Time-Cost Trade-Off
(Time Reduction = Time Compression)
(Time Shortening)
Reasons to Reduce Project Durations:
To avoid late penalties (or avoid damaging the company’s relationship)
To realize incentive pay (monetary incentives for timely or early competition
of a project
To beat the competition to the market to fit within the contractually required
time (influences Bid Price)
To free resources for use on other projects (complete a project early & move
on to another project)
To reduce the indirect costs associated with running the project
To complete a project when weather conditions make it less expensive (Avoid
temporary heating, avoid completing site work during raining season)
Many things make crashing a way of life on some projects (i.e. last minutes changes
in client specification, without permission to extend the project deadline by an
appropriate increment)
Shortening the duration is called project crashing
Methods to reduce durations
1.
Overtime: Have the existing crew work overtime. This increase
the labor costs due to increase pay rate and decrease productivity.
2.
Hiring and/or Subcontracting:
a) Bring in additional workers to enlarge crew size. This increases
labor costs due to overcrowding and poor learning curve.
b) Add subcontracted labor to the activity. This almost always
increases the cost of an activity unless the subcontracted labor
is far more efficient.
3.
Use of advanced technology: Use better/more advanced
equipment. This will usually increase costs due to rental and
transport fees. If labor costs (per unit) are reduced, this could
reduce costs.
How project cost is estimated?
Practices to Estimate the Project Cost
 Some company assigns Overhead office as % of direct cost.
 Most companies don’t consider profit as a cost of the job.
 Cost analysis are completed by using:
o [Direct costs + Indirect costs + Company overhead]
vs. Budgeted (estimated costs)
 Similar to each activity, the project as a whole has an ideal
(least expensive) duration.
How project cost is estimated?
How project is crashed?
Specify 2 activity times and 2 costs
o 1st time/cost combination (D, CD)- called Normal [Normal – usual ‘average’ time, resources]
o 2nd time/cost combination (d, Cd)- called Crash [Expedite by applying additional resources]
General 2 Steps in Project Crashing
• In order to speed up any project schedule with a minimum
cost, you need to follow these two general steps:
1. Select the Critical Path Activities.
2. Within the critical path activities, select the lowest
reduction cost rate “Cost Slope” to reduce it.
• In following slides more details on these 2 General Steps
Steps in Project Crashing
Basic Steps
1. Compute the crash cost per time period. If crash costs are linear over
time:
𝑪𝒓𝒂𝒔𝒉 𝑪𝒐𝒔𝒕 𝒑𝒆𝒓 𝒑𝒆𝒓𝒊𝒐𝒅 = 𝑪𝒐𝒔𝒕 𝑺𝒍𝒐𝒑𝒆
=
2.
(𝑪𝒓𝒂𝒔𝒉 𝒄𝒐𝒔𝒕−𝑵𝒐𝒓𝒎𝒂𝒍 𝒄𝒐𝒔𝒕)
(𝑵𝒐𝒓𝒎𝒂𝒍 𝒕𝒊𝒎𝒆−𝑪𝒓𝒂𝒔𝒉 𝒕𝒊𝒎𝒆)
Using current activity times, find the critical path and identify the
critical activities
Steps in Project Crashing
3. If there is only one critical path, then select the activity on this
critical path that (a) can still be crashed, and (b) has the smallest
crash cost per period.
If there is more than one critical path, then select one activity from
each critical path such that (a) each selected activity can still be
crashed, and (b) the total crash cost of all selected activities is the
smallest.
Note that the same activity may be common to more than one
critical path.
4. Update all activity times. If the desired due date has been reached,
stop. If not, return to Step 2.
Example1: Crashing The Project
Critical Path for Milwaukee Paper (A, C, E, G, H)
0 2
2
2 2 4
4 3 7
C
F
2 0 4
10 6 13
A
0
0 0
0 2
0
Start
0 0
0
13 2 15
4 4 8
H
E
13 0 15
4 0 8
0 3
3
B
1 1
4
3
4
7
8 5 13
D
G
4 1 8
8 0 13
Example1: Crashing The Project
CALCULATION OF CRAH COST/PERIOD
Crash and Normal Times and Costs for Activity B
Activity
Cost
Crash
$34,000 —
Crash Cost – Normal Cost
