∈SXs

maxs∈S

f (x , s)

or the maximal regret in terms of optimality loss

minx∈∩s∈SXs

maxs∈S

(f (x , s)− f (x∗s , s))

.

“Open loop” setting: all decisions are made at time zero.


Reactive Scheduling

“Closed loop” setting: decisions are taken for a certain time periodand updated at the next time period with the newly availableinformation.

Rolling horizon procedure (time-based rescheduling)1 At each step t starting with t = 0, consider a time interval It .2 Schedule all activities available in It (such that the release date ri ∈ It).3 The activities such that ri 6∈ It are either ignored or scheduled in a

relaxed or aggrageted way (ignoring resource constraints,. . . )4 Wait till the end of It , update information, set t ← t + 1 and go to

step 1.

Schedule repairing process (event-based rescheduling)1 Initially schedule all the available activities2 Control in real time the schedule feasibility (or optimality)3 As soon as feasibility is compromised perform a repairing procedure.

Go to step 2.


Common issues for robust and reactive scheduling

Robust vs reactive ?

The two approaches need not necessarily be opposed. Find a robustschedule and repair it as soon as it becomes infeasible or notperformant enough.

Need for flexible solutions

For robust scheduling, we must provide a solution feasible for allscenariosFor reactive scheduling, we must provide a solution easily repairable.

A flexible solution is based on the decomposition of the problemdecision variables in two subsets (stages).

First-stage variables are taken at the highest decision levelSecond-stage (recourse) variables have scenario-dependent values andshould be easily adjustable in case of data change.


Flexible Solution

Back to the schedule example

Description of the solution as a start-time vector S1 = S2 = 0,S3 = S4 = 2 and S5 = 3 is not flexible. If the duration of 1 increasesby 1, solution becomes infeasible.

Consider the start time as second-stage recourse variables andintroduce higher level first-stage sequencing variables:→ activity list scheduling→ partial order scheduling


Activity list solution representation

First-stage decision: sort activities in a totally orderedprecedence-compatible list

Second-stage decision : compute start-times by taking the activitiesin the list order and scheduling them as soon as possible (serialscheduling scheme)

2-list 3-scenario example (1 precedes 3)

2

1

3

4

2

1

3

4

2

1

3

4

2

1

3

4

2

1

3 4

2

1

3

4

List 1,2,3,4 List 2,4,1,3

The activity list representation yields flexible but unstable solutions


Partial-order solution representation

First-stage decision: compute an activity partial order that removesall resource conflictsSecond-stage decision : compute start-times by start the activities asearly as possible (ES-policy) w.r.t. to original and partial order-basedprecedence constraints

2-list 3-scenario example (1 precedes 3)

2

1

3

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1

3

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2

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1

3

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3 4

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Par)al order (1,4) (2,3) Par)al order (2,4)

The partial order schedule representation yields flexible and stablesolutions (removing precedence constraints, decreasing durationsand/or resource demands cannot increase the makespan)






Insertion problems in scheduling: generalities

An insertion problem is a scheduling problem in which the set ofactivities A is partitioned into two sets

The set of already scheduled activities A \ I (forming the baselineschedule).

The set of unscheduled activities I (to be inserted in the baselineschedule).

Solving the insertion problem amounts to schedule all the (scheduledand unscheduled) activities

Subject to scheduling constraints (precedence constraints, timewindows, resource-constraints).

Optimizing a scheduling objective function (makespan,...).

Involving constraints or objective of stability: the obtained schedulemust respect in some way the baseline schedule.


Interest of solving insertion problems

Reactive scheduling

Insertion of unexpected activities without performing a globalrescheduling.

Neighborhood search

Neighborhood defined in terms of reinsertion of (critical) activities.

Issues for designing an insertion problem

The insertion algorithm should be fast, preferably polynomial.

The search space should contain a significant number of solutions.


The disjunctive-graph model

Disjunctive graph G = (V ,U ,E )

n nodes V .

Precedence constraints (original and set by the baseline schedule)modeled by directed arcs U.

Each unscheduled activity is linked to any other activity requiring thesame machine through a pair of opposite (disjunctive) arcs. E is theset of all such pairs.

Each arc is valuated by the duration of the origin activity.

Dummy nodes 0 (n + 1) is a predecessor (successor) of all activities A.


Relevant work for disjunctive insertion problems

Vaessens, R.J.M., Generalized Job Shop Scheduling: Complexity and Local Search, Ph.D.thesis, Department of Mathematics and Computing Science, Eindhoven University ofTechnology, 1995.

