In: Operations Management
Question 3
Time |
Estimates |
(weeks) |
|||
Activity |
Description |
Predecessor |
a |
m |
b |
A |
Position Recruiting |
- |
5 |
8 |
17 |
B |
System Development |
- |
3 |
12 |
15 |
C |
System Training |
A |
4 |
7 |
10 |
D |
Equipment training |
A |
5 |
8 |
23 |
E |
Manual System Test |
B, C |
1 |
1 |
1 |
F |
Preliminary System Changeover |
B, C |
1 |
4 |
13 |
G |
Computer-Personnel Interface |
D, E |
3 |
6 |
9 |
H |
Equipment Modification |
D, E |
1 |
2.5 |
7 |
I |
Equipment testing |
H |
1 |
1 |
1 |
J |
System Debugging and Installation |
F, G |
2 |
2 |
2 |
K |
Equipment Changeover |
G, I |
5 |
8 |
11 |
Use the information in the table to answer the following questions
What is the critical path (the critical activities)? (2 marks
b.Management of CBG is not happy with the completion time in Question 2(iii) and wants the project completed earlier than that. You are given the following additional activity information.
Activity |
Crash Time (weeks) |
Activity Cost (GH¢) |
||||
Normal |
Crash |
|||||
A |
7 |
4,800 |
6,300 |
|||
B |
9 |
9,100 |
15,500 |
|||
C |
5 |
3,000 |
4,000 |
|||
D |
8 |
3,600 |
5,000 |
|||
E |
1 |
0 |
0 |
|||
F |
3 |
1,500 |
2,000 |
|||
G |
5 |
1,800 |
2,000 |
|||
H |
3 |
0 |
0 |
|||
I |
1 |
0 |
0 |
|||
J |
2 |
0 |
0 |
|||
K |
6 |
5,000 |
7,000 |
|||
i. Prepare a table of Allowable Crash Time and Crash Cost per Week for each activity. (5 marks)
Management of CBG can offer a maximum of GH¢800 per week to enable the project manager reduce the project completion time:
ii. prepare a table for the crashing procedure. (5.5 marks)
iii. when will the project now be completed? (0.5 mark)
iv. what is the total extra cost to management of CBG? (1 mark)
v. If management of CBG wishes the computerized accounting system to be ready at week 30, what will be the total extra cost to CBG? (2 marks
ANSWER 3 A=
Answer i=
Activity | a | m | b | Completion time | Completion time ( whole number) |
A | 5 | 8 | 17 | (5+4*8+17)/6 | 9.00 |
B | 3 | 12 | 15 | (3+4*12+15)/6 | 11.00 |
C | 4 | 7 | 10 | (4+4*7+10)/6 | 7.00 |
D | 5 | 8 | 23 | (5+4*8+23)/6 | 10.00 |
E | 1 | 1 | 1 | (1+4*1+1)/6 | 1.00 |
F | 1 | 4 | 13 | (1+4*4+13)/6 | 5.00 |
G | 3 | 6 | 9 | (3+4*6+9)/6 | 6.00 |
H | 1 | 2.5 | 7 | (1+4*2.5+7)/6 | 3.00 |
I | 1 | 1 | 1 | (1+4*1+1)/6 | 1.00 |
J | 2 | 2 | 2 | (2+4*2+2)/6 | 2.00 |
K | 5 | 8 | 11 | (5+4*8+11)/6 | 8.00 |
Answer ii-
Activity | ES | EF | LS | LF |
A | 0 | 9 | 0 | 9 |
B | 0 | 11 | 7 | 18 |
C | 9 | 16 | 11 | 18 |
D | 9 | 19 | 9 | 19 |
E | 16 | 17 | 18 | 19 |
F | 16 | 21 | 20 | 25 |
G | 19 | 25 | 19 | 25 |
H | 19 | 22 | 23 | 26 |
I | 22 | 23 | 26 | 27 |
J | 25 | 27 | 25 | 27 |
K | 27 | 35 | 27 | 35 |
Answer iii= The computerized system will be ready to use after 9+10+6+2+8=35 weeks (which is the duration of the critical path)
Answer iiv= The critical path = longest path = A-D-G-J-K
P.S. AS PER CHEEGG ONLY FIRST 4 PARTS CAN BE ANSWERED.