Which of the following are the guidelines for the construction of a network diagram?
1. Each activity is represented by only one arrow in the network.
2. Two activities can be identified by the same beginning and end events.
3. Dangling must be avoided in a network diagram.
4. Dummy activity consumes no time or resource.
Select the correct answer using the codes given below
Match List-I (Term) with List-II (Characteristics) and select the correct answer using the code given below the lists:
Which one of the following statements is not correct?
(A) PERT → Event oriented CPM → Activity oriented
(B) PERT → 3 time estimates are made
to = optimistic time
tm = most likely time
tP = pessimistic time
CPM → only one time estimate
(C) In PERT slack is calculated, CPM floats calculated
(D) Both PERT and CPM are used for project management
If the earliest starting time for an activity is 10 weeks, the latest finish time is 30 weeks and the duration time of the activity is 11 weeks, then the total float is equal to
The earliest occurrence time for event '1' is 8 weeks and the latest occurrence time for event' 1' is 26 weeks. The earliest occurrence time for event '2' is 32 weeks and the latest occurrence time for event '2' is 37 weeks.
If the activity time is 11 weeks, then the total float will be
Consider an activity having a duration time of Tij. E is the earliest occurrence time and L the latest occurrence time (see figure given).
Consider the following statements in this regard:
1. Total float = Lj − Ei − Tij
2. Free float = Ej − Ei − Tij
3. Slack of the tail event = Lj − Ei
Of these statements
In a small engineering project, for an activity, the optimistic time is 2 minutes, the most likely time is 5 minutes and the pessimistic time is 8 minutes. What is the expected time of the activity?
Consider the following project network
The number of merge events are
For the network shown in the given figure, the expected completion time of the project is
Paths are: 1-2-6-8 = 18
1-2-3-4-5-6-8 = 30
1-2-3-4-5-8 = 27
1-2-3-4-7-8 = 26
1-3-4-5-6-8 = 25
1-3-4-5-8 = 22
1-3-4-7-8 = 21
1-4-5-6-8 = 24
1-4-5-8 = 21
1-4-7-8 = 20
The total float associated with the activity shown below is
= 60 − (20 + 24)
Consider the network. Activity times are given in number of days. The earliest expected occurrence time (TE) for event 50 is
Critical path is given by 10 − 20 − 30 − 40 − 50
∴The earliest expected occurrence time (TE) for the event is 25
The variance (V1) for critical path a → b = 4 time units, b → c = 16 time units, c → d = 4 time units, d → e = 1 time unit. The standard deviation ‘d’ of the critical path a → e is
Consider an activity ‘A’ which is having the following values:
NC = Rs 100
NT = 3 days
CC = Rs 150
CT = 2 days
Find the cost of crashing activity A’ by single day
A project consist of following activities
Minimum time required to complete the project in (days) is __________
A‐ D‐ G = 4 + 5 + 8 = 17
A‐C‐E‐G = 4 + 6 + 5 + 8 = 23
B‐ E‐ G = 6 + 5 + 8 = 19
B‐ F = 6 + 6 = 12
A‐ C‐ F = 4 + 6 + 6 = 16
For the given project, to draw a network diagram how many dummy activities are required?
No. of dummy activities required = 2
In the network shown below. The critical path is along
1-2-3-4-8-9 = 21
1-2-3-4-7-8-9 = 22
1-2-3-5-6-7-8-9 = 24
1-2-5-6-7-8-9 = 22
Consider the network shown in figure. The three time estimates for activities are given along the arrows. Then the standard deviation of the project is
Hence critical path is 1-2-3-4-5-6 Standard deviation is σ
Consider the following project find the critical path duration.
A‐ C = 9
A‐ D = 7
B‐ D = 8
The expected project completion time is found to be 34 days. The standard deviation of the critical path is 5 days. Find the probability of completing the project in 39 days.
P = 0.84
For the project network shown below calculate latest finish time of node 4.