Question

Problem 5-10

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

	IMMEDIATE PREDECESSORS	TIMES (DAYS)
ACTIVITY		a	m	b
A	—	1	4	7
B	—	3	5	10
C	A	2	5	8
D	A	4	6	11
E	B	1	2	3
F	C,D	3	5	7
G	D,E	1	2	6
H	F,G	2	5	6

b. What is the critical path?

	B-E-G-H
	A-D-G-H
	A-D-F-H
	A-C-F-H

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

Project completion time             days

d. What is the probability of completing this project within 22 days? (Do not round intermediate calculations. Round your answer to 4 decimal places.)

Answer 1

data: 5-10) Using given •Expected completion (tel = atum tb time 6 2 Variance (V) = bra 6 Now, A B C D = 65 = 1.361 Activity te 1+16+7/6 = 4 (616) = 1 3+20+10/6=55 (776) = 1.361 2+20+816 = 5 (6/6)², 4+24+4116= (2/6)= E 1+8+3/6 = 2 (2/6)= 0.111 3720x7/6=5 (416)°= 0.444 17 87 616 (5/6)= 0.694 2+ 20+616 = 4.667 (1/6)=0-444 Now, network network diagram diagram becomesi- F G =2.5 H JE 4.667 Start .5 2.5

Path ACEH Time u+5+5+4.667 ut 6.5+5+4.667 18.667 ADFH = 20:167 - 17.667 ADGH BEGH → 4+6.5+ 2.5+ 4.667 5.5+24 2.5+ 4.667 = 14.667 b) critical path is with maximum time ie A-D-E-H ©. es Project compection time (E)= 20.167 d) Variance along critical critical path (i) - + 1.361 to 49470.444 3.249 - P ( X 422):- 2- X= 13 22- 20:167 N 1:01 69 3.249 Using Excel's function :- p - NORM.S.DIST (1,0169, TRUE) = 0.8454