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Example on Critical Path and Float Time Calculations:. Problem: Calculate the float for all activities in network. According to the network diagram P1{1,4} needs 20 days P2{1,5} needs 26 days P3{2,3,5} needs 34 days P4{2,3,4} needs 28 day Therefore P3 is the critical path Pcp=34.
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According to the network diagram • P1{1,4} needs 20 days • P2{1,5} needs 26 days • P3{2,3,5} needs 34 days • P4{2,3,4} needs 28 day • Therefore P3 is the critical path Pcp=34. Solution.
Activities 2, 3, and 5 are on the critical path; they have zero float • But activities 1 and 4 are not on the critical path… Hence,
The float for activities 1 is Pcp-P2 = 34-26= 8 days Since activity 1 is on P2(1,5}
But P4 takes longer, so • The float for activity 4 is Pcp-P4 = 34-28 = 6 days And activity 4 is on P1 and P4
Activity 1 can have a slippage of 8 days • Activity 4 can have a slippage of 6 days. Conclusion…