WOOD 492
MODELLING FOR DECISION SUPPORT
Lecture 23
CPM
Review
– Minimum spanning tree
• Goal: find the smallest network that has a path between each two
nodes (application example: railway networks)
• Greedy algorithm solution : start from any node and in each step
pick the closest unconnected node and add it to the network
– Critical path method (CPM) for time-cost trade-off
• Goal: find the optimal plan to expedite some activities within a
project in order to minimize the costs while meeting the project
deadline
• First, we visualize the activities within a project using a network
(each activity is represented with a node in the network)
• Then we find the “critical path”, which is the longest path through
the project network, going from start to finish nodes.
Oct 31, 2012
Wood 492 - Saba Vahid
2
Example 15 – Project network
• “Reliable Constructions Co.” has identified the activities within a
plant construction project
• The deadline is in 40 weeks
• Project budget is $5.4 million
• Based on activity information we create the project network
• How long will the project take?
– We need to find the paths through the project network and their lengths
– Path: one of the routes following the arcs from Start to Finish node
– Length: sum of the estimated durations of the activities along that path
Example 15
Oct 31, 2012
Wood 492 - Saba Vahid
3
Example 15 - Critical Path Method
Start
•
There are six paths for this network
1.
2.
3.
4.
5.
6.
•
Start-A-B-C-D-G-H-M-Finish
Start-A-B-C-E-H-M-Finish
Start-A-B-C-E-F-J-K-N-Finish
Start-A-B-C-E-F-J-L-N-Finish
Start-A-B-C-I-J-K-N-Finish
Start-A-B-C-I-J-L-N-Finish
40 weeks
31 weeks
43 weeks
44 weeks
41 weeks
42 weeks
A
Critical Path
Activities on each path are done sequentially (can not
overlap), so the project length can not be shorter than
the length of a given path, but it might be longer
– e.g. activity H has 2 predecessors (E, and G which is not
on the path). Looking at the duration of activities after C,
we see that activity E only takes 4 weeks, while activity D
and G take 13 weeks combined. Since H needs to wait for
both E and G to finish, the project will take much longer
than the 31 weeks estimated length for the second path
2
B
4
C
10
6
7
D
I
4
E
5
7
G
F
8
9
J
H
4
5
K
2
M
L
N
6
Finish
Oct 31, 2012
Wood 492 - Saba Vahid
4
Critical Path
Start
• Critical path: Start-A-B-C-E-F-J-L-N-Finish
44 weeks
A
2
• The longest path through the project network
• May have more than one critical path (same length)
• Activities on the longest path can be done sequentially
without having to wait for activities on other paths, so the
project length is equal to the length of this path
• All other paths will reach the Finish node sooner than the
critical path
• Project length is 44 wks: longer than the 40 wks deadline
• The management should focus on reducing the length of
activities on the critical path to finish the project by the
deadline
B
4
C
10
6
7
D
I
4
E
5
7
G
F
8
9
J
H
4
5
K
2
M
L
N
6
• Which activities to choose?
Finish
Oct 31, 2012
Wood 492 - Saba Vahid
5
CPM for time-cost trade-offs
• Crashing an activity: using costly measured to reduce its length
• CPM: determines how much (if any) to crash each activity to reduce the
estimated duration of the project
• Time-cost graph : shows the relationship between time and cost of an activity
Cost
in the normal and crash modes
• Assumptions:
– Maximum time reduction and associated crash costs
For each activity can be estimated with certainty
– Partially crashing an activity is possible, the associated
cost and time move along the line segment in the graph
Crash cost
Crash
Normal
Normal cost
Crash time
Normal time
Duration
• Time-cost trade-off data are required for each activity
Oct 31, 2012
Wood 492 - Saba Vahid
6
Cost-time trade-off
•
•
•
Example of cost-time trade-off for an activity
Activity J (putting up the wallboard)
It’s possible to reduce the duration of this activity by two weeks (through
hiring temporary workers and using overtime)
Normal time: 8 weeks
Normal Cost: $430,000
Crash time: 6 weeks
Crash Cost:$490,000
Maximum reduction in time=8 – 6 = 2 weeks
Crash cost per week saved = (490,000 – 430,000)/2 = $30,000
So for each week the company saves in time, $30,000 are required in
extra costs
Example 15
Oct 31, 2012
Wood 492 - Saba Vahid
7
Which activities to crash?
• How to select the activities to crash, while minimizing the costs of
doing so?
– Marginal Cost analysis
– Linear programming
• Marginal Cost Analysis
1. select the longest path through the network
2. Among the activities on that path, select the one with the lowest “crash cost per
week”
3. Reduce its length by one week if possible, if not move to the next lowest crash
cost/week
4. Review the length of all paths after this reduction and repeat steps 1 to 3 until
you reach the desired length of the project
Example 15
Oct 31, 2012
Wood 492 - Saba Vahid
8
Using LP to choose activities to crash
• The objective: minimize the total cost of crashing the activities
• Variables: xj : reduction in activity j duration by crashing (j=A,B,..,N)
• Constraints:
– Total project duration is less than the desired length
– The reductions can not be more than the maximum allowed
reduction
– The reductions must be non-negative
– Precedence relationships between activities must hold (we need
to define extra variables to write these constraints)
Oct 31, 2012
Wood 492 - Saba Vahid
9
Precedence relationships
• Define new variables: Yj = start time of activity j (for j=B,C,…,N)
• The start time of each activity has to be after all its predecessors are
finished, so
Start time of activity j >= (start time + duration) of its immediate
predecessor
• Duration of each activity j= its normal time – xj
• For example, immediate predecessor of activity F is E, so:
YF >= YE + 4 – XE
• For activities with more than one predecessor, more inequalities are
needed
Oct 31, 2012
Wood 492 - Saba Vahid
10
The LP for cost-time trade-off, Example 15
• Minimize Z = Normal Cost +
(crash cost per week of activity j ∗ reduction in activity j duration)
• Subject to:
• 𝑥𝐴 ≤ 1, 𝑥𝐵 ≤ 2, … (maximum reduction constraints)
• 𝑥𝐴 ≥ 0, 𝑥𝐵 ≥ 0,… (non-negativity constraints)
• 𝑦𝐵 ≥ 0 + 2 − 𝑥𝐴
𝑦𝐶 ≥ 𝑦𝐵 + 4 − 𝑥𝐵
𝑦𝐹𝑖𝑛𝑖𝑠ℎ ≥ 𝑦𝑀 + 2 − 𝑥𝑀
𝑦𝐹𝑖𝑛𝑖𝑠ℎ ≥ 𝑦𝑁 + 6 − 𝑥𝑁 …. (start time constraints/precedence)
• 𝑦𝐹𝑖𝑛𝑖𝑠ℎ ≤ 40 (finish time constraint)
Example 15
Oct 31, 2012
Wood 492 - Saba Vahid
11