1. Project Management Time-Cost Model Tutorial
Objectives and Information
Step 1. Normal duration and cost
Step 2. What to crash
Step 3. Crash the activity
Step 3. Repeat
Step 4. The all crash duration and cost
Step 5. The tradeoff graph
Developed by:
Alex J. Ruiz-Torres, Ph.D.

2. Objectives and Information

3. Objectives and Information
The goal of this process is the development of alternative project plans.
Each project plan has a different cost and duration.
As the project duration is reduced the cost increases.
The idea is similar to the options for shipping via UPS,.. 5 days costs $10, 3 days $16, 2 days costs $22, and overnight $40.

4. Objectives and Information
The project information that must be already defined:
Set of activities and precedence relationships.
Activity duration when its performed at its “normal” speed. Called the Normal Time (NT)
Shortest possible duration. Called the Crash Time (CT)
Activity cost when performed at its normal speed. Called the Normal Cost (NC)
Activity cost when performed at its shortest duration. Called the Crash Cost (CC)

5. Objectives and Information
EXAMPLE
Months
Millions $ CT NT NC CC
Pred a 2 4 8 18 b 3 5 13 c
9.1 d 7 6
8.5
b,c e 14 f 1
d,e

6. Step 1. Normal Duration and Cost

7. Step 1. Normal Duration and Cost
We start by considering all activities are performed at their normal time.
No activity is being expedited/ crashed.
Months
Millions $ CT NT NC CC
Pred a 2 4 8 18 b 3 5 13 c
9.1 d 7 6
8.5
b,c e 14 f 1
d,e

8. Step 1. Normal Duration and CostCT NT NC CC
Pred a 2 4 8 18 b 3 5 13 c
9.1 d 7 6
8.5
b,c e 14 f 1
d,e
We determine the duration and CP for the project with all NT’s
4 9
9 17
0 4 b d
17 20 a f c e
Duration = 20 Months
CP = a – b – d – f
11 17
4 8
5 9

9. Step 1. Normal Duration and CostCT NT NC CC
Pred a 2 4 8 18 b 3 5 13 c
9.1 d 7 6
8.5
b,c e 14 f 1
d,e
4 9
9 17
0 4 b d
17 20 a f
Given all activities are at
their NT, the
total cost is the sum of
Normal Costs
= $36 Million c e
11 17
4 8
5 9

10. Step 1. Normal Duration and Cost
Normal Case
Current Duration = 20 Months
Current Costs = $36 Million

11. Step 2. What to Crash?

12. Step 2. What to Crash?
Objective: to determine what activity to crash in order to reduce the project’s duration.
This is done one time unit at a time. In this example, one month at a time.
For each activity we calculate the crash rate
CR = (CC – NC) / (NT – CT)
The crash rate is the cost to reduce that activity by one time unit
For this example = $million/reduce one month

13. Step 2. What to Crash? crash rate CR = (CC – NC) / (NT – CT) Months
Months
Millions $
$M/mo. CT NT NC CC
Rate
Pred a 2 4 8 18 5 b 3 13 c
9.1
1.1 d 7 6
8.5
2.5
b,c e 14
3.5 f 1
d,e

14. Step 2. What to Crash?
mo.
 mo. $M
$M 
$M/mo. CT NT NC CC
Rate
Pred a 2 4 8 18 5 b 3 13 c
9.1
1.1 d 7 6
8.5
2.5
b,c e 14
3.5 f 1
d,e
Activity c has the smallest rate (therefore smallest cost increase), BUT
This activity is not in the CP. Therefore reducing the time of activity c will not reduce the project’s duration

15. Step 2. What to Crash? b d a f c e
If the duration of activity c is three months (instead of 4 as the original ), the project still takes 20 months
4 9
9 17
0 4 b d
17 20 a f c e
11 17
4 7
6 9
new duration = 3

16. Step 2. What to Crash?
We will crash the activity in the CP with the smallest rate. Activity d.
CP = a – b – d – f
SMALLEST
mo.
 mo. $M
$M 
$M/mo. CT NT NC CC
Rate
Pred a 2 4 8 18 5 b 3 13 c
9.1
1.1 d 7 6
8.5
2.5
b,c e 14
3.5 f 1
d,e

17. Step 3. Crash the activity

18. Step 3. Crash the activity
The duration of Activity d is
reduced by one time unit
duration = 7 months (instead of 8)
We will next determine the project’s
completion time.

