# If the optimistic time, most likely time and pessimistic time or activity A are 4, 6 and 8 weeks. Respectively and for activity B are 5, 5.5 and 9 wee

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If the optimistic time, most likely time and pessimistic time or activity A are 4, 6 and 8 weeks. Respectively and for activity B are 5, 5.5 and 9 weeks respectively, then
1. Expected time of activity A is greater than the expected time of activity B
2. Expected time of activity B is greater than the expected time of activity A
3. Expected time of both activities A and B are same
4. Data too inadequate to compute expected times of activities

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Correct Answer - Option 3 : Expected time of both activities A and B are same

Concept:

(i) PERT is based upon probabilistic approach in which three time estimate are made for each activity i.e. optimistic time, most likely time and pessimistic time.

(ii) Based on these three an expected mean time is computed which is weighted average of above three time.

${t_e} = \frac{{{t_o} + 4{t_L} + {t_p}}}{6}$

where, to = Optimistic time

tL = Most likely time

tp = Pessimistic time

Calculation:

For activity A:

Given, to = 4 week, tL = 6 week and tp = 8 week

∴ Expected time for activity A, ${t_e} = \frac{{{t_o} + 4{t_L} + {t_p}}}{6}$

⇒ ${t_e} = \frac{{4 + 4 \times 6 + 8}}{6} = 6\ week$

For activity B:

Given, to = 5 week, tL = 5.5 week and tp = 9 week

∴ Expected time for activity A, ${t_e} = \frac{{{t_o} + 4{t_L} + {t_p}}}{6}$

⇒ ${t_e} = \frac{{5 + 4 \times 5.5 + 9}}{6} = 6\ week$

Hence Expected time for both the activity is same.