F. Werner and A. Winkler, Insertion techniques for the heuristic solution of the job shopproblem, Discrete Applied Mathematics 58 (1995)

P. Brucker and J. Neyer, Tabu-search for the multi-mode job-shop problem, OR Spektrum20 (1998)

Y.N. Sotskov, T. Tautenhahn and F. Werner, On the application of insertion techniquesfor job shop problems with setup times, RAIRO Operations Research. 33 (1999)

C. Artigues and F. Roubellat, An efficient algorithm for operation insertion in amulti-resource job-shop schedule with sequence-dependent setup times, ProductionPlanning and Control 2 (2002)

T. Kis and A. Hertz, A lower bound for the job insertion problem, Discrete AppliedMathematics 128 (2003)

H. Groflin and A. Klinkert, Feasible insertions in job shop scheduling, short cycles andstable sets, 177 (2007)


The (single-)resource-constrained project schedulingproblem with minimum and maximum time lags(RCPSP/max)

Definition

a resource of availability B

n activities.

Each activity must be processed without interruption on biunspecified units of the resource.

The availability of the resource cannot be exceeded at any timeperiod.

Generalized Precedence constraints between activities Sj − Si ≥ lij , apositive or negative time lag.

Objective: makespan minimization.


Insertion problem example

Data

m = 3

n = 5, I = {1}i 0 1 2 3 4 5 6pi 0 6 4 2 4 2 0bi 0 1 2 2 2 3 0

precedence constraints and baseline schedule

From

Neumann el al (2003)


Resource flow network

Definition

a flow fij ≥ 0 from activity i to activity j represents the number ofmachines on j is a direct successor of activity i .

Each activity has a capacity of bi units.

Activity 0 is a source of m units while Activity n + 1 is a sink.

A flow is valid for an activity set V if it satisfies the conservationproperties and the activity capacity on each resource for each activityof V

The following resource flow is valid for A \ {1}: f03 = 2, f34 = 2, f45 = 2,f05 = 2, f52 = 2, f56 = 1, f26 = 1.Fortemps and Hapke (1997), Artigues and Roubellat (2000), Neumann el al (2003)


Flow formulation of the insertion problem (RCPIP/max)

Baseline resource flow

A Baseline resource flow is any flow valid for A \ I such that

There is no flow involving I

Sj < Si + pi in the baseline schedule implies fijk = 0

Graph induced by a flow

The graph induced by a flow is the graph of precedence constraintsaugmented by an arc (i , j) valuated by pi for each flow fij > 0.

RCPIP/max formulation

Given a baseline flow f , find a flow f ′ valid for A such that its inducedgraph has a minimal longest path length between 0 and n + 1 withoutincreasing the flow between two scheduled activities.


Insertion example

Remark

An insertion position can be represented by a subcut (α, β) of the graphinduced by the baseline flow such that

∑i∈α,j∈β fij ≥ bi


Characterization of feasible insertions for the search variantof the RCPIP/max

Search variant

To obtain the search variant of an insertion such that Cmax ≤ v , we setthe value of arc (n + 1, 0) to max(l(n+1)0,−v).

Distance matrix

δvij denotes the length of the longest path from i to j in the graphinduced by the baseline flow for a makespan of at most v .

δvij can be computed in O(n3) by the Floyd-Warshall algorithm.

Necessary condition for feasibility

There is no solution with a makespan of at most v is the problem ofcomputing Matrix δv admits no solution.


Characterization of feasible insertions for the search variantof the RCPIP/max

Inserting x into (α, β) may generated the following cycles (a ∈ α, b ∈ β).

Insertion-induced elementary cycles

a

x

bδvbx

max(pa, δvax)

δvxa

max(px , δvxb)

δvab

δvba

Theorem

The insertion of x in an insertion position (α, β) is feasible if and only ifmax(L1, L2, L3) ≤ 0 with L1 = maxa∈α(pa + δvxa), L2 = maxb∈β(px + δvbx),L3 = max(a,b)∈α×β(pa + px + δvba).


Problem complexity

Theorem

The RCPIP/max is NP hard

Proof

We prove NP-completeness of the decision variant by reduction fromthe independent set problem.


Proof (illustration)


RCPSP with minimum time lags only (optimizationvariant)

insertion-induced elementary cycles and 0, n + 1 longest paths

Consider distance matrix δ∞ij , a ∈ α, b ∈ β

a x

b n+1a x0(path 3)

δ∞0a pa px δ∞b(n+1)

(cycle 3)δ∞ba

x b0 n+1

δ∞0x px δ∞b(n+1)

(cycle 2) δ∞bx

δ∞0a pa δ∞x(n+1)

0 n+1

δ∞xa(cycle 1)

(path 1) (path 2)

The expression of the longest 0, n + 1 paths allow to define dominancerelations between insertion positions.


