- 71 Views
- Uploaded on

Download Presentation
## PowerPoint Slideshow about ' Topic 11' - roary-parsons

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

Project Scheduling

- Jobs subject to precedence constraints
- Job on arc format (most common)

2

5

6

0

1

4

7

2

3

IE 514

Overview

Critical Path

Method (CPM)

Program Evaluation

and Review Technique

(PERT)

No resource

Constraints

Project

Scheduling

Heuristic Resource

Leveling

Integer Programming

Formulations

Resource

Constraints

IE 514

Planning a Concert

IE 514

Job on Arc Network

- Not allowed: no two jobs can have the same starting and ending node!
- Need to introduce a dummy job.

B

A

D

C

B

A

D

C

IE 514

Changing a Tire

IE 514

B

E

D

C

Job on Node Network- No need for a dummy node
- Less used traditionally
- Recently more popular

IE 514

Critical Path Method (CPM)

- Think of unlimited machines in parallel
- … and n jobs with precedence constraints
- Processing times pj as before
- Objective to minimize makespan

IE 514

Critical Path Method

- Forward procedure:
- Starting at time zero, calculate the earliest each job can be started
- The completion time of the last job is the makespan

- Backward procedure
- Starting at time equal to the makespan, calculate the latest each job can be started so that this makespan is realized

IE 514

Forward Procedure

Step 1:

Set at time t = 0 for all jobs j with no predecessors,

Sj’=0 and set Cj’ = pj.

Step 2:

Compute for each job j

Cj’ = Sj’ + pj.

Step 3:

The optimal makespan is

STOP

IE 514

Backward Procedure

Step 1:

Set at time t = Cmax for all jobs j with no successors,

Cj’’= Cmax and set Sj’’ = Cmax - pj.

Step 2:

Compute for each job j

Sj’’ = Cj’’ - pj.

Step 3:

Verify that

STOP

IE 514

Comments

- The forward procedure gives the earliest possible starting time for each job
- The backwards procedures gives the latest possible starting time for each job
- If these are equal the job is a critical job.
- If these are different the job is a slack job, and the difference is the float.
- A critical path is a chain of jobs starting at time 0 and ending at Cmax.

IE 514

2

3

6

9

5

8

4

7

14

13

12

11

10

Forward Procedure5+6=11

11+12=23

23+10=33

33+9=42

5

43+8=51

14+12=26

26+10=36

51+5=56

5+9=14

43+7=50

36+7=43

14+7=21

26+6=32

IE 514

2

3

6

9

5

8

4

7

14

13

12

11

10

Backwards Procedure24-12=12

34-10=24

43-9=34

51-8=43

14-9=5

56-5=51

36-10=26

43-7=36

56

26-12=14

56-5=51

51-8=43

35-10=26

43-7=36

IE 514

Time/Cost Trade-Offs

- Assumed the processing times were fixed
- More money shorter processing time
- Start with linear costs
- Processing time
- Marginal cost

IE 514

Solution Methods

- Objective: minimum cost of project
- Time/Cost Trade-Off Heuristic
- Good schedules
- Works also for non-linear costs

- Linear programming formulation
- Optimal schedules
- Non-linear version not easily solved

IE 514

Time/Cost Trade-Off Heuristic

- Step 1:
- Set all processing times at their maximum
- Determine all critical paths with these processing times
- Construct the graph Gcp of critical paths
- Continue to Step 2

IE 514

Time/Cost Trade-Off Heuristic

- Step 2:
- Determine all minimum cut sets in Gcp
- Consider those sets where all processing times are larger than their minimum
- If no such set STOP; otherwise continue to Step 2

IE 514

- Step 3:
- For each minimum cut set:
- Compute the cost of reducing all processing times by one time unit.
- Take the minimum cut set with the lowest cost
- If this is less than the overhead per time unit go on to Step 4; otherwise STOP

IE 514

- Step 4:
- Reduce all processing times in the minimum cut set by one time units
- Determine the new set of critical paths
- Revise graph Gcp and go back to Step 2

IE 514

IE 514

3

6

9

14

11

12

Critical Path Subgraph (Gcp)C1=7

C12=2

C6=3

C9=4

C14=8

C11=2

C3=4

Cut sets: {1},{3},{6},{9},

{11},{12},{14}.

