CPM - EMSE 154 254 Applications of Linear and Nonlinear...

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EMSE 154 – 254 Applications of Linear and Nonlinear Optimization CPM/PERT Instructor: Hernan Abeledo Source: LINGO documentation 1 EMSE 154-254 Applied Optimization Modeling CPM/PERT Project Networks
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EMSE 154 – 254 Applications of Linear and Nonlinear Optimization CPM/PERT Instructor: Hernan Abeledo Source: LINGO documentation 2 CPM/PERT Networks and LP Critical Path Method (CPM) and Program Evaluation and Review Technique (PERT) are two closely related techniques for planning large projects. A key part of PERT/CPM is calculating the critical path. That is, identifying the subset of the activities that must be performed exactly as planned in order for the project to finish on time . In the table below, we list the activities involved in the simple, but nontrivial, project of building a house. An activity cannot be started until all of its predecessors are finished:
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EMSE 154 – 254 Applications of Linear and Nonlinear Optimization CPM/PERT Instructor: Hernan Abeledo Source: LINGO documentation 3 The figure shows the so-called activity-on-arc (AOA) network for the house project. We would like to calculate the minimum elapsed time to complete this project. Minimum completion time is equal to the length of the longest path , called the critical path , from the start node to the finish nodes. The project can be completed no sooner than the sum of the times of the successive activities on the critical path. Critical path consists of activities DIG , FOUND , WALLS , RAFTERS , ROOF , and FINISH and has length 27.
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EMSE 154 – 254 Applications of Linear and Nonlinear Optimization CPM/PERT Instructor: Hernan Abeledo Source: LINGO documentation 4 There are two apparently unrelated LP formulations for solving this problem. One formulation seeks a maximum length path in the network diagram. Let variables DIG , FOUND , etc. be either 1 or 0 depending upon whether activities DIG , FOUND , etc. are on or not on the critica1 path. MAX = 3*DIG + 4*FOUND + 2*POURB + 3*JOISTS + 5*WALLS + 3*RAFTERS + 4*FLOOR + 6*ROUGH + 7*ROOF + 5*FINISH + 2*SCAPE; - DIG = -1; DIG - FOUND = 0; FOUND - POURB - JOISTS - WALLS = 0; JOISTS - FLOOR = 0; POURB + WALLS - RAFTERS - SCAPE = 0; FLOOR - ROUGH = 0; RAFTERS - ROOF = 0; ROUGH + ROOF - FINISH = 0; FINISH + SCAPE = 1; ! No need to place binary constraints on variables since the LP solution will be integer. Note that each variable appears exactly twice in the conservation of flow constraints, once with a coefficient of +1 and once with a 1. This is a distinguishing feature of a network LP. Any single constraint (say, -DIG = -1; ) is redundant, since it is implied by the others. In any network LP problem we can eliminate one of the conservation of flow equations.
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EMSE 154 – 254 Applications of Linear and Nonlinear Optimization CPM/PERT Instructor: Hernan Abeledo Source: LINGO documentation 5 Global optimal solution found. Objective value:
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This note was uploaded on 01/02/2010 for the course EMSE 254 taught by Professor Hernanabeledo during the Fall '08 term at GWU.

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CPM - EMSE 154 254 Applications of Linear and Nonlinear...

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