Question

Rich Cole Control Devices, Inc., produces custom-built relay devices for auto makers. The most recent project undertaken by Cole requires 14 different activities appropriate data are shown in the following table Time (days) Immediate bPredecessor(s)Activitya Time (days) Immediate Activity a bPredecessor(s) E, F G, H 10 17 B, C 5 6 10 6 L, M a) The expected completion time of the project days (round your response to two decimal places) The activities that represent the project's critical path are b) If the time to complete the activities on the critical path is normally distributed, then the probability that the critical path will be finished in 54 days or less- I fr

Answer 1

First we calculate the most likely time. This is done by

Most likely time = (a + 4m + b)/6

Then we calculate the project variance by

Variance = ((b – a)/6)^2

Once we have these values we shall calculate the early start, finish, and late start, finish values in order to determine the slack and critical path.

The calculated values for each of them are shown below.

1Activity Predecessor most likely variance Slack 0.00 10.50 12.83 10.50 12.83 0.00 4.17 12.83 22.67 12.83 22.67 0.00 13.67 18.50 17.83 22.67 4.17 6.67 22.67 26.17 22.67 26.170.00 26.17 31.00 26.17 31.00 0.00 1.50 31.00 39.83 31.00 39.83 0.00 1.50 0.00 42.17 50.33 42.17 50.33 0.00 ES EF 5.83 2.33 12.83 13.67 LF 5.83 2.33 0.25 0.44 0.00 1.36 6.25 1.36 2.25 0.25 1.36 1.00 0.69 1.00 0.44 0.69 0.00 5.83 5.83 5.83 10.00 17.83 7.83 9.83 4.83 5.83 3.50 4.83 5.33 8.83 4.33 2.33 8.17 5.83 10 17 10 13.67 19.50 20.33 26.17 10 G,H 31.00 36.33 32.50 37.83 12 36.33 40.67 37.83 42.17 39.83 42.17 39.83 42.17 15 16 6 18 20 Sheet731 Sheet733 Sheet734 Sheet735 Sheet736 Shect737 AVERAGE: 1293680556 COUNT: 96 SUM: 1034.944444囲回凹- +130%

a)

Expected completion time of the project is = 50.33 days

The activities that present the critical path are A-C-E-H-I-K-M-N

b)

We can consider the project completion mean time as 50.33 days. The variance along the critical path is 9.94 days (sum up the variance for ACEHIK)

This means the standard deviation is sqrt(9.94) = 3.15 days.

The z value will be

Z = (54-50.33)/3.15 = 1.16

This means the probability of the project finishing in 54 days or less is P(z) = 0.878 or 87.8%

c)

For 99% probability the z value is 2.32. This means we have

2.32 = (X – 50.33)/3.15

X = 57.65 days

At 57.65 days, there will be a 99% probability of completion.