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Resource Scheduling Constraints
Nicolas BeldiceanuSICS
Lägerhyddsvägen 5SE-75237 Uppsala,
Sweden([email protected])
Resource scheduling constraints
1. Overview2. Cumulative3. Cumulatives
Resource scheduling
• Problem: tasks subject to temporal andresource constraints
• Most successful area of CP• LP is usually poor on this type of problems• Two main categories of resources:
- renewable- non-renewable
Resource scheduling constraints
1. Overview2. Cumulative3. Cumulatives
ARGUMENTS :ctr_arguments(cumulative,
[TASKS-collection(origin-dvar, duration-dvar, end-dvar, height dvar),LIMIT-int]).
PURPOSE :The cumulative constraint enforces that at each point in time,the cumulated height of the set of tasks that overlap that point,does not exceed a given limit.
ORIGIN : [AggounBeldiceanu93]
EXAMPLE :cumulative([[origin-1, duration-3 , end-4 , height-1],
[origin-2, duration-9 , end-11, height-2],[origin-3, duration-10, end-13, height-1],[origin-6, duration-6 , end-12, height-1],[origin-7, duration-2 , end-9 , height-3]],8)
Cumulative
Cumulative
INITIAL GRAPH
Cumulative
1 2 3 4 5 6 7 8 9101112
≤≤≤≤ 8
( 3+1+1+2 ≤≤≤≤ 8 )
7
ARGUMENTS :ctr_arguments(coloured_cumulative,
[TASKS-collection(origin-dvar, duration-dvar, end-dvar, colour-dvar),LIMIT-int]).
PURPOSE :The coloured_cumulative constraint enforces that at each point intime, the number of distinct colours of the set of tasks that overlapthat point, do not exceed a given limit.
ORIGIN : Derived from cumulative and nvalues.
EXAMPLE :coloured_cumulative([[origin-1, duration-2 , end-3 , colour-1],
[origin-2, duration-9 , end-11, colour-2],[origin-3, duration-10, end-13, colour-3],[origin-6, duration-6 , end-12, colour-2],[origin-7, duration-2 , end-9 , colour-3]],2)
Coloured_cumulative
Coloured_cumulative
INITIAL GRAPH
Coloured_cumulative
1 2 3 4 5 6 7 8 9101112
≤≤≤≤ 2
3 2 3 2 1
nvalues([3,2,3,2],≤≤≤≤,2)
7
Definition of cumulative
cumulative(Origins,Durations,Heights,Limit)ListDvars ListDvars ListDvars Int
time
quantityof resource
Limit
task
origin
height
duration
Example of cumulative
O1 in 1..2, O2 in 3..3, D1 in 2..3, D2 in 2..2, H1 in 1..4, H2 in 2..3
cumulative([O1,O2],[D1,D2],[H1,H2],4),
labeling([leftmost,up],[O1,D1,H1,O2,D2,H2]).
[1,2,1,3,2,2] [1,3,1,3,2,3][1,2,1,3,2,3] [1,3,2,3,2,2][1,2,2,3,2,2] [2,2,1,3,2,2][1,2,2,3,2,3] [2,2,1,3,2,3][1,2,3,3,2,2] [2,2,2,3,2,2][1,2,3,3,2,3] [2,3,1,3,2,2][1,2,4,3,2,2] [2,3,1,3,2,3][1,2,4,3,2,3] [2,3,2,3,2,2][1,3,1,3,2,2]
Main applications of cumulative
1 2 3
1
Renewable resource
• disjunctive scheduling
• cumulative scheduling
Non renewable resource
• producer consumer
Bin-packing
Necessary condition for placement
Producer-Consumer
Final Stock
Consumed Quantity C
Stock production task
Consumer task
Initial stock
Produced Quantity P
Cumulative for placement
We want to place the 8 followings rectangles in a 12 x 4 rectangle :R1(5×2), R2(8×3), R3(6×1), R4(5×1), R5(2×1), R6(3×1), R7(2×2), R8(1×2) .
There exists a cumulative solution (by cutting R8 into two pieces) but there is no solution with a non-overlapping constraint on rectangles.(proof : M. Hujter, On the dynamic storage allocation problem, Manuscript, Computer and Automation Institute Hung. Acad. Sci. (1990))
4
3
2
1
00 1 2 3 4 5 6 7 8 9 10 11 12
R8
R1 : 5 × 2
R2 : 8 × 2
R1 : 6 × 1
R8
R4 : 5 × 1
R5
R6:3 × 1
R2 : 2 ×××× 2
Resource scheduling constraints
1. Overview2. Cumulative3. Cumulatives
The cumulatives constraint
The original cumulative constraint:
Restrict the resource consumptionat each point in time.
