This list has been set out in this table, so it is easy to see what needs to be done, how long it will take and what has to be done before something else can. The activity list includes all the activities that need to be done in order for the party to run smoothly.
SOLUTION:
From the precedence table of the previous page I will now produce an activity network assuming that infinite resources are available…
After creating this activity network next I will carry out forwards and backwards passes to obtain the early and late event times, which will be needed later on…
I am now going to draw a table and include the early and late finish times for each activity…
INTERPRETATION:
From this table, and the activity network the fastest time it will take to organise a party this way, would be 19 days.
The critical path for the activity network for organising a party is:
A, C, D, E, G, J, L and M.
I am now gong to construct a cascade chart with both the earliest and latest starts and finishes and the critical path to see how many workers would be needed to complete this activity (page 1 of appendix).
The cascade chart with both the earliest and latest starts and the critical path shows that more than two workers will be needed to complete these tasks in the set time.
I am now going to construct a cascade chart using both the earliest and latest starts to try and get the least amount of workers to complete the tasks in the minimum time. However, here I also have to take into consideration the immediate predecessors. This is because I have to make sure that each activity is placed in a place where its immediate predecessors have already happened (page 2 of appendix).
Workers:
From the chart on page 2 of the appendix it is clear that 3 workers are needed to complete the activities in the minimum time.
To make them all fit, I have used the first worker to do the critical path. Because of this the earliest start and finish times were used.
Then for worker two I used the earliest start and finish times for tasks B, F, N and K. I then used the latest start and finish time for tasks O in order for it to be completed by worker 2.
For worker 3 there is only one tasks needed to be done (H) and I used the earliest start time for this task.
I will now take into consideration the fact that some of these activities will need more than one worker at a time to allow the event to be completed on time.
Below is a table of how many workers each activity will need…
Using this table I will now create a cascade charts and then resource histograms to see how many workers will be needed at each time. I will do this for both the earliest and latest starts and finishes…
Earliest Start: (page 3 and 4 of appendix)
From the histogram on page 4 of the appendix it is clear to see that the maximum number of workers needed is 5. For the majority of the time only 4 workers are needed at any one point. This could mean that to organise a party in this manner you will need 4 workers, and then an extra worker for just 3 days.
I will now do the latest start and finish times and see whether the outcome is different…
Latest Start: (page 5 and 6 of appendix)
From the histogram on page 6 of the appendix showing the latest possible times, the maximum number of workers needed is 7.
CONCLUSION:
In conclusion the histograms show that using the earliest possible start times is the most efficient way, as fewer workers are needed to get the tasks completed in the minimum time.
All the work I have completed in this coursework helps lead up to this conclusion. This is a practical way of arranging a party, and throughout the whole of the coursework the immediate predecessors are taken into consideration. This means that my answers are precise, and everything is done in the correct order.
Changes:
If I were to change anything about this coursework I would have researched further whether each of the activities can be completed in a different order, and at a quicker speed. However, I feel my information is correct, and relevant to the task at hand.
The changes that I would make to allow this to be more accurate would be, there would be more time allowed to complete all of the tasks. This way you can make sure everything is done in the correct way and more accurately. Trying to do things in the minimum time using decision maths could sometimes, in reality, make the task be done not up to the correct standard.
