You are creating a customer database for Modesto Nuts minor league baseball team. Draw a project network given the information below. Complete the forward and backward pass, compute activity slack, and identify the critical path.

How long will this project take? How sensitive is the network schedule? Calculate the free slack and total slack for all noncritical activities.

Draw a project network Complete the forward and backward pass, compute activity slack, and identify the critical path
ID Description Predecessor Time (days)
A Systems design NONE 2
B Subsystem A design A 1
C Subsystem B design A 1
D Subsystem C design A 1
E Program A B 2
F Program B C 2
G Program C D 2
H Subsystem A test E 1
I Subsystem B test F 1
J Subsystem C test G 1
K Integration H,I,J 2
L Integration test K 1

To draw a project network, we need to create nodes for each activity and connect them with arrows to represent the dependencies between activities. The nodes will represent the start and end of each activity, and the arrows will show the flow of the project.

Here is the project network for the customer database creation for the Modesto Nuts minor league baseball team:

A (2)
/ \
/ \
/ \
B (1) C (1) D (1)
| | |
| | |
E (2) F (2) G (2)
| | |
H (1) I (1) J (1)
\ | /
\ | /
\ | /
\ | /
K (2)
|
|
L (1)

To complete the forward and backward pass, we need to calculate the earliest start (ES) and earliest finish (EF) times for each activity.

For the forward pass:
- ES(A) = 0 (since it has no predecessors)
- EF(A) = ES(A) + Time(A) = 0 + 2 = 2

- ES(B) = EF(A) = 2
- EF(B) = ES(B) + Time(B) = 2 + 1 = 3

- ES(C) = EF(A) = 2
- EF(C) = ES(C) + Time(C) = 2 + 1 = 3

- ES(D) = EF(A) = 2
- EF(D) = ES(D) + Time(D) = 2 + 1 = 3

- ES(E) = EF(B) = 3
- EF(E) = ES(E) + Time(E) = 3 + 2 = 5

- ES(F) = EF(C) = 3
- EF(F) = ES(F) + Time(F) = 3 + 2 = 5

- ES(G) = EF(D) = 3
- EF(G) = ES(G) + Time(G) = 3 + 2 = 5

- ES(H) = EF(E) = 5
- EF(H) = ES(H) + Time(H) = 5 + 1 = 6

- ES(I) = EF(F) = 5
- EF(I) = ES(I) + Time(I) = 5 + 1 = 6

- ES(J) = EF(G) = 5
- EF(J) = ES(J) + Time(J) = 5 + 1 = 6

- ES(K) = max(EF(H), EF(I), EF(J)) = 6
- EF(K) = ES(K) + Time(K) = 6 + 2 = 8

- ES(L) = EF(K) = 8
- EF(L) = ES(L) + Time(L) = 8 + 1 = 9

For the backward pass, we need to calculate the latest start (LS) and latest finish (LF) times for each activity.

For the backward pass:
- LF(L) = EF(L) = 9
- LS(L) = LF(L) - Time(L) = 9 - 1 = 8

- LF(K) = LS(L) = 8
- LS(K) = LF(K) - Time(K) = 8 - 2 = 6

- LF(J) = LS(K) = 6
- LS(J) = LF(J) - Time(J) = 6 - 1 = 5

- LF(I) = LS(K) = 6
- LS(I) = LF(I) - Time(I) = 6 - 1 = 5

- LF(H) = LS(K) = 6
- LS(H) = LF(H) - Time(H) = 6 - 1 = 5

- LF(G) = LS(J) = 5
- LS(G) = LF(G) - Time(G) = 5 - 2 = 3

- LF(F) = LS(I) = 5
- LS(F) = LF(F) - Time(F) = 5 - 2 = 3

- LF(E) = LS(H) = 5
- LS(E) = LF(E) - Time(E) = 5 - 2 = 3

- LF(D) = LS(G) = 3
- LS(D) = LF(D) - Time(D) = 3 - 1 = 2

- LF(C) = LS(F) = 3
- LS(C) = LF(C) - Time(C) = 3 - 1 = 2

- LF(B) = LS(E) = 3
- LS(B) = LF(B) - Time(B) = 3 - 1 = 2

- LF(A) = min(LS(B), LS(C), LS(D)) = 2
- LS(A) = LF(A) - Time(A) = 2 - 2 = 0

To compute the activity slack, we subtract the earliest start (ES) from the latest start (LS) for each activity.

Activity Slack:
- Slack(A) = LS(A) - ES(A) = 0 - 0 = 0
- Slack(B) = LS(B) - ES(B) = 2 - 2 = 0
- Slack(C) = LS(C) - ES(C) = 2 - 2 = 0
- Slack(D) = LS(D) - ES(D) = 2 - 2 = 0
- Slack(E) = LS(E) - ES(E) = 3 - 3 = 0
- Slack(F) = LS(F) - ES(F) = 3 - 3 = 0
- Slack(G) = LS(G) - ES(G) = 3 - 3 = 0
- Slack(H) = LS(H) - ES(H) = 5 - 5 = 0
- Slack(I) = LS(I) - ES(I) = 5 - 5 = 0
- Slack(J) = LS(J) - ES(J) = 5 - 5 = 0
- Slack(K) = LS(K) - ES(K) = 6 - 6 = 0
- Slack(L) = LS(L) - ES(L) = 8 - 8 = 0

The critical path is the longest path in the project network, which is determined by the activities with zero slack. In this case, the critical path is:

A - B - E - H - K - L

To calculate the project duration, we take the latest finish time (LF) of the last activity on the critical path, which is activity L:

Project Duration = LF(L) + Time(L) = 9 + 1 = 10 days

The network schedule sensitivity can be determined by examining the total slack time for non-critical activities.

Total Slack for non-critical activities:
- Total Slack(A) = LS(A) - ES(A) = 2 - 0 = 2
- Total Slack(B) = LS(B) - ES(B) = 2 - 2 = 0
- Total Slack(C) = LS(C) - ES(C) = 2 - 2 = 0
- Total Slack(D) = LS(D) - ES(D) = 2 - 2 = 0
- Total Slack(E) = LS(E) - ES(E) = 3 - 3 = 0
- Total Slack(F) = LS(F) - ES(F) = 3 - 3 = 0
- Total Slack(G) = LS(G) - ES(G) = 3 - 3 = 0
- Total Slack(H) = LS(H) - ES(H) = 5 - 5 = 0
- Total Slack(I) = LS(I) - ES(I) = 5 - 5 = 0
- Total Slack(J) = LS(J) - ES(J) = 5 - 5 = 0
- Total Slack(K) = LS(K) - ES(K) = 6 - 6 = 0

From the above calculations, we can see that the total slack for all non-critical activities is either zero or positive, indicating that these activities have some flexibility in their scheduling. However, any delay in the critical path will directly affect the project duration, making the network schedule sensitive.