Da Help me solve this | 31 parts remaining X tion time and the critical path shou de Dave Fletcher was able to determine the activity times for constructing his laser scanning machine. Here are the activity times: Activity Time (weeks) Immediate Predecessor(s) |Activity Time (weeks) Immediate Predecessor(s) 4 B 00 B 10 C, E Da O I D D . F Fletcher would like to determine ES, EF, LS, LF, and slack for each activity. The total project completion time and the critical path should also be determined. To find the critical path, we calculate two distinct starting and ending times for each activity. These are defined as follows: Earliest start (ES) = earliest time at which an activity can start, assuming all predecessors have been completed, Earliest finish (EF) = earliest time at which an activity can be finished, Latest start (LS) = latest time at which an activity can start so as to not delay the completion time of the entire project, Latest finish (LF) = latest time by which an activity has to finish so as to not delay the completion time of the entire project.Dave's earliest start (ES) and earliest finish (EF) are Dave's latest start (LS) and latest finish (LF) are. 0 That's incorrect. Before an activity can start, all its immediate predecessors must be finished: > If an activity is an immediate predecessor for just a single activity, its LF equals the LS of the activity that immediately follows it. > If an activity is an immediate predecessor to more than one activity, its LF is the minimum of all LS values of all activities that immediately follow it. That is: LF = Min(LS of all immediate following activities). The latest start time (LS) of an activity is the difference of its latest finish time (LF) and its activity time. That is: L8 = LF - Activity time. Therefore, LF = LS + Activity time. Dave Fletcher was able to determine the activity times for constructing his laser scanning machine. Fletcher would like to determine ES, EF, LS, LF, and slack for each activity. The total project completion time and the critical path should also be determined. Here are the activity times: Activity Time (weeks) Immediate Predecessor(s) Activity Time (weeks) Immediate Predecessor(s) 6 B A B In TIM - W O C, E DD D . F Dave's earliest start (ES) and earliest finish (EF) are: Activity ES EF A 0 O D 7 10 T m 12 G 10 21 H 12 18 Dave's latest start (LS) and latest finish (LF) are: Activity LS LF H 15