Contents college 3 en 4
• Book: Appendix A.1, A.3, A.4, §3.4, §3.5,
§4.1, §4.2, §4.4, §4.6 (not: §3.6 - §3.8, §4.2
- §4.3)
• Extra literature on resource constrained
project scheduling (will be handed out)
1
Planning and scheduling
optimization techniques
•
•
•
•
•
•
•
•
Dispatching Rules
Composite Dispatching Rules
Adaptive search
Dynamic Programming
(Integer) Linear Programming
Cutting plane methods
Branch and Bound
Beam Search
2
Linear programming (LP) model
• LP:
min c1 x1  c 2 x 2  ...  c n x n
objective function
subject to :
a11 x1  a12 x 2  ...  a1n x n  b1
constraints
a21 x1  a22 x 2  ...  a2n x n  b 2

am1 x1  am2 x 2  ...  amn x n  bm
x j  0 ( j  1,  , n)
• Matrix form:
min c T x
Ax  b
x0
where:
x, c: n-vector
A:
m,n-matrix
b:
m-vector
variable
restrictions
3
Linear programming example
max x1  x 2
subject to :
2x1  x 2  4
3x1  4 x 2  12
x j  0 ( j  1,2)
T
or:
1  x1 
max    
1  x 2 
subject to :
2 1   x1 

 x  
3 4   2 
 x1  0
x    
 2  0
4
 
12
4
Linear programming example:
graphical solution (2D)
2x1  x 2  4
x2 6
max x1  x 2
subject to :
2x1  x 2  4
5
4
x1  x 2  6
3
(objective)
3x1  4 x 2  12
2
x j  0 ( j  1,2)
1
solution
space
3x1  4x 2  12
0
1
2
3
4
5
6
x1
5
Linear programming (cont.)
• Solution techniques:
– (dual) simplex method
– interior point methods (e.g. Karmarkar algorithm)
• Commercial solvers, for example:
– CPLEX (ILOG)
– XPRESS-MP (Dash optimization)
– OSL (IBM)
• Modeling software, for example:
– AIMMS
– AMPL
6
Integer programming (IP) models
• Integer variable restriction
– IP: integer variables only
– MIP: part integer, part non-integer variables
– BIP: binary (0-1) variables
• General IP-formulation:
min c T x
Ax  b
x integer ( x  Z  )
• Complex solution space
7
Integer programming example:
graphical solution (2D)
x2 6
max x1  x 2
5
subject to :
4
x1  x 2  6
3
(objective)
2x1  x 2  4
3x1  4 x 2  12

x j  Z ( j  1,2)
2 optimal solutions!
2
1
0
1
2
3
4
5
6
x1
8
Total unimodularity property for
integer programming models
Suppose that all coefficients are integer in the model:
min c T x
i.e. aij , bi   , i, j
Ax  b
x0
Example: transportation problem
if A has the total unimodularity property
(i.e. every square submatrix has determinant 0,1,-1)

there is an optimal integer solution x*
&
the simplex method will find such a solution 9
Integer programming tricks
PROBLEM: x = 0 or x  k
0 , for x  0
use binary indicator variable y=
1 , for x  k
restrictions:
x  M  y (M is an upperbound on x)
x ky
y  0,1
10
Integer programming tricks (2)
PROBLEM: fixed costs: if xi>0 then costs C(xi)
minimize C( x ) where :
Ax  b
x0
for xi  0,
 0
C( x i )  
k i  c i x i for x i  0.
0 , for xi  0
use indicator variable yi= 
1 , for xi  0
xi  M  yi
restrictions (i): C( x i )  k i  y i  c i  x i
y i  0,1
11
(Integer) programming tricks (3)
• Hard vs. soft restrictions
– hard restriction: must hold, otherwise unfeasibility
for example: x1  x 2  5
– soft restriction: may be violated, with a penalty
for example: minimize c T x  Y  100
x1  x 2  5  Y

x  0, Y  0
12
(Integer) programming tricks (4)
• Absolute values: min  y t
goal
j
a x
j
j
 bt  y t
j ,t
variation
x j,t  0, y t free
solution:

t
yt  y  y

t

min  y t  y t

t
 yt  y  y

t

t
a x
j
j ,t

t
 bt  y  y

t
j
x j,t  0, y t  0, y t  0
13
Integer programming tricks (5)
• Conjunctive/disjunctive programming
- conjunctive set of constraints: must all be satisfied
- disjunctive set of constraints: at least one must be satisfied
• example (Appendix A.4):
min  w j x j
min  w j x j
x k  x j  pk
x k  x j  pk  M1  y
or
x j  xk  p j
x j  x k  p j  M2  (1  y )
j

j

y  {0,1}
14
IP example
nonpreemptive single machine, total weighted
completion time (App. A.3)
model definition:
x jt  1, if job j completes at time t, and 0 otherwise
objective function: minimize weighted completion time:
 t  x jt  completion time of job j if x jt  1

