Question:

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?

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In PERT, certainty about job completion is linked with smaller variance $\sigma^2 = \big(\tfrac{t_p - t_o}{6}\big)^2$. Lesser the variance, more reliable the estimate.
Updated On: Sep 24, 2025
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The Correct Option is B

Solution and Explanation


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.

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