Reservation Systems
TimeTabling
1. Reservation Systems without Slack
2. Reservation Systems with Slack
3. Timetabling with Tooling Constraints
4. Timetabling with Recourse Constraints
5. Scheduling flights at the airport
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Relation between these problems
1. Reservation Systems without Slack (timing of job is fixed)
–
m Parallel machines, n jobs (fixed in time; interval scheduling)
2. Reservation Systems with Slack
–
m Parallel machines, n jobs (more flexible in time)
3. Timetabling with Tooling Constraints
–
Enough identical machines in parallel, n jobs, tools (only 1 of each avail.)
4. Timetabling with Resource Constraints
–
Enough identical machines in parallel, n jobs, one resource
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2
Topic 1
Reservation Systems without Slack
(Interval Scheduling)
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Reservation without Slack

Reservation system with m machines, n jobs

Release date rj (integer), due date dj (integer), weight wj

No slack

Can we accept all jobs? Do we need all machines?
p j  d j  rj
If accepted, job starts at time rj
objective1: Maximize number of jobs processed
OR
objective2: Minimize number of machines needed for all jobs
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4
ILP Problem Formulation
e.g. at car-rental: hi pj

Integer Program

Fixed time periods, assume H periods

Binary variables xij (=1 if job j assigned to i th machine, 0 otherwise)
Job requiring
processing in
period l
Jl ={j| rj l  dj }
m
n
 w x
max
i 1 j 1
m
x
i 1
ij
x
jJ l
ij
ij ij
 1, j  1, ..., n
 1, i  1, ..., m,
l  1, ..., H
Exercise 1: How to adapt model for: do all jobs, but on a
minimum number of machines
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5
Maximizing value of processed jobs
m

In general max
n
 w x
i 1 j 1

is NP-hard (when both pj and wij free)
Two special cases with exact algorithms
–

ij ij
each of these algorithms sorts the jobs to increasing rj
Case 1: processing times pj = 1
–
decompose into separate time units
–
assign in each time unit most valuable jobs first
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6
Case 2: Identical Weights and Machines

Mj = {1,2,…,m} with wij = 1

objective: maximize #jobs taken
Pm | rj , d j , p j  d j  rj | U j

(No time decomposition, but still) an exact, simple algorithm:
–
Dispatch jobs, tentatively, in order of increasing release date
r1  r2  ...  rn
–
when next job has no machine while being earlier completed,
select it at the expense of the later completed scheduled job
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7
Algorithm (case 2: Identical Weights and Machines)

Step 1
–

// J as set of accepted jobs
Step 2
–

Set J =  and j = 1
If available at time rj assign job j to a machine, include j in J, go to Step 4
Step 3
–
( rk  pk )
Of scheduled job set J, select j* with latest finish C j*  max
kJ
–
If C j  rj  p j  C j * do not include j in J and go to Step 4
Else delete job j* from J, assign j to freed machine & include in J

Step 4
–
If j = N STOP, select next job j = j+1 and go back to Step 2
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8
Minimizing #machines used

No slack, arbitrary processing times, equal weights, infinitely
many identical machines in parallel

Easily solved, as follows

–
Order jobs as before to earliest release date
–
Assign job 1 to machine 1
–
Suppose first j-1 jobs have been assigned for processing
–
Try to assign j th job to a machine i ( j-1) as already used
–
If not possible assign to a new machine
Special case of node coloring problem
–
Why special case? (to be discussed later..)
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9
Summary
Reservation Systems without Slack
Maximize # of jobs processed:
Minimize # of machines needed:
simple algorithm
simple algorithm
Topic 2
Reservation Systems with Slack
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Reservation with Slack
p j  d j  rj

Now allow slack
(time slots)

Considering three cases with identical machines
and objective: maximize #jobs
Case 0.
All processing times pj = 1
Case 1.
All processing times pj = p  2 , identical weights
Case 2.
General processing times, weights wj
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(trivial solution)
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11
Case 1: Equal Processing Times

