© J. Christopher Beck 20081 Lecture 6: Time/Cost Trade-off in Project Planning

© J. Christopher Beck 2008 1

Lecture 6: Time/Cost Trade-off in Project Planning


Outline Project Planning with

Time/Cost Tradeoffs Linear and non-linear


Readings

P Ch 4.4-4.5 Slides

borrowedfrom Twente & Iowa See Pinedo CD


Time/CostTrade-Offs

What if you could spend money to reduce the job duration More money

shorter processing time

Run machine at higher speed


Linear Costs

Money

Processingtimemax

jp

ajc

bjc

minjp

maxminjjj ppp

minmaxjj

bj

aj

j pp

ccc

Marginal cost

Cost to reduce job j byby 1 time unit


Problem

Spend money to reduce processing times so as to minimize:

n

jjjj

bj ppccCc

1

maxmax0

“Overhead” cost Cost per activity


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


Source (dummy) node

Sink node

Minimal cut set

Cut set

Sources, Sinks, & Cuts


Cuts Sets & Minimal Cut Sets

Given a graph G(V,E) A cut set is a set of nodes

such that in the graph formed by removing the nodes in S from G there is no path from the source to the sink

A minimal cut set, R, is a cut set such that if you remove any node, a, from R, R \ {a} is not a cut set

V is a set of nodesE is set of edges (arcs)

VS


Is This a Cut Set?Is This a Minimal Cut Set?

1

3

6 9

11

12 14

13

2 4 7

10



1

3

6 9

11

12 14

13

2 4 7

10



1

3

6 9

11

12 14

13

2 4 7

10



1

3

6 9

11

12 14

13

2 4 7

10



1

3

6 9

11

12 14

13

2 4 7

10



1

3

6 9

11

12 14

13

2 4 7

10


Step 1:

Set all processing times at their maximum

Determine all critical paths

Construct the graph Gcp of critical paths

maxjj pp

Time/Cost Trade-off Heuristic


Step 2:

Determine all minimal 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

cpjj Gjpp ,min

Time/Cost Trade-off Heuristic


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


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


Example 4.4.2

Jobs 1 2 3 4 5 6 7 8 9 10 11 12

13 14

pmaxj 5 6 9 12 7 12 10 6 10 9 7 8 7 5

pminj 3 5 7 9 5 9 8 3 7 6 4 5 5 4

caj 20 25 20 15 30 40 35 25 30 20 25 3

520 10

cbj 6 22 12 6 22 31 27 13 18 5 19 2

912 2

cj 7 2 4 3 4 3 4 4 4 5 2 2 4 8

Error in text


Warning: Example 4.4.2

Unfortunately, the text is wrong for this example!


1. Step 1: Maximum Processing Times, Find Gcp

1

2

3

6 9

5 8

4 7

11

10

12 14

13


1. Step 1: Maximum Processing Times, Find Gcp

1

2

3

6 9

5 8

4 7

11

10

12 14

13

56max C

Cost = overhead + job costs = co * Cmax + Σcb

j

= 6 * 56 + 224 = 560


1

3

6 9

11

12 14

c1=7

c3=4

c6=3 c9=4

c11=2

c12=2 c14=8

Minimum cutset with lowest cost

Cut sets: {1},{3},{6},{9}, {11},{12},{14}.

Arbitrarily choose 12Reduce p12 from 8 to 7

1. Step 2 & 3: Min. Cut Sets in Gcp & Lowest Cost


2. Step 4 & 1: Reduce p12 by 1

1

2

3

6 9

5 8

4 7

11

10

12 14

13

55max C

Cost = overhead + processing = c0 * Cmax + Σjob costs = 6 * 55 + 226 = 556



1

3

6 9

11

12 14

c1=7

c3=4

c6=3 c9=4

c11=2

c12=2 c14=8

13

Cut sets: {1},{3},{6},{9}, {11},{12,13},{14}

Why isn’t {12} a cut set?What is the cost of {12, 13}?

c13=4



3. Step 4 & 1: Reduce p11 by 1

1

2

3

6 9

5 8

4 7

11

10

12 14

13

54max C




1

3

6 9

11

12 14

c1=7

c3=4

c6=3 c9=4

c11=2

c12=2 c14=8

13

23 Min. cut sets! Where are they?Min cost min cut set: {2,11}: 4Reduce Job 2 to 5 and Job 11 to 5

c13=4


2 4 7

10

c2=2 c4=3 c7=4

c10=4


4. Step 4 & 1: Reduce p2 & p11 by 1

1

2

3

6 9

5 8

4 7

11

10

12 14

13

53max C




1

3

6 9

11

12 14

c1=7

c3=4

c6=3 c9=4

c11=2

c12=2 c14=8

13

Can’t reduce Job 2 anymore. Why?Min cost min cut set: {11,12}: 4Reduce Job 11 to 4, Job 12 to 6

c13=4


2 4 7

10

c2=2 c4=3 c7=4

c10=4


5. Step 4 & 1: Reducep11 and p12 by 1

1

2

3

6 9

5 8

4 7

11

10

12 14

13

52max C




1

3

6 9

11

12 14

c1=7

c3=4

c6=3 c9=4

c11=2

c12=2 c14=8

13

Cutset: {6, 12}: 4Reduce Job 6 to 11, Job 12 to 5

c13=4

2 4 7

10

c2=2 c4=3 c7=4

c10=4


6. Step 4 & 1: Reducep6 and p12 by 1

1

2

3

6 9

5 8

4 7

11

10

12 14

13

51max C




1

3

6 9

11

12 14

c1=7

c3=4

c6=3 c9=4

c11=2

c12=2 c14=8

13

All cut sets either can’t be reduced (a job at pmin) or has a cost ≥ 6.So we’re done.

c13=4

2 4 7

10

c2=2 c4=3 c7=4

c10=4


Example 4.4.2

It would be very useful for you to work through this entire example on your own!

Remember the book is wrong. You should use the slides.


Linear Programming Formulation

The heuristic does not guarantee optimum See example 4.4.3

You are responsible for the LP formulation! see the text!


Can Also HaveNon-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)?

Documents

© J. Christopher Beck 20081 Lecture 6: Time/Cost Trade-off in Project Planning