Minimum cut

set with lowest cost

IE 514

3

6

9

14

11

12

13

Critical Path Subgraph (Gcp)C1=7

C12=2

C6=3

C9=4

C14=8

C11=2

C13=4

C3=4

Cut sets: {1},{3},{6},{9},

{11},{12,13},{14}.

Minimum cut

set with lowest cost

IE 514

3

6

9

14

12

13

11

Critical Path Subgraph (Gcp)

C1=7

C12=2

C6=3

C9=4

C14=8

C11=2

C13=14

C3=4

Reduce processing

time until = 4

Reduce processing time

next on job 6

IE 514

1

3

6

9

7

2

12

13

11

14

10

Critical Path Subgraph (Gcp)C2=2 C4=3 C7=4

C10=5

C1=7

C12=2

C6=3

C9=4

C14=8

C11=2

C13=4

C3=4

IE 514

Linear Programming Formulation

- The heuristic does not guaranteed optimum
- Here total cost is linear
- Want to minimize

IE 514

Program Evaluation and Review Technique (PERT)

- Assumed processing times deterministic
- Processing time of j random with mean mj and variance sj2.
- Want to determine the expected makespan
- Assume we have
pja = most optimistic processing time

pjm = most likely processing time (mode)

pjb = most pessimistic processing time

IE 514

Expected Makespan

- Estimate expected processing time
- Apply CPM with expected processing times
- Let Jcp be a critical path
- Estimate expected makespan

IE 514

Distribution of Makespan

- Estimate the variance of processing times
- and the variance of the makespan
- Assume it is normally distributed

IE 514

Discussion

- Potential problems with PERT:
- Always underestimates project duration
- other paths may delay the project

- Non-critical paths ignored
- critical path probability
- critical activity probability

- Activities are not always independent
- same raw material, weather conditions, etc.

- Estimates by be inaccurate

- Always underestimates project duration

IE 514

Discussion

- No resource constraints:
- Critical Path Method (CPM)
- Simple deterministic
- Time/cost trade-offs
- Linear cost (heuristic or exact)
- Non-linear cost (heuristic)

- Accounting for randomness (PERT)

IE 514

Resource Constraints

- Renewable resources
- Very hard problem
- No LP
- Can develop an IP
- Let job n+1 be dummy job (sink) and

IE 514

In Practice

- The IP cannot be solved
- (Almost) always resource constraints
- Heuristic:
Resource constraint Precedence constraint

- Example: pouring foundation and sidewalk
- both require same cement mixer
- delaying foundation delays building
- precedence constraint: pour foundation first
- not a logical constraint

IE 514

Optimality of Heuristic

- Say n jobs need the same resource
- Could otherwise all be done simultaneously
- Add (artificial) precedence constraints
- Have n! possibilities
- Will we select the optimal sequence?

IE 514

Decision Support

- Resource leveling
- Solve with no resource constraints
- Plot the resource use as a function of time
- If infeasible suggest precedence constraints
- Longest job
- Minimum slack

- User adds constraints
- Start over

IE 514

Job Shop Scheduling

- Have m machines and n jobs
- Each job visits some or all of the machines
- Customer order of small batches
- Wafer fabrication in semiconductor industry
- Hospital

- Very difficult to solve

IE 514

Job Shop Example

- Constraints
- Job follows a specific route
- One job at a time on each machine

Machine 1

(1,1)

(1,2)

(1,3)

Machine 2

(2,3)

(2,1)

(2,2)

Machine 3

(3,1)

(3,3)

Machine 4

(4,3)

(4,2)

IE 514

Graph Representation

Each job follows a specific route through the job shop ...

(1,1)

(2,1)

(3,1)

Sink

Source

(1,2)

(2,2)

(4,2)

(2,3)

(1,3)

(4,3)

(3,3)

(Conjuctive arcs)

IE 514

(2,1)

(3,1)

Sink

Source

(1,2)

(2,2)

(4,2)

(2,3)

(1,3)

(4,3)

(3,3)

Graph Representation

Machine constraints must also be satisfied ...

