Task A
1. Determine the expected completion time for each of the nine project activities.
The PERT equation used in calculation of the expected time as illustrated by Chemuturi & Cagley (2010) is =
E = (O + 4M + P)/6, where
O= optimistic value
M= probability time value
P=pessimistic time value
E= expected time value
From the above formula, the paper used it in calculating the expected time values listed in the table in the embedded excel workbook (double click on the icon “Students_Task_4Vl_MF6_15”). An example of calculation for A is below:
E = (O + 4M + P)/6, where
O=2, M =3.1, P=3.6
E= (2+4(3.1) +3.6)/6
E=3
2. Determine the variance for each of the nine project activities.
The PERT equation used in calculation of the variance as illustrated by Chemuturi & Cagley (2010) is =
Variance= [(O-E) ² + 4(M-E) ² + (P-E) ²]/6, Where:
O= optimistic value
M= probability time value
P=pessimistic time value
E= expected time value
From the above formula, the paper used it in calculating the variance listed in the table in the embedded excel workbook (double click on the icon “Students_Task_4Vl_MF6_15”). An example of calculation for A2 is below
Variance= [(O-E) ² + 4(M-E) ² + (P-E) ²]/6, where
O=2, M =3.1, P=3.6, E=3
Variance= [(2-3)2 +4 (3.1-3)2 + (3.6-3)2]/6
Variance= [-1+0.04+0.36]/6
Variance= -0.1
3. Develop a PERT chart showing the network diagram with the activities and their expected times.
7 weeks
H
6 weeks
G
12 weeks
11.25 weeks D 12 weeks B D I
3weeks 13.9 weeks 26.5weeks A C E 33 weeks F 4.5 weeks I 4.5 weeks I
(a)Identify the critical path
In identifying the critical path, the paper determined the Earliest Start time (ES), the Earliest Finish time (EF), the Latest Start time (LS), and Latest Finish time (LF) for every activity.
1= (0, 0)
2= (3, 3)
3= (14, 18)
4= (17, 17)
5= (36, 44)
6= (26, 30)
7= (43, 43)
8= (32, 36)
9= (39, 44)
10= (48, 48)
From the diagram, the critical path is A-C-E-I- End since it has a slack of zero
4. Determine the value of each of the following:
a). Determine the expected duration of the entire project, showing all of your work or reasoning.
The expected duration of the project is the sum total of all the activities of the critical path. Therefore, from the PERT chart, the expected duration of the project is as follows:
A= 3, C= 13.9, E= 26.5, I=4.5
Total project duration =47.9
b). Determine the slack for project Activity E, showing all of your work or reasoning.
Earliest Start time (ES) = 17
The Earliest Finish time (EF) =43
The Latest Start time (LS) = 17
Latest Finish time (LF) =43
Slack= LF-EF; 43-43=0
c). Determine the slack for project Activity C, showing all of your work or reasoning.
Earliest Start time (ES) = 3
The Earliest Finish time (EF) =17
The Latest Start time (LS) = 3
Latest Finish time (LF) =17
Slack= LF-EF; 17-17=0
d). Determine the earliest week project Activity F is scheduled to start, showing all of your work or reasoning.
Project activity F Is to start when project activity A is finished. Project A needs 3 weeks to finish, so project activity F is scheduled to start in week 3
e). Determine the latest week project Activity G is scheduled to finish, showing all of your work or reasoning.
Project activity G is scheduled to finish after completion of project activities A, B, D. G or A, C, D, G
A, B, D, G=3+11.25+12+6=32.25
A, C, D, G= 3+13.9+12+6=34.9
Therefore, the latest week project activity G is scheduled to finish is 34.9
5. Determine the probability of completing this project in time for the product launch in 48 weeks, showing all of your work
| TASK | Expected Time | Variance (weeks) | Critical Path Total |
| A | 3 | -0.1 | 3 |
| C | 13.9 | 3.15 | 13.9 |
| E | 26.5 | 14.58 | 26.5 |
| I | 4.5 | 0.0833 | 4.5 |
| END | 47.9 | ||
Probability = (48-47.9)/2.55
=0.03922
51.20%
To determine the probability of finishing the project by 48 week, only the variance and the duration of critical path activities is relevant. Form the table, the probability is observed to be 51.20% that the project activity will be finished in less than 48 hours.
