2 Project Network Design (4 pts A project leam has started gathering information
ID: 354018 • Letter: 2
Question
2 Project Network Design (4 pts A project leam has started gathering information nocessary to develop a project network-predecessor activities. The results i. Develop a network (by handi(1 pts n. Find the slack time on activity if there any in the notwork (I pts) iii. Find the critical path( ii. Find the project duration (Ipt) 3. If a peojoct is expected to complote in 34 days with 6 days variation what is the probability the project will complcte in 40 days? (2 pts.) 4. In a project, the probability of a occurrence of a high-priority risk (P) is 70%. "The direct and indirect costs (C) involved in the project should the risk actually occurs are S13,000 and $7,000, respectively. What is the overall risk exposure (RE)? (1 pts) Hints: C direct +indirect cost Overall risk exposure formula is RE-PxC 5. Using these project selection criteria, please rank and select the best project and describe briefly the rationale for the selection. (2 pts) Align with Total Rank Cost-benefit Difficulty Ability organization's score goal Weight 045 .25 0.2 Project A 5 Project B 7 Project C 0.10 6.5Explanation / Answer
PLEASE FIND BELOW ANSWER TO FIRST QUESTION :
The precedence diagram of activities as follows :
1
2
3
4
5
6
7
8
9
10
11
12
13
The duration figures against each activities are not fully visible. However as understood , following durations have been assumed for each activity
Activity
Duration
1
0
2
6
3
3
4
5
5
3
6
2
7
4
8
2
9
3
10
1
11
5
12
2
13
2
The parallel paths and their corresponding durations as follows :
1-3-5-6-7-8-9-11-12-13 = 26
1-3-5-6-7-8-10-11-12-13 = 24
2-3-5-6-7-8-9-11-12-13 =32
2-3-5-6-7-8-10-11-12-13 = 30
2-4-5-6-7-8-9-11-12-13 = 34
2-4-5-6-7-8-10-11-12-13 = 32
Since , 2-4-5-6-7-8-9-11-12-13 has the longest duration it forms the critical path.
Duration of the project is decided by the duration of the critical path . Therefore duration of the project is 34 .
It is to be noted that all activities on the critical path have ZERO slack .
Therefore activities 2, 4 , 5 , 6 , 7 , 8, 9 , 11, 12 , 13 have ZERO slack
Slack of activity 10 = Duration of 2-4-5-6-7-8-9-11-12-13 – Duration of 2-4-5-6-7-8-10-11-12-13 = 34 – 32 = 2
Slack of activity 3 = Duration of 2-4-5-6-7-8-9-11-12-13 – Duration of 2-3-5-6-7- 8-9-11-12-13 = 34 – 32 = 2
Slack of activity 1 = Duration of 2-4-5-6-7-8-9-11-12-13 – Duration 2-3-5-6-7-8-9—11-12-13 = 34 – 26 = 8
1
2
3
4
5
6
7
8
9
10
11
12
13
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