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Network errors; expected time of engineers

Short answer

Errors: looping, dangling, redundancy, wrong logic, duplicate i-j, wrong numbering, multiple start/end events. Expected times: NTC 8.83, NCell 5.50 time units. Choose NCell because shorter expected time and lower standard deviation (0.83 vs 1.17), hence more certain.

List out various errors in drawing a network diagram. Find out the expected time of each engineer mentioned below. Which engineer will you choose and why? Who is more certain in completing the job?

Typetotmtp
NTC Engineer5912
NCell Engineer459

Errors in drawing a network diagram

  1. Looping (cycle): a chain of activities returns to an earlier event, e.g. 1-2-3-1. Time cannot go backward.
  2. Dangling (open end): an activity, other than the last, ends at an event with no successor, or a non-first activity starts at an event with no predecessor.
  3. Redundancy: an unnecessary dummy or a repeated dependence that does not change the logic.
  4. Wrong logic: a wrong predecessor or successor is shown, or a dummy is missing, so dependence is incorrect.
  5. Same event numbers (duplicate i-j): two activities with the same start and end event, without a dummy.
  6. Wrong numbering: the head event number is smaller than the tail event number (i > j).
  7. Multiple start or end events: more than one starting or finishing event.
  8. Unnecessary crossing of arrows, or a dummy shown with a duration (a dummy must have zero duration).
Looping:   (1)->(2)->(3)
            ^         |
            +---------+
Dangling:  (1)->(2)->(3)->(5)
                 \->(4)   (4 has no successor)

Expected time (PERT)

te=to+4tm+tp6t_e = \dfrac{t_o + 4t_m + t_p}{6}, σ=tp−to6\sigma = \dfrac{t_p - t_o}{6}.

NTC Engineer (5,9,125, 9, 12): te=5+36+126=536=8.83t_e = \dfrac{5 + 36 + 12}{6} = \dfrac{53}{6} = 8.83, σ=12−56=1.17\sigma = \dfrac{12 - 5}{6} = 1.17, variance =1.36= 1.36

NCell Engineer (4,5,94, 5, 9): te=4+20+96=336=5.50t_e = \dfrac{4 + 20 + 9}{6} = \dfrac{33}{6} = 5.50, σ=9−46=0.83\sigma = \dfrac{9 - 4}{6} = 0.83, variance =0.69= 0.69

Engineertet_eσ\sigmaVariance
NTC8.831.171.36
NCell5.500.830.69

Choice: the NCell engineer is chosen because the expected completion time (5.50) is much shorter than that of the NTC engineer (8.83). He is also more certain in completing the job since his standard deviation (0.83) and variance (0.69) are smaller.

Answer: tet_e = 8.83 (NTC) and 5.50 (NCell); NCell engineer is better and more certain.

✓ Answer checked by Claude Opus 5.5 (high effort)

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