Question

Activity	Predecessor	Optimistic	Most Likely	Pessimistic
A	--	12	16	25
B	--	7	14	22
C	--	5	10	14
D	A	15	18	25
E	C	8	12	18
F	B	9	12	15
G	F	18	22	29
H	D,E,G	15	19	27

Using the table above, what is the probability that the project will be completed in 65 or fewer weeks?

-greater than 20% but less than or equal to 24%

-less than or equal to 20%

-greater than 28%

-greater than 24% but less than or equal to 28%

Answer 1

1. B. LESS THAN OR EQUAL TO 20%

EXPLANATION:

ACTIVITY	OPTIMISTIC a	MOST LIKELY m	PESSIMISTIC b	EXPECTED	VARIANCE
				(a + (4m) + b) / 6	((b - a) / 6) ^ 2
A	12	16	25	17	4.69
B	7	14	22	14	6.25
C	5	10	14	10	2.25
D	15	18	25	19	2.78
E	8	12	18	12	2.78
F	9	12	15	12	1
G	18	22	29	23	3.36
H	15	19	27	20	4

CPM ANALYSIS

ACTIVITY	DURATION	ES	EF	LS	LF	SLACK
A	17	0	17	13	30	13
B	14	0	14	0	14	0
C	10	0	10	27	37	27
D	19	17	36	30	49	13
E	12	10	22	37	49	27
F	12	14	26	14	26	0
G	23	26	49	26	49	0
H	20	49	69	49	69	0

FORWARD PASS

We calculate the ES and EF values using a forward pass where the ES of an activity is the maximum EF of all the predecessor activities.

BACKWARD PASS

We calculate the LS and LF values using a backward pass where the LF of the activity is the minimum of all the successor activities.

SLACK

Slack is the value which is determined by subtracting EF from the LF or ES from the LS.

CRITICAL PATH

Critical path is the chain in the project network where the slack value of all the activities is 0, what this means is that any delay in these activities would result in delaying the entire project.

CRITICAL PATH & DURATION

B + F + G + H = 69

PROBABILISTIC ESTIMATE

VARIANCE OF CRITICAL PATH = 6.25 + 1 + 3.36 + 4 = 14.61

WITH THE GIVEN DATA

Due time = 65

Expected time = 69

Variance = 14.61

Stdev = sqrt(variance) = sqrt(14.61) = 3.82(ROUNDED)

Z = DUE - EXPECTED / STDEV

Z = 65 - 69 / 3.82 = -1.05

Z value of -1.05 gives a distribution of 0.1469 OR 14.69 % probability of completion of the project by the due time of 65

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