Activity Networks Example 3matematik

7/28/2019 Activity Networks Example 3matematik

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Activity networks Example 3

Task Duration (days) Immediate

predecessors

A 2 -

B 4 -

C 5 A, B

D 3 B

E 6 C

F 3 C

G 8 D

H 2 D, F

The table below shows the tasks involved in a project,

with their durations and immediate predecessors.

Draw an activity network and use it to find the critical

activities and the minimum duration of the project.


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Begin with a start node,

labelled 1.

Activities A and B have no

preceding activities, so can

both begin at the start node.

A(2)

B(4)

1


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Activity C depends on both A

and B, but activity D

depends on B only.

We can deal with this by

adding a dummy activity.

A(2)

B(4)

1

3

2


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Now add activity C, which

depends on both A and B ...

and activity D, which

depends on B only.

A(2)

B(4)

1

3

2

C(5)

D(3)


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Activities E and F depend on

C only

A(2)

B(4)

1

3

2

C(5)

D(3)

and activity G depends on

D only.

4

5

E(6)

F(3)

G(8)


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Activity H depends on

activities D and F.

A(2)

B(4)

1

3

2

C(5)

D(3)

To deal with this, we need a

dummy activity from event 5.

4E(6)

F(3)

G(8)

5

Now activity H can be added.

6

H(2)


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A finish node is now needed.

A(2)

B(4)

1

3

2

C(5)

D(3)

4E(6)

F(3)

G(8)5

6

H(2)

E(6)

G(8)

7


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The next step is to find the

early event times (EETs).

A(2)

B(4)

1

3

2

C(5)

D(3)

4

F(3)

5

6

H(2)

E(6)

G(8)

7


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Event 1 occurs at time zero.

A(2)

B(4)

1

3

2

C(5)

D(3)

4

F(3)

5

6

H(2)

E(6)

G(8)

70


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Event 2 cannot occur until B

has finished, so the earliest

time for event 2 is 4.

A(2)

B(4)

1

3

2

C(5)

D(3)

4

F(3)

5

6

H(2)

E(6)

G(8)

70

4


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Event 3 cannot occur until

both A and B have finished,

so the earliest time for event

3 is 4.

A(2)

B(4)

1

3

2

C(5)

D(3)

4

F(3)

5

6

H(2)

E(6)

G(8)

70

4

4


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Event 4 cannot occur until C


time for event 4 is 4 + 5 = 9.

A(2)

B(4)

1

3

2

C(5)

D(3)

4

F(3)

5

6

H(2)

E(6)

G(8)

70

4

4 9


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Event 5 cannot occur until D


time for event 5 is 4 + 3 = 7.

A(2)

B(4)

1

3

2

C(5)

D(3)

4

F(3)

5

6

H(2)

E(6)

G(8)

70

4

4 9

7


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Event 6 cannot occur until both D and F

have finished. The earliest time that D can

finish is 7, and the earliest time that F can

finish is 9 + 3 = 12, so the earliest time for

event 6 is 12.

A(2)

B(4)

1

3

2

C(5)

D(3)

4

F(3)

5

6

H(2)

E(6)

G(8)

70

4

4 9

7

12


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Event 7 cannot occur until E, G and H have all finished.

The earliest time that E can finish is 9 + 6 = 15, the

earliest time that G can finish is 7 + 8 = 15, and the

earliest time that H can finish is 12 + 2 = 14, so the

earliest time for event 7 is 15.

A(2)

B(4)

1

3

2

C(5)

D(3)

4

F(3)

5

6

H(2)

E(6)

G(8)

70

4

4 9

7

12

15


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The next step is to find the late event

times (LETs), working backwards

through the network.

A(2)

B(4)

1

3

2

C(5)

D(3)

4

F(3)

5

6

H(2)

E(6)

G(8)

70

4

4 9

7

12

15


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The latest time that event 7 can occur

without delaying the project is 15.

A(2)

B(4)

1

3

2

C(5)

D(3)

4

F(3)

5

6

H(2)

E(6)

G(8)

70

4

4 9

7

12

15 15


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The latest time that activity H can start

is 13, so the latest time for event 6 is

13.

A(2)

B(4)

1

3

2

C(5)

D(3)

4

F(3)

5

6

H(2)

E(6)

G(8)

70

4

4 9

7

12

15 15

13


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Activity G and a dummy activity both lead out of

event 5. The latest time that activity G can start is

7, and the latest time that the dummy activity can

start is 13, so the latest time for event 5 is 7.

A(2)

B(4)

1

3

2

C(5)

D(3)

4

F(3)

5

6

H(2)

E(6)

G(8)

70

4

4 9

7

12

15 15

13

7


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Activities E and F both lead out of event 4. The

latest time that activity E can start is 9, and the

latest time that activity F can start is 9, so the

latest time for event 4 is 9.

A(2)

B(4)

1

3

2

C(5)

D(3)

4

F(3)

5

6

H(2)

E(6)

G(8)

70

4

4 9

7

12

15 15

13

7

9


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The latest time that activity C

can start is 4, so the latest time

for event 3 is 4.

A(2)

B(4)

1

3

2

C(5)

D(3)

4

F(3)

5

6

H(2)

E(6)

G(8)

70

4

4 9

7

12

15 15

13

7

94


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Activity D and a dummy activity both lead out of

event 2. The latest time that activity D can start

is 6, and the latest time that the dummy activity

can start is 4, so the latest time for event 2 is 4.

A(2)

B(4)

1

3

2

C(5)

D(3)

4

F(3)

5

6

H(2)

E(6)

G(8)

70

4

4 9

7

12

15 15

13

7

94

4


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Finally, the latest time for event 1 is zero.

A(2)

B(4)

1

3

2

C(5)

D(3)

4

F(3)

5

6

H(2)

E(6)

G(8)

70

4

4 9

7

12

15 15

13

7

94

4

0


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The minimum duration of the project is 15 days.

A(2)

B(4)

C(5)

D(3)

F(3)

6

H(2)

E(6)

G(8)

0

4

4 9

7

12

15 15

13

7

94

4

0

The critical activities are B, C, D, E and G.

2

1

3 4

7

5

Notice that there are two critical paths: BCE

and BDH.

The float in this example

is analysed in the Notes

and Examples.
http://localhost/var/www/apps/conversion/tmp/scratch_10/4.1.pdfhttp://localhost/var/www/apps/conversion/tmp/scratch_10/4.1.pdfhttp://localhost/var/www/apps/conversion/tmp/scratch_10/4.1.pdfhttp://localhost/var/www/apps/conversion/tmp/scratch_10/4.1.pdf

Activity Networks Example 3matematik

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