Question

2 S 3

			TIMES (days)
Activity	Start Node	Finish Node	OPTIMISTIC to	MOST LIKELY tm	PESSIMISTIC tp
A	1	2	24	30	42
B	1	3	20	23	32
C	2	5	15	27	39
D	3	4	25	43	61
E	4	5	15	42	69

Given the Activity on Arrow Diagram and the data found in the table above, calculate each of the following values:


Activity A………t_e = 31
Standard Deviation = 3
Activity B………t_e =

Standard Deviation =

Activity C………t_e =

Standard Deviation =

Activity D………t_e =

Standard Deviation =

Activity E………t_e =

Standard Deviation = 9


Path 1 – 2 – 5
Mean Time =

Standard Deviation =5


Path 1 – 3 – 4 – 5
Mean Time =

Standard Deviation =

Answer 1

Answer:

Activity	Optimistic time-a	Most likely-m	Pessimistic time-b	Expected time= (a+4*m+ b)/6	Variance, (sigma)^2= (b-a/6)^2	Standard deviation= sqrt(variance)
A	24	30	42	31	9	3
B	20	23	32	24	4	2
C	15	27	39	27	16	4
D	25	43	61	43	36	6
E	15	42	69	42	81	9

Paths and standard deviation

	Paths	Duration	Variance	standard deviation= sqrt(variance)
A-C	1-2-5	58	25	5
B-D-E	1-3-4-5	109	121	11