SIM335 Managing Projects Essay Example | Topics and Well Written Essays - 2500 words - 1

SIM335 Managing Projects - Essay ExampleAny con menses activities, which act follows which? The nedeucerk diagram will look as below and boxes represent the following aspects If exercise A has length of 4 years, earlier start time would be day 0 (zero) and the earlier give up would be day 4. The earliest start for activity B would be day 4 and the earliest block would be day 9 as a result of adding. Earliest finish of activity A (4) + continuation of activity B (5) = 9 early finish for activity B In general, earliest finish of previous activity+ duration of current activity= earliest finish of the current activity. (Early finish for earlier tasks becomes the early the start for next activity). For a case same activity G that has several preceding activities, we take the value, which has the largest earlier finish value (F 13). The latest start and latest finish are determined by working backwards through the project. Latest start for activity Q = earliest finish of activity Q because it is the last activity. In general Latest start= latest finish duration The latest start used as the latest finish for the previous activity. For instance, the latest start for activity Q is 935= 88 and this is used as the value for latest finish for activity P. for activities like I & J we use the latest start for activity K. for activity H we take the least value for latest start in this case is 34 (latest start for activity I). ... calculated by Early finish early start duration=Total shoot a line For instance, to calculate total bollocks up between activities G &H =3413(13+8) =0 Critical path is the least numerate of time that can be taken to complete the project. This means that there is no float, no waive time and a insure in the critical path can lead to a delay in project completion. It can be determined by adding the duration of each activity in the taking over where there is no float. In our case, the critical path is from activity A to Q in the sequence A( 4)+F(9)+G(13)+H(8)+I(7)+K(8)+L(13)+M(9)+N(4)+O(4)+P(9)+Q(5) =93. It is very important as it helps in time scheduling of consecutive activities in a project. Events that lead here are known as milestone (Harold, 2009). If the activity begins on Monday January 16th, the project needs a total of 122 days to finish with a five-day working period indeed we find out how umpteen weeks we be will need. If one week has a total of 5 working days then the project will take 122?5=24 weeks and four days, then the project will abolish on Thursday 5th July 2012. In case activity B is delayed by two days, there would be no effect on the total project duration because it has a float of four days. This is however not the case for activities P and O, delaying activity P by two days or activity O by a day will affect the total duration of the project because there is no float for the two activities (Harold, 2009). Limitations of network diagrams The time estimates depend on individualised bias, th ere is no specific formula for determining the duration of the activities and speculation is used. This may affect the unit project just in case any activity is delayed or takes a long-dated duration to finish. As seen from above, activities without float if delayed for any reason can