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D1 how to CONSTRUCT A scheduling diagram directly from the activity network?

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    I dont know how to start and rules to construct directly from the activity network
    Thanks
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    anyone?
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    For some events ei < li the delay is the slack of the event. If ei = li the event has no slack and it is a critical event. The time available for an activity is the gap between its earliest possible start time and its latest possible finishing time. In some cases this time will be greater than the duration and the change in time is the float of the activity. An activity of 0 float is a critical activity. The critical activities in a network form the longest path from the source to the sink. This is the critical path.

    In a Gantt chart activities are shown as bars against a time scale. The critical activities are fixed so they are inserted first in a single row. Each non-critical activity is placed at its earliest start time as a solid bar with the float shown as a dotted bar, these are positioned on separate rows.

    Scheduling is the systematic allocation of workers to activities. If there is a free worker and an activity to be started then the activity should be assigned to him. The time is then advanced until there is a free worker and an activity which can be started, this is repeated until all the activities have been assigned. If there is ever a choice of activities then the most critical is chosen (the one with the smallest latest starting time).
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    with my exam board i think its called a resource histogram so if its the same thing, then i think i can help you
    you plot a sort of graph with no. of people available on y axis and total duration along x-axis
    then plot ALL critical activities on the first person (basically draw activities as boxes as long as they're duration on x-axis) then when there is no more room for activities on that person, draw the rest on the next person (you can do another method but this is the "easiest" way of getting the most efficient diagram, and efficiency is key in D1 :P: )

    you should get a nice diagram with all activities plotted, hope this helps

    Here's some notes, if mine make no sense http://www.meidistance.co.uk/pdf/d1/mei/c/2/d1c2n.pdf
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    Scheduling is the systematic allocation of workers to activities. If there is a free worker and an activity to be started then the activity should be assigned to him. The time is then advanced until there is a free worker and an activity which can be started, this is repeated until all the activities have been assigned. If there is ever a choice of activities then the most critical is chosen (the one with the smallest latest starting time).
    so can u tell me how many workers are needed and which tasks do they need to take and why and is the mist critical is chosen because it has the smallest latest starting time?
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    Here's some notes, if mine make no sense http://www.meidistance.co.uk/pdf/d1/mei/c/2/d1c2n.pdf
    Thanks but I think its method is quite different from what I learn
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    (Original post by prettyboy_bn)
    so can u tell me how many workers are needed and which tasks do they need to take and why and is the mist critical is chosen because it has the smallest latest starting time?
    so first worker = a c f h j
    second worker = b e i
    third worker = d g
    you choose the smallest latest starting time as if it has only a small slack then there is less options for when it can start, eg 2 activities of length 4 have an early start time of 3 if a has a slack of 5 and b has a slack of 1 then the same worker can do a and b but only if b is chosen first
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    so first worker = a c f h j
    second worker = b e i
    third worker = d g
    but the answer only works for minimum 2 workers
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    (Original post by prettyboy_bn)
    but the answer only works for minimum 2 workers
    what answer only works for a minimum of 2 workers? :confused:
    to complete all tasks in the shortest possible time (which i assume would be the question) you need 3 workers as F G and E (as an example) would all have to take place at the same time
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    Because answer in the book only allows 2 workers and it is:
    1:a,c,d,g,h
    2:b,e,f,i,j
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    (Original post by prettyboy_bn)
    Because answer in the book only allows 2 workers and it is:
    1:a,c,d,g,h
    2:b,e,f,i,j
    ok, i didnt actually read the question there
    i think that if your given the workers and told to find the time you cant use the scheduling method i said, dont know how you would do it though, sorry

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Updated: May 20, 2010
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