Question

0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 Lecture Exercise #14 0.0 0.1 0.2 0.3 0.4 .5120 5517 .5910 .6293 .6664 .7019 .7357 .7673 .7967 5199 5596 5987 .6368 .6736 .7088 .7422 .7734 Activit Predecessor Time (Days) 0.6 у .8023 8238 a m .5000 5398 .5793 .6179 .6554 .6915 .7257 .7580 .7881 .8159 .8413 .8643 .8849 9032 .9192 9332 .9452 .9554 .9641 9713 5239 .5636 .6026 .6406 .6772 7123 .7454 .7764 .8051 .8315 .8554 .8770 .8962 .9131 .9279 9406 b 5040 .5438 .5832 .6217 .6591 .6950 .7291 .7611 .7910 8186 .8438 .8665 .8869 9049 9207 9345 9463 9564 9649 .9719 5319 .5714 6103 .6480 .6844 .7190 .7517 .7823 .8106 .8365 .8599 .8810 .8997 9162 9306 .5080 .5478 5871 .6255 .6628 .6985 .7324 -7642 7939 .8212 .8461 .8686 .8888 9066 .9222 .9357 .9474 9573 .9656 .9726 5160 5557 5948 .6331 .6700 .7054 .7389 .7704 .7995 .8264 .8508 .8729 .8925 .9099 9251 9382 9495 9591 .9671 9738 0.7 0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 5279 5675 .6064 .6443 .6808 .7157 .7486 .7794 .8078 .8340 .8577 .8790 .8980 9147 .9292 .9418 .9525 .9616 .9693 .9756 A .5359 .5753 .6141 .6517 .6879 .7224 .7549 .7852 .8133 .8389 .8621 .8830 9015 9177 .9319 .9441 .9545 .9633 9706 .9767 None 9 10 11 .8289 .8531 .8749 .8944 9115 9265 .9394 9505 9599 .9678 .9744 .8485 .8708 .8907 9082 9236 B A 6 12 18 9515 с A 9 10 11 9370 9484 .9582 .9664 .9732 9608 .9686 9750 9429 9535 .9625 9699 9761 D 5 8 11 0 . B, C What is the expected project completion time? What is the variance of the project completion time? What is the standard deviation of the project completion time? • What is the probability of the project finishing in 33 days or shorter? • What is the probability of the project finishing in 28 days or shorter? .

Answer 1

Based on the given data, we find the expected duration and variance for each activity as shown below:

Variance Activity Optimistic Most Likely Pessimistic Activity Predecessor m b Expected duration a + 4m + b 6 10 o² = (a - by? a 11 18 A А B с D 10 12 10 12 9 6 9 5 0.11 4.00 0.11 A 11 10 B,C 8 11 8 1.00

Based on the above expected duration, we prepare a project entowrk diagram as shown below:

Legend EST ID EFT SLK LST DUR LFT 10 B 22 0 10 12 22 10 D 30 0A 0 0 10 22 0 22 10 с 20 8 30 10 2 12 10 22

The above network diagram in form of formulas is shown below for better understanding and reference

A с D E F G H K 1 B Legend ID EFT N 3 4 EST SLK LST DUR LFT B =E4+F6 5 =C6 =E6-E4 =G6-F6 12 =18 60 A=A6+B8 7 =A8-A6 8 =C8-B8 10 =MIN(E6,E10) 9 10 =MAX(G4,G8) D =16+J8 =18-16 =K8-18 8 =K6 =C6 C =E8+F10 =E10-E8 =G10-F10 10 =18

The above project diagram is prepared as per the legend shown in the top-left corner.

EST = Early start time

LST = Late start time

EFT = Early finish time

LFT = Late finish time

SLK = Slack = LST - EST or Slack = LFT - EFT

The EST of an activity = EFT of previous activity

EFT of an activity = EST + Duration

Similarly, LFT & LST are calculated during the backward pass.

There are 2 paths in the diagram. However, path A-B-D is the longest path and has 0 slack. Hence, this is the critical path of the project.

As seen from the network diagram, the expected project completion time = 30 days

The variance of project completion time = Sum of Variances over critical path = Variance for activity A + Variance for Activity B + Variance for activity D = 0.11 + 4 + 1 = 5.11

Std. Deviation of project completion time = $\sigma$ =  
ariance
= $\sqrt{5.11}$ = 2.261

We find the probability of completion in 33 days or less as shown below:

R = 33 days

E = 30 days

Z value = $\frac{R-E}{\sigma }$ = $\frac{33-30}{2.261}$ = 1.33

For Z = 1.33, Probability from Z table chart = 0.9082 = 90.82%

We find the probability of completion in 28 days or less as shown below:

R = 28 days

E = 30 days

Z value = $\frac{R-E}{\sigma }$ = $\frac{28-30}{2.261}$ = (-0.89)

For Z = (-0.89), Probability from Z table chart = (1 - Probability for Z=0.89) = (1 - 0.8133) = 0.1867 = 18.67%

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