Task B
1. Determine the maximum reduction in time for each of the ten project activities. a. Include your answers for part B1 in Table 1.2, showing all of your work for one activity.
The maximum reduction in time for an activity as illustrated by Chemuturi & Cagley (2010) is;
=NT-CT, where
NT= Normal time for activity completion
CT= Crash Time for activity completion
Taking Task L as an example, where
NT=2 weeks
CT= 1.5 weeks
Using the formula of NT-CT= the maximum reduction in time for an activity
2-1.5= 0.5
From the above formula, the paper used it in calculating the maximum reduction in time for an activity values listed in the table in the embedded excel workbook.
2. Determine the crash cost per week for each of the ten project activities. a. Include your answers for part B2 in Table 1.2, showing all of your work for one activity.
Crashing cost per period as illustrated by Chemuturi & Cagley (2010) is
= Where,
CC= crash cost
NC= Normal cost for activity completion
NT= Normal time for activity completion
CT= Crash Time for activity completion
From the above formula, the paper used it in calculating the crash cost per week for each of the ten project activities as listed in the table in the embedded excel workbook. An example of calculation for B2 is below
=3000
C. Determine the following by using your results from part B: 1. Identify the activities, considering all paths, to be crashed to meet the following criteria: • completion of the project within 22 weeks • achievement of the lowest possible cost
To shorten the project, I will crash on the critical activities starting from the least expensive to the most expensive. From the provided network in task B, there are two paths and the times are shown as normal times;
LOP (M) S (Q) U=30.5
NR (MQ) TU=29.5
The activities that need to be crushed to complete the project in 22 weeks at the lowest possible cost is in path NR (MQ) TU. The activities that is prudent to crash are R, M and U because of their relevance to the project duration and low costs.
| TASK | NORMAL WEEKS | CRASH WEEKS | NORMAL COSTS | CRASH COSTS |
| N | 1.5 | 1.5 (not crashed | 11,000 | 0 |
| R | 3 | 2 | 12,750 | 19225 |
| M | 10 | 4 | 11,000 | 24,750 |
| Q | 6 | 6 (not crashed) | 9500 | 0 |
| T | 7 | 7 (not crashed) | 13,750 | 0 |
| U | 2 | 1.5 | 10,600 | 22,500 |
| TOTAL | 29.5 | 22 | 68,600 | 66475 |
2. Determine the number of weeks each of the activities identified in part C1 should be crashed. 3. Determine the sum of the individual costs of crashing each activity identified in part C1. D. When you use sources
Activity R can be crashed for one week (normal weeks- crash weeks= 3-1= 1 week) at a crash cost totaling to $6475 (crash cost-normal cost=19225-12750= $6475). When the total crash cost is divided by the total crash time allowable, it yields the weekly crash cost ($6475/1) = $6475.
Activity M can also be crashed to a total of 6 weeks (normal weeks- crash weeks= 10-4= 6 weeks) at a crash cost totaling to $13, 750 (crash cost-normal cost=24,750-11,000= $13,750). When the total crash cost is divided by the total crash time allowable, it yields the weekly crash cost ($13, 750/6) = $2291.67.
Lastly, activity U can also be crashed to a total of 0.5 weeks (normal weeks- crash weeks= 2- 1.5= 0.5 weeks) at a crash cost totaling to $11, 900 (crash cost-normal cost=22, 500-10,600= $11,900). When the total crash cost is divided by the total crash time allowable, it yields the weekly crash cost ($11, 900/0.5) = $23, 800
References
Chemuturi, M., & Cagley, T. M. (2010). Mastering software project management: Best practices, tools and techniques. Ft. Lauderdale, FL: J. Ross Pub.
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