Lecture 6:
Time/Cost Trade-off in
Project Planning
© J. Christopher Beck 2008
1
Outline

Project Planning with
Time/Cost Tradeoffs

Linear and non-linear
© J. Christopher Beck 2008
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Readings


P Ch 4.4-4.5
Slides borrowed
from Twente &
Iowa

See Pinedo CD
© J. Christopher Beck 2008
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Time/Cost
Trade-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 2008
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Linear Costs
Money
c aj
Cost to reduce job j by
by 1 time unit
max
p min

p

p
j
j
j
c aj  c bj
Marginal c j  max
min
p

p
cost
j
j
cbj
p
min
j
© J. Christopher Beck 2008
p
max
j
Processing
time
5
Problem

Spend money to reduce processing
times so as to minimize:
n


c0Cmax   cbj  c j p max
 pj
j

j 1
“Overhead” cost
© J. Christopher Beck 2008
Cost per activity
6
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 2008
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Sources, Sinks, & Cuts
Cut set
Sink node
Source (dummy) node
© J. Christopher Beck 2008
Minimal cut set
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Cuts Sets & Minimal Cut Sets



V is a set of nodes
E is set of edges (arcs)
Given a graph G(V,E)
A cut set is a set of nodes S  V
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
© J. Christopher Beck 2008
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Is This a Cut Set?
Is This a Minimal Cut Set?
2
4
7
10
1
6
3
© J. Christopher Beck 2008
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12
11
14
13
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Is This a Cut Set?
Is This a Minimal Cut Set?
2
4
7
10
1
6
3
© J. Christopher Beck 2008
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12
11
14
13
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Is This a Cut Set?
Is This a Minimal Cut Set?
2
4
7
10
1
6
3
© J. Christopher Beck 2008
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12
11
14
13
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Is This a Cut Set?
Is This a Minimal Cut Set?
2
4
7
10
1
6
3
© J. Christopher Beck 2008
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12
11
14
13
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Is This a Cut Set?
Is This a Minimal Cut Set?
2
4
7
10
1
6
3
© J. Christopher Beck 2008
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12
11
14
13
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Is This a Cut Set?
Is This a Minimal Cut Set?
2
4
7
10
1
6
3
© J. Christopher Beck 2008
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12
11
14
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Time/Cost Trade-off Heuristic
Step 1:
Set all processing times at their maximum
pj  p
max
j
Determine all critical paths
Construct the graph Gcp of critical paths
© J. Christopher Beck 2008
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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
pj  p
min
j
, j  Gcp
If no such set STOP; otherwise continue to Step 3
© J. Christopher Beck 2008
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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 2008
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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 2008
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Example 4.4.2
Error in text
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
35 20
10
cbj
6
22
12
6
22
31
27
13
18
5
19
29 12
2
cj
7
2
4
3
4
3
4
4
4
5
2
2
8
© J. Christopher Beck 2008
4
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Warning: Example 4.4.2