A polynomial insertion algorithm InsAct

Definitions

γ0 = {i ∈ A|δ∞ix = 0 and δ∞xi = 0} is the set of activities synchronizedwith x .

α0 = {i ∈ V \ γ0|δ∞ix ≥ 0}β0 = V \ (γ0 ∪ α0)

µ(α, β) = {i ∈ α|δ∞0i + pi = maxAa∈α(δ∞0a + pa)}ν(α, β) = {i ∈ β|δ∞i(n+1) = maxb∈β(δ∞b(n+1))}ν ′(α, β) = {i ∈ ν|δ∞xi = −∞}

Algorithm O(n2m)

Set initial cut (α, β) to (α0, β0)

While there remains enough capacity in (α, β) perform evaluaterecursively all subcuts obtained by removing µ from α and set nextcut (α, β) to (α ∪ ν ′, β \ ν).


RCPSP with minimum time lags only Example


Computational experiments: the standard RCPSP

600 Kolisch, Sprecher and Drexl instances with 120 activities

For each instance,

(Greedily) compute Initial schedule and initial flowRandomly generate an unexpected activity (3 parameters durationrange, time window tightness, resource demand level)insert the activity and compute the resulting makespan

3 compared insertion algorithm

ReschIns: proposed insertion algorithmParLft: parallel schedule generation scheme and min Latest FinishingTime priority ruleReschInsActive: proposed insertion algorithm+ 1 active schedulegeneration scheme pass


Computational results: the standard RCPSP


Conclusion

Even inserting a single activity is hard when maximum time lags areconsidered with the resource-flow model.

Can we a find polynomial insertion framework for the RCPSP/max ?

Use the polynomial algorithm without max. time lags as a heuristic ?

Good results for solving insertion problems with minimum time-lagsonly when associated with a fast active schedule generation schemeprocedure

C. Artigues and F. Roubellat, A polynomial activity insertion algorithm in a multi-resourceschedule with cumulative constraints and multiple modes, European Journal of OperationalResearch 127 (2000)C. Artigues, P. Michelon and S. Reusser, Insertion techniques for static and dynamicresource-constrained project scheduling, European Journal of Operational Research 149 (2003)

C. Artigues, C. Briand, Complexity of insertion problems for resource-constrained project

scheduling with minimum and maximum time lags, Journal of Scheduling, 12(5), pp.447-460,

2009






Activity Network with interval data

0

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3

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5

[0,0]

[2,4] [3,5]

[2,2] [1,2]

[0,0]

Data

Network with n activities subject to precedence constraintsduration interval

[dmini , dmax

i

]scenario d ∈ D = Πn

i=1[dmini , dmax

i ]


Basic problem : computing resource-unconstrained activityfloats

Activities attributes

Best earliest/latest starting times → PWorse earliest/latest starting times → PMinimum activity slack → NP − hard

Interests

Criticity of an activity at the worst case among all scenarios

Compute the float fi of the activity i for a given scenario

fi = lsti − esti = l∗0,n+1 − (l∗0,i + l∗i,n+1)esti et lsti : earliest and latest starting timesl∗i,j : longest path from i to j

Problem

mind∈D fi (d), ∀i = 1, . . . , n


Float property

Dubois et al.There exists an extreme scenario d∗ such that fi (d

∗) = mind∈D fi (d)2n scenarios

For each activity i , there exists a path p and its induced extremescenario dmax(p) such that fi (d

max(p)) = mind∈D fi (d)Where dmax(p) fix all the arcs of p set to dmax and others set to dmin

The Dubois et al. Path algorithm parses all the paths and computes allthe activityJ. Fortin, P. Zielinski, D. Dubois and H. Fargier, Criticality analysis ofactivity networks under interval uncertainty Journal of Scheduling,13(6), 609-627 (2010)

RemarkIt may be costly to enumerate all the paths !


