SKEDSOFT

Latest Allowable Occurrence Time

The next one is the Latest Allowable Occurrence time represented by T_L i. This is illustrated by asimple example.

Earliest occurrence time of event 4 = 9 days. As the activities 3 - 4 take 4 days, the latest time by which activity starts is

T_L⁴ – t_E^3–4 = 9 – 4 = 5th day,

this is also the Earliest Occurrence time of event 3. Similarly, Latest Time by which event 2 occurs in T_L 3 – t_E 2–3 = 5 – 3 = 2 and so on. If a node is connected by number of paths then we have to find Latest Allowable Occurrence time as discussed

below.

The earliest occurrence time of event 6 is 21 days. As activities 4 – 6 take 4 days the earliest occurrence time is

T_L ⁶– t_E ⁴⁶ = 21 – 4 = 17 days. But there is another route 6 – 5 – 4.

 If we consider this routeT_L ⁵ = T_L ⁶ – t_E ⁵⁶ = 21 – 3 = 18th day, T_E⁴ = T_E⁵ – t_E ⁵⁴ = 18 – 7 = 11 days. As the latest allowable

occurrence time for event 4 is 17th day and 11th day, the event 4 will not occur until activities 6 – 4 and6 –5 are completed. As 11th day is the smallest, the event 4 occurs on 11th day. Hence the formula for T_Lⁱ= (T_L ⁱ – t_E ^ij) minimum.