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Project Scheduling

Project Scheduling. 1. Project Planning (revisited) 2. Resource Constrained Project Scheduling 3. Parallel Machine Scheduling. Topic 1. Project Planning (revisited). C. H. A. F. dummy. E. G. B. J. D. Project Planning: Jobs on Arcs. node: fase in project

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Project Scheduling

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  1. Project Scheduling 1. Project Planning (revisited) 2. Resource Constrained Project Scheduling 3. Parallel Machine Scheduling

  2. Topic 1 Project Planning (revisited)

  3. C H A F dummy E G B J D Project Planning: Jobs on Arcs node: fase in project (end of 1 or more tasks) every task := arc (dummies for precedence) jprec duration pj A: - 3 B: - 2 C: A 4 D: A, B 3 E: A 2 F: E 5 G: E 6 H: C, F 2 J: D, G 4 Finish: A to J

  4. C 4 2 5 H 2 A 3 E 2 dummy F 5 1 4 7 G 6 B 2 J 4 3 6 D 3 Critical Path Method (CPM) jprec A: - B: - C: A D: A, B E: A F: E G: E H: C, F J: D, G 3 3 10 13 5 5 0 0 15 15 3 8 11 11 ES1 = 0 ESj = maxi { pi + ESi | i  P(j)} LFn = ESn LFi = minj { LFj - pj | j  Si }

  5. 0 3 ? 5 0 15 j = 0 j =n+1 3 5 3 = max{0+2,0+3} 11 0 Alternative Network: Jobs on Nodes jprec A: - B: - C: A D: A, B E: A F: E G: E H: C, F J: D, G A 3 A 3 C 4 C 4 H 2 H 2 E 2 E 2 F 5 F 5 G 6 G 6 B 2 B 2 D 3 D 3 J 4 J 4 • Advantages • no dummy-jobs needed to enforce precedence • length of node may represent duration or arc-length

  6. Jobs on a time axis jprec A: - B: - C: A D: A, B E: A F: E G: E H: C, F J: D, G A 3 jobs B 2 C 4 D 3 E 2 F 5 G 6 H 2 J 4 t 0 3 5 11 15

  7. cut set 5 6 minimal cut set Project Planning with crashing Heuristic 1 4 4 5 8 0 6 6 n+1 Graph of all Critical Path(s)

  8. Topic 2 Resource Constrained Project Scheduling Problem (RCPSP)

  9. Project PlanningasScheduling • Activities within projectare jobs • Criterium: makespan onder volgorde- (en resource-restricties?) P | prec (, rjk, Rk )| Cmax • CPM-solution assumes: always enough resources RCSP notation/assumptions: • Job j uses rj k units of resource k, during pj periods • Renewable resources: At any time one has Rkof resource k available • Determine all finish times Fj (=genoeg om hele schedule weer te geven)

  10. With resource requirements rj on jobs j Example: job A requires 9 workers for 3 days, etc. rjj pj j prec A: - B: - C: A D: A, B E: A F: E G: E H: C, F J: D, G 9A3 8B2 7C4 7D3 3E2 2F5 3G6 8H2 With R=16 workers available Is Cmax=15 days still possible? 3J4 t 0 3 5 11 15

  11. With resource requirements rjon jobs j rjj pj j P(j) A: { } B: { } C: {A} D: {A, B} E: {A} F: {E} G: {E} H: {C, F} J: {D, G} 9A3 8B2 7C4 7D3 3E2 2F5 3G6 8H2 3J4 Feasible, even with 12 workers t 0 3 5 11 15

  12. CPM on (resource-relaxed) RSCP • Generates “earliest” start- ESj and “latest” LFj finish times • The CPM-time is lower bound on RSCP-time: LFn+1 Cmax • Critical path jobs must start at ESj to accomplish Cmax = LFn+1 • Other jobs have “slack” • Their start time is NOT fixed by CPM • Shift non-critical j in [ESj , LFj] smoothing the use of resources: after resource loading, balancing resource requirements • If capacity Rk is still exceeded, then lengthen makespan (let Fj exceedLFj for some j). • Later, we discuss heuristic methods for RCPSP

