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CMA Final · Strategic Cost Management · Network Analysis - PERT, CPM

An activity in a PERT network has optimistic time 10 weeks, most likely time 16 weeks and pessimistic time 28 weeks. What are its expected time and variance respectively?

Expected time is 17 weeks and variance is 9 weeks squared. Expected time is (10 + 4x16 + 28)/6 = 17. The standard deviation is (28 - 10)/6 = 3, and variance is its square, 9. Quoting 3 would confuse standard deviation with variance.

  1. A17 weeks and 9 weeks squaredCorrect
  2. B18 weeks and 9 weeks squared
  3. C17 weeks and 3 weeks squared
  4. D18 weeks and 3 weeks squared

Explanation

Expected time = (10 + 64 + 28)/6 = 102/6 = 17 weeks. Variance = ((b - a)/6)^2 = (18/6)^2 = 3^2 = 9. Option with 3 weeks squared reports the standard deviation as the variance.

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