Normal Time – Crash Time
Crash Cost/Wk =
$33,000 —
Crash
Cost, Cd
=
$32,000 —
=
$31,000 —
$30,000 —
Normal
Cost, CD
$34,000 – $30,000
3–1
$4,000
= $2,000/Wk
2 Wks
Normal
—
|
1
Crash Time, d
|
2
|
3
Normal Time, D
Time (Weeks)
Table 3.16 (Frome Heizer/Render; Operation Management)
Example1: Crashing The Project
Time (Wks)
Activity Normal
A
B
C
D
E
F
G
H
2
3
2
4
4
3
5
2
Crash
1
1
1
2
2
2
2
1
Cost ($)
Normal
22,000
30,000
26,000
48,000
56,000
30,000
80,000
16,000
Crash
22,750
34,000
27,000
49,000
58,000
30,500
84,500
19,000
Crash Cost Critical
Per Wk ($) Path?
750
2,000
1,000
1,000
1,000
500
1,500
3,000
Yes
No
Yes
No
Yes
No
Yes
Yes
Table 3.5 (Frome Heizer/Render; Operation Management)
Example2: Crashing Project
For the small project shown in the table, it is required to reduce the project
duration.
A) Reduce by 2 periods.
B) Reduce by 5 periods.
Normal
Activity Precedence
Crash
Time, day
Cost, $
Time, day
Cost, $
A
-
4
210
3
280
B
-
8
400
6
560
C
A
6
500
4
600
D
A
9
540
7
600
E
B,C
4
500
1
1100
F
C
5
150
4
240
G
E
3
150
3
150
H
D,F
7
600
6
750
Σ
3050
4280
Example2: Crashing Project
Normal
Activity
1- develop network
Crash
Precedence
Time, day
Cost, $
Time, day
Cost, $
A
-
4
210
3
280
B
-
8
400
6
560
C
A
6
500
4
600
D
A
9
540
7
600
E
B,C
4
500
1
1100
F
C
5
150
4
240
G
E
3
150
3
150
H
D,F
7
600
6
750
Σ
3050
4280
9
D
4
6
5
7
A
C
F
H
FINISH
START
8
4
3
B
E
G
Example2: Crashing Project
2- calculate times, find critical path
9
4
13
D
6
0
4
4
4
A
0
0
0
0
6
2
15
10
5
10
C
4
4
0
15
15
F
10
10
0
7
22
H
15
15
0
22
0
START
0 0 0
22
0
22
FINISH
0
8
8
10
B
7
7
4
14
14
E
15
 Project completion time = 22 working days
 Critical Path: A, C, F, H.
15
5
3
17
22
0
22
G
19
Activity
Total float
Free float
19
A
0
0
5
22
B C D
7 0 2
2 0 2
E F G H
5 0 5 0
0 0 5 0
Example2: Crashing Project
3- calculate cost slope
Activity Precedence
Time, day
Cost, $
Time, day
Cost, $
Cost Slope,
$/day
Normal
Crash
A
-
4
210
3
280
70
B
-
8
400
6
560
80
C
A
6
500
4
600
50
D
A
9
540
7
600
30
E
B,C
4
500
1
1100
200
F
C
5
150
4
240
90
G
E
3
150
3
150
**
H
D,F
7
600
6
750
150
Σ
3050
4280
Remarks
1- [G can not expedite]
2- lowest slope and can be expedited on critical path is activity (C) WITH 2 periods
Example2: Crashing Project
Normal
Activity
Reduce 2 periods of activity (C) with
increase of cost (2*50) =100
ES
LS
Activity
EF
TF
LF
9
4
Crash limit (d @ cost)
4
4
4
A
0
0
0
0
4
0
8
Cost, $
Time, day
Cost, $
$/day
A
-
4
210
3
280
70
B
-
8
400
6
560
80
C
A
6
500
4
600
50
D
A
9
540
7
600
30
E
B,C
4
500
1
1100
200
F
C
5
150
4
240
90
G
E
3
150
3
150
**
H
D,F
7
600
6
750
150
Σ
4
0
8
5
13
13
F
8
3050
8
0
7
20
H
13
13
0
20
0
START
0 0 0
4280
13
C
4
Cost Slope,
Time, day
D
4
0
13
Crash
Precedence
20
0
20
FINISH
0
8
8
8
B
5
5
4
12
12
E
13
13
 Project completion time = 20 working days
 Critical Path: A, C, F, H. & A, D, H
5
3
15
20
0
20
G
17
17
5
20
Activity
Total float
Free float
A
0
0
B C D
5 0 0
0 0 0
E F G H
5 0 5 0
0 0 5 0
Example2: Crashing Project
Normal
Activity
Reduce 1 period of activity (A) with
increase of cost =70
9
3
12
0
3
3
3
A
0
0
0
0
4
0
7
Cost, $
Time, day
Cost, $
$/day
A
-
4
210
3
280
70
B
-
8
400
6
560
80
C
A
6
500
4
600
50
D
A
9
540
7
600
30
E
B,C
4
500
1
1100
200
F
C
5
150
4
240
90
G
E
3
150
3
150
**
H
D,F
7
600
6
750
150
Σ
3
0
7
5
12
12
F
7
3050
7
0
7
19
H
12
12
0
19
0
START
0 0 0
4280
12
C
3
Cost Slope,
Time, day
D
3
Crash
Precedence
19
0
19
FINISH
0
8
8
7
B
8
4
4
11
11
E
12
12
 Project completion time = 19 working days
 Critical Path: A, C, F, H. & A, D, H
5
3
14
19
0
19
G
16
16
5
19
Activity
Total float
Free float
A
0
0
B C D
4 0 0
0 0 0
E F G H
5 0 5 0
0 0 5 0
Example2: Crashing Project
Normal
Activity
Reduce farther 1 period of 2 activities
(D,F) with increase of cost (30+90) =120
3
8
11
0
3
3
3
A
0
0
0
0
4
0
Cost, $
Time, day
Cost, $
$/day
A
-
4
210
3
280
70
B
-
8
400
6
560
80
C
A
6
500
4
600
50
D
A
9
540
7
600
30
E
B,C
4
500
1
1100
200
F
C
5
150
4
240
90
G
E
3
150
3
150
**
H
D,F
7
600