The generalized cumulatives constraint:
A pool of cumulative resources ,Height of a task can be negative ,Maximum or minimum resource consumption ,Holds for time-points crossed byat least one task.
1 2 3 4 5 601234
12
3 4
time
resource consumption ≤≤≤≤ 4
time
resource consumption
1 2 3 4 5 6-2-1012-101
Cumulatedprofiles
2
13
45
76 ≥≥≥≥ 0
≥≥≥≥ 0
Parameters• Tasks:
• Origin• Duration• End• Height• Machine
1 2 3 4 5 601231
time
Origin EndDuration
HeightMachine
ConsumptionResource 2
Resource 1
Origin+Duration=End
0 0
• Constraint:
• upper• lower 1
231
≤ 1231
≥
• MachineId• Limit
01231
LimitIdentifier
• Resources:
task(O,D,E,H,M)
machine(M,L)
bound(Ctr)
DEFINITION
For all tasks :
For all time-points crossed by task :
Let be the machine where task is assigned ,
For all tasks which both cross point , and are assigned to :
The sum of the height of tasks is not greater (less)
than the capacity of machine .
tt
ti
ms i m
ms
time
resource consumption
-2-1012-101
3
5
76 ≥≥≥≥ 0
≥≥≥≥ 0
Machine 1
Machine 2
1
2
4
1 2 3 4 5 6
EXAMPLE
Original cumulative constraint
1 2 3 4 5 6
01234
12
3 4
time
resource consumption
Maximum capacity
({machine-1 origin-2 duration-2 height-2machine-1 origin-4 duration-2 height-1machine-1 origin-3 duration-1 height-1machine-1 origin-5 duration-1 height-3},{id-1 capacity-4},
≤≤≤≤)
cumulatives
cumulative constraint withmachine choice
({machine-1 origin-2 duration-3 height-1,machine-1 origin-3 duration-2 height-2,machine-2 origin-2 duration-4 height-1,machine-1 origin-6 duration-1 height-2},{id-1 capacity-3,id-2 capacity-1},
≤)
cumulatives
resource consumption
Machine 2 with capacity 1
Machine 1 with capacity 3
1 2 3 4 5 601231
time
1
2
3
4
Uninterupted minimum levelof resource consumption
({machine-1 origin-2 duration-4 height-0,machine-1 origin-2 duration-4 height-1,machine-1 origin-2 duration-2 height-1,machine-1 origin-4 duration-2 height-2,machine-1 origin-4 duration-1 height-1},{id-1 capacity-2},≥≥≥≥)
cumulatives
1 2 3 4 5 601234
time
resource consumption
23
45
1
Minimum consumption is 2
First use Last use
Minimum level ofresource consumption
({machine-1 origin-2 duration-2 height-1,machine-1 origin-2 duration-1 height-2,machine-1 origin-3 duration-1 height-1,machine-1 origin-5 duration-2 height-2},{id-1 capacity-2},≥≥≥≥)
cumulatives
1 2 3 4 5 601234
time
resource consumption
No constraint at this point
31
2
4
Minimum consumption is 2
Producer-consumer
cumulatives machine-1 origin-1 duration-3 height-2,machine-1 origin-1 duration-4 height-3,machine-1 origin-1 duration-6 height-1,machine-1 origin-5 duration-2 height-4},
({
{id-1 capacity-7},
≤)
time
resource consumption
No negative stock
1 2 3 4 5 601234567
1 4
2
3
Stock at thispoint in time
Producer-Consumer withmultiple resources
cumulatives
{id-1 capacity-4,id-2 capacity-3},≤ )
time
resource consumption
No negative stock on resource 2
1 2 3 4 5 601234123
Stocks at thispoints in time
3
1
2
4
5 No negative stock on resource 1
machine-1 origin-1 duration-1 height-1,machine-1 origin-1 duration-3 height-2,
machine-2 origin-1 duration-3 height-1,
machine-2 origin-3 duration-4 height-2}
machine-1 origin-3 duration-4 height-1,
,
({
Workload covering
cumulatives
{id-1 capacity-0},≥)
time
resource consumption
Green tasks”cover” yellow tasks
1 2 3 4 5 6-4-3-2-10123
Cumulatedprofile
machine-1 origin-2 duration-2 height-(-2),machine-1 origin-4 duration-2 height-(-3),
({
1
3
2
45
6
machine-1 origin-1 duration-4 height-1,machine-1 origin-5 duration-2 height-1,machine-1 origin-2 duration-4 height-1,machine-1 origin-3 duration-3 height-1},
Workloads covering
cumulatives
{id-1 capacity-0,id-2 capacity-0}≥ )
({
time
resource consumption
1 2 3 4 5 6-2-1012-101
Cumulatedprofiles
2
13
45
76
machine-1 origin-2 duration-2 height-(-2),
machine-1 origin-1 duration-4 height-1,
machine-1 origin-4 duration-2 height-(-1),
machine-1 origin-2 duration-3 height-2,machine-1 origin-5 duration-2 height-2,
machine-2 origin-3 duration-2 height-(-1),
machine-2 origin-1 duration-4 height-1},
Green tasks”cover” yellow tasks on
both resources
Exercise: project scheduling
Each task requires a given number of persons.Knowning that you have at most 7 persons,find the earliest end.
