G 3 (J2 jJ to tp tm Activity C 2 7 3 3.5 0.6944 D 2 6 3 3.5 0.4444 J 3 6 5 4.8 0.25 (b) Details of the EST and LST and floats are given below. Activity EST EFT LFT LST Critical? A 0 2 2 0 0 Y B 2 5 16.5 13.5 11.5 N C 2 5.5 5.5 2 0 Y D 2 5.3 5.5 2.2 0.2 N E 5.5 6.5 6.5 5.5 0 Y F 6.5 16.5 16.5 6.5 0 Y G 0 3 16.5 13.5 13.5 N H 16.5 22.5 22.5 16.5 0 Y I 22.5 24.5 24.5 22.5 0 Y J Y 24.5 29.3 29.3 24.5 0 Activities A, C, E, F, H, I are critical. Expected time before the national campaign can be launched is 29.3 days. (c) The variance of the critical path is 0.6944 + 0.25 = 0.9444 (Note: Activity D is not on the critical path), so the standard deviation is 0.9718 To find the probability that expected time is greater than 32 days we need to calculate the Z value which is = 2.77 From tables this give a probability of 0.0028 So there is a 0.28% chance that the expected time will exceed 32 days, which means that there is a probability of 100 0.28 = 99.72% that – the launch can take place within 32 days.
G 3 Activity to tp tm jJ (J2 C D J 2 2 3 7 6 6 3 3 5 3.5 3.5 4.8 0.6944 0.4444 0.25 (b) Details of the EST and LST and floats are given below. Activity EST EFT LFT LST Critical? A B C D E F G H I J 0 2 2 2 5.5 6.5 0 16.5 22.5 24.5 2 5 5.5 5.3 6.5 16.5 3 22.5 24.5 29.3 2 16.5 5.5 5.5 6.5 16.5 16.5 22.5 24.5 29.3 0 13.5 2 2.2 5.5 6.5 13.5 16.5 22.5 24.5 0 11.5 0 0.2 0 0 13.5 0 0 0 Y N Y N Y Y N Y Y Y Activities A, C, E, F, H, I are critical. Expected time before the national campaign can be launched is 29.3 days. (c) The variance of the critical path is 0.6944 + 0.25 = 0.9444 (Note: Activity D is not on the critical path), so the standard deviation is 0.9718 To find the probability that expected time is greater than 32 days we need to calculate the Z value which is = 2.77 From tables this give a probability of 0.0028 So there is a 0.28% chance that the expected time will exceed 32 days, which means that there is a probability of 100 – 0.28 = 99.72% that the launch can take place within 32 days.
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