6
750
150
Σ
3
0
3050
7
7
4
11
11
F
7
7
0
7
18
H
11
11
0
18
0
START
0 0 0
4280
11
C
3
Cost Slope,
Time, day
D
3
Crash
Precedence
18
0
18
FINISH
0
8
8
7
B
3
3
4
11
11
E
11
11
 Project completion time = 19 working days
 Critical Path: A, C, F, H. & A, D, H
4
3
14
18
0
18
G
15
15
4
18
Activity
Total float
Free float
A
0
0
B C D
3 0 0
0 0 0
E F G H
4 0 4 0
0 0 4 0
Example2: Crashing Project
Normal
Activity
Reduce farther 1 period of activity (H)
with increase of cost =150
3
8
11
0
3
3
3
A
0
0
0
0
4
0
Cost, $
Time, day
Cost, $
$/day
A
-
4
210
3
280
70
B
-
8
400
6
560
80
C
A
6
500
4
600
50
D
A
9
540
7
600
30
E
B,C
4
500
1
1100
200
F
C
5
150
4
240
90
G
E
3
150
3
150
**
H
D,F
7
600
6
750
150
Σ
3
0
3050
7
7
4
11
11
F
7
7
0
6
17
H
11
11
0
17
0
START
0 0 0
4280
11
C
3
Cost Slope,
Time, day
D
3
Crash
Precedence
17
0
17
FINISH
0
8
8
7
B
2
2
4
11
11
E
10
10
 Project completion time = 19 working days
 Critical Path: A, C, F, H. & A, D, H
3
3
14
17
0
17
G
14
14
3
17
Activity
Total float
Free float
A
0
0
B C D
2 0 0
0 0 0
E F G H
3 0 3 0
0 0 3 0
Example2: Crashing Project
Normal
Reduction from 22 to 17 increase to 3490
Cycle
Activity
0
-
1
C
Time
cost
2
Activity
Total cost
Duration
3050
22
3150
20
100
2
A
1
70
3220
19
3
D,F
1
30+90
3340
18
4
H
1
150
3490
17
Crash
Time, day
Cost, $
Time, day
Cost, $
$/day
A
-
4
210
3
280
70
B
-
8
400
6
560
80
C
A
6
500
4
600
50
D
A
9
540
7
600
30
E
B,C
4
500
1
1100
200
F
C
5
150
4
240
90
G
E
3
150
3
150
**
H
D,F
7
600
6
750
150
Σ
3050
3600
3500
Cost
3400
3300
3200
3100
3000
16
17
18
19
20
Project Duration
Cost Slope,
Precedence
21
22
23
4280
Example2: Crashing Project
Accelerating the Critical and Noncritical path
Normal
Activity
Crash
Cost
Slope,
Precedence
Time,
day
Cost,
$
Time,
day
Cost,
$
$/day
A
-
4
210
3
280
70
B
-
8
400
6
560
80
C
A
6
500
4
600
50
D
A
9
540
7
600
30
E
B,C
4
500
1
1100
200
F
C
5
150
4
240
90
G
E
3
150
3
150
**
H
D,F
7
600
6
750
150
Σ
3050
4280
How project is crashed?
Network Compression Algorithm
 Determine Normal project duration, and cost.
 Identify Normal duration Critical Path.
 For large network, using CRITICALITY THEOREM to
eliminate the noncritical paths that do not need to be crashed.
 Eliminate Activities with having “TF > the required project
reduction time”.
 Compute the cost slop
𝑪𝒓𝒂𝒔𝒉 𝑪𝒐𝒔𝒕 𝒑𝒆𝒓 𝒑𝒆𝒓𝒊𝒐𝒅 = 𝑪𝒐𝒔𝒕 𝑺𝒍𝒐𝒑𝒆
=
(𝑪𝒓𝒂𝒔𝒉 𝒄𝒐𝒔𝒕−𝑵𝒐𝒓𝒎𝒂𝒍 𝒄𝒐𝒔𝒕)
(𝑵𝒐𝒓𝒎𝒂𝒍 𝒕𝒊𝒎𝒆−𝑪𝒓𝒂𝒔𝒉 𝒕𝒊𝒎𝒆)
How project is crashed?
Network Compression Algorithm
 Shortening the CRITICAL ACTIVITIES beginning with the
activity having the lowest cost-slope
 Determine the compression limit (Nil)
 - Crash Limit, or
Nil = Min  - Free Float of any of the non critical activities
in the parallel paths competing for critical path.*
 Organize the data as in the following table:
How project is crashed?
Network Compression Algorithm
 Update the project network
 When a new Critical Path is formed:
- Shorten the combination of activity which Falls on Both
Critical Paths, OR
- Shorten one activity from each of the critical paths. Use
the combined cost of shortening both activities when
determining if it is cost-effective to shorten the project.
 At each shortening cycle, compute the new project duration
and project cost
 Continue until no further shortening is possible
How project is crashed?
Network Compression Algorithm
 Tabulate and Plot the Indirect project Cost on the same timecost graph
 Add direct and indirect cost to find the project cost at each
duration.
 Use the total project cost-time curve to find the optimum time.
Example 3
38470
Example 3
20 23 43
D
24 4 47
0
0 12 12
A
0 0 12
2@100
Cycle #
0
1
We have one critical path, A-C-G- I.
Either crash A
at cost SR 100/week or
crash C
at cost SR 200/week or
crash G
at cost SR 60/week or
crash I
at cost SR 75/week.
4
12 8 20
B
14 2 22
2@150
20 5 25
E
22 2 27
1@50
27 20 47
G
27 0 47
5@60
12 15 27
C
12 0 27
3@200
27 5 32
F
29 2 34
1@300
32 13 45
H
34 2 47
2@40
Activity to
Can Be
Shorten
Shortened


