Need all Otherwise negative Review don't touch if not able to solve all

2011 - Nov [6] (a) The number of days of total float (TF), earliest start times (EST) and duration in days are given for some of the following activities. Activity TF EST Duration 1-2 0 0 1-3 0 1-4 5 2-4 0 4 2-5 1 5 3-6 2 12 4-6 0 12 5-7 3 6 - 7 23 6 - 8 2 7-8 0 23 8-9 30 6 (i) Draw the network. (ii) List the paths with their corresponding durations and state when the project can be completed. (10 marks) (CAFG - II NN 2012 - May [5] (b) The following network and table are presented to you: W7 Y V TA Activity Normal Normal Crash Crash Duration Cost (5) Duration Cost (5) (Days) (Days) T 8 2,250 6 2,750 U 16 1,875 11 2,750 V 14 2,250 9 3,000 W 12 3,000 9 3,750 15 1,000 14 2,500 Y 10 2,500 8 2,860 Perform step by step crashing and reduce the project duration by 11 days while minimizing the crashing cost. What would be the cost of the crashing exercise? (8 marks) [CAFG - II] 2012 - May [7] Answer the following questions : (a) The following is a part of a network. P. S Q R What are activities P and Q called? How would you rectify the situation? (4 marks) [CAFG - II] 2015 - Dec [10] (b) The activities involved in a PERT project are detailed in the following table: Job (1 - i) DURATION TIME (DAYS) most optimistic time most likely time most pessimistic time 1-2 3 6 15 2-3 6 12 30 3-5 5 11 17 7-8 4 19 28 5-8 1 4 7 6-7 3 9 27 15 3 6 4-5 1-6 2-4 5 NIN 14 8 5 (0) Draw a network diagram. (ii) Find the critical path, expected time for completion of the project and project variance after estimating the earliest and latest event times for all nodes. (iii) Find the probability of completing the project before 31 days. (iv) What is the chance of project duration exceeding to 46 days? (v) What will be the effect on the current critical path if the most likely time of activity 3 - 5 gets revised to 14 instead of 11 days (given in the table)? [Given that the probability of standard normal curve between): (a) Z = 0 and Z = 2 is 0.4772, and (b) Z= 0 and Z = -0.1 is 0.3413 (11 marks)