Question

The following represents a project that should be scheduled using CPM:

                             IMMEDIATE                                TIMES (DAYS)

                            PREDECESSORS

ACTIVITY                                                                  a           m         b

A                                   —                                          1           4          7

B                                   —                                          3           5          10

C                                  A 2          5          8

D                                  A                                            4           6          11

E                                   B                                           1            2          3

F                                 C,D 3          5          7

G                                D,E                                         1            2          6

H                                F,G                                         2            5          6

b. What is the critical path?

            A-D-F-H

            B-E-G-H

            A-C-F-H

            A-D-G-H

c. What is the expected project completion time?

Project completion time __________ days

d. What is the probability of completing this project within 22 days?

Probability _________          

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	(1 + (4 * 4) + 7) / 6 = 4	((7 - 1) / 6)^2 = 1
B	(3 + (4 * 5) + 10) / 6 = 5.5	((10 - 3) / 6)^2 = 1.3611
C	(2 + (4 * 5) + 8) / 6 = 5	((8 - 2) / 6)^2 = 1
D	(4 + (4 * 6) + 11) / 6 = 6.5	((11 - 4) / 6)^2 = 1.3611
E	(1 + (4 * 2) + 3) / 6 = 2	((3 - 1) / 6)^2 = 0.1111
F	(3 + (4 * 5) + 7) / 6 = 5	((7 - 3) / 6)^2 = 0.4444
G	(1 + (4 * 2) + 6) / 6 = 2.5	((6 - 1) / 6)^2 = 0.6944
H	(2 + (4 * 5) + 6) / 6 = 4.67	((6 - 2) / 6)^2 = 0.4444

CPM

ACTIVITY	DURATION	ES	EF	LS	LF	SLACK
A	4	0	4	0	4	0
B	5.5	0	5.5	5.5	11	5.5
C	5	4	9	5.5	10.5	1.5
D	6.5	4	10.5	4	10.5	0
E	2	5.5	7.5	11	13	5.5
F	5	10.5	15.5	10.5	15.5	0
G	2.5	10.5	13	13	15.5	2.5
H	4.67	15.5	20.17	15.5	20.17	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.

B. CRITICAL PATH = A-D-F-H

C. DURATION OF PROJECT = 20.17

VARIANCE OF CRITICAL PATH = 1 + 1.3611 + 0.4444 + 0.4444 = 3.2499

STANDARD DEVIATION = SQRT(VARIANCE) = SQRT(3.2499) = 1.802748

EXPECTED TIME = 20.17

DUE TIME = 22

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

Z = (22 - 20.17) / 1.802748 = 1.02

D. PROBABILITY FOR A Z VALUE OF 1.02 = NORMSDIST(1.02) = 0.8461

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