Project Scheduling
1. Project Planning (revisited)
2. Resource Constrained Project Scheduling
3. Parallel Machine Scheduling
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Topic 1
Project Planning (revisited)
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Project Planning: Jobs on Arcs
node:
j prec duration pj
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
(end of 1 or more tasks)
every task := arc
(dummies for precedence)
C
H
A
dummy
A: -
fase in project
F
E
G
B
J
D
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Critical Path Method (CPM)
j prec
3 3
10 13
C4
2
A: -
5
A3
B: -
D: A, B
E: A
F: E
dummy
C: A
E2
0 0
1
B2
J: D, G
15 15
F5
4
7
G6
3 8
J4
3
6
D3
G: E
H: C, F
5 5
H2
ES1 =
ESj =
LFn =
LFi =
11 11
0
maxi { pi + ESi | i  P(j)}
ESn
minj { LFj - pj | j  Si }
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Alternative Network: Jobs on Nodes
j prec
0
A: -
3
A3
B: -
?
C4
H2
5
C:0A
E2
D: A, B
3
j=0
G: E
15
G6
j =n+1
5
E: A
F: E
F5
B2
0
D3
3
= max{0+2,0+3}
J4
11
H: C, F
J: D, G
Advantages
- no dummy-jobs needed to enforce precedence
- length of node may represent duration or arc-length
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Jobs on a time axis
j prec
A3
jobs
A: -
B2
B: -
C4
C: A
D3
D: A, B
E2
E: A
F5
F: E
G6
G: E
H: C, F
H2
J: D, G
J4
0
3
5
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15
t
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Project Planning with crashing
Heuristic
cut set
1
4
5
8
0
6
minimal cut set
n+1
Graph of all Critical Path(s)
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Topic 2
Resource Constrained Project Scheduling
Problem (RCPSP)
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Project Planning as Scheduling

Activities within project are 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 Rk of resource k available

Determine all finish times Fj (=genoeg om hele schedule weer te geven)
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With resource requirements rj on jobs j
r j j pj
Example: job A requires 9 workers for 3 days, etc.
j prec
9A3
A: -
8B2
B: -
7C4
C: A
7D3
D: A, B
3E2
E: A
2F5
F: E
G: E
3G6
With R=16 workers available
Is Cmax=15 days still possible?
0
3
5
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H: C, F
8H2
J: D, G
3J4
11
15
t
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With resource requirements rj on jobs j
r j j pj
j P(j)
9A3
A: { }
8B2
B: { }
7C4
C: {A}
7D3
D: {A, B}
3E2
E: {A}
2F5
F: {E}
G: {E}
3G6
8H2
0
3
5
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J: {D, G}
3J4
Feasible, even with 12 workers
11
H: {C, F}
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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 exceed LFj for some j ).

Later, we discuss heuristic methods for RCPSP
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Resource loading and leveling
19
loading: jobs start at ESj
D
leveling: shift jobs in time
13
10
C DD
A
E E
7
B
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C
F
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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
A(t) = { j | Fj – pj < t  Fj }
jA(t) rjk  Rk
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t
 resources k ,  t
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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], periods t = 1,2, ..H
For j J: Fj  {1, …, H}


t
X jt  1
( Choosing H large enough)
E.g.
Fj  t t X jt
Fj = 4 en pj = 2
Xj4= 1
Use extra binary variables
–
Aj3= 1
Xjt = 1 if task j ends in period t;
Ajt   s t
t  p j 1
X jt
0
–
Aj4= 1
1
2
3
4
Ajt = 1 if task j is active in period t;
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Resulting ILP model for RCPSP
Min
s.t.
Fn 1
(1)
 j  J,  hP(j)
Fh  Fj  p j
Precedence restrictions
(2)
(3)
Fj   t t X jt
X
t
jt
jJ
1
jJ
Determine one last period
(4)
Ajt   s t
t  p j 1
X jt
 j  J,  t = 1, …, H
Determining active periods of tasks
(5)

Ajt rjk  Rk
j
 t = 1, …, H,  k  K
Consumption of scarce resources
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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
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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
rj j pj
8/16
00
00
0/10
4/12
6/16
3D4
4E2
2F4
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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
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Parallel SGS (on example)
tg
finished R(tg) schedulable Dg

0
4
j select R(tg) Fj
{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
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Parallel SGS (formal)

Notation (sets in bold)
–
–
–
–

Cg = activities: scheduled and finished;
Ag = activities: scheduled and still active;
C0 = {0};
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
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Serial SGS (informal)
Each stage selects a job  n stages

set of jobs already selected and planned: scheduled jobs (S )

decision set (D ): jobs with all predecessors in S

remaining set of jobs: these will first enter D and then 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')
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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 } Dg
0.
{0/4}
{A, D}
r j j pj
Sj
Fj
updates t / R(t)
3D4
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}
4E2
4
6
R(4)=2, R(6)=4
(idem, E has lower slack)
2.
{0/1,4/2,6/4}
{A, F}
2A3
6
9
R(6)=2, R(9)=4
6
10
R(6)=0, R(9)=2, R(10)=4
(idem)
4.
{0/1,4/2,6/2,9/4}
{B, F}
4F2
(idem)
5.
{0/1,4/2,6/0,9/2,10/4} {B}
4B2
10
12
R(10)=0, R(12)=4
6.
{.. 10/0, 12/4}
3C 1
12
13
R(12)=1
{C}
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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};
–
Rk(t) = remaining capacity of resource k at time t
F0 = 0;
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 *)
–
Determine Dg and take j Dg
(* Dg set of decidable jobs *)
// schedule job j :
–
Determine ESj = max { Fh | hP(j) }
–
Determine Fj = pj + min { t [ESj, LSj]  Fg | Rk(t) is feasible for j }
–
Update Rk(t)
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(* respecting precedence *)
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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’
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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.)
and Dg = 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 from D with the highest priority, let it start at time tg
4. If jobs are still unscheduled then goto step 2
else stop
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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
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Topic 3
Parallel Machine Scheduling Problems
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Parallel Machine Models

Lets say we have
–
Multiple machines (m), where
–
the makespan should be minimized (Cmax)
Pm || Cmax

We denote this problem as

Problem already NP-hard for m = 2 (partitioning problem)

LPT-rule good heuristic
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Worst case behaviour LPT

Maximal deviation from optimal value
Cmax ( LPT ) 4 1
 
Cmax (OPT ) 3 3m

Example: four parallel machines, nine jobs
jobs
p(j)
1
7
Cmax(opt)=12
2
7
3
6
but
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6
5
5
6
5
7
4
8
4
9
4
Cmax(LPT)=15
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More makespan problems
1| prec | Cmax
easy
Pm | prec | Cmax
NP-hard
P | prec | Cmax
easy
Pm | p j  1, M j | Cmax
Least Flexible Job first is optimal
if Mj is nested
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Allowing preemptions

Lets say we have
–
Multiple machines (m)
–
preemption is allowed at any point in time
–
and makespan is to be minimized (Cmax)
Pm | prmp | Cmax

We denote this problem as

LP model gives Cmax but not a schedule
–
only total time spent on machine
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Heuristic solution

Longest Remaining Processing Time First (LRPT-rule)
–
Academic solution
–
Not practical
Pm | prmp | Cmax

LRPT-rule optimal for

Example: two parallel machines, three jobs with p = (8, 7, 6)
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in discrete time
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One final model

Lets say we have
–
Multiple machines (m), where
–
the total completion time should be minimized (Cj)
Pm ||  C j

We denote this problem as

SPT-rule gives optimal schedule

For the weighted case: WSPT does not give optimal schedule
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