G
5
Nil

2
Weeks Cost per
Shortened Week


2
60
47 12 59
I
47 0 59
2@75
2
ES D EF
Activity
LS TF LF
Crash limit
Cost for
Project
Total Cost
Cycle
Duration
59

36,500
120
57
36,620
-
Example 3
20 23 43
D
22 2 45
0
0 12 12
A
0 0 12
2@100
Cycle #
0
1
2
Now we have two critical paths:
A-C-F-H-I and A-C-G- I.
Either crash A
at cost SR 100/week or
crash C
at cost SR 200/week or
crash I
at cost SR 75/week or
crash F and G
at cost SR 360/week or
crash H and G
at cost SR 100/week.
2
12 8 20
B
14 2 22
2@150
20 5 25
E
22 2 27
1@50
27 18 45
G
27 0 47
3@60
12 15 27
C
12 0 27
3@200
27 5 32
F
27 0 32
1@300
32 13 45
H
32 0 45
2@40
Activity to
Can Be
Shorten
Shortened


G
5
I
2
Nil

2

Weeks Cost per
Shortened Week


2
60
2
75
45 12 57
I
45 0 55
2@75
ES D EF
Activity
LS TF LF
Crash limit
Cost for
Project
Total Cost
Cycle
Duration
36,500
59

120
36,620
57
150
36,770
55
-
Example 3
20 23 43
D
22 2 45
0
0 12 12
A
0 0 12
2@100
Cycle #
0
1
2
3
Now we have two critical paths:
A-C-F-H-I and A-C-G- I.
Either crash A
at cost SR 100/week or
crash C
at cost SR 200/week or
crash F and G
at cost SR 360/week or
crash H and G
at cost SR 100/week.
2
12 8 20
B
14 2 22
2@150
20 5 25
E
22 2 27
1@50
27 18 45
G
27 0 45
3@60
12 15 27
C
12 0 27
3@200
27 5 32
F
27 0 32
1@300
32 13 45
H
32 0 45
2@40
Activity to
Can Be
Shorten Shortened


