For an activity, the estimates for optimistic, pessimistic and most likely times are 2, 12 and 5 days. The expected time and variance of this activity are

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BPSC AE Paper 4 (General Engineering Science) 10 Nov 2022 Official Paper
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  1. 5.7 days and 2.8 respectively assuming beta - distribution for time estimates
  2. 6.7 days and 4.5 respectively assuming beta - distribution for time estimates
  3. 5 days and 3.5 respectively assuming beta - distribution for time estimates
  4. 6 days and 4 respectively assuming beta - distribution for time estimates

Answer (Detailed Solution Below)

Option 1 : 5.7 days and 2.8 respectively assuming beta - distribution for time estimates
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Detailed Solution

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Concept:

The PERT (Project Evaluation and Review Technique) technique is used, when activity time estimates are stochastic in nature. For each activity, three values of time (optimistic, most likely, pessimistic) are estimated.

  • Optimistic time (to) estimate is the shortest possible time required for the completion of the activity
  • Most likely time (tm) estimate is the time required for the completion of activity under normal circumstances
  • Pessimistic time (tp) estimate is the longest possible time required for the completion of the activity


In PERT expected time of an activity is determined by using the below-given formula:

\({t_e} = \frac{{\left( {{t_o} + 4{t_m} + {t_p}} \right)}}{6}\)

Variance, σ2 = \({\left[ {\frac{{{t_P} - {t_0}}}{6}} \right]^2}\)

Calculation:

to = 2, tp = 12, and tm = 5

Expected time

\({{\text{t}}_{\text{E}}}=\frac{{{\text{t}}_{\text{o}}}+4{{\text{t}}_{\text{m}}}+{{\text{t}}_{\text{p}}}}{6}=\frac{2+4\left( 5 \right)+12}{6}=5.7\text{ }\!\!~\!\!\text{ days}\)

Variance, σ2 = \({\left[ {\frac{{{t_P} - {t_0}}}{6}} \right]^2}\) = \({\left[ {\frac{{{12} - {2}}}{6}} \right]^2}\)= 2.8
Note: In official paper the given options are not match with correct values, so we updated the option 1.

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