T12-1
T21-3
T34-3
T42-2
T53-4
T61-6
T70-0
Exercise: project scheduling (load)
Each task requires a given load (duration x needed resource).Knowning that you have at most 7 persons, modify your previous programin order to find the earliest end.
T11
T23
T312
T44
T512
T66
T70
Exercise: perfect square
Write a program for finding a solution for placing21 squares of respective size [50,42,37,35,33,29,27,25,24,19,18,17,16,15,11,9,8,7,6,4,2]) into asquare of size 112 without overlap.
Exercise: machine scheduling
Each task has to run on a given machinechosen from a set of machines. Each machinecan only run one task at a time and each taskshould not be interrupted. Find the earliest end.
T12-{1,2}
T21-{1,3}
T34-{1,3}
T42-{2,3}
T53-{1}
T61-{1,3}
T70-{}
Exercise: machine scheduling(quotient labeling)
Create a specialized labeling which, for eachtask, creates a compulsory part (intersection of the task positioned at its earliest start andof the task positioned at its latest start) ofsize 1. Justify the use of this heuristics.
Exercise: machine scheduling(variable speed)
The effective duration of a task is now computedby multiplying the ”duration” by a fixed coefficient(2 of M1, 1 of M2, 3 for M3).Modify your program to handle this factor.
The Sweep Algorithm: Fixed Tasks
resource
timeEvent pointsMinimum level to reach
for those instants wherethere is at least one task
Current positionof the sweep line: 3
Sweep-linestatus
01234
Task 1
Task 2 Task 3
• nb_task=1• sum_height=2
1 2 4 5
Sweep line status:number of tasks
sum of the height
•
•which overlap the sweep line
Event points:start of each task
end of each task
•
•
• nb_task=2• sum_height=3
•••• nb_task=0•••• sum_height=0
• nb_task=1• sum_height=4
• nb_task=0• sum_height=0
Not Yet Fixed Tasks
Preconditions for a Check Event
max(origin[t])<min(end[t])
min(machine[t])=max(machine[t])=r
max(height[t])<max(0,Limit[m])
•
•
•
machine[t]
: nb_task=nb_task+1
: nb_task=nb_task−−−−1
Start
End
QUESTIONANSWER
ACTION
: which instants to check ?: those instants for which there is a task which can cause a problem
: count the number of such tasks which overlap the sweep line
Not Yet Fixed Tasks (continued)
Preconditions for a Bad Profile Event
max(origin[t]) < min(end[t])
min(machine[t]) = max(machine[t]) = r
max(height[t]) < 0
•
•
•
r
: sum_height=sum_height+max(height[t])
: sum_height=sum_height −−−−max(height[t])
Start
End
QUESTIONANSWER
ACTION
: how to build an ”optimistic” resource consumption profile ?: consider the ”bad” and ”good” tasks
: sum up the height of such tasks which overlap the sweep line
Preconditions for a Good Profile Eventr ∈∈∈∈ dom(machine[t])
max(height[t]) > 0
•
•
r
: sum_height=sum_height+max(height[t])
: sum_height=sum_height −−−−max(height[t])
Start
End
10. if nb_task≠0 and sum_height<limit to reach then return fail
1. Scan the tasks and generate the check and profile events
2. Sort the events in increasing order
3. Set nb_task and sum_height to 0 and d to the smallest date
The Sweep Algorithm: Main Loop(simplified version: one resource, no pruning)
4. while there still exist an event E do
9. Extract the next event E (if it exists)
INITIALIZATION
CHECK
MODIFY SWEEP STATUS
CHECK
5. if d≠date of E then
6. if nb_task≠0 and sum_height<limit to reach then return fail
7. Set d to date of E
8. if type of E=check then add E.inc to nb_task else add E.inc to sum_height
Principle of the Generalization Algorithm
1 2 3 4 5 60
1
2
3
1
time
1
2
3
4
Machine 1
Machine 2
1 2 3 4 5 6
?