Path algorithm

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[0,0]

[2,4] [3,5]

[2,2] [1,2]

[0,0]

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[0]

[4] [3]

[2] [2]

[0]

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[0]

[4] [5]

[2] [1]

[0,0]

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[0,0]

[2] [3]

[2] [2]

[0,0]

0 0

3 1

0 0

4 6

0 0

1 1

[dura0on] float


Activities-Independent MIP

Q(k) :

min z(k) = Sn+1 −∑i

dmaxi xi (1)

subject to

z(k) ≥ z(k − 1), (2)

Sj ≥ Si + dmini + (dmax

i − dmini )xi ∀(i , j) ∈ A, (3)

xi ≤∑j∈Γ−1

i

xj ≤ 1 ∀i 6= 0, n + 1, (4)

xi ≤∑j∈Γi

xj ≤ 1 ∀i 6= 0, n + 1, (5)

x0 = xn+1 = 1, (6)∑i∈Uk

xi ≥ 1, (7)

xi ∈ {0, 1} ∀i ∈ V . (8)


Activities-dependent MIP

P(k) :

min fk = Sn+1 −∑i

dmaxi xi (9)

subject to

Sj ≥ Si + dmini + (dmax

i − dmini )xi ∀(i , j) ∈ A, (10)

xi ≤∑j∈Γ−1

i

xj ≤ 1 ∀i 6= 0, n + 1, (11)

xi ≤∑j∈Γi

xj ≤ 1 ∀i 6= 0, n + 1, (12)

x0 = xn+1 = 1, (13)

xk = 1, (14)

xi ∈ {0, 1} ∀i ∈ V . (15)


Float properties

For an extreme scenario, there are at most 2 points where the longestpath l∗0,n+1 and the longest path p traversing i diverge

There exists (a, b) ∈ p such that :

l∗0,n+1 = l∗0,a + l∗a,b + l∗b,n+1

l∗0,i = l∗0,a + l∗a,il∗i,n+1 = l∗i,b + l∗b,n+1

l∗a,b = l∗a,b(dmin) + dmaxa − dmin

a

The float is valid for all activities of p between a and b


A new branch and bound algorithm

Branch and bound for each activity i

Root node: a partial path reduced to {i}Left extension of node {a, . . . , i , . . . , b} generated by inserting apredecessor of a at the beginning of the partial pathRight extension of node {a, . . . , i , . . . , b} generated by appending asuccessor of b at the end of the partial pathChild nodes are either all possible left etensions of all possible rightextensionsUse of lower bound and dominance rules based on path structures


Instances

Density of G = (V ,A). Order strength = |A+||V |(|V |−1)/2

120-activities PSPLIB-instances. 0, 1 ≤ OS ≥ 0, 5 anddurations ∈ [d ; 1, 2d ]

100-activities instances. 0, 1 ≤ OS ≤ 0, 9 and durations ∈ [d ; 1, 2d ]

Modified : durations ∈ [d ∈ [10; 50α]; [d ; (1 + β)d ]].α, β ∈ {0, 3; 0, 6; 0, 9}

Run on an Intel 2.6 GHz


Original instances

serie average time sec. maximum timePath MIPd BB Path MIPd BB

PSPLIB

OS ∈ [0.1, 0.5] < 0.005 4.55 < 0.005 0.01 23.02 0.01

RG

OS = 0.1 < 0.005 0.25 < 0.005 0.01 0.33 0.01OS = 0.2 < 0.005 0.54 < 0.005 0.01 0.69 0.01OS = 0.3 0.02 1.08 < 0.005 0.02 1.61 0.01OS = 0.4 0.04 2.29 < 0.005 0.05 3.40 0.01OS = 0.5 0.12 4.99 0.01 0.19 7.71 0.02OS = 0.6 0.43 9.22 0.01 0.62 16.51 0.02OS = 0.7 2.50 14.95 0.02 3.79 21.13 0.04OS = 0.8 33.96 23.12 0.02 55.99 98.80 0.04OS = 0.9 > 1800 17.43 0.02 > 1800 111.45 0.03


