Question

1. Consider a project having the following seven activities:

                                                                        Optimistic        Most likely      Pessimistic

                                          Immediate              Time (o)          Time (m)         Time (p)

            Activity                 Predecessors         (weeks)           (weeks)           (weeks)

                  A                     none                            4                      7                      9

                  B                     none                            3                      5                      8

                  C                     A                               2                      4                      6

                  D                     A, B                            6                      9                      9

                  E                      C, D                            5                      5                      5

                  F                      B                               2                      5                      9

                  G                     E, F                             3                      3                      6

(a) What is the expected time for each activity?

(b) What is the expected project completion time?

(c) What is the critical path?

(d) What are the expected time and variance of each path?

(e) What is the probability that the project will be completed in less than 22 weeks?

Answer 1

EXPECTED TIME = (OPTIMISTIC TIME + (4 * MOST LIKELY TIME) + PESSIMISTIC TIME) / 6

VARIANCE = ((PESSIMISTIC TIME - OPTIMISTIC TIME) / 6)^2

ACTIVITY	EXPECTED TIME	VARIANCE
A	(4 + (4 * 7) + 9) / 6 = 6.83	((9 - 4) / 6)^2 = 0.6944
B	(3 + (4 * 5) + 8) / 6 = 5.17	((8 - 3) / 6)^2 = 0.6944
C	(2 + (4 * 4) + 6) / 6 = 4	((6 - 2) / 6)^2 = 0.4444
D	(6 + (4 * 9) + 9) / 6 = 8.5	((9 - 6) / 6)^2 = 0.25
E	(5 + (4 * 5) + 5) / 6 = 5	((5 - 5) / 6)^2 = 0
F	(2 + (4 * 5) + 9) / 6 = 5.17	((9 - 2) / 6)^2 = 1.3611
G	(3 + (4 * 3) + 6) / 6 = 3.5	((6 - 3) / 6)^2 = 0.25

CPM

ACTIVITY	DURATION	ES	EF	LS	LF	SLACK
A	6.83	0	6.83	0	6.83	0
B	5.17	0	5.17	1.66	6.83	1.66
C	4	6.83	10.83	11.33	15.33	4.5
D	8.5	6.83	15.33	6.83	15.33	0
E	5	15.33	20.33	15.33	20.33	0
F	5.17	5.17	10.34	15.16	20.33	9.99
G	3.5	20.33	23.83	20.33	23.83	0

FORWARD PASS: ES = MAXIMUM EF OF ALL PREDECESSOR ACTIVITIES; 0 IF NO PREDECESSORS ARE PRESENT. EF = ES + DURATION OF THE ACTIVITY.

BACKWARD PASS: LF = MINIMUM LS OF ALL SUCCESSOR ACTIVITIES; COMPLETION TIME OF THE PROJECT IF NO SUCCESSORS ARE PRESENT. LS = LF - DURATION OF THE ACTIVITY.

SLACK = LF - EF, OR, LS - ES

CRITICAL PATH = PATH WITH THE LONGEST COMBINED DURATION VALUE AND 0 SLACK.

CRITICAL PATH = A-D-E-G

DURATION OF PROJECT = 23.83

PROBABILITY

VARIANCE = 0.6944 + .25 + 0 + 0.25 = 1.1944

STANDARD DEVIATION = SQRT(VARIANCE) = SQRT(1.1944) = 1.092886

EXPECTED TIME = 23.83

DUE TIME = 22

Z = (DUE TIME - EXPECTED TIME) / STANDARD DEVIATION OF CRITICAL PATH)

Z = (22 - 23.83) / 1.092886 = -1.67

PROBABILITY FOR A Z VALUE OF -1.67 = NORMSDIST(-1.67) = 0.0475