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# Activity PowerPoint PPT Presentation

Activity. Duration. Predecessor. Activity. A. 3. -. B. 5. -. C. 4. A. D. 7. B. E. 10. B. F. 3. E. G. 8. C, D. Example AOA (Activity On Arrow).

Activity

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Activity

Duration

Predecessor

Activity

A

3

-

B

5

-

C

4

A

D

7

B

E

10

B

F

3

E

G

8

C, D

Example AOA (Activity On Arrow)

In this example we will carry out a calculation by using AOA-network, on the basis of information given in the table. The Project consists of 7 activities. Durations and sequences are shown in the table.

1

Node nr.

Activity

Duration

Predecessor

activity

A

3

-

B

5

-

C

4

A

Earliest

Latest

D

7

B

Time point

Time point

E

10

B

F

3

E

G

8

C, D

For being able to start the project, we should have a start node(event).

For each node, we have divided the circle into upper and lover halves. The upper half represents the activity number.

We have divided the lower part into halves as well. In the left half, we will write the earliest start and in the right half, the latest start that the activity can happen.

2

5

C (4)

G (8)

D (7)

1

3

6

E (10)

F (3)

4

Activity

Duration

Predecessor

activity

A

3

-

B

5

-

C

4

A

D

7

B

E

10

B

F

3

E

G

8

C, D

A (3)

Accordingly it is illustrated for activities C,D,E,F and G. In this network, we will carry out a calculation.

Activity B has a duration of five units and lies between node 1 and 3. Both activities of A and B have predecessor activities and begin from start-node 1.

By using information in the table, we can draw a network. Lines In the network represent activities given in the table. The durations which stand in the table are given in the parenthesis in the network as well.

Activity A which is shown between node 1 and 2, has a duration of 3 units.

B (5)

2

5

C (4)

G (8)

A (3)

D (7)

1

3

6

B (5)

E (10)

F (3)

4

Activity

Duration

Predecessor

activity

A

3

-

B

5

-

Node nr.

C

4

A

D

7

B

E

10

B

F

3

E

G

8

C, D

Earliest

Latest

Time point

Time point

The calculation is carried out in two steps. First we go forwards in the network. In this step we will determine earliest start for all nodes.

In the second step, we go backwards from final node in the network to the start node.

In this step we determine the latest time for each activity in the network.

2

5

C (4)

G (8)

A (3)

D (7)

1

3

6

B (5)

Activity

Duration

Predecessor

activity

A

3

-

E (10)

F (3)

B

5

-

4

C

4

A

D

7

B

E

10

B

F

3

E

G

8

C, D

We can reach Node(event) 4, by passing through activity E.

Activity B has a duration of 5 so the earliest start for event 3 becomes 5.

We will go on with forward calculation in the network. The only way for reaching to node 3 is to accomplish activity B.

The duration of activity A is 3. Therefore event 2 can happen earliest at 3.

If event 1 happens earliest at time 0, event 2 can not happen before end of activity A.

In our project, we have not said anything on startpoint of the project but we will calculate relatively and say that the project starts at time 0.

The Duration of activity E is 10 so the earliest start for the event 4 is 10 pluss earliest start for event 3 which is 5. this gives 15 as the earliest start for event 4.

3

0+3=3

0+5=5

5

0

15

5+10=15

2

5

C (4)

G (8)

A (3)

D (7)

1

3

6

B (5)

Activity

Duration

Predecessor

activity

A

3

-

E (10)

F (3)

B

5

-

4

C

4

A

D

7

B

E

10

B

F

3

E

G

8

C, D

3+4=7

12

3

Now we have two different values and should choose the largest of them. Therefore the earliest start for activity 6 becomes 6.

Now we have determinedall the earliest starts for activities.

We have two different values, but both activities should be finished before event 5 occurrence.

It results in the need to choose one of two and we choose the larger value.so the earliest start for event 5 becomes 12.

The same applies to event 6 where we have two alternative ways for reaching to it: via activity F or Activity G.

If we go via activity C the earliest start for event 5 becomes 3 pluss the duration of activity C equal to 4 that gives 7.

By passing through activity F we gain 15 pluss 3 ,the duration of F, that becomes 18.

By going through, the activity G we obtain 12 pluss 8 which is the duration of G, so the earliest start for event 6 becomes 20.

But there was another way for reaching event 5. By going through activity D, the earliest event time becomes five pluss the duration of activity D which is 7. So we will have the earliest time of event 5 equal to 12.

For event 5, there are two possibilities for reaching to: via activity C or D.

