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PERT/CPM

CONSTRUCTION

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ACT PRECEDES DUR

A D,F 2

D E 6

E H 7

H END 2

F G 3

G END 5

B I 2

I J 2

J K 3K END 4

C M 5

M N 1

N O 6

O END 8

Draw the network diagram and determine the critical path.

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1 A2

2

D6

F

3

3 E7 4 H

2

5G

5

6

B

2

I

27

J

3 8 K4

C

59

M

110

N

6

O

811

END

ACT PRECEDES DUR

A D,F 2

D E 6

E H 7

H END 2

F G 3

G END 5

B I 2

I J 2

J K 3

K END 4C M 5

M N 1

N O 6

O END 8

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1 A2

2

D6

F3

3 E7 4 H

2

5G

5

6

B

2

I

27

J

38

K

4

C

5 9M

110

N

6

O

811

END

ACT PRECEDES DUR

A D,F 2

D E 6

E H 7

H END 2

F G 3

G END 5

B I 2

I J 2

J K 3

K END 4

C M 5

M N 1

N O 6

O END 8

0

2 4 7

112

8

15

17

5 10

5 6 12

200

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1 A2

2

D6

F3

3 E7 4 H

2

5G

5

6

B

2

I

27

J

38

K

4

C

5 9M

110

N

6

O

811

END

0

2 4 7

202

8

15

20

5 20

5 6 12

200

161311

91811

5

1512

3

1265

0

ACT PRECEDES DUR

A D,F 2

D E 6

E H 7

H END 2

F G 3

G END 5

B I 2

I J 2

J K 3

K END 4C M 5

M N 1

N O 6

O END 8

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1 A2

2

D6

F

3

3 E7 4 H

2

5G

5

6

B

2

I

27

J

38

K

4

C

59

M

110

N

6

O

811

END

ACT PRECEDES DUR

A D,F 2

D E 6

E H 7

H END 2

F G 3

G END 5

B I, E 2

I J, H 2

J K 3

K END 4

C M, G 5

M N 1

N O 6

O END 8

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1 A2

2

D6

F3

3 E7 4 H

2

5G

5

6

B

2

I

27

J3 8

K

4

C

5 9M

110

N

6

O

811

END

0

ACT PRECEDES DUR

A D,F 2

D E 6

E H 7

H END 2

F G 3

G END 5

B I, E 2

I J, H 2

J K 3

K END 4

C M, G 5

M N 1

N O 6

O END 8

2 4 7

24

112

815

17

5

5

0

5 6 12

20

10

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1 A2

2

D6

F3

3 E7 4 H

2

5G

5

6

B

2

I

27

J3 8

K

4

C

5 9M

110

N

6

O

811

END

0

ACT PRECEDES DUR

A D,F 2

D E 6

E H 7

H END 2

F G 3

G END 5

B I, E 2

I J, H 2

J K 3

K END 4

C M, G 5

M N 1

N O 6

O END 8

2 4 7

24

112

815

17

5

5

0

5 6 12

20

10

12

15

18

16

6

5

12

13

18

11

11

5

11

9

3

0

15

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1 A2

2

D6

F3

3 E7 4 H

2

5G

5

6

B

2

I

27

J3 8

K

4

C

5 9M

110

N

6

O

811

END

0

ACT PRECEDES DUR

A D,F 2

D E 6

E H 7

H END 2

F G 3

G END 5

B I, E 2

I J, H 2

J K 3

K END 4

C M, G 5

M N 1

N O 6

O END 8

2 4 7

24

112

815

17

5

5

0

5 6 12

20

10

12

15

18

16

6

5

12

13

18

11

11

5

11

9

3

0

15

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Definition of Terms

Critical Path the longest route in the network of activities representing the project- duration of CP = duration of the project

- there maybe more than one CP in a project and may shift as the projectmoves towards completion

Forward Pass

Earliest Start Time (ES)

earliest time an activity can start

ES = maximum EF of immediate predecessors Earliest finish time (EF)

earliest time an activity can finish

earliest start time plus activity time

EF= ES + t

Latest Start Time (LS)Latest time an activity can start without delaying critical path time

