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### Lecture 5: Project Planning 2

© J. Christopher Beck 2005

Outline

- Time/Cost Tradeoffs
- Linear and non-linear
- Adding Workforce Constraints
- Slides borrowed from Twente & Iowa
- See Pinedo CD

© J. Christopher Beck 2005

Time/CostTrade-Offs

- What if you could spend money to reduce the job duration
- More money shorter processing time
- Run machine at higher speed

© J. Christopher Beck 2005

Problem

- Spend money to reduce processing times so as to minimize:

Cost per activity

“Overhead” cost

© J. Christopher Beck 2005

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

© J. Christopher Beck 2005

Time/Cost Trade-off Heuristic

- Step 1:
- Set all processing times at their maximum
- Determine all critical paths
- Construct the graph Gcp of critical paths

© J. Christopher Beck 2005

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 3

© J. Christopher Beck 2005

Time/Cost Trade-Off Heuristic

- 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

© J. Christopher Beck 2005

Time/Cost Trade-Off Heuristic

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

© J. Christopher Beck 2005

8

4

5

9

7

3

2

1

6

13

11

10

12

14

Step 1: Maximum Processing Times, Find GcpCost = overhead + job costs

= co * Cmax + Σcaj

= 6 * 56 + 350

= 686

© J. Christopher Beck 2005

1

3

6

9

11

12

14

Step 2 & 3: Min. Cut Sets in Gcp & Lowest Costc1=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

© J. Christopher Beck 2005

8

4

5

9

7

3

2

1

6

13

11

10

12

14

Step 4 & 1: Reduce Processing Time for Each Job by 1Cost = overhead + processing

= c0 * Cmax + Σjob costs

= 6 * 55 + 352

= 682

© J. Christopher Beck 2005

1

3

6

9

14

11

12

13

Step 2 & 3: Min. Cut Sets in Gcp & Lowest Costc1=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

© J. Christopher Beck 2005

1

3

6

9

11

12

14

13

Next 3 Iterations

c1=7

c12=2

c6=3

c9=4

c14=8

c11=2

c13= 4

c3=4

Next 3 iterations

reduce processing

time from 7 to 4

Cost = overhead + processing

= co * Cmax + Σjob costs

= 6 * 52 + 355

= 667

© J. Christopher Beck 2005

1

3

6

9

11

12

14

13

Step 1,2, & 3

c1=7

c12=2

c6=3

c9=4

c14=8

c11=2

c13= 4

c3=4

Reduce processing time

next on job 6

Q: why not 12?

© J. Christopher Beck 2005

4

1

3

6

9

7

2

12

13

11

14

10

After More Iterations …c2=2 c4=3 c7=4

c10=5

c1=7

c12=2

c6=3

c9=4

c14=8

c11=2

c13= 4

c3=4

© J. Christopher Beck 2005

Linear Programming Formulation

- The heuristic does not guarantee optimum
- See example 4.4.3
- Here total cost is linear so use LP
- Want to minimize

© J. Christopher Beck 2005

Linear Program

Minimize

subject to

earliest start

time of job k

processing

time of job k

© J. Christopher Beck 2005

Can Also Have Non-linear Costs

- Arbitrary function cj(pj) → cost of setting job j to processing time pj
- LP doesn’t work!
- See Section 4.5
- A question I like:
- Given processing times and cj(pj), which algorithm would you use (heuristic or LP)?

© J. Christopher Beck 2005

What If Jobs Require Resources?

- Back to fixed durations
- Without resources → easy
- With resources → hard
- Resource Constraint Project Scheduling Problem (RCPSP)

© J. Christopher Beck 2005

RCPSP

- n: jobs j=1,…,n
- N: resources i=1,…,N
- Rk: availability of resource k
- pj: duration of job j
- Rkj: requirement of job j for resource k
- Pj: (immediate) predecessors of job j
- Minimize Cmax

© J. Christopher Beck 2005

RCPSP Example

© J. Christopher Beck 2005

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