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

In a PERT network the critical path has an expected duration of 50 days. The variances of its four activities are 4, 9, 4 and 8 days squared. Assuming a normal distribution, what is the probability of completing the project within 55 days? (Normal table: Z = 0.5 gives 0.6915; Z = 1.0 gives 0.8413; Z = 2.0 gives 0.9772.)

The probability is 0.8413. The critical path variance is 25, so the standard deviation is 5 days. The Z value is (55 − 50)/5, which equals 1.0, and the normal table gives 0.8413. The value 0.1587 is the chance of overrunning 55 days.

  1. A0.6915
  2. B0.9772
  3. C0.1587
  4. D0.8413Correct

Explanation

Project variance = 4 + 9 + 4 + 8 = 25, so the standard deviation is 5 days. Z = (55 − 50) / 5 = 1.0, giving a probability of 0.8413. Using 0.6915 would mean a standard deviation of 10, which is wrong. The value 0.1587 is the probability of exceeding 55 days, not of finishing within it.

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