Question

The estimated times and immediate predecessors for activities in a project  are given below in the table. Assume that the activity time  are independent.

Task                            Predecessor                   Optimistic   Most Likely   Pessemistic     

                                                                          Time (wks)    time (wks)     time (wks)

Architectural design                           --           10                  30                    45

Software                      Architectural design     13                  15                    23

Compilg                      Software                          2                    3                      4

Hardware                    Architectural design     15                  20                    25

Integration                  Software & Hardware   20                  22                    24

Bug fixes                     Integration                    3                    4                      11

            

a)  Draw an activity-on-node network diagram for this project.

b) Identify the critical path and estimate the expected time to complete the project.

c)  What is the probability of finishing this project in 77 weeks?

Answer 1

EXPECTED TIME = (OPTIMISTIC TIME + (4 * MOST LIKELY TIME) + PESSIMISTIC TIME) / 6

VARIANCE = ((PESSIMISTIC TIME - OPTIMISTIC TIME) / 6)^2

ACTIVITY	EXPECTED TIME	VARIANCE
A	(10 + (4 * 30) + 45) / 6 = 29.17	((45 - 10) / 6)^2 = 34.0278
B	(13 + (4 * 15) + 23) / 6 = 16	((23 - 13) / 6)^2 = 2.7778
C	(2 + (4 * 3) + 4) / 6 = 3	((4 - 2) / 6)^2 = 0.1111
D	(15 + (4 * 20) + 25) / 6 = 20	((25 - 15) / 6)^2 = 2.7778
E	(20 + (4 * 22) + 24) / 6 = 22	((24 - 20) / 6)^2 = 0.4444
F	(3 + (4 * 4) + 11) / 6 = 5	((11 - 3) / 6)^2 = 1.7778

CPM

ACTIVITY	DURATION	ES	EF	LS	LF	SLACK
A	29.17	0	29.17	0	29.17	0
B	16	29.17	45.17	33.17	49.17	4
C	3	45.17	48.17	73.17	76.17	28
D	20	29.17	49.17	29.17	49.17	0
E	22	49.17	71.17	49.17	71.17	0
F	5	71.17	76.17	71.17	76.17	0

ES = MAX(EF OF ALL PREDECESSORS); 0 FOR FIRST ACTIVITY

EF = ES + DURATION

LF = MIN(LS OF ALL SUCCESSOR ACTIVITIES); MAX EF AS LF FOR LAST ACTIVITY

LS = LF - DURATION

SLACK = LF- EF

CRITICAL PATH = LONGEST PATH WITH 0 SLACK:

CRITICAL PATH = ADEF

DURATION OF PROJECT = 76.17

PROBABILITY OF FINISHING IN 77 WEEKS

VARIANCE OF CRITICAL PATH = SIGMA(VARIANCE OF THE CRITICAL ACTIVITIES) = 39.0278

EXPECTED TIME = DURATION OF THE PROJECT = CRITICAL PATH = 76.17

DUE TIME = 77

STANDARD DEVIATION OF CRITICAL PATH = SQRT(VARIANCE)

Z = (DUE TIME - EXPECTED TIME) / STANDARD DEVIATION OF CRITICAL PATH)

Z = (77 - 76.17) / SQRT(39.0278) = 0.13

PROBABILITY FOR A Z VALUE OF 0.13 = 0.5517

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