  13. Resource loading and leveling 19 D loading: jobs start at ESj leveling: shift jobs in time 13 C D D 10 A E E F 7 B C

  14. Conceptual Model • Notation • Decision variable: Fj (finish time) for j  J • P(j) = set of predecessors of job j • A(t) set of active jobs j in period t • Model Min Fn+1 s.t. Fh Fj - pj j  J,  hP(j) Fj  0  j  J jA(t) rjk  Rk  resources k ,  t A(t)={ j | Fj – pj < t  Fj }  t

  15. 3 0 1 2 4 Making an ILP Model for RCPSP To achieve an ILP formulation: How can we linearize A(t)={ j | j active at t } and Fj? • Change to discrete time-horizon [0,H], periodst = 1,2, ..H For j J: Fj  {1, …, H} ( Choosing H large enough) • Use extra binary variables • Xjt = 1 if task j ends in periodt; • Ajt = 1 if task j is active in periodt; E.g. Fj = 4 en pj = 2 Xj4= 1 Aj3= 1 Aj4= 1

  16. Resulting ILP model for RCPSP Min s.t. (1)  j  J,  hP(j) Precedence restrictions (2) j  J (3)  j  J Determine one last period (4)  j  J,  t = 1, …, H Determining active periods of tasks (5)  t = 1, …, H,  k  K Consumption of scarce resources

  17. Remarks • Aiding variables Fj en Ajt can be eliminated: see book • Exact ILP methods useful only for benchmarking other methods on small problem sizes • RCPSP is generalization of job shop problem

  18. 0/13 3/15 5/16 2A3 4B2 3C1 00 00 0/10 4/12 6/16 3D4 4E2 2F4 Heuristic methods for RCPSP • Two Schedule Generation Schemes can construct a schedule • Serial Method (task oriëntated): fixes in each of n iterations one task at earliest feasible starting time • Parallel Method (time oriëntated): at next finish time of active task start one or more tasks • Example 0/13 3/15 5/16 2A3 4B2 3C1 ESj / LSj 8/16 rj j pj 00 00 0/10 4/12 6/16 3D4 4E2 2F4

  19. Parallel SGS (informal) • At most n stages • Each stage g <= n represents: 1. Schedule time tg(=next finish time of scheduled job) 2. A partial schedule, consisting of four disjoint sets of jobs: • Completed, scheduled jobs at time tg • Active jobs: scheduled jobs, not yet completed at time tg • Decidable jobs: unscheduled, schedulable jobs (that can be chosen to start at time tn) • remaining set: unscheduled jobs that cannot start at time tn

  20. Parallel SGS (on example) tg finished R(tg) schedulable Dg j select R(tg) Fj 0  4 {A, D} 3D4 1 4 (3D) {} - (* not A,E ; both require > 1 *) 4 3D 4 {2A3, 4E2} 4E2 0 6 (4E) 6 4E 4 {2A3, 2F4} 2A3 2 9 (2A) {F} 2F4 0 10 (2F) 9 2A 2 { }({B}) - (* 2 short for B *) 10 2F 4 {B} 4B2 0 12 (4B) 12 4B 4 {C} 3C1 1 13

  21. Parallel SGS (formal) • Notation (sets in bold) • Cg = activities: scheduled and finished; C0 = {0}; • Ag = activities: scheduled and still active; A0 = {}; • Dg = activities: schedulable according to prec.- AND cap. restrictions • Rk(t) = remaining capacity of resource k at time t Rk(0) = Rk • Algorithm While some jobs  Cg  Ag do(* i.e., unscheduled jobs *) • g = g+1; • tg = minjA(g) {Fj} , jg=arg (* next finish time active task *) • Update Cg,Ag, Rk(tg), Dg(* for jg scheduled and finished *) • While Dg   do(* also Rk(tg) sufficient *) • Take jDg, set Fj = tg + pj en update Ag, Rk(tg), Dg

  22. Serial SGS (informal) Each stage selects a job  n stages • set of jobs already selected and planned: scheduled jobs (S) • decision set (D): jobs withall predecessors in S • remaining set of jobs: these will first enter Dandthen S Procedure 1. Start with scheduled set S=empty; j=0 2. Add j to S; Update schedulable set D (:= {j | P(j)S} ); 3. Select job j from decision set D(with highest priority), and (using {t/R(t) : finish times t} ) set j’s start/finish time as early as possible; 4. If |S|<n then go to step 2 (otherwise be happy with an 'active schedule')

