1. Project Management 2017 Dr. Mahmoud Abbas Mahmoud
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Project Management
Project Time-Cost Trade-Off
مالحظاتعن مهمةتعجيلالمشاريع
Dr. Mahmoud Abbas Mahmoud
Assistant Professor
2017
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Normal Network Crash Network
Project completion time = 13 days Project completion time = 10 days
Cost = 5800 $ Cost = 9400 $
CP is (1 – 2 – 4 – 5) CP is (1 – 2 – 4 – 5)
Project crash limit = Project completion time (with normal times) - Project
completion time (with crashed times)
Project crash limit = 13 – 10 = 3 days
Now we start compression of the project times (Normal times). From (Normal
Network), there is only one critical path (1 – 2 – 4 – 5). We will select one of the
critical activities.
Activity (4, 5) is selected because it has the smallest slope.
The time of this activity will be reduced from 3 to 2 days
The new network will be as follows
The new:
Project completion time = 12 days
Cost = 5800 + (1 × 200) = 6000 $
CP is (1 – 2 – 4 – 5)
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Now we will start new compression of the project times. We will select a new
critical activity with smallest slope. Therefore, we will select the activity (2, 4).
The time of this activity will be reduced from 2 to 1 day.
The new network will be as shown.
The new:
Project completion time = 11 days
Cost = 6000 + (1 × 400) = 6400 $
CP is (1 – 2 – 4 – 5)
Now we will start new compression of the project times. We will select a new
critical activity with smallest slope.
Therefore, we will select the activity (1, 2)
The time of this activity will be reduced from 8 to 7 days
The new network will be as shown.
Now the new:
Project completion time = 10 days
Cost = 6400 + (1 × 1000) = 7400 $
CP is (1 – 2 – 4 – 5)
We can see that the new project completion time (10 days) is equal to the (crash)
completion time. Therefore, it is no longer possible to reduce the time of the
project, and the total cost will be 7400 $.
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EXAMPLE- 2
A project consists of the following activities. The normal and crash points for each
activity are given in Table 2 below. It is required to complete the project with
minimum possible time and cost.
TABLE 2
Activity
(i, j)
Normal Crash
Time (days) Cost ($) Time (days) Cost ($)
(1, 2) 8 100 6 200
(1, 3) 4 150 2 350
(2, 4) 2 50 1 90
(2, 5) 10 100 5 400
(3, 4) 5 100 1 200
(4, 5) 3 80 1 100
Solution
Calculate the slope for each activity.
Activity
(i, j)
Normal Crash
ΔC =
Cc - Cn
ΔD =
Dn - Dc
Slope =
ΔC/ ΔD
$/day
Time
(days)
Cost
($)
Time
(days)
Cost
($)
(1, 2) 8 100 6 200 100 2 50
(1, 3) 4 150 2 350 200 2 100
(2, 4) 2 50 1 90 40 1 40
(2, 5) 10 100 5 400 300 5 60
(3, 4) 5 100 1 200 100 4 25
(4, 5) 3 80 1 100 20 2 10
∑ = 580 ∑ = 1340
Normal Network Crash Network
Project completion time = 18 days Project completion time = 11 days
Cost = 580 $ Cost = 1340 $
CP is (1 – 2 – 5) CP is (1 – 2 – 5)
Crash limit = 18 – 11 = 7 days
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Selected critical activity (1, 2) [smallest slope]. The time of this activity can be
reduced from 8 to 6 days.
Compression limit = Min. [(8 - 6), Max. FF of noncritical activities]
From the previous (Normal Network) we can calculate the free floats of the
noncritical activities as follows.
The Max. FF is of activity (4, 5) = 5
Compression limit = Min. [2, 5] = 2
Therefore, we will reduce the time of activity (1, 2) from 8 days to 6 days.
The result will be as shown
The new:
Project completion time = 16 days
Cost = 580 + 2(50) = 680 $
CP is (1 – 2 – 5)
Select a new critical activity (2, 5) [smallest slope]. The time of this activity can
be reduced from 10 to 5 days.
Compression limit = Min. [(10 - 5), Max. FF of noncritical activities]
The new Max. FF is of activity (4, 5) = 4
Compression limit = Min. [5, 4] = 4
Therefore, we will reduce the time of activity (2, 5) only from 10 days to 6 days.
The result will be as shown
The new:
Project completion time = 12 days
Cost = 680 + 4(60) = 920 $
There are two critical paths, they are:
CP1 is (1 – 2 – 5) and CP2 is (1 – 3 – 4 – 5)
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Now referring to the new network we can see that in order to make any useful
decrease to the activity (2 – 5), the activity (1 – 3) or activity (3 – 4) or activity (4
– 5) must be decreased also.
The best choice is which it has the lowest cost.
(2 – 5), (1 – 3) the cost = 60 + 100 = 160 $
(2 – 5), (3 – 4) the cost = 60 + 25 = 85 $
(2 – 5), (4 – 5) the cost = 60 + 10 = 70 $ ……. The lowest cost
Therefore, we will decrease the activities (2 – 5) and (4 – 5) together and the
cost of this step is = 70 $
The new:
Project completion time = 11 days
Cost = 920 + 70 = 990 $
Critical paths are: CP1 is (1 – 2– 5) and CP2 is (1 – 3 – 4 – 5)
We can see that the new project completion time (11 days) is equal to the (crash)
completion time. Therefore, it is no longer possible to reduce the time of the
project, and the total cost is 990 $.
EXERCISE
For the information of a project consists of the activities listed in Table 3. It is
required to complete the project with minimum possible time and cost.
TABLE 3
Activity
(i, j)
Normal Crash
Time
(days)
Cost
($)
Time
(days)
Cost
($)
(1, 2) 3 15 2 20
(2, 3) 3 15 1 35
(2, 4) 6 10 3 40
(2, 5) 15 70 13 90
(3, 4) 2 10 1 25
(4, 5) 4 10 2 30