LS= LF - t

Latest finish time (LF)

latest time an activity can be completed without delaying critical path time

LS = minimum LS of immediate predecessors

Backward Pass

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1 A2

2

D6

F3

3 E7 4 H

2

5G

5

6B

2

I2 7

J3 8

K

4

C

5 9M

110

N

6

O

811

END

0

ACT D ES EF

A 2

D 6

E 7

H 2

F 3

G 5

B 2

I 2

DM1 0

J 3

DM2 0

K 4

C 5

DM3 0

M 1

N 6

O 8

2 4 7

24

112

815

17

5

5

0

5 6 12

20

10

0 2

2 8

0 2

2 4

2 2

8 15

4 7

4 4

15 17

7 11

2 5

0 5

5 5

5 10

5 6

6 12

12 20

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ACT D LF LS

A 2

D 6

E 7

H 2

F 3

G 5

B 2

I 2

DM1 0

J 3

DM2 0

K 4

C 5

DM3 0

M 1

N 6

O 8

1 A2

2

D6

F3

3 E7 4 H

2

5G

5

6B

2

I2 7

J3 8

K

4

C

5 9M

110

N

6

O

811

END

20

12

15

18

16

6

5

12

13

18

11

11

5

11

9

3

0

15

20 12

12 6

6 5

20 15

15 12

15 15

5 0

20 18

18 11

11 5

5 3

20 16

18 18

16 13

11 11

13 11

11 9

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ACT D ES EF LF LS Float

TF

A 2 0 2 5 3

D 6 2 8 11 5

E 7 8 15 18 11

H 2 15 17 20 18

F 3 2 5 15 12

G 5 5 10 20 15

B 2 0 2 11 9

I 2 2 4 13 11

DM1 0 2 2 11 11

J 3 4 7 16 13

DM2 0 4 4 18 18

K 4 7 11 20 16

C 5 0 5 5 0

DM3 0 5 5 15 15

M 1 5 6 6 5

N 6 6 12 12 6

O 8 12 20 20 12

Total Float is the maximum amount of time that this activity

can be delay in its completion before it becomes a critical

activity, i.e., delays completion of the project

TF = LS - ES

3

3

3

3

10

10

9

9

9

9

14

3

0

10

0

0

0

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ACT PRECEDES DURA C, D, F 4

B F, D 7

C E, G, I 5

D E, G, I 10

E H 2

F G, I 6

G H 4

H END 8

I END 9

Construct a Network Diagram and Deter mine the Critical Path

ACT PRECEDES DUR

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ACT PRECEDES DUR

A C, D, F 4

B D, F 7

C E, G, I 5

D E, G, I 10

E H 2

F G, I 6

G H 4

H END 8

I END 9

1

2A4

C5

4

3B7

F6

5

E2

6

7

H8

I9

G 4D 10 END

START

J 0 K 0

ACT PRECEDES DUR

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ACT PRECEDES DUR

A C, D, F 4

B D, F 7

C E, G, I 5

D E, G, I 10

E H 2

F G, I 6

G H 4

H END 8

I END 9

1

2A4

C5

4

3B7

F6

5

E2

6

7

H8

I9

G 4D 10 ENDSTART J 0 K 0

0

4

7

4

9

17

19

17

13

2129

26

0

ACT PRECEDES DUR

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ACT PRECEDES DUR

A C, D, F 4

B D, F 7

C E, G, I 5

D E, G, I 10

E H 2

F G, I 6

G H 4

H END 8

I END 9

1

2A4

C5

4

3B7

F6

5

E2

6

7

H8

I9

G 4D 10 ENDSTART J 0 K 0

0

4

7

4

9

17

19

17

13

21 29

26

29

21

17

20

19

17

11

7

12

73

029

0

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ACT PRECEDES DUR

A C, D, F 4

B D, F 7

C E, G, I 5

D E, G, I 10

E H 2

F G, I 6

G H 4

H END 8

I END 9

1

2A4

C5

4

3B7

F6

5

E2

6

7

H8

I9

G 4D 10 ENDSTART J 0 K 0

0

4

7

4

9

17

19

17

13

21 29

26

29

21

17

20

19

17

11

7

12

73

00

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Reducing Time and Cost

1. To avoid penalties for not completing the project on time.2. To take advantage of monetary incentives for completing the project on or before

the target date.3. To free the resources such as money, equipment, and men for use on other

project.4. Reduce indirect cost associated with the project such as:

1. Facilities and equipment cost2. Supervision cost3. Labor Cost4. Personnel Cost

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Crashing- spending more money to get something done more quickly.- The key is to attain maximum decrease in schedule time with

minimum cost.

