Inspection interval with BS

[Rui Assis]

How does BlockSim determine the inspection intervals in condition monitoring? What is the rational behind it?

Thank you in advance.

[tarik]

If you are referring to determining the optimum inspection interval, BlockSim doesn’t provide such calculation. I recommend that you use different scenarios in your model (for ex, inspecting every 1month, 2 months..) and determining which inspection policy ensures that your system stay above a certain availability requirement or under a cost requirement (cost associated to performing inspection, corrective action, downtime). You can check the following example
http://www.weibull.com/hotwire/issue28/hottopics28.htm

If you are referring to the optimum preventive maintenance interval calculations?
It‘s based on a model that describes the associated costs and risks. The model assumes that if the unit fails before time t, a corrective action will occur and if it does not fail by time t, a preventive action will occur. In other words, the unit is replaced upon failure or after a time of operation, t, whichever occurs first.
Thus, the optimum replacement time can be found by minimizing the cost per unit time, CPUT(t). CPUT(t) is given by:

CPUT(t) = Total Expected Replacement Cost per Cycle/ Expected Cycle Length

Because I can’t type the equations here, please refer to http://www.weibull.com/SystemRelWeb/preventive_maintenance.htm
(specifically read the ‘Determining Preventive Replacement Time’ and ‘Optimum Age Replacement Policy’ sections)

[Rui Assis]

Thank you Tarik for your answer.

I was referring to determining the optimum inspection interval in condition monitoring of a single item. The example in hotwire/issue28 treats a block of items and doesn’t match my problem. My problem is, what schedule of inspections to choose when you are expecting a potential failure showing up sooner or later and want to prevent it from becoming an operational failure. I know that the inspection intervals must become shorter and shorter as the probability of a failure increases with time. I suppose that such a schedule has to be found by trading off two opposite costs: risk of choosing intervals too large and miss the detection of the potential failure in course on one hand, and costs of performing inspections too frequently that also are costly on the other.

Do you have an answer for this or know related literature?

Thank you.