The time estimates obtained from four contractors (P, Q, R and S) for executing a particular job are as under:
\[\begin{array}{|c|c|c|c|} \hline \textbf{Contractor} & \textbf{Optimistic time, $t_o$} & \textbf{Most likely time, $t_m$} & \textbf{Pessimistic time, $t_p$} \\ \hline \text{P} & 5 & 10 & 13 \\ \hline \text{Q} & 6 & 9 & 12 \\ \hline \text{R} & 5 & 10 & 14 \\ \hline \text{S} & 4 & 10 & 13 \\ \hline \end{array}\]
Which of these contractors is more certain about completing the job in time?
Step 1: Formula for variance in PERT analysis.
In PERT (Program Evaluation and Review Technique), the variance of activity time is given by:
\[
\sigma^2 = \left(\frac{t_p - t_o}{6}\right)^2
\]
A smaller variance means the contractor is more certain about the job completion time.
Step 2: Calculate variance for each contractor.
- For P:
\[
\sigma^2 = \left(\frac{13 - 5}{6}\right)^2 = \left(\frac{8}{6}\right)^2 = 1.78
\]
- For Q:
\[
\sigma^2 = \left(\frac{12 - 6}{6}\right)^2 = \left(\frac{6}{6}\right)^2 = 1.00
\]
- For R:
\[
\sigma^2 = \left(\frac{14 - 5}{6}\right)^2 = \left(\frac{9}{6}\right)^2 = 2.25
\]
- For S:
\[
\sigma^2 = \left(\frac{13 - 4}{6}\right)^2 = \left(\frac{9}{6}\right)^2 = 2.25
\]
Step 3: Comparison.
- P → Variance = 1.78
- Q → Variance = 1.00 (minimum)
- R → Variance = 2.25
- S → Variance = 2.25
Step 4: Conclusion.
Since contractor Q has the minimum variance, Q is more certain about completing the job in time.
Which of the following statements (pertaining to CPM network analysis) are correct?
A. It is an event-oriented method.
B. It is an activity-oriented method.
C. Time and cost are controlling factors.
D. Time alone is the controlling factor.
Choose the most appropriate answer from the options given below:
A weight of $500\,$N is held on a smooth plane inclined at $30^\circ$ to the horizontal by a force $P$ acting at $30^\circ$ to the inclined plane as shown. Then the value of force $P$ is:
A steel wire of $20$ mm diameter is bent into a circular shape of $10$ m radius. If modulus of elasticity of wire is $2\times10^{5}\ \text{N/mm}^2$, then the maximum bending stress induced in wire is: