Question

QUESTION 2. The following table explain the logical relationship of activities of a given project, including the duration in weeks of each activity.

Activity	Follows	Duration (Weeks)
M	G, F	4
F	Start	6
G	Start	3
D	Start	7
K	D, G	5
L	K, J	8
J	D	3
S	L, M, F	4
Finish	S

1. Draw the network
2. Perform the forward path as determine the project duration (5 Marks)
3. Perform the back path and determine the critical path (5 Marks)

Answer 1

a. Network Diagram

b. Project Duration : 24 Days

ES = Earliest Start, EF = Earliest Finish, LS = Latest Start, LF = Latest Finish

Forward Path:

Activities F,G,D has no any predecessor so all these activities start on Day 0.

ES =0, EF = Project Duration

For all other activities

E: (ES, EF)
ES = Latest Earliest Finish time of predecessors.

EF = ES + Duration

c. Critical path: D-K-L-S

Backward Path:

L: (LS, LF)

LF = Lowest Latest Start time (LS)  of Following activities.

LS = LF - Project Duration

Activity S is the last activity in the project so for S, LS= ES and LF = EF.

S has three predecessors F, M and L

LF for M = LS for S  = 20

LS for M = 20 - 4 = 16

LF for F = min (LS for S or LS for M) = min (20,16) = 16

LS for F = 16-6 =10

LF for L = LS for S = 20

LS for L = 20-12 =8

Now, L had two predecessors K and J

LF for K = LS for L = 12

LS for K = 12-5 = 7

LF for J = LS for L =12

LS for J = 12-3 =9

Now for remaining two activities D & G.

LF for D = min (LS for K, LS for J) = 7

LS for D = 7-7 =0

LF for G = min(LS for M, LS for K) = 7

LS for G = 7-3 =4.

All activities are critical for which ES = LS.

In this project, D K L and S are critical activities.

And Critical Path = D-K-L-S