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

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

a) The expected (estimated) time for activity [tex]$C$[/tex] is [tex]$\square$[/tex] weeks. (Round your response to two decimal places.)

Sagot :

GinnyAnswer
To find the expected (estimated) time for activity \( C \), we use the PERT (Program Evaluation and Review Technique) formula for the expected time:

[tex]\[ TE = \frac{a + 4m + b}{6} \][/tex]

where
- \( a \) is the optimistic time,
- \( m \) is the most likely time, and
- \( b \) is the pessimistic time for the activity.

For activity \( C \), the given times are:
- \( a = 9 \) weeks (optimistic time),
- \( m = 12 \) weeks (most likely time),
- \( b = 18 \) weeks (pessimistic time).

Plugging these values into the PERT formula, we get:

[tex]\[ TE = \frac{9 + 4(12) + 18}{6} \][/tex]

First, calculate the values inside the parentheses:

[tex]\[ 9 + 4(12) + 18 = 9 + 48 + 18 = 75 \][/tex]

Next, divide the sum by 6:

[tex]\[ TE = \frac{75}{6} \approx 12.5 \][/tex]

Hence, the expected (estimated) time for activity [tex]\( C \)[/tex] is [tex]\( 12.5 \)[/tex] weeks.
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