Given an activity's optimistic, most likely, and pessimistic time estimates of 5, 10, and 15 days respectively, compute the PERT variance for this activity
Given an activity's optimistic, most likely, and pessimistic time estimates of 5, 10, and 15 days...
Given an activity's optimistic, most likely, and pessimistic time estimates of 5, 7, and 15 days respectively, compute the PERT expected activity time (t) and variance of activity completion time (Var) for this activity. 1. t = 8, and Var = 2.78 2. t = 7, and Var = 1.78 3. t = 14, and Var = 2.78 4. t = 8, and Var = 1.78
Given an activity's optimistic, most likely, and pessimistic time estimates of 6, 12, and 18 days respectively, compute the PERT variance for this activity
An activity on a PERT network has these time estimates: optimistic = 2, most likely = 3, and pessimistic = 10. What is its expected activity time and variance? A. 5; 1.78 B. 4; 1.78 C. 5; 10.67 D. 4; 10.67 E. None of the above.
Consider the following precedence chart: ACTIVITY PRECEDING ACTIVITIES OPTIMISTIC TIME (days) PESSIMISTIC TIME days) MOST LIKELY TIME (days) 10 10 10 B.C DE The total slack for activity D is O 2 O O O
Given the following activity list and times in days: (15 points) Activity Optimistic Most Likely Pessimistic Time Time Time A 7 10 12 B 6 8 14 C 5 9 12 D 12 14 21 E 10 12 15 F 4 5 8 G 1 3 8 H 12 15 17 NOTE 1: ASSUME THE CRITTCAL PATH IS “B-E-H”. Calculate the schedule duration of the critical path in days. What is the probability of the project being completed...
An activity on a PERT network has an optimistic time of 10 hours, a most likely time of 13 hours, and a pessimistic time of 28 hours. Find its VARIANCE. 20.25 hour2 25 hour? 9 hour2 16 hour2 12.25 hour2
Project task E has the following estimates: (1) optimistic time - 10 days, (2) Pessimistic time - 20 days, and (3) Most likely time - 12 Days. The standard deviation for time using the beta statistical distribution is ________ days. 2.33 1.5 02 1.67
Activity D on a PERT network has an optimistic duration of 3 days, a most likely duration of 5 days and a pessimistic duration of 13 days. What is the expected duration of activity D? Formula: Expected Activity time: t = (a + 4m + b)/6 a. 5 days b. 6 days c. 7 days d. 10 days e. 12 days
123) Given (Time in weeks): Activity Optimistic Most Likely Pessimistic A 3 4 5 B 6 7 14 C 2 3 10 D 6 9 12 E 4 5 12 F 1 3 11 G 1 2 9 H 2 5 8 I 1 4 7 Determine: (a) the critical path. ( HOW DO I DO THIS WITH OUT A PREDECESSOR?) (b) the probability that the project will be completed in 22 weeks.
TIMES (days) Activity Start Node Finish Node OPTIMISTIC to MOST LIKELY tm PESSIMISTIC tp A 1 2 24 30 42 B 1 3 20 23 32 C 2 5 15 27 39 D 3 4 25 43 61 E 4 5 15 42 69 Given the Activity on Arrow Diagram and the data found in the table above, calculate each of the following values: Activity A………te = 31 Standard Deviation = 3 Activity B………te = Standard Deviation = Activity C………te...