Question

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

	Intermediate	Times(Days)
Activity	Predecessors	a	m	b
A	-----	1	2	6
B	-----	1	2	9
C	A	2	2	11
D	A	1	8	9
E	B	1	2	3
F	C,D	3	4	11
G	D,E	1	3	11
H	F,G	2	3	11

What is the critical path?

What is the expected project completion time? (Round your answer to 3 decimal places.)

Project completion time ????? days

What is the probability of completing this project within 19 days? (Do not round intermediate calculations. Round your answer to 4 decimal places.)

Probability ???

Answer 1

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 * 2) + 6) / 6 = 2.5	((6 - 1) / 6)^2 = 0.6944444
B	(1 + (4 * 2) + 9) / 6 = 3	((9 - 1) / 6)^2 = 1.7777778
C	(2 + (4 * 2) + 11) / 6 = 3.5	((11 - 2) / 6)^2 = 2.25
D	(1 + (4 * 8) + 9) / 6 = 7	((9 - 1) / 6)^2 = 1.7777778
E	(1 + (4 * 2) + 3) / 6 = 2	((3 - 1) / 6)^2 = 0.1111111
F	(3 + (4 * 4) + 11) / 6 = 5	((11 - 3) / 6)^2 = 1.7777778
G	(1 + (4 * 3) + 11) / 6 = 4	((11 - 1) / 6)^2 = 2.7777778
H	(2 + (4 * 3) + 11) / 6 = 4.1666667	((11 - 2) / 6)^2 = 2.25

CPM

ACTIVITY	DURATION	ES	EF	LS	LF	SLACK
A	2.5	0	2.5	0	2.5	0
B	3	0	3	5.5	8.5	5.5
C	3.5	2.5	6	6	9.5	3.5
D	7	2.5	9.5	2.5	9.5	0
E	2	3	5	8.5	10.5	5.5
F	5	9.5	14.5	9.5	14.5	0
G	4	9.5	13.5	10.5	14.5	1
H	4.1666667	14.5	18.6666667	14.5	18.6666667	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 = ADFH

#DURATION OF PROJECT = 18.667

#PROBABILITY

VARIANCE OF CRITICAL PATH = SIGMA(VARIANCE OF THE CRITICAL PATH) = 6.5

DUE TIME = 19

ESTIMATED TIME = LENGTH OF CRITICAL PATH = 18.6666667

STDEV = SQRT(VARIANCE) = SQRT(6.5) = 2.54950976

Z = (DUE - ESTIMATED) / STDEV = (19 - 18.6666667) / 2.54950976 = 0.13074408

#PROBABILITY = NORMSDIST(0.13074408) = 0.552

**Z VALUE HAS NOT BEEN ROUNDED

Do leave a like for the answer if you found it helpful.