C max-1
t x
t 0
jt
 completion time of job j
n C max-1
 min imize 
j1
w
t 0
j
 t  x jt (objective function)
15
IP example (cont.)
Restriction: all jobs must be completed once:
C max-1
x
t 0
jt
1
Restriction: only one job per time t:
if job j is in process during t, it must be completed
somewhere during [t,t+pj]

t p j
x
s t
js
 1 (if job j is in process during t)
n t p j
   x js  1 (restricti on : exactly one job per time t)
j1 s  t
16
IP example (cont.)
Complete IP-model:
n C max-1
minimize 
j1
w
t 0
j
 t  x jt
subject to :
C max-1
x
t 0
jt
 1 (for j  1, , n)
n t p j
x
j 1 s  t
js
 1 (for t  0, , Cmax - 1)
x jt  0,1 (for j  1, , n, t  0, , Cmax - 1)
nCmax
integer
variables
17
IP example (cont.)
Additional restriction: precedence constraints
Model definition: SUCC(j) = successors of job j
 job j must be completed before all jobs in SUCC(j):
C max 1
 tx
t 0
jt
 completion time of job j
kt
 pk  start time of job k
C max 1
 tx
t 0

C max 1
 tx
t 0
jt

C max 1
 tx
t 0
kt
(for k  SUCC(j), j  1, , n)
18
Integer programming
solution techniques
• Heuristic vs. explicit approach:
– trade-off between solution quality and computation time
– trade-off between implementation effort/costs and yield
(i.e. profits gained from solution quality improvement)
• Heuristic methods; for example:
–
–
–
–
–
local search (e.g. simulated annealing, tabu search, k-opt)
(composite) dispatching rules (e.g. EDD, SPT, MS)
adaptive search
rounding fractional solutions
19
beam search
Integer programming
solution techniques (cont.)
• Explicit methods; 3 categories:
1. dynamic programming
2. cutting plane (polyhedral) methods
3. branch and bound
or: hybrid methods (combination of the above)
• Commercial IP solvers usually use a
combination of heuristics and 2, 3
20
Dynamic programming
• Problem divided into stages x t (t  0, , T)
• Each stage can have various states it
• A recursive objective function is used to
iterate through all states and all stages
(forwards or backwards)
F0 (i0 )  c 0 (constant)
Ft (it )  min{c t (it , x t )  Ft 1 (it 1 (it , x t ))}
xt
21
Cutting plane methods
STEP 0:
STEP 1:
STEP 2:
Create a relaxation of the problem by
omitting restrictions
(e.g. the integrality restrictions)
Solve the current problem
If solution is infeasible then generate
a restriction that cuts of the solution,
and add it to the problem  STEP 1
Otherwise: DONE
22
Branch and bound
• Enumeration in a search tree
root node
Level 0
Level 1
child nodes
...
child nodes
...
Level 2
• each node is a partial solution, i.e. a part of
23
the solution space
Branch and bound example 1
Disjunctive programming (appendix A.4):
disjunctive set of constraints: at least one must be satisfied
xj = completion time of job j
restriction: either x k  x j  pk or x j  x k  p j (j, k  I)
solve LP without
disjunctive restrictions
(= LP relaxation)
if disjunct. restr.
violated for j & k
Level 0
Level 1
x k  x j  pk
x j  xk  p j
...
24
Branch and bound (cont.)
• Upper bound: e.g. a feasible solution
• Lower bound:
e.g. a solution to an “easier” problem
• Node elimination (fathom/discard nodes):
when lower bound >= upper bound
25
Branch and bound (cont.)
• Branching strategy:
how to partition solution space
• Node selection strategy:
– sequence of exploring nodes:
• depth first (tries to obtain a solution fast)
• breadth/best bound first (tries to find the best solution)
– which nodes to explore (filter and beam width)
• filter width: #nodes selected for thorough evaluation
• beam width: #nodes that are branched on ( filter width)
 Beam search
26
Branch and bound example 2
• Single machine, maximum lateness, release
and due dates
Jobs
p(j)
r(j)
d(j)
1
4
0
8
2
2
1
12
3
6
3
11
4
5
5
10
(1,?,?,?)
Level 0
(?,?,?,?)
(2,?,?,?)
(3,?,?,?)
(4,?,?,?)
Level 1
lower bound: EDD + preemption
27
Branch and bound example 2
• Lower bound for: (1,?,?,?)
r(2)
r(3)
1
r(4)
3
Jobs
p(j)
r(j)
d(j)
4
1
4
0
8
2
2
1
12
3
6
3
11
4
5
5
10
3
2
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
t
d(4)<d(3)
d(3)<d(2)
Lower bound: Lmax = max(0,17-12,15-11,0)=5
28
Branch and bound example 2 (cont.)
Level 0
(?,?,?,?)
LB=5
infeasible:
(1,?,?,?)
(1,3,4,3,2)
LB=6*
=UB (1,2,?,?)
(1,2,4,3)
(1,2,4,3)
LB=7*
=UB (2,?,?,?) (3,?,?,?)
LB=5*=UB
(1,3,?,?) (1,3,4,2)
DONE
(4,?,?,?) Level 1
Jobs
p(j)
r(j)
d(j)
1
4
0
8
2
2
1
12
3
6
3
11
(1,3,4,2)
29
4
5
5
10
Branch and bound example 3
LP solution: x1  0.8,
x 2  2.4
x1  0
x1  1
x1  0,
x1  1,
x2  3
x2  2
obj: 3
x2 6
obj: 3
5
4
x1  x 2  6
3
(objective)
2
x2  2
x2  3
x1  1,
x1  0,
x2  2
x2  3
obj: 3
obj: 3
1
0
1
2
3
4
5
6
x1
30
Beam search example 1
single-machine, total weighted tardiness
Upper bound: ATC rule (apparent tardiness cost):
• schedule 1 job at a time
• every time a machine comes available, determine
ranking of jobs:
I j (t) 
wj
pj
e
WSPT
rule
 max( d j p j  t ,0 ) 