?
Removes 3,4 from origin of task 4
Task Interval
DEFINITION: At each instant: ΣΣΣΣ heigths ≥≥≥≥ Capacity
NECESSARY CONDITION: ΣΣΣΣ surfaces ≥≥≥≥ utilisation of interval x Capacity
IDEA: sum up over each instant where we have at least one task
time1 2 3 4 5 6-4-3-2-10123
≥≥≥≥ 1
time
resource consumption
1 2 3 4 5 6-4-3-2-10123
13
2
456
≥≥≥≥ 1
interval ≥≥≥≥ (1+1+1)X1(3-1)+(1+1+1-2)+(1+1+1-2)
Task Interval (atleast case)
• For tasks with negative height
• For tasks with positive height
• If non-negative capacity
• If negative capacity
: minimum intersection with interval
: maximum intersection with interval
: minimum utilization in time of interval
: maximum utilization in time of interval
CHECKING THE NECESSARY CONDITION
• Forbid to cover more than a given surface in an interval
• Force to cover at least a given surface in an interval
PRUNING ACCORDING TO THE NECESSARY CONDITION
Minimum Intersection of a Task with an Interval
interval
T at its earliest startand minimum duration
T at its latest endand minimum duration
2 1
Compute the size of the intersection withthe interval of these two instances,
and take the minimum
Result is 1
Key ideas of most algorithms forresource scheduling: compulsory part
DEFINITION : The Compulsory Part CP (A. Lahrichi PhDthesis 1979) is the intersection of all feasibleinstances of the task T
INTEREST : No need to wait until all the variables(start, duration, resource consumption) of task T are fixed to deduce some information about resource consumption of T over time.The other tasks will not have access to that resource amount � pruning
MAIN QUESTION : How to efficiently compute the CP?
To compute CP, we need to compute only the intersection of the two followings instances of the rectangle task :
Imin= task at its earliest date with minimum duration Imax= task at its latest date
Compulsory part
Startmin Startmax Startmin + Durationmin
Durationmin
Durationmax
CP of a rectangle task
3TTT CPCPCP21
∪∪Startmax
Startmin
T1 T2T3
T1 T2T3
321 TTTCP ∪∪
∅=1TCP
∅=2TCP
CP of the “global” task
CP of a sequence of rectangles
Scheduling Problem under Labour Constraints
resource
time0 1 2 3 4 5 6 7 8 9 10 11 12
6543210
Compulsory part of T2
� T1 can not start at positions 0,1,2 and 6
T2T1• Starting time of of T2 within [1, 3]
• No restriction on the starting time of T1
• Resource capacity equal to 6
Use of CP for pruning
Plane task (LT)
Smax Emin
∞
Smin Smax Emin Emax
Earliest startImin
Latest endImax
Task
CP of a task with valleys
CP = Imin∩ Imax ∩ LT
Key ideas of most algorithms forresource scheduling: task interval
IDEA : Compute the intersection of all feasible instancesof a task T with a given interval and take the minimum.Repeat the previous step for all tasksCumulate all the minimum intersections and checkthat it does not exceed the available surface
MAIN QUESTION : How to efficiently compute the minimumintersection with an interval?
Exercise: maximum intersection
Consider a cumulative scheduling problem where one needs to use almostall the resource availability (i.e. the sum of the surface of the tasks is almostequal to the resource availability).
Question 1: consider a fixed interval low..up and a task t characterizedby its origin, duration and end (3 domain variables).Come up with a method for computing the maximum intersectionin surface (or a lower approximation) of task t with intervallow..up.
Question 2: how could you use the result of question 1 to perform somepruning in the context of the cumulative constraint ?
