Question

A future construction project has the following activity relationships and activity time estimates (in days):

Activity	Immediate Predecessor	To	Tm	Tp
A	None	4	5	6
B	None	3	6	21
C	B	1	4	7
D	A	2	5	8
E	C, D	3	9	27
F	C	3	6	15
G	E, F	2	4	6
H	F, G	2	2	2

• Leave 2-decimal if not exact, do not round up, for example, 4.3268... will be kept as 4.32

1. Please calculate the “expected finish time” and “variance” for each activity. What is the expected finish time of activity C?   What is the activity variance of activity D?  
2. Use “expected finish time” of each activity to construct the network. What is the critical path of the project? (no space permitted between letters, acceptable format: ABCD) What is the expected project completion time in days?  What is the variance of project completion time?  
3. Use “expected finish time” to decide ES, EF, LS, LF of each activity. What is the LS time of activity F?   What is the slack time of activity D?

Answer 1

Answer Table:

Activity	Immediate Predecessor	To	Tm	Tp	Te	Variance	ES	EF	LS	LF	Slack
A	None	4	5	6	5	0.11	0	5	2	7	2
B	None	3	6	21	8	9.00	0	8	0	8	0
C	B	1	4	7	4	1.00	8	12	8	12	0
D	A	2	5	8	5	1.00	5	10	7	12	2
E	C, D	3	9	27	11	16.00	12	23	12	23	0
F	C	3	6	15	7	4.00	12	19	16	23	4
G	E, F	2	4	6	4	0.44	23	27	23	27	0
H	F, G	2	2	2	2	0.00	27	29	27	29	0

Expected Finish time (Te) of C = 4

Activity variance of activity D = 1

Network:

Critical path = BCEGH

Expected project completion time = 29 days

Variance of the project completion time = Sum of variances of critical activities = 26.44

LS time of activity F is 16 days

Slack time of activity D = 2 days

Explanation:

For each activity:

Expected time, Te = (To+4*Tm+Tp)/6

Variance = (Tp - To)/6

ES, EF, LS, LF of each activity

ES or Earliest Start Time = EF or Earliest Finish time of the predecessor of the corresponding activity

if an activity has more than 1 predecessor than the maximum of the EF of the predecessors is considered as the ES for the corresponding activity.

EF = ES+ Expected time

The ES and EF are calculated from the start to end activities of the network.

Example: Activities A and B do not have any predecessors, hence can be started at 0 time. So the ES is 0 for both of them.

Activity C has one predecessor which is B. So, the ES = EF of B = 8

E has two predecessors, C and D.The ES of activity E = Max of (EF of Activity C and D) = MAX(12,10) = 12

LS or Late Start time = LF or Late Finish Time - Duration of the corresponding project.

LF = LS of the successor

The LF and LS are calculated from the end to start the activity. Hence the activity with the highest EF becomes the end activity and the EF becomes the LF.

If an activity has more than one successor, the minimum of the LS values is considered for the LF.

Example: Activity H has the highest EF time. This means this activity finishes last. Hence the latest finish for the project is 29 days.

LF of Activity H = 29 days

Activity H has two predecessors F and G. Hence the LS of H becomes the LF of both F and G which is 29-expected time of H = 29-2 = 27

But Activity F is the predecessor of two activities G and H. So, the LF = Min (LS of G, LS of H) = Min (23,27) = 23

Slack = LF - EF or LS - ES

If slack is 0, the activity is a critical activity and all the zero slack activities constitute the critical path.