Answer :
Sure, let's address each part of the question step-by-step:
a) The expected (estimated) time for activity \( C \):
The expected time \( t_e \) for activity \( C \) is calculated using the PERT (Program Evaluation and Review Technique) formula:
[tex]\[ t_e = \frac{a + 4m + b}{6} \][/tex]
Given the values:
- \( a = 9 \)
- \( m = 12 \)
- \( b = 18 \)
Using the formula:
[tex]\[ t_e = \frac{9 + 4(12) + 18}{6} \][/tex]
[tex]\[ t_e = \frac{9 + 48 + 18}{6} \][/tex]
[tex]\[ t_e = \frac{75}{6} \][/tex]
[tex]\[ t_e = 12.5 \][/tex]
So, the expected time for activity \( C \) is 12.5 weeks.
b) The variance for activity \( C \):
The variance \( \sigma^2 \) for activity \( C \) is calculated using the formula:
[tex]\[ \sigma^2 = \left(\frac{b - a}{6}\right)^2 \][/tex]
Given the values:
- \( a = 9 \)
- \( b = 18 \)
Using the formula:
[tex]\[ \sigma^2 = \left(\frac{18 - 9}{6}\right)^2 \][/tex]
[tex]\[ \sigma^2 = \left(\frac{9}{6}\right)^2 \][/tex]
[tex]\[ \sigma^2 = \left(1.5\right)^2 \][/tex]
[tex]\[ \sigma^2 = 2.25 \][/tex]
So, the variance for activity \( C \) is 2.25 weeks.
c) Based on the calculation of the estimated times, the critical path is \( A-B-E-F-H-J-K \).
d) The estimated time for the critical path:
To determine the estimated time for the critical path \( A-B-E-F-H-J-K \), we sum the expected times \( t_e \) of each activity on this path.
Given the expected times:
- \( t_e(A) = 9 \text{ weeks} \)
- \( t_e(B) = 10.67 \text{ weeks} \)
- \( t_e(E) = 2.5 \text{ weeks} \)
- \( t_e(F) = 9.5 \text{ weeks} \)
- \( t_e(H) = 2 \text{ weeks} \)
- \( t_e(J) = 9 \text{ weeks} \)
- \( t_e(K) = 1.83 \text{ weeks} \)
Using the provided calculations directly,
The total estimated time for the critical path \( A-B-E-F-H-J-K \) is:
[tex]\[ 9 + 10.67 + 2.5 + 9.5 + 2 + 9 + 1.83 = 42.5 \][/tex]
So, the estimated time for the critical path is 42.5 weeks.
a) The expected (estimated) time for activity \( C \):
The expected time \( t_e \) for activity \( C \) is calculated using the PERT (Program Evaluation and Review Technique) formula:
[tex]\[ t_e = \frac{a + 4m + b}{6} \][/tex]
Given the values:
- \( a = 9 \)
- \( m = 12 \)
- \( b = 18 \)
Using the formula:
[tex]\[ t_e = \frac{9 + 4(12) + 18}{6} \][/tex]
[tex]\[ t_e = \frac{9 + 48 + 18}{6} \][/tex]
[tex]\[ t_e = \frac{75}{6} \][/tex]
[tex]\[ t_e = 12.5 \][/tex]
So, the expected time for activity \( C \) is 12.5 weeks.
b) The variance for activity \( C \):
The variance \( \sigma^2 \) for activity \( C \) is calculated using the formula:
[tex]\[ \sigma^2 = \left(\frac{b - a}{6}\right)^2 \][/tex]
Given the values:
- \( a = 9 \)
- \( b = 18 \)
Using the formula:
[tex]\[ \sigma^2 = \left(\frac{18 - 9}{6}\right)^2 \][/tex]
[tex]\[ \sigma^2 = \left(\frac{9}{6}\right)^2 \][/tex]
[tex]\[ \sigma^2 = \left(1.5\right)^2 \][/tex]
[tex]\[ \sigma^2 = 2.25 \][/tex]
So, the variance for activity \( C \) is 2.25 weeks.
c) Based on the calculation of the estimated times, the critical path is \( A-B-E-F-H-J-K \).
d) The estimated time for the critical path:
To determine the estimated time for the critical path \( A-B-E-F-H-J-K \), we sum the expected times \( t_e \) of each activity on this path.
Given the expected times:
- \( t_e(A) = 9 \text{ weeks} \)
- \( t_e(B) = 10.67 \text{ weeks} \)
- \( t_e(E) = 2.5 \text{ weeks} \)
- \( t_e(F) = 9.5 \text{ weeks} \)
- \( t_e(H) = 2 \text{ weeks} \)
- \( t_e(J) = 9 \text{ weeks} \)
- \( t_e(K) = 1.83 \text{ weeks} \)
Using the provided calculations directly,
The total estimated time for the critical path \( A-B-E-F-H-J-K \) is:
[tex]\[ 9 + 10.67 + 2.5 + 9.5 + 2 + 9 + 1.83 = 42.5 \][/tex]
So, the estimated time for the critical path is 42.5 weeks.
