Question

Assume a project information is given in the table below:

Activity	Immediate Processor	Duration (Days)
A	---	5
B	---	5
C	---	5
D	A, B	4
E	B	3
F	C	8
G	D, E	6

1. What is the minimum number of activities that have to be delayed for the project to get delayed?

1. 1
2. 2
3. 3
4. 4
5. 5
6. None of the above (What is your answer?)

1. What is the minimum number of non-critical activities that have to be delayed for the project to get delayed?

1. 1
2. 2
3. 3
4. 4
5. 5
6. None of the above (What is your answer?)

1. What is the maximum number of activities that can be delayed for the project not to get delayed?

1. 1
2. 2
3. 3
4. 4
5. 5
6. None of the above (What is your answer?)

1. What is the maximum number of activities that can be delayed for the project not to have additional critical activities?

1. 1
2. 2
3. 3
4. 4
5. 5
6. None of the above (What is your answer?)

Answer 1

1. A. 1; ANY CRITICAL ACTIVITY CAN BE DELAYED TO DELAY THE PROJECT

2. A. 1, E IS A NON CRITICAL ACTIVITY WITH A SLACK OF 1, DELAYING IT ONCE WILL MAKE IT CRITICAL.

3. B. 2, C AND F CAN BE DELAYED ONCE WITH NO CHANGE TO THE CRITICAL PATH.

4. B. 2. C AND F

ACTIVITY	DURATION	ES	EF	LS	LF	SLACK	CRITICAL
A	5	0	5	0	5	5 - 5 = 0	YES
B	5	0	5	0	5	5 - 5 = 0	YES
C	5	0	5	2	7	7 - 5 = 2
D	4	5	9	5	9	9 - 9 = 0	YES
E	3	5	8	6	9	9 - 8 = 1
F	8	5	13	7	15	15 - 13 = 2
G	6	9	15	9	15	15 - 15 = 0	YES

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 = A---D--G & B--D--G(2 CRITICAL PATHS)

DURATION OF PROJECT = 15

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