Modified instances

serie average time sec. maximum timePath MIPd BB Path MIPd BB

α− β = 0.3OS = 0.1 < 0.005 0.24 < 0.005 0.01 0.33 0.01OS = 0.2 < 0.005 0.51 < 0.005 0.01 0.67 0.01OS = 0.3 0.02 1.03 < 0.005 0.02 1.58 0.01OS = 0.4 0.04 2.17 < 0.005 0.06 3.6 0.01OS = 0.5 0.12 4.86 0.01 0.22 8.64 0.02OS = 0.6 0.44 9.04 0.01 0.62 14.07 0.02OS = 0.7 2.51 15.65 0.02 3.83 24.78 0.04OS = 0.8 34.37 24.62 0.02 58.01 73.29 0.06OS = 0.9 > 1800 17.12 0.02 > 1800 78.14 0.04α− β = 0.6OS = 0.1 < 0.005 0.26 < 0.005 0.01 0.35 < 0.005OS = 0.2 < 0.005 0.56 < 0.005 0.01 0.72 0.01OS = 0.3 0.02 1.15 < 0.005 0.02 1.71 0.01OS = 0.4 0.04 2.37 < 0.005 0.05 3.67 0.01OS = 0.5 0.12 5.34 0.01 0.17 8.08 0.02OS = 0.6 0.43 10.71 0.01 0.65 17.49 0.02OS = 0.7 2.5 23.43 0.02 4.03 52.17 0.04OS = 0.8 34.35 63.89 0.02 56.45 157.07 0.06OS = 0.9 > 1800 94.61 0.02 > 1800 590.17 0.04α− β = 0.9OS = 0.1 < 0.005 0.27 < 0.005 < 0.005 0.34 0.01OS = 0.2 < 0.005 0.6 < 0.005 0.01 0.82 0.01OS = 0.3 0.02 1.25 < 0.005 0.02 1.85 0.01OS = 0.4 0.04 2.55 < 0.005 0.05 3.82 0.01OS = 0.5 0.12 5.77 0.01 0.22 9.49 0.02OS = 0.6 0.43 12.12 0.01 0.66 23.89 0.02OS = 0.7 2.52 34.76 0.01 3.87 84.68 0.03OS = 0.8 34.38 124.22 0.02 57.71 267.88 0.05OS = 0.9 > 1800 338.98 0.01 > 1800 1776.73 0.08






Project schedulingRCPSP, Res.-Constr. Project Scheduling Problem


i dur . res. usage

1 2 12 2 23 2 14 1 25 2 1

Schedule


ES policies


i poss. durs. res. usage

1 {1, 2, 8} 12 {2, 3} 23 {2} 14 {1, 3} 25 {1, 2} 1

Possible resource allocations


ES policies → resource flows

Original network

Resource flows for previous two allocations


Definitions

V = {0, 1, . . . , n, n + 1} = set of activities (project) with 0 dummmystart activity and n + 1 dummy end activity

R set of resource types

ak availability of resource type k ∈ R

rik demand of activity i for resource type k ∈ R

E ⊂ V × V : precedence constraints

Pi ⊂ R+: set of possible durations of activity i ∈ V , withP0 = Pn+1 = {0} (either interval or discrete set)

p = (p0, p1, . . . , pn, pn+1) with pi ∈ Pi : sample / scenario

P = P0 × P1 × . . .× Pn+1: all scenarios


Definitions (2)

Decision variables in RCPSP: start times

Decision variables for ES policies: selection X ⊂ V × V

Feasibility of a selection:1 G (V ,E ∪ X ) acyclic2 no more forbidden sets for E ∪ X , with a forbidden set a set of

incomparable activities whose resource consumption jointly exceed theresource availability.For the example: {1, 2, 5}, {1, 3, 4}, {2, 4}, {3, 4, 5}

When each |Pi | = 1, finding an ES policy with lowest longest pathlength in G (V ,E ∪ X ) solves the (deterministic) RCPSP(ES policies generalize the disjunctive-graph representation for jobshops)

There is a one-to-many relation between feasible selections andresource flows


Evaluating absolute robustness

sn+1(X ,p) =longest path length in graph G (V ,E ∪ X )

with node weights p

X =set of all feasible selections

s∗n+1(p) = minX∈X

sn+1(X ,p) (RCPSP)

regret ρ(X ,p) =sn+1(X ,p)− s∗n+1(p)

Maximum regret for feasible selection X is

ρmax(X ) = maxp∈P

ρ(X ,p) = maxp∈P

{sn+1(X ,p)− s∗n+1(p)

}Computing ρ(X ,p) and ρmax(X ) is NP-hard.

Let an extreme scenario have pi = maxPi or = minPi , ∀i ∈ V .Theorem: the max absolute regret can always be reached in an extremescenario.


Absolute minimax regret optimization

The optimization problem AR-RCPSP that we wish to solve can now bestated as follows:

ρ∗ = minX∈X

ρmax(X ) = minX∈X

maxp∈P

ρ(X ,p)

= minX∈X

min ρsubject toρ ≥ sn+1(X ,p)− s∗n+1(p) ∀p ∈ P

This does require prior knowledge of s∗n+1(p).

Let P = {p1, . . . ,p|P|}.


