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There are four consecutive activities in a simple linear network, each with mean duration μ and first two with '2k'as the standard deviation, third with 'k' as the standard deviation and fourth has ‘zero’ standard deviation. The overall project duration through these activities is likely to be in the range
  • a)
  • b)
  • c)
  • d)
Correct answer is option 'B'. Can you explain this answer?
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And standard deviation . The expected duration of the entire network is given by:

E(T) = E(A) + E(B) + E(C) + E(D)

Since each activity has the same mean duration, we can simplify the equation to:

E(T) = 4E(A)

To find the variance of the entire network, we use the formula:

Var(T) = Var(A) + Var(B) + Var(C) + Var(D) + 2Cov(A,B) + 2Cov(B,C) + 2Cov(C,D) + 2Cov(A,C) + 2Cov(B,D)

Since each activity has the same standard deviation, we can simplify the equation to:

Var(T) = 4Var(A) + 4Cov(A,B) + 4Cov(B,C) + 4Cov(C,D) + 6Cov(A,C) + 6Cov(B,D)

Assuming that the activities are independent, we can set all the covariances to zero, giving us:

Var(T) = 4Var(A)

Therefore, the standard deviation of the entire network is:

SD(T) = sqrt(Var(T)) = 2sqrt(Var(A)) = 2sigma

where sigma is the standard deviation of each activity.
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There are four consecutive activities in a simple linear network, each with mean duration μ and first two with '2k'as the standard deviation, third with 'k' as the standard deviation and fourth has ‘zero’ standard deviation. The overall project duration through these activities is likely to be in the rangea)b)c)d)Correct answer is option 'B'. Can you explain this answer?
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