Answer to Question #237097 in Operations Research for opr

Question

A small project is composed of 7 activities whose time estimates in weeks are listed below:

Activity Predecessors Optimistic Most likely Pessimistic

A - 1 2 4

B - 5 6 7

C - 2 4 5

D A 1 3 4

E C 4 5 7

F A 3 4 5

G B,D,E 1 2 3

i) Draw the network.

ii) Calculate the expected duration and variance of every task.

iii) Determine the critical path.

iv) Calculate the expected project duration and the variance of the project duration based on network analysis.

v) Calculate the probability that the project will be completed on or before a deadline of 10 weeks

Answer 1

Project scheduling with uncertain activity times (Optimistic, Most likely, Pessimistic)

Expected time of each activity,

The earliest and latest expected time for each activity is calculated by considering the expected time $t_e$

Edge and it's preceded and succeeded node

(i) The network diagram for the project, along with activity time, is

Forward Pass Method

$E_1=0\\ E_2=E_1+t_{1,2} [t_{1,2}=A=2.17]=0+2.17=2.17\\ E_3=E_1+t_{1,3} [t_{1,3}=C=3.83]=0+3.83=3.83\\$

$E_4=Max[{Ei+t_{i,4}}][i=12,3,]\\ =Max[{E_1+t_1,4;E_2+t_2,4;E_3+t_{3,4}}]\\ =Max[{0+6;2.17+2.83;3.83+5.17}]\\ =Max[{6;5;9}]\\ =9\\ E_5=Max[{E_i+t_{i,5}}][i=24,]\\ =Max[{E_2+t_2,5;E_4+t_{4,5}}]\\ =Max[{2.17+4;9+2}]\\ =Max[{6.17;11}] =11 \\ Backward \ Pass\ Method\\ L_5=E_5=11\\ L_4=L_5-t_{4,5} [t_{4,5}=G=2]=11-2=9\\ L_3=L_4-t_{3,4} [t_{3,4}=E=5.17]=9-5.17=3.83\\ L_2=Min[{L_j-t_{2,j}}][j=54,]\\ =Min[{L_5-t_{2,5};L_4-t_{2,4}}]\\ =Min[{11-4;9-2.83}]\\ =Min[{7;6.17}]\\ =6.17$

$L_1=Min[{L_j-t_{1,j}]}[j=43,2,]\\ =Min[{L_4-t_{1,4};L_3-t_{1,3};L_2-t_{1,2}}]\\ =Min[{9-6;3.83-3.83;6.17-2.17}]\\ =Min[{3;0;4}]\\ =0$

(iii) The critical path in the network diagram has been shown. This has been done by double lines by joining all those events where E-values and L-values are equal.

The critical path of the project is : 1-3-4-5 and critical activities are C,E,G

(iv) The total project time is 11

The network diagram for the project, along with E-values and L-values, is

(ii) Variance & Probability:

(v) Z = (12 − 26.000) / 1.333 ^0.5 = -12.126 (Z is negative here)

Since, the overall expected time of project completion is 26 weeks as calculated above, therefore it is not possible for a project to get finish in 10 weeks.

Probability of Completion: 0.000%