FormulationScenario-independent constraints

∑j∈V ,j 6=i

fjik =∑

j∈V ,j 6=i

fijk = rik ∀i ∈ V \ {0, n + 1},∀k ∈ R

∑j∈V ,j 6=0

f0jk =∑

j∈V ,j 6=(n+1)

fj,n+1,k = ak ∀k ∈ R

fijk ≥ 0 ∀(i , j) ∈ V × V ,∀k ∈ R

0 ≤ fijk ≤ Mxij ∀(i , j) ∈ V × V ,∀k ∈ R

xij = 1 ∀(i , j) ∈ E

xij ∈ {0, 1} ∀(i , j) ∈ V × V

Variables:fijk = number of resources type k going directly from act i to jxij = 1 if (i , j) ∈ E ∪ X , with X the selection to be made


FormulationScenario-dependent constraints

ρ∗ = min ρ

subject to

ρ ≥ Shn+1 − s∗n+1(ph) h = 1, . . . , |P|

Shj ≥ Sh

i + phi −M(1− xij)∀(i , j) ∈ V × V , i 6= j ,

h = 1, . . . , |P|Shi ≥ 0 ∀i ∈ V , h = 1, . . . , |P|

with Shj = (earliest) starting time of activity j in scenario h


Full formulation (“master”)

ρ∗ = min ρ

subject to

ρ ≥ Shn+1 − s∗n+1(ph) h = 1, . . . , |P|

Shj ≥ Sh

i + phi −M(1− xij)∀(i , j) ∈ V × V , i 6= j ,

h = 1, . . . , |P|Shi ≥ 0 ∀i ∈ V , h = 1, . . . , |P|∑

j∈V ,j 6=i

fjik =∑

j∈V ,j 6=i

fijk = rik ∀i ∈ V \ {0, n + 1},∀k ∈ R

∑j∈V ,j 6=0

f0jk =∑

j∈V ,j 6=(n+1)

fj,n+1,k = ak ∀k ∈ R

fijk ≥ 0 ∀(i , j) ∈ V × V ,∀k ∈ R

0 ≤ fijk ≤ Mxij ∀(i , j) ∈ V × V ,∀k ∈ R

xij = 1 ∀(i , j) ∈ E

xij ∈ {0, 1} ∀(i , j) ∈ V × V


Scenario relaxation1

Replace P by P; optimal objective value is ρ∗(P).

1: initialisationP1 = p1, q = 1, LB = 0, UB = +∞. Compute s∗n+1(p1).

2: master problemSolve the restricted master to obtain LB = ρ∗(Pq) and ES policy Xq. IfLB = UB then stop.

3: maximum-regret computationEvaluate the maximum regret ρmax(Xq) of policy Xq against all scenarios andobtain worst-case duration vector pq+1 and makespan s∗n+1(pq+1). SetUB = min{ρmax(Xq);UB}.

4: optimality checkIf LB = UB then stop; else set q = q + 1, Pq = Pq−1 ∪ {pq} and go to Step2.

1T. Assavapokee, M.J. Realff, and J.C. Ammons. A new min-max regret robustoptimization approach for interval data uncertainty. JOTA, 137:297–316, 2008.


Step 3: evaluate selection XSimultaneous computation of the maximal regret and of new scenario + RCPSP solution


∑

(i,j)∈E∪Xpiφij

− S∗(p)

subject to ∑

(i,j)∈E∪Xφij =

∑(j,i)∈E∪X

φji ∀i ∈ V \ {0, n + 1}

∑(0,j)∈E∪X

φ0j =∑

(j,n+1)∈E∪Xφj,n+1 = 1

φij ≥ 0 ∀(i , j) ∈ E ∪ X

S∗(p) = minY∈X

Sn+1

subject to Sj ≥ Si + piS0 = 0

∀(i , j) ∈ E ∪ Y

The variables φij route a unit flow through the network G (V ,E ∪ X )


Step 3 (ii)integrate two levels


∑

(i,j)∈E∪Xpiφij

− Sn+1

subject to ∑

(i,j)∈E∪Xφij =

∑(j,i)∈E∪X

φji ∀i ∈ V \ {0, n + 1}

∑(0,j)∈E∪X

φ0j =∑

(j,n+1)∈E∪Xφj,n+1 = 1

φij ≥ 0 ∀(i , j) ∈ E ∪ X

Sj ≥ Si + pi ∀(i , j) ∈ E ∪ Y

S0 = 0

Y ∈ X

For Y ∈ X : resource-flow formulation.Still non-linearity in the objective. . .


Step 3 (iii)remove non-linearity

We need only consider two values pmini = minPi and pmax

i = maxPi for pi(i ∈ V \ {0, n + 1}).

New variables ai (i = 1, . . . , n): replace pi by (1− ai )pmini + aip

maxi .