Which Method is Best?

Increasing Your Resources - this essentially means decreasing the time it takes toperform individual activities by increasing the number of people working on thoseactivities.

Pros: This makes sense, at first glance. For example, if it takes Junior 4 hoursto complete an activity, it would logically take Bob and Thirdy 2 hours tocomplete the same activity.

Cons:Adding resources isnt always the best solution, though. Sometimes itends up taking moretime in the long run.

Consider the following:New resources aren't going to be familiar with the tasks at hand, so they willprobably be less productive than current team members.Who will guide the new members up the learning curve? Usually it will bethe most productive members of the team, who could themselves be working

to get the task finished more quickly.

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Which Method is Best?

Fast Tracking - which involves over-lapping tasks which were initiallyscheduled sequentially. Or you might be able to optimize your schedule inother areas.

For example, maybe you can split long tasks into smaller chunks to

squeeze more work into a shorter period of time

reduce lag times between tasks

or reduce the scope to eliminate less important tasks.

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When NOT to Crash

The key to project crashing is attaining maximum reduction in schedule time withminimum cost. Quite simply, the time to stop crashing is when it no longer becomescost effective. A simple guideline is:

Crash only activities that are critical.

Crash from the least expensive to most expensive.

Crash an activity only until:It reaches its maximum time reduction.It causes another path to also become critical.

It becomes more expensive to crash than not to crash.

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Procedure in Crashing Project Time

1. Obtain an estimate of regular and crash time plus the cost of each activity.

2. Determine the length of all path and their float time.3. Determine which activities are on the critical path.4. Crash the critical activities in the order of increasing costs as long as crashing

cost do not exceeds the benefits.

Requirements for Crashing Time

1. Regular time and crash time estimates for each activity2. Regular cost and crash estimates for each activity.3. A list of activities that are on the critical path.

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From the following data, develop an optimum time cost plan assuming thattotal crashing cost is P15, 000.

Act Normal

Time

Crash

Time

Cost/day

to CrashA 8 8

B 12 10 P6,000

C 7 6 2,000

D 6 2 6,000

E 9 7 4,500

F 4 2 7,000

1

2

3 4

5 6

A8

B12

C7

D6

E9

F4

Path 1-2-5-6 or A-B-F = 8+12+4 = 24days

Path 1-3-4-5-6 or C-D-E-F = 7+6+9+4= 26days Critical Path

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Rank the Critical Path activities in order of lowest crashing cost.

Act NormalTime CrashTime AvailableDays Cost/day toCrash

C 7 6 1 2,000

E 9 7 2 4,500

D 6 2 4 6,000

F 4 2 2 7,000

A-B-F = 24daysC-D-E-F = 26days Critical Path

Shorten the project one day and then check after each reduction to see which path is critical

Shorten Activity C by 1 day at P2,000/dayCP1 = 26 1 =25 days

Shorten Activity Eby 1 day at P4,500/dayCP2 = 25 1 =24 days

A-B-F = 24daysCritical Path

C-D-E-F = 24days

1

2

3 4

5 6

A8

B12

C7

D6

E9

F4

1

2

3 4

5 6

A8

B12

C7

D6

E9

F4

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Path Activity Crash Cost/day

A-B-F

A No more Reduction

B 6,000

F 7,000

C-D-E-F

C No More Reduction

D 6,000

E 4,500

F 7,000

A-B-F = 24daysCritical Path

C-D-E-F = 24days

Act NormalTime

CrashTime

AvailableDays

Cost/dayto Crash

C 7 6 1 2,000

E 9 7 2 4,500

B 12 10 2 6,000

D 6 2 4 6,000

F 4 2 2 7,000

Try shorten B and E23 days, Cost = 10, 500Shorten F,23 days, Cost = 7,000

Thus, Shortening F is more practical.Results, 23 daysProject Duration, Total Crashing Cost = 2000 + 4,500 + 7, 000 = 13, 500

1

2

3 4

5 6

A8

B12

C

6

D

6

E

8

F4