Now assume processing times all equal to some p  2
–
NOTE: slack can be different

Interaction between time units

Method: Barriers algorithm
–
wait for critical jobs to be released
–
start the job with the earliest deadline

Slot number l = l th job to start

S(l) starting time of l th slot
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12
Barriers Algorithm



Barriers Algorithm (again open, whether this algorithm is exact!)
Barrier
Slot
number
Release
time
–
ordered pair (l, r)
–
Expresses waiting constraint
Approach: Starting with a k-partial schedule
–
construct a (k+1)-partial schedule
–
or add a new barrier to the barrier list Lb and start over from scratch
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13
Earliest Deadline with Barriers

Selects machine h for job in (k+1)th slot
m
k  1 multiple of m

h
otherwise
(k  1) mod m

Compute the starting time
  max t1 , t2 , t3 , t4 
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14
Computing the Starting Time

 is defined as the earliest time that the next job can start

Claim:
  max t1 , t2 , t3 , t4 
t1  max S (1),..., S (k )
Minimum
release date
of jobs left
0
k 1  m

t3  
S (k  1  m)  p otherwise
t4  max r : (k  1, r )  Lb 
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15
Job in (k+1)th slot, can be a crisis job

Choose for (k+1)th slot: among available jobs, j ’ with earliest dj’

The creation of (k+1)-partial-schedule is successful if d j '    p

Otherwise, job j ’ is a ‘crisis job’ and one calls upon the
Crisis Routine

Backtrack l=k, k-1, .. seeking 'pull-job' jl with dj(l) > dj'
(most recently scheduled with later deadline than job j' )


If no pull-job is found, then exclude the crisis job
Otherwise:
– Add barrier (l, r*) in Lb, with r* min. release time in
restricted set Jr , of jobs in part. sched. after pull job
– Start anew with the updated barrier list
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16
Barriers Algorithm: an example (all pj=10)
Jobs
1
2
3
4
rj
0
2
5
5
dj
20
30
19
30
d 3  19
1
Machine 1
Crisis job
3
Pull job (d2 > d3)
2
Machine 2
0
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10
20
30
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17
An example (cont’d)

Slot k=3 gives crisis job j=3 with pull job 2 in slot l=2

Setup restricted set: Jr={3}

Include new barrier: (l, r*) = (2,5)  Lb

Restart with updated list Lb from scratch (with extra wait command)
(min. release date r* = r3= 5)
Crisis job
M1
M2
Now at k=2, wait till
1
3
1
2
3
2
Pull job (d2 > d3)
0
10
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20
  t4  5
4
Optimal schedule
0

10
20
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18
CASE 2: General processing times

Reservation with slack: processing times pj and weights wj
NP-hard 
No efficient algorithm 
Heuristic needed

Composite dispatching rule
–
Preprocessing: determine flexibility of jobs and machines
–
Dispatch least flexible job first on the least flexible machine, etc
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19
Defining priority indices

yik := # candidate jobs on machine i in time slot [k-1,k]

|Mj|= # machines suitable for job j

Priority index Ij for jobs (lower index =higher priority):
–
more value | less proc. | less machines  lower index Ij
e.g., I j  ( p j /w j )* M j

// increases in parameters p j /w j , M j
Priority index gi k j of using machine i for job j in interval [k, k+pj]

e.g., gikj  max y i ,k 1 ,y i ,k  2 ,...,y i ,k  p j
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
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20
Algorithm
1: - Calculate both priority indices Ij and yik
- Order jobs according to job priority index (Ij)
- Set j = 1
2: - Among available machines and time slots assign j to the
combination <machine i , [k,k+pj] > with lowest value gikj
- Discard job j if it cannot be processed at all
3: - If j = n then STOP, otherwise go back to Step 2 with j = j+1
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21
Summary
Reservation Systems with Slack
Equal processing times:
General processing times :
Barrier algorithm
Composite dispatching rule
Topic 3
Timetabling with Tooling Constraints
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Timetabling