Replace φij in the constraints by φminij + φmax

ij , in which φminij fulfills the role

of φij when ai = 0 and φmaxij functions as φij in the cases where ai = 1. This

is achieved by adding :∑(i,j)∈E∪X

φmaxij ≤ ai ∀i ∈ V \ {0, n + 1}

∑(i,j)∈E∪X

φminij ≤ 1− ai ∀i ∈ V \ {0, n + 1}

Replace piφij by pmini φmin

ij + pmaxi φmax

ij .


Step 3 (iv)“bounded contribution”

Step 3 absorbs most of the running time.

Let z = obj subproblem; we look for pq+1 such that z ≥ LB + 1.

→ Replace the objective function of Step 3 by

b∗(X ) = min b

and add constraints

z + b ≥ LB + 1

z ≤

∑(i,j)∈E∪X

pmini φmin

ij + pmaxi φmax

ij

− Sn+1

z , b ≥ 0

This model has a solution b∗(X ) = 0 iff exists z ≥ LB + 1.

Drawback: value S∗n+1 may not be the optimal makespan. We need to solvea standard RCPSP instance to obtain s∗n+1(p∗) (in Step 3) before adding p∗

to P (in Step 4).C. Artigues TU Munchen- Technology-Based Services and Operations Management, July 20, 2011 59 / 67

Heuristic

1: initialisationP1 = {p1}, q = 1 and compute s∗n+1(p1).

2: generate new solutionUse s∗n+1(pq) and Pq to produce a new approximate solution Xq.

3: generate new duration vectorsGenerate one or more new duration vectors pq+1 that represent scenariosunder which policy Xq performs badly, together with an RCPSP upper bounds∗n+1(pq+1).

4: iterateSet q = q + 1, Pq = Pq−1 ∪ {pq} and go to Step 2.

Step 2: list representation of optimal schedule for p1 and “path relinking”with current best → schedule → resource flow, and choices by computingregret only for P.

Step 3: upper bound on regret of X via maxp∈P {sn+1(X ,p)− sn+1(∅,p)}

Cycling → other scenario-generation options. . .


Scenario relaxation, exact algorithmn = 10, 10 instances2 per cell

parametersstandard bounded contribution

timeit

timeit

OS RF RC St 2 St 3 tot St 2 St 3 tot

.4.45

.3 0.03 0.04 0.07 2.7 0.02 0.02 0.04 2.7

.6 0.12 1.15 1.27 2.1 0.12 0.71 0.83 2.1

.9.3 11.30 15.27 26.57 9.0 10.23 9.15 19.38 15.40.6 16.48 882.59 899.07 1.6(2) 19.06 734.84 753.90 4.25(1)

.6.45

.3 0.01 0.02 0.03 2.1 0.01 0.02 0.03 2.8

.6 0.06 0.12 0.18 2.1 0.06 0.07 0.13 2.8

.9.3 0.15 0.40 0.55 3.4 0.17 0.24 0.41 4.3.6 0.20 11.17 11.37 1.2 0.22 14.77 14.99 1.7

.8.45

.3 0.01 0.02 0.03 1.2 0.01 0.00 0.01 1.2

.6 0.01 0.02 0.03 1.2 0.01 0.02 0.03 1.2

.9.3 0.03 0.05 0.08 1.8 0.03 0.04 0.07 1.8.6 0.05 0.14 0.19 1 0.05 0.14 0.19 1

2Data generated by RanGen, pmin and pmax generated randomly(E. Demeulemeester, M. Vanhoucke, and W. Herroelen. A random network generatorfor activity-on-the-node networks. JOS, 6:13–34, 2003).


Comparison exact / heuristicn = 10

parameters exact heuristicOS RF RC time gap∗ opt. time gap∗ opt.

0.40.45

0.3 0.04 0.00 10 2.03 17.78 50.6 0.83 0.00 10 2.53 31.37 5

0.90.3 19.38 0.00 10 3.88 44.87 40.6 753.90 0.00 9 2.58 25.00 5

0.60.45

0.3 0.03 0.00 10 1.36 28.21 70.6 0.13 0.00 10 2.44 25.00 7

0.90.3 0.41 0.00 10 2.24 30.51 40.6 14.99 0.00 10 1.86 11.90 8

0.80.45

0.3 0.01 0.00 10 0.21 0.00 100.6 0.03 0.00 10 0.00 0.00 10

0.90.3 0.07 0.00 10 1.22 17.50 70.6 0.19 0.00 10 0.73 0.00 10

“opt.” based on CPU bound of 30 min for exact algorithm

Stopping criteria heuristic: CPU time (30 min), max nr of scenarios (100)and max nr consecutive cycling occurrences (10).

gap∗ = 100× f (alg)−f (opt)δ , with δ = 1

n

∑ni=1

(pmaxi − pmin

i

).


