Question

A project schedule has the following characteristics

Activity	1 - 2	1 - 3	2 - 4	3 - 4	3 - 5	4 - 9	5 - 6	5 - 7	6 - 8	7 - 8	8 - 10	9 - 10
Time	4	1	1	1	6	5	4	8	1	2	5	7

Construct the network and calculate the earliest start time, earliest finish time, latest start time and latest finish time of each activity and determine the Critical path of the project and duration to complete the project.

Answer 1

E₁ = 0

E₂ = 0 + 4 = 4

E₃ = 0 + 1 = 1

E₄ = 4 + 1 = 5

E₅ = 1 + 6 = 7

E₆ = 7 + 4 = 11

E₇ = 8 + 7 = 15

E₈ = 15 + 2 = 7

E₉ = 5 + 5 = 10

E₁₀ = 17 + 5 = 22

L₇ = 31

L₁₀ = 22

L₉ = 22 – 7 = 15

L₈ = 22 – 5 = 17

L₇ = 17 – 2 = 15

L₆ = 17 – 1 = 16

L₅ = (16 – 4) or (15 – 8)

whichever is minimum = 7

L₄ = 15 – 5 = 10

L₃ = (10 – 1) or (7 – 6)

whichever is minimum = 1

L₂ = 10 – 1 = 9

L₁ = 0

Activity	Duration tij	EST	EFT = EST + tij	LST = LFT – tij	LFT
1 - 2	4	0	4	9 – 4 = 5	9
1 - 3	1	0	1	1 – 1 = 0	1
2 - 4	1	4	5	10 – 1 = 9	10
3 - 4	1	1	2	10 – 1 = 9	10
3 - 5	6	1	7	7 – 6 =1	7
4 - 9	5	5	10	15 – 5 =10	15
5 - 6	4	7	11	16  4 = 12	16
5 - 7	8	7	15	15 – 8 = 7	15
6 - 8	1	11	12	17 – 1 = 16	17
7 - 8	2	15	17	17 – 2 = 15	17
8 - 10	5	17	22	22 – 5 = 17	22
9 - 10	7	10	17	22 – 7 = 15	22

Since EFT and LFT is same on 1 - 3, 3 - 5, 5 - 7 and 7 - 8 and 8 - 10 the critical path is 1 - 3 - 5 - 7 - 8 - 10 and the duration is 22 time units.