Time-Cost Trade Off s
Time-cost trade-offs involve the balance between how long a task takes and how
much it costs. In many situations, speeding up a task can increase costs, as
more resources or efforts are needed. On the other hand, reducing costs might
slow down the task, as fewer resources are used or more time is needed to
complete it.
This concept is important in project management, where the goal is to find the
right balance to meet deadlines and stay within budget without sacrificing quality.
Time-Cost Trade Off s
Another important aspect of project management is about reducing the overall project time at an acceptable
cost. Most activities within a project can be reduced at some cost, but there are some activities that cannot be
reduced in terms of time. The process of reducing the time necessary to complete a project by adding resources
is called crashing. Crashing an activity refers to taking special costly measures to minimize the activity time in its
normal value.Some special measures to reduce time are using overtime, hiring additional temporary employees,
obtaining special equipment, adopting a new method of working, etc. Crashing the project refers to crashing a
number of activities to reduce the time of the project below its normal value. The minimum possible duration for
an activity is known as the crash duration.
y Figure 7.21
Since cost is of prime importance in deciding whether or not to take advantage of feasible
time reductions. It will greatly affect our decision what activities, to crash. Figure 7.21 shows Crash Cost Crash
the relationship of the normal and crash points. The figure shows that whenever an activity
reduced the time of completion, the cost increases, in other words at the time decreases, Normal Cost Normal
the cost increases. The normal point on the time cost graph shows the time and the cost of
activity in a normal way. The rash point shows the time and cost when the activity is fully
crashed. For us to understand fully how to apply crashing ofactivities let us consider the next x
example. Crash Normal
Time Time
The two-SUV maintenance project consists of five activities. The mechanics have extensive experience
with similar car problems. The times for the maintenance activities are shown in the table.
Activity Description Immediste Time Normal Crash duration Crash
Predecessor (days) Cost (days) Cost
A Overhaul SUV 1 -- 9 P 1,500 4 P 2,000
B Adjust SUV 1 A 5 1,000 3 1,400
C Overhaul SUV 2 -- 8 1,600 4 2,200
D Adjust SUV 2 C 4 900 2 1,150
E Test SUV B,D 3 800 1 1,300
Suppose the current project can be completed within 15 days, what activities must be crashed to
meet the target completion time?
Solution:
Step 1: Develop the project network.
A B
Figure 7.22
Start E Finish
C D
Step 2: Determine the completion time per activity.
Figure 7.23
A B
9 5
E
Start Finish
C D 3
8 4
Step 3: Established the earliest start time and time using forward pass.
Figure 7.24
A 0 9 B 9 14
9 5
E 14 17
Start Finish
3
C 0 8 D 8 12
8 4
Step 4: Develop the latest finish time and latest start time using backward pass.
Figure 7.25
0 9 9 14
14 17
2 10 10 14
Step 5: Compute the slack of all activites using the formula provided below and identify the zero slack to establish the critical path.
The formula is Slack = LS - ES = LF - EF.
Activity ES LS EF LF Slack Critical Path
A 0 0 9 9 0 Yes
B 9 9 14 14 0 Yes
C 0 2 8 10 2 No
D 8 10 12 14 2 No
E 14 14 17 17 0 Yes
Therefore, the critical path is A B E with 17 days finish the project.
Step 6: Determine each actvity that can be crashed and the crashing process costs.
Let Eᵢ = expected time for activity i.
We must identify the following information: Eᵢ' = time for activity i under maximum crashing.
1. Estimated activity cost in normal or expected time. Mᵢ = maximum reduction in time due to crashing
(Mᵢ = Eᵢ - Eᵢ')
2. Estimated time to complete the activity under maximum crashing.
3. Estimated activity cost under maximum crashing.
Let Cᵢ = estimated cost for activity i in normal time.
Cᵢ' = estimated cost for activity i under maximum crashing.
Kᵢ = crashing cost for activity
(Kᵢ = (Cᵢ' - Cᵢ) / Mᵢ)
Let Eᵢ = expected time for activity i.
Eᵢ' = time for activity i under maximum crashing.
Mᵢ = maximum reduction in time due to crashing
(Mᵢ = Eᵢ - Eᵢ')
Let Cᵢ = estimated cost for activity i in normal time.
Cᵢ' = estimated cost for activity i under maximum crashing.
Kᵢ = crashing cost for activity
Time (days) Total Cost Maximum Reduction Cash Cost
Activity
Normal Crash Normal (Cᵢ) Crash (Cᵢ') in Time (Mᵢ) per Day (Kᵢ)
A 9 4 P 1,500 P 2,000 5 P 100
B 5 3 P 1,000 P 1,400 2 P 200
C 8 4 P 1,600 P 2,200 4 P 150
D 4 2 P 900 P 1,150 2 P 125
E 3 1 P 800 P 1,300 2 P 250
P 5,800 P 8,050
Recall that the critical path is A → B → E, these are activities that we can crash to minimize the time to 15 days. We
need to select the lowest crash cost of P100 for activity A. Selecting activity A would lead to additional costs of P100
per day, thus the new cost for the project would be: New Cost = 5,800 + 100(2) = 5,800 + 200 = ₽6,000.
