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Hello student welcome.
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It is given in this question a network consists of the activities in the following list and times are given in weeks.
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We are asked to draw the network diagram calculate the es, ef, lslf and slack for each activity please make a note that here esef lslf represent early start early finish late start and late finish and also we need to calculate the project completion time now first let us draw the network diagram for a given table.
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Let us take the starting one as start and the last turn as end.
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Now from the table we can observe that a and b do not have any predecessors.
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Hence the first circle would be a and this one would be b.
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Since a and b do not have any predecessors, let us mark the arrows from start to a and b.
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Now coming to c, c is a follower of a.
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So, we would mark c as follows.
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And coming to d, d is a follower of both a and b.
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Hence, there would be arrows pointed towards d from both a and b.
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Now coming to e, e is a follower of c.
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So, e would be here and there would be an arrow pointing from c to e.
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And f is a follower of d now both e and f point towards the end this is the required network diagram that we were asked to draw now coming to the second part calculate the es ef ls ls lf and slack for each activity let us mark the given time duration for each activity here for eight for b the time duration is three here also make a note that the time are given in weeks, so it would be three weeks, and for c it is seven weeks, for e it is four weeks, for f it is six weeks, and for d it is three weeks.
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Now coming to the calculation of e s, ef, ls and lf, now this is part b of the question.
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Let us fill this table to find es, ef, ls, lf and slack values.
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See off from the network diagram.
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First let us find out e s and ef, and we know a and b do not have any predecessors hence the es and ef values for a and b both would be 0.
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Now coming to ef values, ef is 0 added to the duration time of a, that is 8.
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Hence 0 plus 8 becomes 8.
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Similarly for b, 0 plus 3 becomes 3.
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Now coming to activity c, c is followed by a.
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Hence the early start of c would be early finish of a and 7 added to 8 begins 15.
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The early finish of c is the duration time of c added to the early start.
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Now coming to d, b is followed by both a and b but let us take the higher value that is 8.
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Hence the early start of b also would be 8 and coming to the early finish 8 added to 3.
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Would become 11.
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Now coming to e is followed by c.
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So the early start of e would be the early finish of c 15 and 15 added to 4 becomes 19.
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Now f is followed by d so the early start of f would be the early finish of d that is 11.
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11 added to 6 becomes 70.
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So these are early start and early finish.
04:41
So while calculating early stand and early finish we have followed the forward pass that is we calculated the activities from left to right which is known as forward pass...