Comparison exact / heuristicn = 20

parameters exact heuristicOS RF RC time opt. time opt.

0.40.45

0.3 542.31 8 2.00 10.6 1947.81 1 3.22 0

0.90.3 2247.11 0 3.59 00.6 1830.47 1 5.18 0

0.60.45

0.3 61.00 10 1.63 20.6 1559.89 0 2.57 1

0.90.3 1466.95 0 2.58 30.6 2081.57 1 1.87 0

0.80.45

0.3 0.68 10 0.84 50.6 133.15 10 1.10 4

0.90.3 177.56 10 1.41 20.6 1808.54 7 0.29 0


Other comparisons?n = 20

We compute

LB(X ,Y ) = maxp∈P

{sn+1

(X , p

)− sn+1

(Y , p

)},

via MIP. This is a lower bound for ρmax(X ) for any feasible Y .

parametersLB(exact, heur) LB(heur, exact)

OS RF RC

0.40.45

0.3 0.00 31.480.6 49.29 33.12

0.90.3 80.63 38.100.6 89.45 4.56

0.60.45

0.3 0.00 36.210.6 79.05 42.92

0.90.3 77.81 42.650.6 51.26 33.64

0.80.45

0.3 0.00 33.290.6 0.00 75.42

0.90.3 0.00 47.380.6 18.04 68.00


Other heuristics

Heuristic 2

1: determine the average duration pi = b pmini +pmax

i

2 c for each activity i2: solve the corresponding deterministic RCPSP3: impose a resource flow on this deterministic schedule4: output the solution found

Heuristic 3

1: ignore the activity durations2: find a solution (ES policy) with a transitive closure having the minimum

number of arcs via MIP


Comparison heuristics n = 30

parameters Scen-Rel-Heur Heuristic 2 Heuristic 3OS RF RC time opt. time opt. time opt.

0.40.45

0.3 4.99 2 0.00 0 0.27 00.6 6.89 1 0.00 0 105.54 1

0.90.3 25.79 0 0.00 0 703.85 00.6 17.59 0 0.00 0 1800.17 0

0.60.45

0.3 21.15 1 0.00 1 976.51 00.6 17.30 1 0.00 0 1800.06 0

0.90.3 52.18 0 0.01 0 1800.03 00.6 53.07 0 0.00 0 1800.04 0

0.80.45

0.3 14.74 2 0.01 0 1753.83 00.6 11.62 1 0.01 0 1800.07 0

0.90.3 31.81 1 1.19 0 1800.05 00.6 32.46 0 0.00 0 1800.02 0

parameters SRH H2 H3OS RF RC H2 H3 SRH H3 SRH H2

0.40.45

0.3 36.90 41.65 150.30 57.24 239.00 77.620.6 39.20 43.70 403.50 108.31 344.00 94.87

0.90.3 15.50 18.80 381.80 163.44 310.10 124.640.6 55.20 58.80 420.10 122.18 390.70 210.23

0.60.45

0.3 34.20 39.40 244.40 132.52 197.00 97.560.6 39.80 54.90 332.00 142.10 304.60 56.78

0.90.3 34.60 56.50 354.80 180.30 365.10 169.230.6 73.70 86.90 467.30 156.34 444.80 274.59

0.80.45

0.3 18.50 28.10 206.60 93.24 169.30 73.520.6 57.00 86.40 376.00 196.03 358.80 216.76

0.90.3 59.70 90.50 406.70 215.73 400.80 91.08


Summary and conclusions

Robust ES-policies for resource-constrained project scheduling underuncertainty

Scenario-relaxation algorithm and heuristic

Only small instances. . .

Evaluation? Simulation?Number of scenarios is very high.

Dependence in durations?‘Less’ uncertainty? (e.g., Bertsimas & Sim)Is it at all appropriate to model duration uncertainty in this way?

Robustness / regret objectives for NP-hard problems? In our case,evaluation comes down to solving a multi-mode RCPSP.

Artigues, C., Leus, R. and Talla Nobibon, F. (2010). Robust optimization for

resource-constrained project scheduling with uncertain activity durations.

Research report 1034, Department of Decision Sciences and Information

Management, FBE, KULeuvenC. Artigues TU Munchen- Technology-Based Services and Operations Management, July 20, 2011 67 / 67


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