Unfortunately, the text is wrong for this
example!
© J. Christopher Beck 2008
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1. Step 1: Maximum Processing
Times, Find Gcp
2
4
7
10
1
6
9
3
11
5
© J. Christopher Beck 2008
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14
13
8
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1. Step 1: Maximum Processing
Times, Find Gcp
Cmax  56
2
4
7
10
1
6
9
3
Cost = overhead + job costs
= co * Cmax + Σcbj
= 6 * 56 + 224
= 560
© J. Christopher Beck 2008
12
11
5
14
13
8
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1. Step 2 & 3: Min. Cut Sets in
Gcp & Lowest Cost
c1=7
c6=3
6
1
c9=4
9
3
c3=4
Cut sets: {1},{3},{6},{9},
{11},{12},{14}.
Arbitrarily choose 12
Reduce p12 from 8 to 7
© J. Christopher Beck 2008
c12=2
c14=8
12
14
11
c11=2
Minimum cut
set with lowest cost
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2. Step 4 & 1: Reduce p12 by 1
Cmax  55
2
4
7
10
1
6
9
3
Cost = overhead + processing
= c0 * Cmax + Σjob costs
= 6 * 55 + 226
= 556
© J. Christopher Beck 2008
12
11
5
14
13
8
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2. Step 2 & 3: Min. Cut Sets in
Gcp & Lowest Cost
c1=7
c6=3
6
1
c9=4
9
3
c3=4
Cut sets: {1},{3},{6},{9},
{11},{12,13},{14}
Why isn’t {12} a cut set?
What is the cost of {12, 13}?
© J. Christopher Beck 2008
c12=2
c14=8
12
14
11
13
c11=2
c13=4
Minimum cut
set with lowest cost
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3. Step 4 & 1: Reduce p11 by 1
Cmax  54
2
4
7
10
1
6
9
3
Cost = overhead + processing
= c0 * Cmax + Σjob costs
= 6 * 54 + 228
= 552
© J. Christopher Beck 2008
12
11
5
14
13
8
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3. Step 2 & 3: Min. Cut Sets in
Gcp & Lowest Cost
c2=2
c4=3
2
4
c7=4
7
c1=7
c6=3
6
1
c9=4
9
3
c3=4
23 Min. cut sets! Where are they?
Min cost min cut set: {2,11}: 4
Reduce Job 2 to 5 and Job 11 to 5
© J. Christopher Beck 2008
c10=4
10
c12=2
c14=8
12
14
11
13
c11=2
c13=4
Minimum cut
set with lowest cost
28
4. Step 4 & 1: Reduce
p2 & p11 by 1
Cmax  53
2
4
7
10
1
6
9
3
Cost = overhead + processing
= c0 * Cmax + Σjob costs
= 6 * 53 + 232
= 550
© J. Christopher Beck 2008
12
11
5
14
13
8
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4. Step 2 & 3: Min. Cut Sets in
Gcp & Lowest Cost
c2=2
c4=3
2
4
c7=4
7
c1=7
c6=3
6
1
c9=4
9
3
c3=4
Can’t reduce Job 2 anymore. Why?
Min cost min cut set: {11,12}: 4
Reduce Job 11 to 4, Job 12 to 6
© J. Christopher Beck 2008
c10=4
10
c12=2
c14=8
12
14
11
13
c11=2
c13=4
Minimum cut
set with lowest cost
30
5. Step 4 & 1: Reduce
p11 and p12 by 1
2
4
Cmax  52
7
10
1
6
9
3
Cost = overhead + processing
= c0 * Cmax + Σjob costs
= 6 * 52 + 236
= 548
© J. Christopher Beck 2008
12
11
5
14
13
8
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5. Step 2 & 3: Min. Cut Sets in
Gcp & Lowest Cost
c2=2
c4=3
2
4
c7=4
7
c1=7
c6=3
6
1
3
c3=4
c9=4
9
c10=4
10
c12=2
c14=8
12
14
11
13
c11=2
c13=4
Cutset: {6, 12}: 4
Reduce Job 6 to 11, Job 12 to 5
© J. Christopher Beck 2008
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6. Step 4 & 1: Reduce
p6 and p12 by 1
Cmax  51
2
4
7
10
1
6
9
3
Cost = overhead + processing
= c0 * Cmax + Σjob costs
= 6 * 51 + 241
= 547
© J. Christopher Beck 2008
12
11
5
14
13
8
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6. Step 2 & 3: Min. Cut Sets in
Gcp & Lowest Cost
c2=2
c4=3
2
4
c7=4
7
c1=7
c6=3
6
1
c9=4
9
3
c3=4
c10=4
10
c12=2
c14=8
12
14
11
13
c11=2
c13=4
All cut sets either can’t be reduced
(a job at pmin) or has a cost ≥ 6.
So we’re done.
© J. Christopher Beck 2008
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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.
© J. Christopher Beck 2008
35
Linear Programming Formulation

The heuristic does not guarantee optimum


See example 4.4.3
You are responsible for the LP formulation!

see the text!
© J. Christopher Beck 2008
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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 2008
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