(Disjunctive arcs)

IE 514

Solving the Problem

- Select one of each pair of disjunctive arcs
- The longest path in this graph G(D) determines the makespan

(1,1)

(2,1)

(3,1)

Sink

Source

(1,2)

(2,2)

(4,2)

(2,3)

(1,3)

(4,3)

(3,3)

IE 514

Feasibility of the Schedule

- Are all selections feasible?

(1,1)

(2,1)

(3,1)

Sink

Source

(1,2)

(2,2)

(4,2)

(2,3)

(1,3)

(4,3)

(3,3)

IE 514

Solution Methods

- Exact solution
- Branch and Bound
- 20 machines and 20 jobs

- Dispatching rules (16+)
- Shifting Bottleneck

- Search heuristics
- Tabu, SA, GA, etc.

IE 514

Definitions

- A schedule is nondelay if no machine is idled when there is an operation available
- A schedule is called active if no operation can be completed earlier by altering the sequence on machines and not delaying other operations
- For “regular” objectives the optimal schedule is always active but not necessarily nondelay

IE 514

Nonactive Schedule

Machine 1

(1,1)

(2,1)

Machine 2

(2,3)

(2,2)

(2,1)

Machine 3

(3,2)

0 2 4 6 8

IE 514

Active Schedule, not Nondelay

Machine 1

(1,1)

Machine 2

(2,3)

(2,2)

(2,1)

Machine 3

(3,2)

0 2 4 6 8

IE 514

Branch and Bound

- Operation (i,j) with duration pij
- Minimize makespan
- Branch by generating all active schedules
- Notation
- Let W denote operations whose predecessors have been scheduled
- Let rij be the earliest possible starting time of (i,j) in W.

IE 514

Generating Active Schedules

Step 1. (Initialize)

Let W contain the first operation of each job

Let rij = 0 for all (i,j)W.

Step 2. (Machine selection)

Compute the current partial schedule

and let i* denote the machine where minimum achieved

IE 514

Generating Active Schedules

Step 3. (Branching)

Let W’ denote all operations on machine i such that

For each operation in W’ consider a partial schedule with that operation next on i*

For each partial schedule, delete operation from W and include immediate follower in W. Go back to Step 2.

IE 514

Example

10

8

(1,1)

(2,1)

(3,1)

4

0

0

8

3

5

6

Source

(2,2)

(1,2)

(4,2)

(3,2)

Sink

0

3

7

(1,3)

4

(2,3)

(4,3)

IE 514

Level 1: select (1,1)

10

8

(1,1)

(2,1)

(3,1)

4

10

0

10

0

8

3

5

6

Source

(2,2)

(1,2)

(4,2)

(3,2)

Sink

0

3

7

(1,3)

4

(2,3)

(4,3)

Disjunctive Arcs

IE 514

10

8

(1,1)

(2,1)

(3,1)

4

0

4

0

8

3

5

6

Source

(2,2)

(1,2)

(4,2)

(3,2)

Sink

0

4

3

7

(1,3)

4

(2,3)

(4,3)

Disjunctive Arcs

IE 514

Branching Tree

No disjunctive arcs

(1,1) scheduled first

on machine 1

LB = 24

(1,3) scheduled first

on machine 1

LB = 26

IE 514

Level 2: Select (2,2)

10

8

(1,1)

(2,1)

(3,1)

4

10

0

10

0

8

3

5

6

Source

(2,2)

(1,2)

(4,2)

(3,2)

Sink

0

3

7

(1,3)

4

(2,3)

(4,3)

Disjunctive Arcs

IE 514

No disjunctive arcs

(1,1) scheduled first

on machine 1

LB = 24

(1,3) scheduled first

on machine 1

LB = 26

(1,1) first on M1 and

(2,2) first on M2

IE 514

Lower Bounds

- Lower bounds
- The length of the critical path in G(D’)
- Quick but not very tight

