Project Management: PERT Analysis and Scheduling

Understanding PERT Calculations

PERT (Program Evaluation and Review Technique) is a project management tool used to schedule, organize, and coordinate tasks within a project. It is particularly useful for projects where the time required to complete individual activities is uncertain. The core of PERT involves estimating three time durations for each activity:

• Most Optimistic Time (a): The shortest possible time in which an activity can be completed, assuming everything goes exceptionally well.
• Most Likely Time (m): The most probable time required to complete an activity under normal conditions.
• Most Pessimistic Time (b): The longest possible time an activity might take if unforeseen difficulties arise.

From these estimates, the expected time (Te) and variance (σ²) for each activity are calculated using the following formulas:

• Expected Time (Te) = (a + 4m + b) / 6
• Variance (σ²) = ((b - a) / 6)²

(i) Network Diagram, Critical Path, and Expected Project Completion Time

First, let's calculate the expected time (Te) and variance (σ²) for each activity:

Activity	a	m	b	Te = (a + 4m + b) / 6	Variance (σ²) = ((b - a) / 6)²
A	5	7	9	(5 + 4*7 + 9) / 6 = 42 / 6 = 7	((9 - 5) / 6)² = (4 / 6)² = (2/3)² = 0.44
B	6	8	16	(6 + 4*8 + 16) / 6 = 54 / 6 = 9	((16 - 6) / 6)² = (10 / 6)² = (5/3)² = 2.78
C	5	9	13	(5 + 4*9 + 13) / 6 = 54 / 6 = 9	((13 - 5) / 6)² = (8 / 6)² = (4/3)² = 1.78
D	16	21	26	(16 + 4*21 + 26) / 6 = 126 / 6 = 21	((26 - 16) / 6)² = (10 / 6)² = (5/3)² = 2.78
E	11	19	27	(11 + 4*19 + 27) / 6 = 114 / 6 = 19	((27 - 11) / 6)² = (16 / 6)² = (8/3)² = 7.11
F	9	10	17	(9 + 4*10 + 17) / 6 = 66 / 6 = 11	((17 - 9) / 6)² = (8 / 6)² = (4/3)² = 1.78
G	5	9	13	(5 + 4*9 + 13) / 6 = 54 / 6 = 9	((13 - 5) / 6)² = (8 / 6)² = (4/3)² = 1.78
H	2	3	4	(2 + 4*3 + 4) / 6 = 18 / 6 = 3	((4 - 2) / 6)² = (2 / 6)² = (1/3)² = 0.11
I	7	8	9	(7 + 4*8 + 9) / 6 = 48 / 6 = 8	((9 - 7) / 6)² = (2 / 6)² = (1/3)² = 0.11

Network Diagram:

The network diagram illustrates the sequence of activities and their dependencies. Each node represents an event, and arrows represent activities.

Start --> A (7) --> B (9) --> D (21) --> H (3) --> I (8) --> End
          |          |                 /
          |          |                /
          |          --> E (19) --> G (9) --> I (8)
          |                           ^
          --> C (9) --> F (11) --------

Paths and their Durations:

We need to identify all possible paths from start to end and calculate their total expected durations:

• Path 1: A-B-D-H-I
  • Duration = Te(A) + Te(B) + Te(D) + Te(H) + Te(I) = 7 + 9 + 21 + 3 + 8 = 48 days
  • Variance = Var(A) + Var(B) + Var(D) + Var(H) + Var(I) = 0.44 + 2.78 + 2.78 + 0.11 + 0.11 = 6.22
• Path 2: A-B-E-G-I
  • Duration = Te(A) + Te(B) + Te(E) + Te(G) + Te(I) = 7 + 9 + 19 + 9 + 8 = 52 days
  • Variance = Var(A) + Var(B) + Var(E) + Var(G) + Var(I) = 0.44 + 2.78 + 7.11 + 1.78 + 0.11 = 12.22
• Path 3: A-C-F-H-I
  • Duration = Te(A) + Te(C) + Te(F) + Te(H) + Te(I) = 7 + 9 + 11 + 3 + 8 = 38 days
  • Variance = Var(A) + Var(C) + Var(F) + Var(H) + Var(I) = 0.44 + 1.78 + 1.78 + 0.11 + 0.11 = 4.22

Critical Path and Expected Project Completion Time:

The Critical Path is the longest path through the network, determining the minimum time required to complete the project. From the calculations above:

• Path 1: 48 days
• Path 2: 52 days
• Path 3: 38 days

The longest path is A-B-E-G-I with an expected duration of 52 days.

Therefore, the Critical Path is A-B-E-G-I, and the Expected Project Completion Time (Te_project) is 52 days.

The variance of the critical path (σ²_project) = 12.22

The standard deviation of the critical path (σ_project) = sqrt(12.22) ≈ 3.496 days.

(ii) Probability that the project will be completed in 61 days

To calculate this probability, we use the Z-score formula, assuming the project completion time follows a normal distribution:

Z = (Ts - Te_project) / σ_project

Where:

• Ts = Target completion time = 61 days
• Te_project = Expected project completion time = 52 days
• σ_project = Standard deviation of the critical path = 3.496 days

Z = (61 - 52) / 3.496 = 9 / 3.496 ≈ 2.574

Now, we need to find the probability associated with a Z-score of 2.574 from the standard normal distribution table. (Using a standard Z-table or calculator, the cumulative probability for Z = 2.57 is approximately 0.9949 or 99.49%).

Therefore, the probability that the project will be completed in 61 days is approximately 99.49%.

(iii) When the company wants to be 95% sure that the system will be installed by a certain date, how many days prior to that should it start the work?

This question asks for the target completion time (Ts) such that there is a 95% probability of project completion. First, we need to find the Z-score corresponding to a 95% cumulative probability (0.95) from the standard normal distribution table. For a cumulative probability of 0.95, the Z-score is approximately 1.645.

Now, we use the Z-score formula rearranged to solve for Ts:

Ts = Te_project + Z * σ_project

Where:

• Te_project = 52 days
• Z = 1.645
• σ_project = 3.496 days

Ts = 52 + 1.645 * 3.496

Ts = 52 + 5.751 ≈ 57.751 days

Rounding up to the next whole day to ensure 95% certainty, the project should be scheduled for approximately 58 days.

Therefore, to be 95% sure that the system will be installed by a certain date, Quick Industries should plan to start the work approximately 58 days prior to that desired completion date.