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Answer 1

1)

Step 1 Draw the project network diagram B >E —> F A G >H L C->D J -> Step 2 For each activity, find the expected time TE = (a + 4m + b)/6, and Variance o? = [(b-a)/6)? Find the Earliest start and finish and Latest start and finish times and Slack for each activity. Earliest Start is calculated by forward pass method as latest of the predecessors Earliest Finish. Latest Finish is calculated using reverse pass method as earliest of the successors Latest Start. Then Earliest Finish is calculated by the formula, EF = Earliest Start + Activity Duration And, Latest Start is calculated by the formula, LS = Latest Finish - Activity Duration Slack = Latest Finish - Earliest Finish, or Latest Start - Earliest Start Activity Predecesso ID a m b ES LS EF LF Slack 30 0 30 0 A B 20 20 0 60 30 120 A 65 30 с А 30 95 50 30 40 80 100 100 35 0.1 30 30 95 90 60 50 30 0.1 D с 0 0 30 30 25 Expected variance (TE) (6) 30 11.1 60 100.0 65 69.4 55 136.1 30 2.8 0 0.0 30 2.8 20 11.1 30 44.4 10 0.4 0 0.0 30 44.4 E F B E Success Critical or Activity B,C Critical E D Critical G Critical F G H Critical 1, Critical L Critical 0.1 25 G D,F 30 35 90 95 150 120 120 180 200 230 210 210 260 0 H 10 30 95 120 150 150 180 200 220 230 230 0 150 150 150 180 200 230 230 230 260 - H 120 150 180 200 200 210 230 20 60 0 .ااا 20 25 10 0.1 25 H 8 J K K 12 0.1 60 20 20 0.1 20 L L IK 0 Critical Step 3 Find the critical path. Critical path is that path in the network diagram, which has maximum length (duration) as per expected time (u) The expected duration of critical path is the sum of expected duration of each activity on critical path. Similarly, variance of critical path is the sum of individual variance of each activity on critical path. Critical path, Expected duration, variance and standard deviation of Critical path Critical Path A-C-D-G-H-I-L Expected Completion Time (TE) = 260 days Sum of Variances along CP, 02 = 319.44 days Standard Deviation = vv = 0 17.87 days

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2)

For target time of 265 days, z value = (265-260)/17.87 = 0.28

P(z) = NORMSDIST(0.28) = 0.61

Probability of completing the project in less than 265 days = 0.61 or 61 %

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3)

There are four paths in the network.

These paths and their respective duration are listed below:

Path	Duration
A-B-E-F-G-H-I-L	230.1
A-B-E-F-G-H-J-K-L	210.2
A-C-D-G-H-I-L	260
A-C-D-G-H-J-K-L	240.1

Activity A is on critical path and has the lowest crash cost/day.

Crash time of activity A = 30-20 = 10 days, So, activity A is crashed by 10 days at a cost of $ 1500 per day

Activity D on critical path has the next lowest crash cost/day. Crash time of activity D = 55-30 = 25 days. So, activity D is crashed by 25 days at a cost of $ 1900 per day

Now, the project duration has been shortened by 10+25 = 35 days

Activity H or I has the next lowest crash cost/day. So, either of these activities is crashed by 5 days.

Let's say activity H is crashed by 5 days at a cost of $ 2000 per day

Summary: Activity A, D and H needs to be crashed by 10, 25 and 5 days respectively.

Project duration after crashing = 220 days

Total crash cost = 1500*10+1900*25+2000*5 = $ 72,500

Total savings in penalty = 10000*40 = $ 400,000