Question

Development of a new deluxe version of a particular software product is being considered. The activities necessary for the completion of this are listed in the table below (Time in weeks).

Activity	Immediate Predecessor	a	m	b
A	--	2	5	8
B	A	3	6	9
C	A	4	7	10
D	B	2	5	14
E	D	3	3	3
F	C	6	8	10
G	E, F	1	1	1
H	C	6	10	14
I	G, H	3	4	5

(a) If you wish to reduce the time required to complete this project by two weeks, which activity(ies) should be crashed and why?

Answer 1

We will select activities = A, C, H, I

EXPLANATION:

In order to find the expected time, we will use the following formula

T = (a + 4m + b) / 6

A. (2 + (4 * 5 ) + 8) / 6 = 5

B. (3 + (4 * 6 ) + 9) / 6 = 6

C. (4 + (4 * 7 ) + 10) / 6 = 7

D. (2 + (4 * 5 ) + 14) / 6 = 6

E. (3 + (4 * 3 ) + 3) / 6 = 3

F. (6 + (4 * 8 ) + 10) / 6 = 8

G. (1 + (4 * 1 ) + 1) / 6 = 1

H. (6 + (4 * 10 ) + 14) / 6 = 10

I. (3 + (4 * 4 ) + 5) / 6 = 4

USING THE EXPECTED TIME, WE CAN FIND THE CRITICAL PATH:

ACTIVITY	DURATION	PRE 1	PRE 2	ES	EF	LS	LF	SLACK
A	5			0	5	0	5	0
B	6	A		5	11	6	12	1
C	7	A		5	12	5	12	0
D	6	B		11	17	12	18	1
E	3	D		17	20	18	21	1
F	8	C		12	20	13	21	1
G	1	E	F	20	21	21	22	1
H	10	C		12	22	12	22	0
I	4	G	H	22	26	22	26	0

The ES, EF values are calculated using a forward pass where the ES of the next activity is the maximum EF of all the predecessor activities.

The LS, LF values are calculated using a backward pass where the LS of the next activity is the minimum LF of all the predecessor activities.

The critical path represents all the activities in a particular chain which have the largest duration and 0 slack for all activities, and due to this fact, any delay in the critical activities can cause a delay for the entire project.

A - C - H - I = 26

If we have to crash an activity to reduce the project time, the critical activities would be selected in order to bring about a decrease in the project time and therefore, for crashing, we will select activities = A, C, H, I