5+7=12

12+8=20

5

20

0

15+3=18

15

2

5

C (4)

12

3

G (8)

A (3)

D (7)

1

3

6

B (5)

Activity

Duration

Predecessor

5

20

0

activity

A

3

-

E (10)

F (3)

B

5

-

4

C

4

A

D

7

B

15

E

10

B

F

3

E

G

8

C, D

we will calculate beckwards in the network.

In the case that a finish date had been determined for the project, we would insert this value in the right half of event 6.but no date is given so we assume that project should be finished as soon as possible. Therefore we say that the latest timefor event 6 is equal to its earliest timeand we set 20 as the latest time for that.

The latest time for event 4 becomes 20 minus 3 that gives 17.

For reaching to event 5, we can just go through activity G with a duration of 8.

For event 4, we go from event 6 through activity F with a duration of 3.

Therefore the latest time for event 5 becomes 20 minus 8 equal to 12.

If event 5 happens later than 12, event 6 will not have 20 as the latest start.

We will now carry out a similar process used for caculating ealiest time points but this time we will go bavkwards whole the way by subtracting

12

20-8=12

20

17

20-3=17

2

5

C (4)

12

3

G (8)

A (3)

D (7)

1

3

6

B (5)

Activity

Duration

Predecessor

5

20

0

activity

A

3

-

E (10)

F (3)

B

5

-

4

C

4

A

D

7

B

15

E

10

B

F

3

E

G

8

C, D

12-4=8

12-7=5

8-3=5

For event 1 we have a case with two alternatives as well. From event 2 we have 8 and the duration of activity A is 3, so we gain 5. but we need to look at other alternative as well.

We can reach to event 2 just by going through activity C. By this way we gain 12 minus the duration of C which is 4. This results in a latest time equal to 8 for event 2.

But we should look at the other alternative as well. We gain 12 minus the duration of activity D equal to 7 that gives 5.

Then we have values of 5 and 7 and should choose the lower value. So the earliest time for event 3 is 5.

Event 3 is slightly similar to event 5 and 6 when we calculate forwardly. The latest time for event 3 should be chosen between two alternatives because there are two acctivities branching from event 3.

Initially we go from event 4 and gain 17 minus 10 which is the duration of activity E. This gives the latest time of 7 for event 3.

Again we have two values, 0 and 5 and we should choose the smaller value.

By starting from event 3 with 5 and subtracting the duration of activity B(5), we gain the latest time for event 1 that is equal to 0.

12

8

5

20

0

5-5=0

17

17-10=7

Activity

Duration

ES

EF

LS

LF

FL

A

3

0

3

5

8

5

B

5

0

5

0

5

0

C

4

3

7

8

12

5

D

7

5

12

5

12

0

E

10

5

15

7

17

2

F

3

15

18

17

20

2

G

8

12

20

12

20

0

Now we have carried out calculations of all of project nodes. we will now look at calculation of activities we will do this in the table.

From network, we see that activity A and B have event 1 as their start event, where the earliest time is 0.In other words, the start time for activities A and B is 0.

Activity F has event 4 as its start node. Therefore the earliest start for activity F becomes 15.

Activity C has event 2 as its start-event. Therefore the earliest start for activity C becomes 3.

Both activity D and E have node 3 as their start node where the earliest start is 5.

We can extract the earliest start of all activities by looking at earliest start of start-node of each activity.

Now we will calculate the earliest start of all the activities. This value is demonstrated by ES(earliest start)

First we list all activities in a column in extreme left. In its adjacent column, the respective durations are shown. For example activity A has a duration of 3.

Similarly, Activity G has the event 5 as start-event and it gives 12 as the earliest start for activity G.

Activity

Duration

ES

EF

LS

LF

FL

A

3

0

3

5

8

5

B

5

0

5

0

5

0

C

4

3

7

8

12

5

D

7

5

12

5

12

0

E

10

5

15

7

17

2

F

3

15

18

17

20

2

G

8

12

20

12

20

0

For activity B, EF is equal to 5 pluss 0 that gives 5

Activity C has the event 2 as its start event therefore the earliest start for activity C becomes 3.

Calculations of all remaining activities are carried out similarly.

For activity A we gain 0 plus 3 that gives an EF equal to 3.

For finding this value for each activity, we take the earliest time of its start event and add to the duration of that activity. In other words since the earliest start of activities is equal to earliest time of its start event, we can find EF by aggregating ES and the duration of activity.

Now we will calculate the earliest time points at which activities can end. These time points are supplied in the column under EF(Earliest Finish)

EFi=ESi+ti

Activity

Duration

ES

EF

LS

LF

FL

A

3

0

3

5

8

5

B

5

0

5

0

5

0

C

4

3

7

8

12

5

D

7

5

12

5

12

0

E

10

5

15

7

17

2

F

3

15

18

17

20

2

G

8

12

20

12

20

0

We can find this value directly from the network diagram as well. The latest time that an activity can finish is same as the latest time of the end-event of that activity.