(All given) n jobs must be processed

enough (e.g. >= n) identical machines in parallel

Minimizing Makespan

Tooling Constraints
–
Set of different available tools k (say in set K ):
of each k, only one available (for use by one job at a time)
–

Job j requires tools of subset Kj during its processing
Timetabling is Special Case of RCPSP :
–
no precedence constraints
–
Rk = 1; rjk=1 or 0 depending on k being part of Kj
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23
Why Timetabling = Node coloring
Example: all jobs take one time unit
jobs
Tool 1
Tool 2
Tool 3
Tool 4
1
1
1
0
1
2
0
1
0
0
3
0
1
1
1
4
1
0
0
1
5
1
0
1
1
1
7
2
6
0
0
1
0
7
1
0
0
0

Job = node;

Jobs need same tool  arc
between nodes

6
3
5
4
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Makespan <= H time units ?
 Can graph be colored
with  H colors ?
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24
Heuristic for Node Coloring


Terminology
–
degree of node = number of arcs connected to node
–
saturation level = number of colors connected to node
Intuition
–
Color high degree nodes first
–
Color high saturation level nodes first
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26
Heuristic Algorithm for Node Coloring
Step 1:
Order nodes in decreasing order of degree
Step 2:
Use color 1 for first node
Step 3:
Choose uncolored node with maximum saturation
level, breaking ties according to degree
Step 4:
Color using the lowest possible color-number
Step 5:
If all nodes colored, STOP;
otherwise go back to Step 3
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27
Example book
Nodes
Degree
1
5
2
2
3
5
4
4
5
5
6
2
7
3
1
7
2
6
3
5
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4
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28
Example book
Saturation + Degree in uncolored subgraph = Degree original graph
Nodes
Degree sub
Saturation
1
4
1
2
2
0
3
4
1
4
3
1
5
-
6
1
1
7
2
1
1
7
2
6
3
5
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4
Operations Scheduling
29
Example book
Nodes
Degree sub
Saturation
1
3
2
2
1
1
3
-
4
2
2
5
-
6
0
2
7
2
1
1
7
2
6
3
5
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4
Operations Scheduling
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Example book
Nodes
Degree sub
Saturation
1
-
2
0
2
3
-
4
1
3
5
-
6
0
2
7
1
2
1
7
2
6
3
5
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4
Operations Scheduling
31
Example book
Nodes
Degree sub
Saturation
1
-
2
0
2
3
-
4
-
5
-
1
6
0
2
7
0
3
Result 4 colors
7
2
e.g.
7: blue
2: red or green
6
3
5
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6: yellow or green
4
Operations Scheduling
32
Worse: a reverse method
Nodes
Degree sub
1
5
2
2
3
5
4
4
5
5
4
6
2
7
3
11
77
22
66
33
5
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Operations Scheduling
33
Example of Timetabling with tools

For solution: see previous node coloring example.
jobs
Tool 1
Tool 2
Tool 3
Tool 4
1
1
1
0
1
2
0
1
0
0
3
0
1
1
1
4
1
0
0
1
5
1
0
1
1
6
0
0
1
0
7
1
0
0
0
Job 1: yellow
Job 3: blue
Job 7: blue
Job 4: green
Job 5: red
Job 2: red or green
Job 6: yellow or green
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34
Relation to Reservation Models

Closely related to reservation problem with zero slack and
arbitrary processing times (interval scheduling)

Interval scheduling = Special case of timetabling problem
–
jobs with common time periods  unary (pj=1) jobs with common tools
–
new color = new machine  new color = extra period
–
job occupies adjacent time periods (creating easy solvable color problem)

jobs occupy tools ‘that need not be adjacent’
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35
Example

