If an activity has its optimistic, most likely and pessimistic times as 2, 3 and 7 respectively, then its expected time and variance are respectively
A. 3.5 and $$\frac{5}{6}$$
B. 5 and $$\frac{{25}}{{36}}$$
C. 3.5 and $$\frac{{25}}{{36}}$$
D. 4 and $$\frac{5}{6}$$
Answer: Option C
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Comments ( 2 )
The normal time required for the completion of project in the above problem is
A. 9 days
B. 13 days
C. 14 days
D. 19 days
A. $$\frac{{{{\text{t}}_{\text{o}}} + 3{{\text{t}}_{\text{m}}} + {{\text{t}}_{\text{p}}}}}{2}$$
B. $$\frac{{{{\text{t}}_{\text{o}}} + 3{{\text{t}}_{\text{m}}} + {{\text{t}}_{\text{p}}}}}{3}$$
C. $$\frac{{{{\text{t}}_{\text{o}}} + 4{{\text{t}}_{\text{m}}} + {{\text{t}}_{\text{p}}}}}{4}$$
D. $$\frac{{{{\text{t}}_{\text{o}}} + 4{{\text{t}}_{\text{m}}} + {{\text{t}}_{\text{p}}}}}{5}$$
E. $$\frac{{{{\text{t}}_{\text{o}}} + 4{{\text{t}}_{\text{m}}} + {{\text{t}}_{\text{p}}}}}{6}$$
A construction schedule is prepared after collecting
A. Number of operations
B. Output of labour
C. Output of machinery
D. All the above
A. 3.5 and $$\frac{5}{6}$$
B. 5 and $$\frac{{25}}{{36}}$$
C. 3.5 and $$\frac{{25}}{{36}}$$
D. 4 and $$\frac{5}{6}$$
But In CPM method is shown the longest duration that means answer should be Maximum time
expected time = [2 + 7 + 4(3) ]/6 = 3.5
Variance = [(tp - to)/6]^2
= [(7-2)/6]^2 = (5/6) ^ 2 = 25/36
so right option is C