Question

Here is an Activity on Arrow (AOA) network diagram for a project. The lettered A through F represents an activity, as shown in the accompanying precedence diagram.

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The probabilistic time for each activity is as follows

Activity	Time	Activity	Time
A	4-5-6	D	4-5-6
B	2-5-6	E	2-5-6
C	3-4-5	F	3-4-6

What is the expected time and standard deviation of path A-C-D?

What percentage will path A-C-D finish within 13 weeks?

Please type your answer and the steps of computation

A D E LL

Answer 1

To find the expected time, we will use the following formula.

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

A. (4 + (4 * 5 ) + 6) / 6 = 5

B. (2 + (4 * 5 ) + 6) / 6 = 5

C. (3 + (4 * 4 ) + 5) / 6 = 4

D. (4 + (4 * 5 ) + 6) / 6 = 5

E. (2 + (4 * 5 ) + 6) / 6 = 5

F. (3 + (4 * 4 ) + 6) / 6 = 4

To find the variance, we will use the following formula.

Variance = [(b - a) / 6] ^2

A. [(6 -4) / 6 ] ^ 2 = 0.11

B. [(6 -2) / 6 ] ^ 2 = 0.44

C. [(5 -3) / 6 ] ^ 2 = 0.11

D. [(6 -4) / 6 ] ^ 2 = 0.11

E. [(6 -2) / 6 ] ^ 2 = 0.44

F. [(6 -3) / 6 ] ^ 2 = 0.25

ACTIVITY	OPTIMISTIC	MOST LIKELY	PESSIMISTIC	EXPECTED TIME	VARIANCE
	a	m	b	t = (a + 4 m + b) / 6	[(b - a) / 6] ^2
A	4	5	6	5	0.11
B	2	5	6	5	0.44
C	3	4	5	4	0.11
D	4	5	6	5	0.11
E	2	5	6	5	0.44
F	3	4	6	4	0.25

CPM

The ES, EF values are calculated using a forward pass where the ES of the next activity is the maximum EF of all the predecessor activities.

The LS, LF values are calculated using a backward pass where the LS of the next activity is the minimum LF of all the Predecessor activities.

The critical path represents all the activities in a particular chain which have the largest duration and 0 slack for all activities, and due to this fact, any delay in the critical activities can cause a delay for the entire project.

ACTIVITY	DURATION	PRE 1	PRE 2	ES	EF	LS	LF	SLACK
A	5			0	5	0	5	0
B	5			0	5	4	9	4
C	4	A		5	9	5	9	0
D	5	C		9	14	9	14	0
E	5	B		5	10	9	14	4
F	4	B		5	9	10	14	5

CRITICAL PATH & DURATION:

A - C - D = 14

PROBABILITY

CRITICAL VARIANCE = A - C - D = 0.11 + 0.11 + 0.11 = 0.33

STANDARD DEVIATION = SQRT(CRITICAL VARIANCE) = SQRT(0.33) = 0.57

EXPECTED = 13 DAYS

Z = DUE - EXPECTED / STANDARD DEVIATION

Z = 14 - 13 / 0.57 = 1.754

THE Z VALUE OF 1.754 GIVES 0.9603 OR 96.03% PROBABILITY OF COMPLETION.