Question

There is a 95.05% chance the project below can be completed in X days or less. What is X? In the space provided below type in the values for each activity’s expected time, variance, list of critical activities, Project duration and value of X. Draw the network diagram (diagram required only in the pdf file). Activity ----- Predecessors----------- Optimistic (days)----------- Most likely (days) ------------pessimistic(days) ------A ------------none ---------------------------1 --------------------------------4 --------------------------------------- 7 ------B------------none ----------------------------2 --------------------------------2 --------------------------------------- 2 ------C---------------A-------------------------------2 --------------------------------5 --------------------------------------- 8 ------D---------------A-------------------------------3 --------------------------------4 --------------------------------------- 5 ------E--------------B,C -----------------------------4 --------------------------------6 --------------------------------------- 8 ------F--------------B,C------------------------------0---------------------------------0 --------------------------------------- 6 ------G--------------D,E -----------------------------3 --------------------------------6 --------------------------------------- 9

1. Expected value for each activity: BLANK-1

2. Variance for each activity: BLANK-2

3. Critical activities: BLANK-3

4. Project duration: BLANK-4

5. X = BLANK-5

Answer 1

X = 24.1 OR 24

EXPLANATION:

EXPECTED TIME = (A + (4M) + B) / 6; WHERE A = OPTIMISTIC TIME, M = MOST LIKELY TIME, B = PESSIMISTIC TIME

VARIANCE = ((B - A) / 6)**2

ACTIVITY	EXPECTED TIME	VARIANCE
A	(1 + (4 * 4) + 7) / 6 = 4	((7 - 1) / 6)^2 = 1
B	(2 + (4 * 2) + 2) / 6 = 2	((2 - 2) / 6)^2 = 0
C	(2 + (4 * 5) + 8) / 6 = 5	((8 - 2) / 6)^2 = 1
D	(3 + (4 * 4) + 5) / 6 = 4	((5 - 3) / 6)^2 = 0.1111
E	(4 + (4 * 6) + 8) / 6 = 6	((8 - 4) / 6)^2 = 0.4444
F	(0 + (4 * 0) + 6) / 6 = 1	((6 - 0) / 6)^2 = 1
G	(3 + (4 * 6) + 9) / 6 = 6	((9 - 3) / 6)^2 = 1

CPM

ACTIVITY	DURATION	ES	EF	LS	LF	SLACK
A	4	0	4	0	4	0
B	2	0	2	7	9	7
C	5	4	9	4	9	0
D	4	4	8	11	15	7
E	6	9	15	9	15	0
F	1	9	10	20	21	11
G	6	15	21	15	21	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

The 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 = A-C-E-G

DURATION OF PROJECT = 21

CRITICAL PATH VARIANCE = SUM OF VARIANCE OF ACTIVITIES ON THE CRITICAL PATH = 3.4444

STDEV = SQRT(VARIANCE) = SQRT(3.4444) = 1.855909

DUE DATE = EXPECTED COMPLETION TIME + (Z * STANDARD DEVIATION OF CRITICAL PATH)

CONFIDENCE INTERVAL = 95.05

Z VALUE = NORMSINV(95.05 / 100) = 1.65

DUE TIME = 21 + (1.65 * 1.855909) = 24.1