Question

Given is a CPM project network diagram as shown below.

Activity	Start	A	B	C	D	E	F	G	H	End
days	0	2	6	8	3	4	2	3	2	0

a) The Project Completion time = ____ days.

b) The Earliest Start time, ES, of Activity G = ____ days.

c) The Latest Finish time, LF, of Acitivity C = ____ days.

d) The critical activities are = ____ (ex. Fill in answer as: ABCD)

Answer 1

a) Solution for Project Completion time

Path & Duration are:

ACFH = 5+9+8+4 = 26

ACEGH = 5+9+9+5+4 = 32

BDGH = 6+6+5+4 = 21

Project completion time will be maximum of path duration = 32days

ES, the earliest time activity can start, assuming all preceding activities start as early as possible.

EF, the earliest time the activity can finish.

LS, the latest time the activity can start and not delay the project.

LF, the latest time the activity can finish and not delay the project

Computation of earliest starting and finishing times is aided by two simple rules:

1. The earliest finish time for any activity is equal to its earliest start time plus it's expected duration, t:

EF = ES + t

2. ES for activities at nodes with one entering arrow is equal to EF of the entering arrow. ES for activities leaving nodes with multiple entering arrows is equal to the largest EF of the entering arrow.

Computation of the latest starting and finishing times is aided by the use of two rules:

1. The latest starting time for each activity is equal to its latest finishing time minus its expected duration:

LS = LF - t

2. For nodes with one leaving arrow, LF for arrows entering that node equals the LS of the leaving arrow. For nodes with multiple leaving arrows, LF for arrows entering that node equals the smallest LS of leaving arrows.

and therefore;

b) ES of G = 23days

c) LF of C = 14days

d) Critical activities are those in the critical path, ACEGH