19. Step 3. Crash the activityCT NT NC CC
Pred a 2 4 8 18 b 3 5 13 c
9.1 d 7 6
8.5
b,c e 14 f 1
d,e
We calculate the completion time and critical path with d = 7 9
9 16
new duration = 7
0 4 b d 19 a f
Duration = 19 Months
Thus crashing d by one
time unit reduced the
project by one time unit. c e 15 8

20. Step 3. Crash the activity
When expediting activity d by one month the project cost increases by the crash rate for that activity
Therefore the project cost at 19 months is $38.5 Million ( )

21. Step 4. Repeat
This means steps 2 – 3 will be repeated X times or until no alternatives are available.
Back to step 2. What to crash?

22. Step 2. What to Crash? b d a f c e CP = a – b – d – f (The CP did notCT NT NC CC
Pred a 2 4 8 18 b 3 5 13 c
9.1 d 7 6
8.5
b,c e 14 f 1
d,e
We need to determine the critical path
4 9
9 16
0 4 b d
16 19 a f c e
CP = a – b – d – f
(The CP did not
CHANGE)
10 16
4 8
5 9

23. Step 2. What to Crash? The critical path is CP = a – b – d – f
d cannot be crashed again. Is at its minimum possible duration (already at 7 months).
The CT is the smallest possible duration..
Next smallest is b.
available LOWEST
mo.
 mo. $M
$M 
$M/mo. CT NT NC CC
Rate
Pred a 2 4 8 18 5 b 3 13 c
9.1
1.1 d 7 6
8.5
2.5
b,c e 14
3.5 f 1
d,e

24. Step 3. Crash the activityCT NT NC CC
Pred a 2 4 8 18 b 3 5 13 c
9.1 d 7 6
8.5
b,c e 14 f 1
d,e
We calculate the project’s duration with b = 4 months
4 8
8 15
0 4 b d
15 18 a f c e
Duration = 18 Months
8 14 8

25. Step 3. Crash the activity
Activity b had a crash rate of 4
Reducing the time of an activity increases the cost of the project by its crash rate.
Project cost at 18 Months is
= $42.5 Million

26. Step 4. Repeat
Back to step 2. What to crash?

27. Step 2. What to Crash? b d a f c e We determine the critical pathCT NT NC CC
Pred a 2 4 8 18 b 3 5 13 c
9.1 d 7 6
8.5
b,c e 14 f 1
d,e
We determine the critical path
4 8
8 15
all activities but e
have 0 slack
0 4 b d
15 18 a f
Duration = 18 Months
CP = a – b – d – f
CP = a – c – d – f
(we now have 2 CPs) c e
8 14
9 15
4 8

28. Step 2. What to Crash? We now have 2 CPs
CP = a – b – d – f
CP = a – c – d – f
To reduce the project to 17 months we could do one of the following
Crash a
Crash both b and c
Crash f
Lowest cost = a
$M/mo.
Rate
Pred a 5 b 4 c
1.1 d
2.5
b,c e
3.5 f 6
d,e

29. Step 2. What to Crash? b d a f c e
Why not just c? If we only crashed c, the project’s completion time would not be reduced. Same if we only crashed b. They are in parallel, so both must be reduced simulateneosly to change the project’s completion time.
4 8
8 15
0 4 b d
15 18 a f
CP = a – b – d – f
CP = a – c – d – f c e
8 14
9 15
4 7
5 8 29

30. Step 3. Crash the activity
Activity a is selected. Now has a duration of 3 months.
3 7
7 14
0 3 b d
14 17 a f c e
Duration = 17 Months
7 13
3 7 30

31. Step 3. Crash the activity
Activity a had a crash rate of 5
Reducing the time of an activity increases the cost of the project by its crash rate.
Project cost at 17 Months is
= $47.5 Million
Step 4.We will stop here,
X times = 3 times

32. Step 5. The all crash duration and cost

33. Step 5. The all crash duration and costCT NT NC CC
Pred a 2 4 8 18 b 3 5 13 c
9.1 d 7 6
8.5
b,c e 14 f 1
d,e
We calculate the completion time and critical path with all durations at crash time
2 5
5 12
0 2 b d
12 13 a f c e
Duration = 13 Months
5 9
2 5

34. Step 5. The all crash duration and costCT NT NC CC
Pred a 2 4 8 18 b 3 5 13 c
9.1 d 7 6
8.5
b,c e 14 f 1
d,e
As we assumed all the activities are crashed to their maximum, the project’s cost is the sum all the crash costs.
Project Cost = $76.6 Million.
This is a rough estimate. For example, activity e can be un-crashed back to 6 months and the project would still have a duration of 13 months.

35. Step 6. The Tradeoff Plot

36. Step 6. The Tradeoff Plot Alternatives Generated Duration Cost 20 3619
38.5 18
42.5 17
47.5 13
76.6
Alternatives Generated