Single-Variable Optimization Numerical

Single-Variable Optimization Numerical

<p>Nonlinear Optimization MSC 523</p><p>Turn-in Problem 2 Single-Variable Optimization – Numerical Inspect and Repair Availability Problem</p><p>For some restorable systems, such as alarm systems and standby systems, failures are not observed until the system is required to operate. If dormant (undetected) failures may occur while in a non-operating state, then the system availability can be influenced by the frequency in which the systems are inspected and repaired if found inoperable. Dormant failures are often caused by corrosion or mechanical fracture but the dominant cause of dormant failures can often be latent manufacturing defects. If inspection requires some downtime, then there is a trade-off between inspection downtime and the detection and restoration of failures. The following model assumes idealized repair that restores the system to as good as new condition. It is based upon maximizing a steady-state availability. Let R(t) be the dormant failure distribution,</p><p>R(t) = e-t for the exponential failure distribution,</p><p> t1 = inspection time,</p><p> t2 = repair time if necessary</p><p>T = time between inspections (the decision variable).</p><p>If the dormant failure distribution is exponential, then the expected available time (up- time) during a cycle is given by:</p><p>1 eT A(T)  T [T  t1  t2 (1 e )]</p><p>A nuclear power plant has several sophisticated radiation detectors throughout the facility. These detectors historically have had a constant failure rate (exponential failure distribution). Periodically, these detectors can be removed and brought into a lab for testing. If found defective, they are replaced. What is the optimum time between inspections, T*? Nonlinear Optimization MSC 523</p><p>Name: ______</p><p>Turn-in Problem 2 Single-Variable Optimization – Numerical Inspect and Repair Availability Problem</p><p>Answer the following using a different single-variable search algorithm for each case:</p><p>Case 1 2 3 Failure rate  .0002 .001 .0008 Inspection time t1 16 8 24 Repair time t2 48 72 12 (a) What is the optimal inspection interval, T*, to the nearest hour?</p><p>(b) What is the corresponding availability, A(T*)?</p><p>(c) Algorithm used for optimization?</p><p>Note: All time units are in hours</p><p>Attach your work showing the results of each iteration:</p>

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