# Please assist in detailed explanation on the following question in Project Scheduling:1) - A company decides to plan a project, with activities, precedence and deterministic durations given in the following table:Activity iDuration(days)Immediate PredecessorsA32noneB21noneC30noneD45AE26A, BF28CG20E, FQuestions:1.1) What method should be used? Using only one pseudo-activity, draw the project network, with arcs labeled by the activity name and duration.1.2) Compute and display in a table the earliest and latest starting and finishing times, as well as slack, for each activity, and identify the critical path(s).1.3) Draw a Gantt chart of the schedule, based on earliest starting times1.4) Which activities must be completed by day 60 to ensure that the project is not delayed?

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Please assist in detailed explanation on the following question in Project Scheduling:

1) - A company decides to plan a project, with activities, precedence and deterministic durations given in the following table:

 Activity i Duration (days) Immediate Predecessors A 32 none B 21 none C 30 none D 45 A E 26 A, B F 28 C G 20 E, F

Questions:

1.1) What method should be used? Using only one pseudo-activity, draw the project network, with arcs labeled by the activity name and duration.

1.2) Compute and display in a table the earliest and latest starting and finishing times, as well as slack, for each activity, and identify the critical path(s).

1.3) Draw a Gantt chart of the schedule, based on earliest starting times

1.4) Which activities must be completed by day 60 to ensure that the project is not delayed?

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Step 1

Since it has been mentioned that the arcs need to be labelled with activity name and duration it is implied we use AOA (Activity-on-Arc) convention. Given the activity and predecessor relationship we develop a AOA representation using a dummy activity to capture the dependency of both the activities D and E on activity A. The dummy activity is represented by dotted lines.

Step 2

Now using a forward pass to calculate the ES and EF times for the different activities between the nodes as given below we get:

1-2: ES = 0 EF = 32

1-3: ES = 0 EF = 21

1-4: ES = 0 EF = 0

2-3: ES = 32 EF = 32 (dummy activity)

3-5: ES = max (EF2-3, EF1-3) = max (32,21) = 32 EF = 32 + 26 = 58

4-5: ES = EF1-4 = 30 EF = 30 + 28 = 58

5-6: ES = max (EF3-5,EF4-5) = max(58.58) = 58 EF = 58+20 = 78

2-6: ES = EF1-2 = 32 ES = 32 + 45 = 77

At node 6 the ES = max(EF2-6,EF5-6) = max (77,78)=78 and since this is the ending node the values of LS and ES are also the same.

Now using a reverse pass to calculate the LS and ES we get:

2-6: LF = LS6 = 78 LS = 78-45 = 33

5-6: LF = LS6 = 78 LS = 78-20 = 58

4-5: LF = LS5-6 = 58 LS = 58 -28 = 30

3-5: LF = LS5-6 = 58 LS = 58 – 26 = 32

2-3: LF= LS3-5 = 32 LS = 32 (dummy activity)

1-4: LF = LS4-5 = 30 LS = 30-30 =0

1-3: LF = LS3-5 = 32 LS = 32 – 21 =11

1-2: LF = min(LS2-3,LS2-6) = min (32,33) = 32 LS = 32 – 32 = 0

Step 3

Now to find the slack we need to find the values of LS-ES (or LF-EF) for each of the activities as shown below. The activities with value of LS – ES = 0 are activities with zero slack (or critical activities). We get,

1-2: LS – ES = 0

We skip the dummy activity as it is of zero duration.

1-4: LS – ES = 0

1-3: LS – ES = 11-...

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