ProjmanagementXmp

ProjmanagementXmp - The Problem Job No. Immediate...

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The Problem eo b e Job No. Immediate predecessors amb a. Construct the appropriate network diagram. . dicate the critical path. 1 - 2 3 4 2 1 123 3 14 5 1 2 b. Indicate the critical path. c. What is the expected completion time for the project? d. You can accomplish any one of the 4 13 4 1 1 5 2 135 6 3 following at an additional cost of $1500: 1. Reduce J5 by 2 days. 2. Reduce J3 by 2 days. 7 4 189 8 5 , 6 246 9 82 4 1 2 3. Reduce J7 by 2 days. If you will save $1000 for each day that the earliest ompletion time is reduced 10 7 345 11 9,10 5 7 8 completion time is reduced, which action, if any, would you choose? e. What is the probability that the roject will take more than 30 days 1 project will take more than 30 days to complete?
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Project Management – the network (How about starting and ending nodes?) J5 J2 J8 J1 J3 J6 J9 J4 J11 J7 J10 2
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Finding the expected finishing time d g t e e pected s g t e • The expected time can be easily calculated by this formula or using spread sheet 6 b 4m a ET + + = 3
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Expected completion time pected co p et o t e Job No. Predecessors a m b ET 1 -2 3 4 3 2 11 2 3 2 3 14 5 1 2 6 4 13 4 1 1 5 3 5 2 1 3 5 6 31 2 3 2 7 41 8 9 7 8 5,6 2 4 6 4 9 82 4 1 2 5 10 73 4 5 4 4 9,10 5 7 8 6.83
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Project Management – the network J5 3 J2 2 J8 4 J1 3 J3 6 J6 2 J9 5 J4 J11 J7 7 J10 4 5 6.83 5
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Project management he critical path oject a age e t t e c t ca pat • We can use J1 as the starting node and J11 as the ending node. To find the critical path is equivalent to finding the longest path from the starting node to the ending node . • The network is simple. We can see that there are only 3 possible paths from the starting node to the ending node: 1 2 5 8 9 11 (23 83 days) J1 - J2 - J5 - J8 - J9 -J11 (23.83 days) – J1 – J3 – J6 – J8 – J9 – J11 (26.83 days) – J1 – J4 – J7 – J10 – J11 (25.83 days) • The critical path is given by ( J1 – J3 – J6 – J8 – J9 – J11 ) with a total completion time of 26.83 days . 6
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Project Management – the project critical path J5 3 J2 2 J8 4 J1 3 J3 6 J6 2 J9 5 J4 J11 J7 7 J10 4 5 6.83 7
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ritical paths C t ca pat s • Remark: It is very rare to have just a few paths (from starting node to ending node) in a project network. If that is the case, we can identify all the paths and evaluate e completion time of each one the completion time of each one. • The critical path is the one with the longest completion time . • Otherwise, we have to apply a more systematic method to identify the critical path. The approach we introduce ere is called the Critical Path Method (CPM) here is called the Critical Path Method. (CPM) 8
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The CPM – Finding start time and finish time Finding the earliest start time (ES) and earliest finish time (EF) of an activity or job. ES EF J?? 5 Let t be the ET of an activity or job. Then the ES of an activity is given by the following expression ES = max {EF of all its immediate predecessors} 9 EF = ES + t
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Project Management – the CPM S a F of all its immediate predecessors} S = ax F of all its immediate predecessors} J5 3 58 ES = max {EF of all its immediate predecessors} Cannot find the ES of J8 because we do
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This note was uploaded on 09/21/2011 for the course COMP 4631 taught by Professor Ding during the Fall '11 term at HKUST.

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ProjmanagementXmp - The Problem Job No. Immediate...

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