Suppose now we have a reservation system
Job

pj
1
2
2
3
3
1
rj
0
2
3
dj
2
5
4
Here we transform to time-tabling:
Job 1 must
be processed
in time [0,2]
Job 2 must
be processed
in time [2,5]
Job
1
2
3
Tool 1
1
0
0
Tool 2
1
0
0
Tool 3
0
1
0
Tool 4
0
1
1
Tool 5
0
1
0
Toool 5
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36
Other example

Adjusted example of Timetabling with tools
jobs
Tool 1
Tool 2
Tool 3
Tool 4

1
1
1
0
1
2
0
1
0
0
3
1
1
1
1
4
11
0
0
1
5
1
0
1
1
6
0
0
1
0
7
1
0
0
0
4
0
1
1
0
5
1
1
1
0
6
1
0
0
0
7
0
0
1
0
Rearrange and rename tools
jobs
Tool 3 1
Period
Tool 4 2
Period
Tool 1 3
Period
Period
Tool 2 4
1
0
1
1
1
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2
0
0
0
1
3
1
1
1
1
Operations Scheduling
37
Summary
Timetabling with Tooling Constraints
Node Coloring heuristic
Topic 4
Timetabling with Resource Constraints
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Operations Scheduling
Timetabling with single resource

Unlimited number of identical machines in parallel

All n jobs must be processed

Resource constraints
–
Single resource of total quantity R
–
Required is a certain amount for each job
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39
Resource Constraints

One type of tool but R units of it (resource)

Job j needs Rj units of this resource

Clearly, if Rj + Rk > R then job j and k cannot be processed at the
same time, etc

Applications

–
scheduling a construction project (R = crew size)
–
exam scheduling (R = number of seats)
Special case of RCPSP with one resource, pi  1, no precedence
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40
Special case: Bin-Packing

Assume equal processing times pj = 1

Unlimited number of machines

Minimize makespan

Equivalent to bin-packing problem
–
each bin has capacity R
–
item of size rj
–
pack into the minimum number of bins
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41
Solving the Bin-Packing Problem

Known to be NP-hard

Many heuristics developed

First fit (FF) heuristic
–
Always put an item in the first bin it fits into
–
Cmax (FF) 
Know that
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17
Cmax (OPT)  2
10
Operations Scheduling
42
Example



Assume 18 items and R = 2100
–
Jobs 1-6 require 301 resources
–
Jobs 7-12 require 701 resources
–
Jobs 13-18 require 1051 resources
FF heuristic:
–
We assign the first 6 jobs to one interval (3016=1806)
–
We then assign jobs two at a time to next 3 intervals (7012=1402)
–
We then assign just one of the remaining jobs to each interval
Poor performance when jobs are assigned in arbitrary order
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43
First Fit Decreasing (FFD)

An improvement of FF:

Order jobs in decreasing order first
Cmax (FFD) 
11
Cmax (OPT)  4
9

Know that

FF and FFD can be extended to different release dates
R = 12
6 jobs of size 6, 4, 4, 4, 3, 3
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44
Discussion

