Consider the given project network, where numbers along various activities represent normal time. The free float on activity 4 - 6 and the project duration, respectively, are
A
-2 and 13
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B
2 and 12
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C
0 and 12
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D
2 and 13
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Solution
The correct option is D 2 and 13
Critical path is 1 - 2 - 5 - 6 - 7
Project duration, tE=3+2+3+5=13
Free float (4−6)=Ej−(Ei+tijE)
=8−(4+2)=2
Points to Remember:
Free float is that part of total float which can be utilised without affecting the float of successive activities, i.e. if an activity is delayed by this value then also the successor activity starts at it's earliest start time.