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Use the Triple Time Estimate table below to determine the critical path and comp

ID: 432970 • Letter: U

Question

Use the Triple Time Estimate table below to determine the critical path and completion time for the project

(note: times are in weeks).

Task description

optimistic

Likely

Pessimistic

Period 1

Period 2

A

Task a

1

3

5

B

Task b

2

5

11

A

C

Task c

4

7

13

A

D

Task d

5

10

12

B

c

E

Task e

4

5

6

C

f

Task f

3

3

3

d

__________15a) What steps comprise the “critical path”?

__________15b) What is the minimum completion time?

__________15c) What is the probability of completing the project in less than 25 weeks?

__________15d) If TASK B can be completed in half the time for a small fee, would it be worthwhile

Task description

optimistic

Likely

Pessimistic

Period 1

Period 2

A

Task a

1

3

5

B

Task b

2

5

11

A

C

Task c

4

7

13

A

D

Task d

5

10

12

B

c

E

Task e

4

5

6

C

f

Task f

3

3

3

d

Explanation / Answer

The precedence diagram of activities as follows :

                                                                                                                   A

                               B

                                                                             C

                                                                                     D

                       E

                                                                                   F

Please find below table highlighting expected durations as well as Variances of each activity.

Expected duration = ( Optimistic time + 4 x Most likely time + Pessimistic time ) / 6

Variance = ( Pessimistic time – Optimistic time ) ^2/36

Activity

To

tm

tp

te = ( to + 4.tm+tp)/6

Variance ( tp-to)^2/36

A

1

3

5

3

0.44

B

2

5

11

5.5

2.25

C

4

7

13

7.5

2.25

D

5

10

12

9.5

1.36

E

4

5

6

5

0.11

F

3

3

3

3

0.00

The expected durations of all parallel paths as follows :

A-B-D-F = 3 + 5.5 + 9.5 + 3 = 21

A-C-D-F = 3 + 7.5 + 9.5 + 3 = 23

A-C-E = 3 + 7.5 + 5 = 15.5

Out of above , A-C-D-F has the longest duration and hence forms the critical path.

Since minimum completion time of a project will be same as duration of the critical path, minimum completion time will be 23 weeks

STEPS COMPRISING CRITICAL PATH = ACDF

MINIMUM COMPLETION TIME = 23 WEEKS

Variance of the critical path’s duration

= Sum of variances of A,C,D,F

= 0.44 + 2.25 + 1.36+ 0

= 4.05 weeks

Therefore,

Standard deviation of duration of critical path = Square root ( Variance ) = Square root ( 4.05) = 2.01

Let z value corresponding to probability of completing the project in maximum 25 weeks =Z1

Therefore ,

Expected duration of critical path + Z1 x Standard deviation of duration of critical path = 25

Or, 23 + 2.01.Z1 = 25

Or, Z1 = 2/2.01 = 1 ( approx.)

Corresponding probability for Z= 1 as derived from standard normal distribution table is 0.84134.

PROBABILITY OF COMPLETING THE PROJECT FOR LESS THAN 25 WEEKS = 0.84134

Since task B does not lie on the critical path, shortening its duration will not help to reduce minimum completion time of the project. Therefore, it is not worthwhile to pay a small fee to complete task B in half the time.

                                                                                                                   A

                               B

                                                                             C

                                                                                     D

                       E

                                                                                   F