This chapter only considered very simple models

In practice:
–
Dynamic reservation systems
–
Price considerations (yield management)
–
Other requirements (such as in final topic..)
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45
Extra Topic:
Scheduling flights at the airport,
A reservation problem with no slack :
When planning check-in desks
(as adjacent tools)
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Operations Scheduling
Planning Check-in Desks
period
desk 1
desk 2
desk 3
desk 4
desk 5
desk 6
desk 7
desk 8
desk 9
desk 10
desk 11
1
2
3
4
5
6
7
flight 1
8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
flight 3
flight 8
flight 6
flight 4
flight 9
flight 2
flight 7
flight 5
flight 10
rj = # desks required during (given) time-interval I(j) for flights j=1,2,..n
Optimization Problem
- Minimize total #desks R* for all flight-desk -assignments
Distinguishing feature:
- Assigned desks must be adjacent
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47
Algorithm: Earliest Release Date or First-Fit
(analogous to algorithm for minimum #machines, topic 1 )
flight
starting period
ending periode
# desks required
d\t
1
2
3
4
1
1
1
1
d\t
1
2
3
4
1
1
1
1
2
1
1
1
2
1
1
1
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1
1
3
3
3
1
1
1
2
4
3
3
3
1
1
1
2
4
2
3
3
2
2
3
5
1
3
4
6
2
4
5
7
1
5
7
9
3
5
3
3
4
2
6
3
3
4
7
8
5
4
3
3
2
6
4
3
3
7
4
5
5
5
9
Flight 5
4does not fit
8
9
5
5
5
5
5
5
Operations Scheduling
48
Another Example
Scheduling flights (j=1,2,..8) in 30 min. periods (t=1,2,..9)
Data:
flights desks
j
r j
SAS
2
KLM 3
CAN 3
AF
4
ALIT 3
BA
4
NW
1
LH
1
A feasible, but non-optimal solution:
periods
I(j)
2-4
2-3
1-5
1-1
4-4
6-8
5-8
8-9
Period t =1 2 3
desk 1
SAS
2
3
KLM
4
5
6 CAN
7
8
9
10 AF
11
12
4
5
6
7
8 9=T
LH
BA
LH
ALIT
NW
Find better solutions with R < 11!
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Operations Scheduling
49
First attempt
1
2 3
SAS
1
2
3
KLM
4 AF
5
6 CAN
7
8
9
4
5
6
SASNW
7
8
LH
NW
9
ALIT
BA
BA
Define R(t) as sum of rj’s over jobs j with tI(j)
Then R* >= max t R(t) (a lower bound called Rlow)
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50
Optimal Assignment with R*=8 desks
period> 1 2 3
1 AF KLM
2
3
4
SAS
5
6 CAN
7
8
9
4
5
6
8
LH
9
ALIT
NW
BA
Here Rlow=8 and
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7
R* = Rlow
Operations Scheduling
51
Question: Always R* = Rlow ?
j 1
2
3
4
5
6
7
8
9
10
r(j) 4
4
3
2
5
7
1
1
1
1
I(j) 1-1 1-1 2-2 2-2 4-4 5-5 1-3 3-4 2-4 2-4
r\t
1
1
2
3
4
5
6
7
8
9
10
11
1
2
4
11
10
3
10
3
9
4
4
9
5
5
11
6
1
2
2
7
6
7
5
8
12
R(t)
3
11
1
2-5
9
9
5
8
9
9
Is R* = Rlow ?
Exercise 2: find a schedule using only #desk-rows 9=Rlow
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52
Notation for an ILP model
Ij (= [aj .. bj ]) Check-in interval
rj
# of desks required
dj
Largest number of adjacent desks as occupied by j
desks
df
flight f
dg
flight g
1
af
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ag
bf
bg
period
Operations Scheduling
55
Exercise 3: Finish this ILP model
min D

With integer variables:
D : Total number of desks required
dj : Largest desk number assigned to flight j

With parameters:
rj : Number of desks required for the check-in process of flight j
Ij : Check-in time interval of flight j
Hint: disjunctive model, comparable to that for job-shop-scheduling
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Operations Scheduling
56
An alternative ILP model
 Binaries yj t r (=1 if j occupies in period t largest desk r; 0 otherwise)
 Flight j require n(j,t) adjacent desks in period t
min D
N

r n( j ,t )
N

r n( j ,t )
r y j, t, r  D
j , t  I (j )
y j, t, r  1
j , t  I (j )
no overlap?
r ,  t
Exercise 4: model the missing constraint-set
Hint: similar to the technique used in the RCSP model
Operational Research & Management
Operations Scheduling
57
With n(j,t) desks for a flight
t
d 4
 4
4

Static

Dynamic (8 possible shapes)
4
4
4
4
4
4
4
4
or
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4
4
4
4
4
4
4
4
4
4
4
4
4
4
or
4
4
4
4
4
4
4
4
or …
Operations Scheduling
58