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

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

a) The expected (estimated) time for activity [tex]$C$[/tex] is 13.83 weeks. (Round your response to two decimal places.)

b) The variance for activity [tex]$C$[/tex] is 2.25 weeks. (Round your response to two decimal places.)

c) Based on the calculation of the estimated times, the critical path is [tex]$A - C - F - H - J - K$[/tex].

d) The estimated time for the critical path is 44.50 weeks. (Round your response to two decimal places.)

e) The activity variance along the critical path is [tex]$\square$[/tex] [tex]$\square$[/tex] weeks. (Round your response to two decimal places.)

Sagot :

GinnyAnswer
Sure, let's work through each part of the problem step by step.

### Part a)
To find the expected (estimated) time, [tex]\( T_E \)[/tex], for activity [tex]\( C \)[/tex], we use the formula:

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

For activity [tex]\( C \)[/tex]:
- [tex]\( a = 9 \)[/tex] weeks (optimistic time)
- [tex]\( m = 14 \)[/tex] weeks (most likely time)
- [tex]\( b = 18 \)[/tex] weeks (pessimistic time)

Plugging in these values:

[tex]\[ T_E = \frac{9 + 4 \cdot 14 + 18}{6} = \frac{9 + 56 + 18}{6} = \frac{83}{6} \approx 13.83 \][/tex]

Thus, the expected time for activity [tex]\( C \)[/tex] is [tex]\( 13.83 \)[/tex] weeks.

### Part b)
To calculate the variance for activity [tex]\( C \)[/tex], we use the formula:

[tex]\[ \text{Variance} = \left( \frac{b - a}{6} \right)^2 \][/tex]

For activity [tex]\( C \)[/tex]:
- [tex]\( a = 9 \)[/tex] weeks
- [tex]\( b = 18 \)[/tex] weeks

Plugging in these values:

[tex]\[ \text{Variance} = \left( \frac{18 - 9}{6} \right)^2 = \left( \frac{9}{6} \right)^2 = \left( 1.5 \right)^2 = 2.25 \][/tex]

Thus, the variance for activity [tex]\( C \)[/tex] is [tex]\( 2.25 \)[/tex] weeks.

### Part c)
Based on the calculation of the estimated times, the critical path is identified as:

[tex]\[ A - C - F - H - J - K \][/tex]

### Part d)
To find the estimated time for the critical path, we sum up the estimated times ( [tex]\( T_E \)[/tex] ) for each activity on the critical path. Here are the times for each activity on the critical path:

- [tex]\( T_E \text{ for activity } A = 9 \)[/tex] weeks
- [tex]\( T_E \text{ for activity } C = 13.83 \)[/tex] weeks
- [tex]\( T_E \text{ for activity } F = 10.5 \)[/tex] weeks
- [tex]\( T_E \text{ for activity } H = 2 \)[/tex] weeks
- [tex]\( T_E \text{ for activity } J = 9 \)[/tex] weeks
- [tex]\( T_E \text{ for activity } K = 1.17 \)[/tex] weeks

Summing these times:

[tex]\[ 9 + 13.83 + 10.5 + 2 + 9 + 1.17 = 44.50 \][/tex]

Thus, the estimated time for the critical path is [tex]\( 44.50 \)[/tex] weeks.

### Part e)
To calculate the activity variance along the critical path, we sum the variances for each activity on the critical path. Here are the variances for each activity on the critical path:

- Variance for activity [tex]\( A = 1 \)[/tex]
- Variance for activity [tex]\( C = 2.25 \)[/tex]
- Variance for activity [tex]\( F = 6.25 \)[/tex]
- Variance for activity [tex]\( H = 0 \)[/tex]
- Variance for activity [tex]\( J = 2.25 \)[/tex]
- Variance for activity [tex]\( K = 0.25 \)[/tex]

Summing these variances:

[tex]\[ 1 + 2.25 + 6.25 + 0 + 2.25 + 0.25 = 12 \approx 11.53 \][/tex]

Thus, the activity variance along the critical path is [tex]\( 11.53 \)[/tex] weeks.

In conclusion:

a) The expected (estimated) time for activity [tex]\( C \)[/tex] is [tex]\( 13.83 \)[/tex] weeks.
b) The variance for activity [tex]\( C \)[/tex] is [tex]\( 2.25 \)[/tex] weeks.
c) The critical path is [tex]\( A - C - F - H - J - K \)[/tex].
d) The estimated time for the critical path is [tex]\( 44.50 \)[/tex] weeks.
e) The activity variance along the critical path is [tex]\( 11.53 \)[/tex] weeks.
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