Transcribed image text: Bearing failure in the blower used in diesel engines in tractors has been believed to occur according to a Weibull distribution with a mean life of 205000 km and a standard deviation of 6500 km. A sudden failure of the bearing results in costly repairs, and in total, a failure replacement is 11 times as expensive as a preventive replacement. Use the Glasser's graphs in the posted Lecture Notes 04 to answer the following questions. (1) Determine the optimal preventive replacement interval of the constant interval policy) to minimize the expected total cost per km. (2) What is the expected cost saving associated with your optimal value in (1) over a replace-only-on-failure policy? (3) Given that the cost of a failure replacement is $2050, what is the cost per km associated with your optimal policy given in (1)? (4) What if you use the age based preventive replacement policy? Answer all three sub-questions above. Your answers for (1): The optimal preventive replacement interval (of the constant interval policy) is: tp = Your answers for (2): The expected cost saving associated with your optimal value in (1) over a replace-only-on-failure policy is: Your answers for (3): The cost per km associated with your optimal policy given in (1) is: Your answers for (4) (Note: use the age based preventive replacement policy.): (4.1) The optimal preventive replacement interval (of the constant interval policy) is: tp = (4.2) The expected cost saving associated with your optimal value in (1) over a replace-only-on-failure policy is: (4.3) The cost per km associated with your optimal policy given in (1) is: Transcribed image text: Bearing failure in the blower used in diesel engines in tractors has been believed to occur according to a Weibull distribution with a mean life of 205000 km and a standard deviation of 6500 km. A sudden failure of the bearing results in costly repairs, and in total, a failure replacement is 11 times as expensive as a preventive replacement. Use the Glasser's graphs in the posted Lecture Notes 04 to answer the following questions. (1) Determine the optimal preventive replacement interval of the constant interval policy) to minimize the expected total cost per km. (2) What is the expected cost saving associated with your optimal value in (1) over a replace-only-on-failure policy? (3) Given that the cost of a failure replacement is $2050, what is the cost per km associated with your optimal policy given in (1)? (4) What if you use the age based preventive replacement policy? Answer all three sub-questions above. Your answers for (1): The optimal preventive replacement interval (of the constant interval policy) is: tp = Your answers for (2): The expected cost saving associated with your optimal value in (1) over a replace-only-on-failure policy is: Your answers for (3): The cost per km associated with your optimal policy given in (1) is: Your answers for (4) (Note: use the age based preventive replacement policy.): (4.1) The optimal preventive replacement interval (of the constant interval policy) is: tp = (4.2) The expected cost saving associated with your optimal value in (1) over a replace-only-on-failure policy is: (4.3) The cost per km associated with your optimal policy given in (1) is