Question

Recently, you were assigned to manage a project for your company. You have constructed a network diagram depicting the various activities in the project (shown below to the right). In addition, you have asked your team to estimate the amount of time that they would expect each of the activities to take. Their responses are shown in the following table: ơ Time Estimates (days) Optimistic Most Likely Pessimistic Activity ơ ال 7 کره ) | Start Finish 6 ای یا نه w co a. What is the expected completion time of the project? The expected completion time is days. (Enter your response as an integer.)

Answer 1

1. EXPECTED TIME(ROUNDED)

FORMULA:

(a + (4m) + b) / 6

A = (5 + (4 * 7) + 10) / 6 = 7

B = (3 + (4 * 8) + 12) / 6 = 8

C = (5 + (4 * 6) + 8) / 6 = 6

D = (2 + (4 * 3) + 6) / 6 = 3

E = (3 + (4 * 8) + 10) / 6 = 8

VARIANCE(ROUNDED 4 PLACES)

FORMULA:

((b - a) / 6) ^ 2

A = (10 - 5) / 6) ^ 2 = 0.6944

B = (12 - 3) / 6) ^ 2 = 2.25

C = (8 - 5) / 6) ^ 2 = 0.25

D = (6 - 2) / 6) ^ 2 = 0.4444

E = (10 - 3) / 6) ^ 2 = 1.3611

A. CPM

After forwards and backwards pass, we get the following value based on the precedence relationship.

ACTIVITY	DURATION	ES	EF	LS	LF	SLACK
A	7	0	7	3	10	3
B	8	0	8	0	8	0
C	6	7	13	10	16	3
D	3	7	10	13	16	6
E	8	8	16	8	16	0

CRITICAL PATH = B + E = 16

DURATION = 16

B. VARIANCE OF CRITICAL PATH = B + E = 2.25 + 1.3611 = 3.6111

STDEV = SQRT(CRITICAL VARIANCE) = 1.9003

Z = DUE - EXPECTED / STDEV

PROBABILITY OF COMPLETION IN 22

EXPECTED = 16

DUE = 22

Z = 22 - 16 / 1.9003 = 3.16

Z value that corresponds to 3.16 = 0.9992 or 99.92%

C. PROBABILITY OF COMPLETION IN 18

EXPECTED = 16

DUE = 18

Z = 18 - 16 / 1.9003 = 1.05

Z value that corresponds to 1.05 = 0.8531 or 85.31%