Question

In: Statistics and Probability

Sager Products has been in the business of manufacturing and marketing toys for toddlers for the...

Sager Products has been in the business of manufacturing and marketing toys for toddlers for the past two decades. Jim Sager, president of the firm, is considering the development of a new manufacturing line to allow it to produce high-quality plastic toys at reasonable prices. The development process is long and complex. Jim estimates that there are five phases involved and multiple activities for each phase.

Phase 1 of the development process involves the completion of four activities. These activities have no immediate predecessors. Activity A has an optimistic completion time of 2 weeks, a probable completion time of 3 weeks, and a pessimistic completion time of 4 weeks. Activity B has estimated completion times of 5, 6, and 8 weeks; these represent optimistic, probable, and pessimistic time estimates. Similarly, activity C has estimated completion times of 1 week, 1 week, and 2 weeks; and activity D has expected completion times of 8 weeks, 9 weeks, and 11 weeks.

Phase 2 involves six separate activities. Activity E has activity A as an immediate predecessor. Time estimates are 1 week, 1 week, and 4 weeks. Activity F and activity G both have activity B as their immediate predecessor. For activity F, the time estimates are 3 weeks, 3 weeks, and 4 weeks. For activity G, the time estimates are 1 week, 2 weeks, and 2 weeks. The only immediate predecessor for activity H is activity C. Time estimates for activity H are 5 weeks, 5 weeks, and 6 weeks. Activity D must be performed before activity I and activity J can be started. Activity I has estimated completion times of 9 weeks, 10 weeks, and 11 weeks. Activity J has estimated completion times of 1 week, 2 weeks, and 2 weeks.

Phase 3 is the most difficult and complex of the entire development project. It also consists of six separate activities. Activity K has three time estimates of 2 weeks, 2 weeks, and 3 weeks. The immediate predecessor for this activity is activity E. The immediate predecessor for activity L is activity F. The time estimates for activity L are 3 weeks, 4 weeks, and 6 weeks. Activity M has 2 weeks, 2 weeks, and 4 weeks for the estimates of the optimistic, probable and pessimistic time estimates. The immediate predecessor for activity M is activity G. Activities N and O both have activity I as their immediate predecessor. Activity N has 8 weeks, 9 weeks, and 11 weeks for its three time estimates. Activity O has 1 week, 1 week, and 3 weeks as its time estimates. Finally, activity P has time estimates of 4 weeks, 4 weeks, and 8 weeks. Activity J is its only immediate predecessor.

Phase 4 involves five activities. Activity Q requires activity K to be completed before it can be started. The three time estimates for activity Q are 6 weeks, 6 weeks, and 7 weeks. Activity R requires that both activity L and activity M be completed first. The three time estimates for activity R are 1, 2, and 4 weeks. Activity S requires activity N to be completed first. Its time estimates are 6 weeks, 6 weeks, and 7 weeks. Activity T requires that activity O be completed. The time estimates for activity T are 3 weeks, 3 weeks, and 4 weeks. The final activity for phase 4 is activity U. The time estimates for this activity are 1 week, 2 weeks, and 3 weeks. Activity P must be completed before activity U can be started.

Phase 5 is the final phase of the development project. It consists of only two activities. Activity V requires that activity Q and activity R be completed before it can be started. Time estimates for this activity are 9 weeks, 10 weeks, and 11 weeks. Activity W is the final activity of the process. It requires three activities to be completed before it can be started. These are activities S, T, and U. The estimated completion times for activity W are 2 weeks, 4 weeks, and 5 weeks.

  1. Develop a table showing the activities of the project, the duration of each activity, the ES, EF, LS, LF and slack times.
  2. Determine the expected completion time for the entire process. Show the variance for each activity and compute for the total project variance.
  3. Determine the critical path and identify the activities on the critical path.

Solutions

Expert Solution

Solution :

Activity Immediate predecessor Optimistic Most probable Pessimistic Expected time Variance
A - 2 3 4 3 0.6667
B - 5 6 8 6.166666667 1.5
C - 1 1 2 1.166666667 0.1667
D - 8 9 11 9.166666667 1.5
E A 1 1 4 1.5 1.5
F B 3 3 4 3.166666667 0.1667
G B 1 2 2 1.833333333 0.1667
H C 5 5 6 5.166666667 0.1667
I D 9 10 11 10 0.6667
J D 1 2 2 1.833333333 0.1667
K E 2 2 3 2.166666667 0.1667
L F 3 4 6 4.166666667 1.5
M G 2 2 4 2.333333333 0.6667
N I 8 9 11 9.166666667 1.5
O I 1 1 3 1.333333333 0.6667
P J 4 4 8 4.666666667 2.6667
Q K 6 6 7 6.166666667 0.1667
R L,M 1 2 4 2.166666667 1.5
S N 6 6 7 6.166666667 0.1667
T O 3 3 4 3.166666667 0.1667
U P 1 2 3 2 0.6667
V Q,R 9 10 11 10 0.6667
W S,T,U 2 4 5 3.833333333 1.5
Activity Immediate predecessor Expected time ES EF LS LF Slack
A - 3 0 3 15.5 18.5 15.5
B - 6.166666667 0 6.166667 12.66667 18.83333 12.66667
C - 1.166666667 0 1.166667 32 33.16667 32
D - 9.166666667 0 9.166667 0 9.166667 0
E A 1.5 3 4.5 18.5 20 15.5
F B 3.166666667 6.166667 9.333333 18.83333 22 12.66667
G B 1.833333333 6.166667 8 22 23.83333 15.83333
H C 5.166666667 1.166667 6.333333 33.16667 38.33333 32
I D 10 9.166667 19.16667 9.166667 19.16667 0
J D 1.833333333 9.166667 11 26 27.83333 16.83333
K E 2.166666667 4.5 6.666667 20 22.16667 15.5
L F 4.166666667 9.333333 13.5 22 26.16667 12.66667
M G 2.333333333 8 10.33333 23.83333 26.16667 15.83333
N I 9.166666667 19.16667 28.33333 19.16667 28.33333 0
O I 1.333333333 19.16667 20.5 30 31.33333 10.83333
P J 4.666666667 11 15.66667 27.83333 32.5 16.83333
Q K 6.166666667 6.666667 12.83333 22.16667 28.33333 15.5
R L,M 2.166666667 13.5 15.66667 26.16667 28.33333 12.66667
S N 6.166666667 28.33333 34.5 28.33333 34.5 0
T O 3.166666667 20.5 23.66667 31.33333 34.5 10.83333
U P 2 15.66667 17.66667 32.5 34.5 16.83333
V Q,R 10 15.66667 25.66667 28.33333 38.33333 12.66667
W S,T,U 3.833333333 34.5 38.33333 34.5 38.33333 0

expected completion time = 38.33333

the total project variance = 1.5 + 0.6667 + 1.5 + 0.1667 + 1.5 = 5.3333

The critical path avoids project delays by identifying the tasks that are critical for the project schedule. By establishing the duration of each activity on the critical path you can determine the project completion time.

Critical Path = D - I - N - S - W


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