4. An R & D project has a list of tasks to be performed whose time estimates are given in the table
R & D project Time estimates
Activity
i - j
Activity
Name
Time
(days)
To
Time
(days)
tm
Time
(days)
tp
1-2 A 4 6 8
1-3 B 2 3 10
1-4 C 6 8 16
2-4 D 1 2 3
3-4 E 6 7 8
3-5 F 6 7 14
4-6 G 3 5 7
4-7 H 4 11 12
5-7 I 2 4 6
6-7 J 2 9 10
a. Draw the network diagram for the R&D project
b. Find the critical path and calculate the project duration
c. Find the probability that the project is completed in 19 days.
2
3
5
7
6
4
1
A
B
C
D
E
F
G
H
I
J
a. The network diagram for the R&D project is shown below:
b. The critical path is 1-2-4-6-7-J with a project duration of 19 days.
c. The probability that the project is completed in 19 days is 1, since the critical path is 19 days.
a. To draw the network diagram for the R&D project, we need to connect the activities based on their dependencies and show their time estimates.
The network diagram for the R&D project is as follows:
```
A(4) C(6) G(3) I(2)
| | | |
| | | |
v v v v
Start(0) -> B(2) -> D(1) -> E(6) -> J(2) -> Finish(19)
| | | |
| | | |
v v v v
------------------> F(6) --------------------------------
```
b. To find the critical path and calculate the project duration, we need to calculate the earliest start time (ES), earliest finish time (EF), latest start time (LS), and latest finish time (LF) for each activity.
Using the table provided, we can calculate these values as follows:
```
Activity ES EF LS LF Slack
1-2(A) 0 4 0 4 0
1-3(B) 4+1 7 4 7 0
1-4(C) 0+2 8 4 10 2
2-4(D) 4 5 8 9 4
3-4(E) 7 13 10 16 3
3-5(F) 13 19 16 22 3
4-6(G) 5 8 9 12 4
4-7(H) 5+4 13 9 17 4
5-7(I) 19+2 21 19 21 0
6-7(J) 8+2 12 12 16 4
```
The critical path is the longest path through the network diagram where the total duration of the project is equal to the sum of the durations of all activities on that path. In this case, the critical path is:
Start -> 1-4 -> 3-5 -> Finish
The project duration is the total duration of the activities on the critical path. In this case, the project duration is 19 days.
c. To find the probability that the project is completed in 19 days, we need to calculate the probability of completing each activity within its time estimate.
Using the provided time estimates, we can calculate the probabilities as follows:
P(A <= 4 days) = (4 - 6) / (8 - 6) = 0.5
P(B <= 2 days) = (2 - 3) / (10 - 3) = 0.125
P(C <= 6 days) = (6 - 8) / (16 - 8) = 0.5
P(D <= 1 day) = (1 - 2) / (3 - 2) = 0.5
P(E <= 6 days) = (6 - 7) / (8 - 7) = 1
P(F <= 6 days) = (6 - 7) / (14 - 7) = 1
P(G <= 3 days) = (3 - 5) / (7 - 5) = 0
P(H <= 4 days) = (4 - 11) / (12 - 11) = 0
P(I <= 2 days) = (2 - 4) / (6 - 4) = 0
P(J <= 2 days) = (2 - 9) / (10 - 9) = 0
To find the probability that the project is completed in 19 days, we multiply the probabilities of completing all the activities within their time estimates:
P(Project completed in 19 days) = P(A <= 4 days) * P(B <= 2 days) * P(C <= 6 days) * P(D <= 1 day) * P(E <= 6 days) * P(F <= 6 days) * P(G <= 3 days) * P(H <= 4 days) * P(I <= 2 days) * P(J <= 2 days)
= 0.5 * 0.125 * 0.5 * 0.5 * 1 * 1 * 0 * 0 * 0 * 0
= 0
Therefore, the probability that the project is completed in 19 days is 0.
a. To draw the network diagram for the R&D project, we need to use the table provided to identify the activities and their dependencies. The activities are represented by the numbers in the first column, and the dependencies are indicated by the numbers in the second column.
Here is the network diagram for the R&D project:
1
/ \
/ \
A(4) B(2)
| |
2 3
| |
D(1) E(6)
\ /
\ /
4
/ \
/ \
C(6) F(6)
| |
6 5
| |
G(3) I(2)
| |
7 7
\ /
\ /
7
|
J(2)
|
7
The numbers in parentheses represent the time estimates for each activity.
b. To find the critical path and calculate the project duration, we need to determine the earliest start time (ES), earliest finish time (EF), latest start time (LS), and latest finish time (LF) for each activity.
1. Calculate ES and EF:
- ES for activity 1 is 0, as it is the starting activity.
- EF for activity 1 is the time estimate for activity 1, which is 4.
2. Calculate LS and LF:
- LF for the last activity, activity 7, is the same as its EF, which is 19.
- LS for the last activity, activity 7, is LF - time estimate, which is 19 - 2 = 17.
3. Calculate LS and LF for the remaining activities by moving backwards through the network diagram:
- LF for activity 6 is the same as its EF, which is 16.
- LS for activity 6 is LF - time estimate, which is 16 - 3 = 13.
- LF for activity 5 is ES of activity 5, which is 16.
- LS for activity 5 is LF - time estimate, which is 16 - 6 = 10.
- LF for activity 4 is the same as its EF, which is 10.
- LS for activity 4 is LF - time estimate, which is 10 - 1 = 9.
- LF for activity 3 is ES of activity 3, which is 9.
- LS for activity 3 is LF - time estimate, which is 9 - 2 = 7.
- LF for activity 2 is LS of activity 2, which is 7.
- LS for activity 2 is LF - time estimate, which is 7 - 4 = 3.
- LF for activity 1 is LS of activity 1, which is 3.
- LS for activity 1 is LF - time estimate, which is 3 - 4 = -1 (not feasible).
- Therefore, the critical path is: 1-3-5-7
- The project duration is the LF of the last activity, which is 19.
c. To find the probability that the project is completed in 19 days, we need to calculate the cumulative probability of completing each activity within its time estimate.
1. For each activity, calculate the estimated time (tp) minus the minimum time (tm):
- For activity 1, the time difference is 8 - 4 = 4.
- For activity 2, the time difference is 3 - 1 = 2.
- For activity 3, the time difference is 10 - 2 = 8.
- For activity 4, the time difference is 3 - 1 = 2.
- For activity 5, the time difference is 7 - 6 = 1.
- For activity 6, the time difference is 7 - 3 = 4.
- For activity 7, the time difference is 12 - 9 = 3.
2. Calculate the cumulative probability by multiplying the time differences for each activity:
- Probability = (4/4) * (2/2) * (8/8) * (2/2) * (1/1) * (4/4) * (3/3) = 1
Therefore, the probability that the project is completed in 19 days is 100%.