G
5
I
2
A
2
Nil

2


Weeks Cost per
Shortened Week


2
60
2
75
2
100
45 10 55
I
45 0 55
0
ES D EF
Activity
LS TF LF
Crash limit
Cost for
Project
Total Cost
Cycle
Duration
59
36,500

120
57
36,620
150
55
36,770
200
53
36,970
-
Example 3
18 23 41
D
20 2 43
0
0 10 10
A
0 0 10
0
Cycle #
0
1
2
3
4
Activity to
Shorten

G
I
A
H, G
Now we have two critical paths:
A-C-F-H-I and A-C-G- I.
Either crash C
at cost SR 200/week or
crash F and G
at cost SR 360/week or
crash H and G
at cost SR 100/week.
2
10 8 18
B
12 2 20
2@150
18 5 23
E
20 2 25
1@50
25 18 43
G
25 0 43
3@60
10 15 25
C
10 0 25
3@200
25 5 30
F
25 0 30
1@300
30 13 43
H
30 0 43
2@40
Can Be
Shortened

5
2
2
2
Nil

2


2
Weeks Cost per
Shortened Week


2
60
2
75
2
100
2
60+40
43 10 53
I
43 0 53
0
ES D EF
Activity
LS TF LF
Crash limit
Cost for
Project
Total Cost
Cycle
Duration
36,500
59

120
36,620
57
150
36,770
55
200
36,970
53
200
37,170
51
Example 3
Now we have three critical paths;
A-C-F-H-I, A-C-G- I, and A-B-D-I.
Either crash C and B
at cost SR 350/week or
crash F, G and B
at cost SR 510/week.
18 23 41
D
18 0 41
0
0 10 10
A
0 0 10
0
18 5 23
E
20 2 25
1@50
25 16 41
G
25 0 41
1@60
25 5 30
F
25 0 30
1@300
30 11 41
H
30 0 41
0
10 8 18
B
10 0 18
2@150
-
0
10 15 25
C
10 0 25
3@200
Cycle #
0
1
2
3
4
5
Activity to
Can Be
Shorten Shortened


G
5
I
2
A
2
G, H
2
B, C
2
Nil

2


2

Weeks
Cost per
Shortened
Week


2
60
2
75
2
100
2
60+40
2
150+200
Cost for
Cycle

120
150
200
200
700
41 10 51
I
41 0 51
0
ES D EF
Activity
LS TF LF
Crash limit
Project
Total Cost
Duration
59
36,500
57
36,620
55
36,770
53
36,970
51
37,170
49
37,870
Example 3 – Final Results
16 23 39
D
16 0 39
0
0 10 10
A
0 0 10
0
10 6 16
B
10 0 16
0
16 5 21
E
18 2 23
1@50
23 16 39
G
23 0 39
1@60
10 13 23
C
10 0 23
1@200
23 5 28
F
23 0 28
1@300
28 11 39
H
28 0 39
0
Cycle #
0
1
2
3
4
5
Project Direct Cost
Duration
59
36500
57
36620
55
36770
53
36970
51
37170
49
37870
39 10 49
I
39 0 49
0
ES D EF
Activity
LS TF LF
Crash
limit
Indirect
Cost
Total Cost
7375
7125
6875
6625
6375
6125
43875
43745
43645
43595
43545
43995
Example 3 – Final Results
Project Optimal Duration
50000
45000
40000
Cost (SR)
35000
Direct
Cost
Indirect
cost
Total
Cost
30000
25000
20000
15000
10000
5000
0
48
49
50
51
52
53
54
55
56
Project Duration (Week)
57
58
59
60
CASE WORK
Data on small maintenance project is given as below: On completion, the project
will give a return of SR110/day. Using time-cost trade-off method, how much
would you like to compress the project for maximizing the return (ignore the
Indirect cost effect)? Show all calculations.
Activity
A
B
C
D
E
F
G
H
I
Depends
on



A
B
B
C
D, E
F, G
Normal
Time
Cost
6 days
SR700
4 days
400
5 days
650
8 days
625
10 days
200
7 days
500
3 days
600
6 days
300
7 days
350
Crash
Time
4 days
4 days
4 days
5 days
7 days
5 days
3 days
5 days
4 days
Cost
SR800
400
700
700
350
700
600
400
425