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Chapter 3
Question 3, Problem 3.17b

Ross Hopkins, president of Hopkins Hospitality, has developed the tasks, durations, and predecessor relationships in the following table for building new motels.

[tex]\[
\begin{array}{|c|c|c|c|c|c|c|c|c|c|}
\hline
\multirow{2}{\ \textless \ em\ \textgreater \ }{Activity} & \multicolumn{3}{|c|}{Time (weeks)} & \multirow{2}{\ \textless \ /em\ \textgreater \ }{Immediate Predecessor(s)} & \multirow{2}{\ \textless \ em\ \textgreater \ }{Activity} & \multicolumn{3}{|c|}{Time (weeks)} & \multirow{2}{\ \textless \ /em\ \textgreater \ }{Immediate Predecessor(s)} \\
\cline{2-4} \cline{7-9}
& a & m & b & & & a & m & b & \\
\hline
A & 4 & 9 & 10 & - & G & 3 & 3 & 4 & C, E \\
\hline
B & 2 & 9 & 24 & A & H & 2 & 2 & 2 & F \\
\hline
C & 9 & 12 & 18 & A & I & 5 & 5 & 5 & F \\
\hline
D & 4 & 7 & 10 & A & J & 6 & 7 & 14 & D, G, H \\
\hline
E & 1 & 3 & 4 & B & K & 1 & 1 & 4 & I, J \\
\hline
F & 5 & 8 & 20 & C, E & & & & & \\
\hline
\end{array}
\][/tex]

a) The expected (estimated) time for activity \(C\) is 12.5 weeks. (Round your response to two decimal places.)

b) The variance for activity \(C\) is 2.25 weeks. (Round your response to two decimal places.)

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 is [tex]\(\square\)[/tex] weeks. (Round your response to two decimal places.)



Answer :

GinnyAnswer
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.

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