Use the following network answer the following questions: F, 3 B, 6 E, 5 H, 6 G,
ID: 339289 • Letter: U
Question
Use the following network answer the following questions: F, 3 B, 6 E, 5 H, 6 G, 4 D, 7 a(4). Find the critical path and the project completion time. b(6). Make a table of ES, EF, LS, and LF for each activity. c(3). Calculate the slack for activity D d(7). The entrepreneur is considering reducing the time for completion of the project. He can hire an additional worker to help reduce the times required for activities B, C, F, and G; by one day each only. ) Which activity should the additional worker perform to reduce the project completion time by one day? ii) Which activities should the worker perform to reduce the project completion time by two days? Please provide complete logic for your activity selection for full credit.Explanation / Answer
a.
Paths and duration
Path Duration
A-B-E-F-H 4+6+5+3+6 = 24
A-C-E-F-H 4+3+5+3+6 = 21
A-C-E-G-H 4+3+5+4+6 = 22
A-D-G-H 4+7+4+6 = 21
Path with longest duration is A-B-E-F-H, with duration of 24 periods.
Thus, A-B-E-F-H path is critical path.
Project completion time = 24
b) Schedule for activities:
Activity
Preceding Activity
time
ES
EF
LS
LF
Slack
ti
ESi
EFi
LSi
LFi
Slack
ES = max(EF of preceding activities)
EF = ES + ti
LS = LF – ti
LF = min(LS of succeeding activities)
(LF-EF)
A
-
4
0
4
0
4
0
B
A
6
4
10
4
10
0
C
A
3
4
7
7
10
3
D
A
7
4
11
13
20
9
E
B, C
5
10
15
10
15
0
F
E
3
15
18
15
18
0
G
D, E
4
15
19
20
24
5
H
F
6
18
24
18
24
0
c) slack for activity D: LF – EF = 20 – 11 = 9 days
d)
According to crashing principle always select critical path activities to reduce the project completion time. Since the critical path duration determines the project completion time.
Critical Activities: A, B, E, F, H
i)
Additional workers are available for activities B, C, F, and G. Out of these activities only B and F are critical activities. Thus, either B or F should be selected for additional worker to reduce the project completion time by one day.
ii)
Same B or F can be selected to reduce the project completion time depending on the allowable crashing limits.
Activity
Preceding Activity
time
ES
EF
LS
LF
Slack
ti
ESi
EFi
LSi
LFi
Slack
ES = max(EF of preceding activities)
EF = ES + ti
LS = LF – ti
LF = min(LS of succeeding activities)
(LF-EF)
A
-
4
0
4
0
4
0
B
A
6
4
10
4
10
0
C
A
3
4
7
7
10
3
D
A
7
4
11
13
20
9
E
B, C
5
10
15
10
15
0
F
E
3
15
18
15
18
0
G
D, E
4
15
19
20
24
5
H
F
6
18
24
18
24
0