Prepared For

Sayeedul Karim (SDK)

MGT 314.5

Prepared By
Name: ID#
Sadaf Islam 0930418030
Tarek- Al- Zobaer 1010052 030
Md. Saiful Islam 1010051030
Mushfique Enam Ratul 1010852030
Sabbir Islam 092195530

March 2, 2014

Activity | Predecessors | Normal Time | Crash Time | Max. Time Red. (NT-CT) | Cost Increase/Period | A | None | 7 | 6 | 1 | 200 | B | None | 12 | 9 | 3 | 250 | C | A | 8 | 6 | 2 | 250 | D | A | 6 | 5 | 1 | 300 | E | B | 1 | 1 | 0 | | F | C,D | 1 | 1 | 0 | | G | D,E | 3 | 1 | 2 | 200 | H | F | 3 | 2 | 1 | 350 | I | G | 2 | 2 | 0 | |

AON Network Diagrams:

C

F

H
A

D

Start

Finish

I
G
B

E

From Start to Finish there are 4 paths

Path Length of Path

Start-A-C-F-H-Finish 0+7+8+1+3+0 19
Start-A-D-F-H-Finish 0+7+6+1+3+0 17
Start-A-D-G-I-Finish 0+7+6+3+2+0 18
Start-B-E-G-I-Finish 0+12+1+3+2+0 18

To find out the Total cost after crashing we have to follow some steps. They are:

1. Finding the critical path:
After observing the length of path we can say that ‘A-C-F-H’ is the critical path because it will take the greatest time to complete.

A-C-F-H 7+8+1+3 19 weeks

2. From the critical path, finding the lowest crash cost/week:
We have the information that – Activity | Cost Increase/ Period ($) | A | 200 | C | 250 | F | - | H | 350 |

So we can say that, A has the lowest crash cost/Week.

3. Reducing the time for ‘A’ until- a. It cannot be further reduced:
We can crash ‘A’ up to 1 week not more than 1 week because A’s maximum time red is given 1 week.

b. Another path become critical:
From the chart we can see that activity ‘A’ is present in 3 paths. If we crash one week then all of the three paths will be reduced by one