  23. Serial SGS (on example) Sj=min {t  Fg | t  ESj and k, t’ [t, t+pj)  Fg : Rk(t’)  rj } step{ t / R(t) | t in Fg } Dgrj j pj SjFjupdates t / R(t) 0. {0/4} {A, D} 3 D 4 0 4 R(0)=1, R(4)=4 (e.g. as priority rule: D has lower slack 8=10-(0+2)) 1. {0/1,4/4} {A, E} 4 E 2 4 6 R(4)=2, R(6)=4 (idem, E has lower slack) 2. {0/1,4/2,6/4} {A, F} 2 A 3 6 9 R(6)=2, R(9)=4 (idem) 4. {0/1,4/2,6/2,9/4} {B, F} 4 F 2 6 10 R(6)=0, R(9)=2, R(10)=4 (idem) 5. {0/1,4/2,6/0,9/2,10/4} {B} 4 B 2 10 12 R(10)=0, R(12)=4 6. {.. 10/0, 12/4} {C} 3 C 1 12 13 R(12)=1

  24. Serial SGS (formal) • Notations (sets in bold) • Sg = activities already scheduled (in steps <=g-1) S0 = { 0 (dummy)}; • Dg = activities (precedence-)schedulable (i.e, predecessors in Sg) • Fg = { finishtimes of jobs j  Sg } F0= {0}; F0 = 0; • Rk(t) = remaining capacity of resource k at time t • Repeat n times (i.e., “for g := 0 to n do”) • Determine : Fg, Rk(t) voor tFg(*Initially F0= {0}, Rk(0)=full resource k *) • DetermineDgand take j Dg(* Dg set of decidable jobs *) // schedule job j : • Determine ESj = max { Fh | hP(j) } (* respecting precedence *) • Determine Fj = pj+ min { t [ESj, LSj]  Fg | Rk(t) is feasible for j } • Update Rk(t)

  25. Comments on Serial SGS • Constructs a feasible schedule • When resources are ample: schedule is optimal • Leads to schedule in class of ‘active schedules’, • I.e., no operation can be earlier without others finishing later • Class of ‘active schedules’ contains an optimal schedule • One can specify selection in “Take j in Dg” by giving: • priority rule(s) • an a priori list // and this explais the name ‘list scheduling’

  26. Comments on Parallel Method Algorithm (informal) 1. Start with finish time of active job 0 (scheduled with finish time 0) 2. In step g: tg= earliest time (that unscheduled jobs may start.) andDg= collection of jobs that may start at time tg Normally, when Dg empty: tg = first finish time of active job j 3. Select the job fromD with the highest priority, let it start at time tg 4. If jobs are still unscheduled thengoto step 2 elsestop

  27. Comments on Parallel SGS • If ample resources then schedule is optimal • Schedule will be a non-delay schedule: • Exercise: why is that so? • Class of ‘Non-Delay schedules’ contains not always an optimal schedule • For serial en parallel: • Single pass: 1 SGS combined with one Priority Rule • Multi pass • 1 SGS with all PR’s • Forward-backward scheduling • Sampling method: “random”, “biased” or “regret based biased” • Metaheuristics

  28. Topic 3 Parallel Machine Scheduling Problems

  29. Parallel Machine Models • Lets say we have • Multiple machines (m), where • the makespan should be minimized (Cmax) • We denote this problem as • Problem already NP-hard for m = 2 (partitioning problem) • LPT-rule good heuristic

  30. Worst case behaviour LPT • Maximal deviation from optimal value • Example: four parallel machines, nine jobs Cmax(opt)=12 but Cmax(LPT)=15

  31. More makespan problems easy NP-hard easy Least Flexible Job first is optimal if Mj is nested

  32. Allowing preemptions • Lets say we have • Multiple machines (m) • preemption is allowed at any point in time • and makespan is to be minimized (Cmax) • We denote this problem as • LP model gives Cmax but not a schedule • only total time spent on machine

  33. Heuristic solution • Longest Remaining Processing Time First (LRPT-rule) • Academic solution • Not practical • LRPT-rule optimal for in discrete time • Example: two parallel machines, three jobs with p = (8, 7, 6)

  34. One final model • Lets say we have • Multiple machines (m), where • the total completion time should be minimized (SCj) • We denote this problem as • SPT-rule gives optimal schedule • For the weighted case: WSPT does not give optimal schedule

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