- Linear programming relaxation
- A maximum lateness problem (see book)

- The length of the critical path in G(D’)

IE 514

Shifting Bottleneck

- Minimize makespan in a job shop
- Let M denote the set of machines
- Let M0 M be machines for which disjunctive arcs have been selected
- Basic idea:
- Select a machine in M - M0 to be included in M0
- Sequence the operations on this machine

IE 514

Example

IE 514

Iteration 1

10

8

(1,1)

(2,1)

(3,1)

4

0

0

8

3

5

6

Source

(2,2)

(1,2)

(4,2)

(3,2)

Sink

0

3

7

(1,3)

4

(2,3)

(4,3)

IE 514

Selecting a Machine

Set up a nonpreemptive single machine maximum lateness

problem for Machine 1:

Optimum sequence is 1,2,3 with Lmax(1)=5

IE 514

Set up a nonpreemptive single machine maximum lateness

problem for Machine 2:

Optimum sequence is 2,3,1 with Lmax(2)=5

IE 514

10

8

(1,1)

(2,1)

(3,1)

4

0

10

0

8

3

5

6

Source

(2,2)

(1,2)

(4,2)

(3,2)

Sink

0

3

3

7

(1,3)

4

(2,3)

(4,3)

IE 514

Set up a nonpreemptive single machine maximum lateness

problem for Machine 2:

Optimum sequence is 2,1,3 with Lmax(2)=1

IE 514

10

8

(1,1)

(2,1)

(3,1)

4

10

0

8

8

0

8

3

5

6

Source

(2,2)

(1,2)

(4,2)

(3,2)

Sink

0

3

3

7

(1,3)

4

(2,3)

(4,3)

IE 514

Discussion

- Procedure continues until all the disjunctive arcs have been added
- Extremely effective
- Fast
- Good solutions

- ‘Just a heuristic’
- No guarantee of optimum

IE 514

Solving the Maximum Lateness Problem

(•,•,•)

(1,•,•)

(2,•,•)

(3,•,•)

(1,2,3)

(1,3,2)

(3,1,2)

(3,2,1)

IE 514

Discussion

- The solution is actually a little bit more complicated than before
- Precedence constraints because of other (already scheduled) machines
- Delay precedence constraints
(see example in book)

IE 514

Discussion

- Shifting bottleneck can be applied generally
- Basic idea
- Solve problem “one variable at a time”
- Determine the “most important” variable
- Find the best value of that variable
- Move on to the “second most important” ….
- Here we treat each machine as a variable

IE 514

Total Weighted Tardiness

- We now apply a shifted bottleneck procedure to a job shop with total weighted tardiness objective
- Need n sinks in disjunctive graph
- Machines scheduled one at a time
- Given current graph calculate completion time of each job
- Some n due dates for each operation
- Piecewise linear penalty function

IE 514

Cost Function for Operation (i,j)

IE 514

Machine Selection

- Machine criticality
- Solve a single machine problem
- Piecewise linear cost function
- May have delayed precedence constraints
- Generalizes single-machine with n jobs, precedence constraints, and total weighted tardiness objective
- ATC rule

IE 514

Generalized ATC Rule

Earliest time machine can be used

Ranks jobs

good schedule

Scaling constant

IE 514

Criticality of Machines

- Criticality = subproblem objective function
- Simple

- More effective ways, e.g.
- Add disjunctive arcs for each machine
- Calculate new completion times and

IE 514

Example

5

10

4

Sink

(1,1)

(2,1)

(3,1)

5

0

4

5

6

Source

(3,2)

(1,2)

(2,2)

Sink

0

3

7

Sink

(3,3)

5

(2,3)

(1,3)

IE 514

Subproblem: Machine 1

IE 514

IE 514

IE 514

Subproblem Solutions

- Solve using dispatching rule
- Use K=0.1
- Have t = 4,

- For machine 1 this results in

Schedule

first

IE 514

5

10

4

Sink

(1,1)

(2,1)

(3,1)

5

5

0

4

5

6

Source

(3,2)

(1,2)

