Question

The following represents a project that should be scheduled using CPM: TIMES (DAYS) IMMEDIATE b ACTIVITY PREDECESSORS m a 1 4 7 3 5 10 B 2 5 4 6 11 D B 1 2 3 E C,D D,E F,G 3 5 7 F G 1 2 6 2 5 b. What is the critical path? A-D-F-H A-D-G-H B-E-G-H A-C-F-H

Answer 1

b.

Activity	a	m	b	Expected completion time	variance
A	1	4	7	4	1
B	3	5	10	5.5	1.361111
C	2	5	8	5	1
D	4	6	11	6.5	1.361111
E	1	2	3	2	0.111111
F	3	5	7	5	0.444444
G	1	2	6	2.5	0.694444
H	2	5	6	4.666666667	0.444444

Path	Duration	Variance
A-C-F-H	18.66667
A-D-F-H	20.16667	3.25
A-D-G-H	17.66667
B-E-G-H	14.66667

Critical path is the longest path which is A-D-F-H

Formula

C15 2 Activity 3 A 5c 6o 7 E 8 F 9 G 10 H Expected completion time =(B3+D3+4*C3)/6 =(B4+D4+4*C4)/6 =(B5+D5+4*C5)/6 =(B6+D6+4*C6)/6 =(B7+D7+4*07)/6 = (B8+D8+4*C8)/6 =(B9+D9+4*C9)/6 =(B10+D10+4*C10)/6 variance =((D3-B3)/6)^2 =((D4-B4)/6)^2 =((D5-B5)/6)^2 =((D6-B6)/6)^2 =((D7-B7)/6)^2 =((D8-B8)/6)^2 =((D9-B9)/6)^2 =((D10-B10)/6)^2 12 Variance 13 Path 14 A-C-F-H 15 A-D-F-H 16 A-D-G-H 17 B-E-G-H Duration =E3+E5+E8+E10 =E3+E6+E8+E10 =E3+E6+E9+E10 =E4+E7+E9+E10 =F3+F6+F8+F10 19

c. Project completion time = Duration of critical path = 20.167 days(Rounded to 3 decimals)

d.

Project variance = Sum of variances of critical activities = 1+1.361111+0.444444+0.444444 = 3.25

Project standard deviation = sqrt(Project variance ) = sqrt(3.25) = 1.802775638

We know,

z = (x-expected completion time)/project standard deviation

= (22-20.167)/1.802775638

= 1.01676546 = 1.02 (Rounded to 2 decimal places)

corresponding probability = NORM.S.DIST(1.02,TRUE) = 0.84613577 = 0.8461(Rounded to 4 decimal places)