For activity B, event 3 with latest start of 5 is the final event. So the latest finish for activity B becomes 5.

Event 5 is the final event for both activities C and D and this gives latest finish of 12 for both activities.

For activity E, event 4 is the final event and the latest finish becomes 17.

Activity A has event 2 as its final event and the latest time that the event 2 can happen is 8. Therefore the latest finish of A becomes 8.

Activities F and G have event 6 as their end-event so the latest finish for both activities is equal to 20.

Now we have accomplished the calculation of earliest start and earliest finish for all activities. From now on we will look at latest start and finish times. First we look at latest finish(LF).

Activity

Duration

ES

EF

LS

LF

FL

A

3

0

3

5

8

5

B

5

0

5

0

5

0

C

4

3

7

8

12

5

D

7

5

12

5

12

0

E

10

5

15

7

17

2

F

3

15

18

17

20

2

G

8

12

20

12

20

0

For remaining activities, same calculations are carried out as illustrated in the table.

Activity B has a latest finish equal to 5 and a duration of 5. This gives a latest start of 0.

The latest start for activity C becomes 8 or 12 minus 4.

For activity A it becomes 8 minus 3 equal to 5. 8 is the latest finish and 3 is the duration of activity.

The latest start is the latest finish minus the duration of each activity.

Now it remains to calculate the latest start of each activity. In the table it is demonstrated by LS(latest start)

LSi=LFi-ti

Activity

Duration

ES

EF

LS

LF

FL

A

3

0

3

5

8

5

B

5

0

5

0

5

0

C

4

3

7

8

12

5

D

7

5

12

5

12

0

E

10

5

15

7

17

2

F

3

15

18

17

20

2

G

8

12

20

12

20

0

We can determine this by looking at the float of different activities.In order to calculate float of each activity, we take the latest time of its end-node and subtract from that the earliest time of its start-node. Now we have the total period in which the activity will be carried out. From this value, we subtract duration of same activity. The remaining value is called float. The float is the surplus time and represents the amount that activities have freedom for planning.

Which activities influence in this end date?

Which activities have importance for determining the end date.

In our case, the FL(Float) is equal to LF minus ES minus the duration of activity 1.

Now we have calculated sufficiently in order to know when earliest and latest times of each event occurs. We also know when each activity can start and finish at earliest and latest.we have found out that the project becomes accomplished in 20 days.

Fli =LFi-Esi -ti

=LFi-(Esi +ti)

=LFi-Efi

Fli =(LFi-ti)-ESi

=LSi-ESi

Activity

Duration

ES

EF

LS

LF

FL

A

3

0

3

5

8

5

B

5

0

5

0

5

0

C

4

3

7

8

12

5

D

7

5

12

5

12

0

E

10

5

15

7

17

2

F

3

15

18

17

20

2

G

8

12

20

12

20

0

Similarly it is carried out for the remaining activities.

For activity B, the float is 5 minus 5=0.

Activity C has a float of 5, when LF is 12 and EF is 7.

Now we have two equivalent formulas for calculation of the float. In this example We use LF minus EF.

For activity A LF is equal to 8 minus EF equal to 3 that gives a float of 5.

Fli =LFi-Esi -ti

=LFi-(Esi +ti)

=LFi-Efi

Fli =(LFi-ti)-ESi

=LSi-ESi

2

5

C (4)

12

3

G (8)

A (3)

D (7)

1

3

6

B (5)

5

20

Activity

Duration

ES

EF

LS

LF

FL

0

A

3

0

3

5

8

5

E (10)

F (3)

B

5

0

5

0

5

0

4

C

4

3

7

8

12

5

15

D

7

5

12

5

12

0

E

10

5

15

7

17

2

F

3

15

18

17

20

2

G

8

12

20

12

20

0

12

8

Some of activities have float of zero. It means that there is no freedom in planning of such activities, in other words they can not be extended. We call this type of activities, critical activities. In contrast, activities with positive float have possibility to be extended without affecting the finish date of network.

Activities B,D, and G which all have float of zero are critical activities.

We can draw a continous chain of critical activities from first event to the last event. This chain includes activities B, D and G and we call it critical path.

Pay attention that the float that we are speaking about belongs to whole chain in the network. For instance activities E and F lie in a chain between event 3 and 6. They both have a float of 2 days; but we can not simultanously use 2 days float for each of them.In other words these two activities altogether have a float of two days. 2 days for E, 2 days for F or 1 day for each of them.

The same applies to the chain in which A and C lie. In this case these two, altogether share a float of five days.

5

20

0

17