(2,2)

Sink

5

0

3

7

Sink

(3,3)

5

(2,3)

(1,3)

IE 514

IE 514

IE 514

5

10

4

Sink

(1,1)

(2,1)

(3,1)

5

5

10

0

4

5

6

Source

(3,2)

(1,2)

(2,2)

Sink

5

3

0

3

7

Sink

(3,3)

5

(2,3)

(1,3)

IE 514

IE 514

5

10

4

Sink

(1,1)

(2,1)

(3,1)

5

5

10

4

0

4

5

6

Source

(3,2)

(1,2)

(2,2)

Sink

5

5

3

0

3

7

Sink

(3,3)

5

(2,3)

(1,3)

IE 514

Final Schedule

Machine 1

1,1

1,2

1,3

Machine 2

2,3

2,1

2,2

Machine 3

3,3

3,2

3,1

0 5 10 15 20 25 30

IE 514

Performance

- Objective function
- Try finding a better schedule using LEKIN
(optimal value = 18)

IE 514

Random Search Methods

- Popular to use genetic algorithms, simulated annealing, tabu search, etc.
- Do not work very well
- Problems defining the neighborhood
- Do not exploit special structure

IE 514

Defining the Neighborhood

- Approximately nm neighbors!
- Simply too inefficient

5

10

4

Sink

(1,1)

(2,1)

(3,1)

5

5

10

4

0

4

5

6

Source

(3,2)

(1,2)

(2,2)

Sink

5

5

3

0

3

7

Sink

(3,3)

5

(2,3)

(1,3)

IE 514

Job Shop with Makespan

- Random search methods can be applied
- Use ‘critical path’ neighborhood
- Can eliminate many neighbors immediately

- Specialized methods usually better
- Random search = ‘giving up’ !
- Traveling Salesman Problem (TSP)
- Very well studied
- Lin-Kernighan type heuristics (1970)
- Order of 1000 times faster than random search

IE 514

The Nested Partitions Method

- Partitioning
- by scheduling the bottleneck machine first

- Random sampling
- using randomized dispatching rules

- Calculating the promising index
- incorporating local improvement heuristic

- Can incorporate any special structure!

IE 514

Applications

- Very common in applications:
- Paper mills
- Steel lines
- Bottling lines
- Food processing lines

IE 514

Classical Literature

- Exact solutions
- Simple flow shop with makespan criterion
- Two machine case (Johnson’s rule)

- Realistic problems require heuristic approaches

IE 514

Objectives

- Multiple objectives usual
- Meet due dates
- Maximize throughput
- Minimize work-in-process (WIP)

Setting for job

j on Machine i

IE 514

Generating Schedules

- Identify bottlenecks
- Compute time windows at bottleneck stage
- Compute machine capacity at bottleneck
- Schedule bottleneck stage
- Schedule non-bottlenecks

IE 514

Identifying Bottlenecks

- In practice usually known
- Schedule downstream bottleneck first
- Determining the bottleneck
- loading
- number of shifts
- downtime due to setups

- Bottleneck assumed fixed

IE 514

Identifying Time Window

- Due date
- Shipping day
- Multiply remaining processing times with a safety factor

- Release date
- Status sj of job j
- Release date if sj = l

Decreasing

function -

determined

empirically

IE 514

Computing Capacity

- Capacity of each machine at bottleneck
- Speed
- Number of shifts
- Setups

- Two cases:
- Identical machines
- Non-identical machines

IE 514

Scheduling Bottleneck

- Jobs selected one at a time
- Setup time
- Due date
- Capacity

- For example ATCS rule

IE 514

Schedule Remaining Jobs

- Determined by sequence at bottleneck stage
- Minor adjustments
- Adjacent pairwise interchanges to reduce setup

IE 514

Summary: Job Shops

- Representation: graph w/disjoint arcs
- Solution methods
- Branch-and-bound
- Shifted bottleneck heuristic
- Beam search
- Can incorporate bottleneck idea & dispatching rules

- Random search methods

- Special case: flow shops

IE 514

Download Presentation

Connecting to Server..