In: Operations Management
t
he activities described by the following table are given for the Howard Corporation in Kansas:
Activity Time (in weeks) Immediate
Predecessor(s) Activity Time (in
weeks) Immediate Predecessor(s)
Activity Time (in weeks) Immediate
Predecessor(s)
A 10 - D 6 B G 6 E, F
B 6 A E 10 B H 6 D
C 3 A F 10 C I 2 G, H
b) The activities on the critical path are A-B-E-G-I
Activity | Duration, D | Early start, ES=Max of early finish of preceding activities | Early finish, EF = ES + D | Late finish, LF= Min of LS of successor activities | Late start, LS= LF - D | Total slack= LF-EF | Critical activity, activities with 0 slack time |
A | 10 | 0 | 10 | 10 | 0 | 0 | Yes |
B | 6 | 10 | 16 | 16 | 10 | 0 | Yes |
C | 3 | 10 | 13 | 16 | 13 | 3 | No |
D | 6 | 16 | 22 | 26 | 20 | 4 | No |
E | 10 | 16 | 26 | 26 | 16 | 0 | Yes |
F | 10 | 13 | 23 | 26 | 16 | 3 | No |
G | 6 | 26 | 32 | 32 | 26 | 0 | Yes |
H | 6 | 16 | 22 | 32 | 26 | 10 | No |
I | 2 | 32 | 34 | 34 | 32 | 0 | Yes |
c) what is the total project completion time for Howard Corporation in weeks?
(Enter your response as a whole number.)
(b)
Activity | Duration | Predecessors | ES | EF | LF | LS | Total slack | Critical? |
A | 10 | - | 0 | 10 | 10 | 0 | 0 | Y |
B | 6 | A | 10 | 16 | 16 | 10 | 0 | Y |
C | 3 | A | 10 | 13 | 16 | 13 | 3 | N |
D | 6 | B | 16 | 22 | 26 | 20 | 4 | N |
E | 10 | B | 16 | 26 | 26 | 16 | 0 | Y |
F | 10 | C | 13 | 23 | 26 | 16 | 3 | N |
G | 6 | E, F | 26 | 32 | 32 | 26 | 0 | Y |
H | 6 | D | 22 | 28 | 32 | 26 | 4 | N |
I | 2 | G, H | 32 | 34 | 34 | 32 | 0 | Y |
The activities on the critical path are A-B-E-G-I (Zero total slack)
c)
The total project completion time = LF of the last activity 'I' = 34 weeks