Simulation Project 2 - Bearing Replacement [PDF]

  • 0 0 0
  • Suka dengan makalah ini dan mengunduhnya? Anda bisa menerbitkan file PDF Anda sendiri secara online secara gratis dalam beberapa menit saja! Sign Up
File loading please wait...
Citation preview

Simulation Project 2 – Bearing Replacement A milling machine has three different bearings that fail in service. The distribution of the life of each bearing is identical, as shown in the following table: Bearing Life (Hours) Probability



1000 1100 1200 1300 1400 1500 1600 1700 1800 1900 0.10



0.13



0.25



0.13



0.09



0.12



0.02



0.06



0.05



0.05



When a bearing fails, the mill stops, a repairperson is called, and a new bearing is installed. The delay time of the repairperson's arriving at the milling machine is also a random variable having the distribution given in the following table: Delay Time (Minutes) 5 10 15 Probability 0.60 0.30 0.10 Downtime cost for the mill is estimated at $10 per minute. The direct on-site cost of the repairperson is $30 per hour. It takes 20 minutes to change one bearing, 30 minutes to change two bearings, and 40 minutes to change three bearings. The cost of each new bearing is $32. A proposal has been made to replace all three bearings whenever a bearing fails. Management needs an evaluation of the proposal. The total cost per 10,000 bearing-hours will be used as the measure of performance for current and proposed methods. Design and perform a simulation of 15 bearing changes for each bearing under the current method of operation and proposed method. Replicate the simulation 100 times. Based on simulation results, prepare a report for management containing the answers to following questions: a) What is the (long term) average life of each bearing? Compare with theoretical value. b) What is the (long term) average delay time until the arrival of repairperson? Compare with theoretical value. c) What is the total cost per 10,000 bearing-hours under two methods? Should they implement the new procedure? d) What is the amount of saving, if any, per 10 000 hours of bearing life and also per year if the machine is run 8 hours a day for 300 days in a year? Now suppose that the distribution of the life time of each bearing is has exponential distribution with a mean life of 1400 hours. Run the simulation again and compare your results with previous ones.