K p


MS rule
look-ahead parameter:
K = 4.5 + R (R  0.5)
K = 6 - 2R (R  0.5)
R  dmax  dmin  / Cmax = due date range factor
31
Beam search example 1 (cont.)
single-machine, total weighted tardiness
(?,?,?,?)
Jobs
p(j)
d(j)
w(j)
w(j)/p(j)
1
10
4
14
1.4
2
10
2
12
1.2
3
13
1
1
0.1
4
4
12
12
3
(1,?,?,?)
(2,?,?,?)
(3,?,?,?)
(4,?,?,?)
Upper bound by ATC rule: max(d j  p j  t ,0)  0 ( j  1,2,3)
 I j (t) 
wj
pj
e
 max( d j p j  t ,0 ) 




K p



wj
pj
(j  1,2,3)
32
Beam search example 1 (cont.)
single-machine, total weighted tardiness
Jobs
p(j)
d(j)
w(j)
w(j)/p(j)
1
10
4
14
1.4
2
10
2
12
1.2
3
13
1
1
0.1
4
4
12
12
3
(?,?,?,?)
(1,?,?,?)
Upper bound Jobs C(j)
1 10
by ATC rule:
2 24
3 37
4 14
(2,?,?,?)
d(j) T(j) w(j)*T(j)
4 6
84
2 22
264
1 36
36
12 2
24
Total = 408
(3,?,?,?)
(4,?,?,?)
33
Beam search example 1 (cont.)
single-machine, total weighted tardiness
(?,?,?,?)
(1,?,?,?)
(2,?,?,?)
(3,?,?,?)
(4,?,?,?)
UB=408 UB=436 UB=814 UB=440
explored
further
(beam width = 2)
4 nodes
analyzed
(filter width=4)
discarded
34
Beam search example 1 (cont.)
(?,?,?,?)
UB=408
436
(1,?,?,?)
UB=480
(1,2,?,?)
(2,?,?,?)
706
(1,3,?,?)
440
814
(3,?,?,?)
408
(1,4,?,?)
UB=408
(1,4,2,3)
best solution
(4,?,?,?)
436
(2,1,?,?)
(2,3,?,?)
554
436
(1,4,3,2)
(2,4,?,?)
(2,4,1,3)
608
(2,4,3,1)
35