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- At the beginning of each week, a machine is in one of four conditions: 1 = excellent; 2 = good; 3 = average; 4 = bad. The weekly revenue earned by a machine in state 1, 2, 3, or 4 is 100, 90, 50, or 10, respectively. After observing the condition of the machine at the beginning of the week, the company has the option, for a cost of 200, of instantaneously replacing the machine with an excellent machine. The quality of the machine deteriorates over time, as shown in the file P10 41.xlsx. Four maintenance policies are under consideration: Policy 1: Never replace a machine. Policy 2: Immediately replace a bad machine. Policy 3: Immediately replace a bad or average machine. Policy 4: Immediately replace a bad, average, or good machine. Simulate each of these policies for 50 weeks (using at least 250 iterations each) to determine the policy that maximizes expected weekly profit. Assume that the machine at the beginning of week 1 is excellent.It costs a pharmaceutical company 75,000 to produce a 1000-pound batch of a drug. The average yield from a batch is unknown but the best case is 90% yield (that is, 900 pounds of good drug will be produced), the most likely case is 85% yield, and the worst case is 70% yield. The annual demand for the drug is unknown, with the best case being 20,000 pounds, the most likely case 17,500 pounds, and the worst case 10,000 pounds. The drug sells for 125 per pound and leftover amounts of the drug can be sold for 30 per pound. To maximize annual expected profit, how many batches of the drug should the company produce? You can assume that it will produce the batches only once, before demand for the drug is known.Telephone calls come into an office at random. The time between calls is the following discrete distribution: Service time Probability 6 .40 7 .30 8 .20 9 .10 When the receptionist is busy, a telephone message tells the caller that the call is important and to please wait on the line until the next reservation agent becomes available. We can assume that customers don’t hang up once they join the call queue. The service time for call follows a normal distribution with a mean of 8 mins and standard deviation of 2.5 mins Simulate the operation of the office telephone system for 1100 customers. Discard the first 100 customers and use data over the next 1000 customers to answer the following questions: What is the average waiting time for customers? What is the utilization of the receptionist? What is the probability that a customer waits for longer than 1 min before their call is answered?
- The average time between the arrivals of the taxis arriving at the airport to pick up passengers has an exponential distribution, with an average of 10 minutes. a) What is the probability that a passenger will wait for the taxi less than 15 minutes? b) What is the probability that a passenger will wait for the taxi between 20 and 30 minutes? Solve using the cumulative distribution function H1(Physics: acceleration) Average acceleration is defined as the change of velocity divided by the time taken to make the change, as shown in the following formula: a = (v1 - v0) / t Here, v0 is the starting velocity in meters/second, v1 is the ending velocity in meters/second, and t is the time span in seconds. Assume v0 is 5.6, v1 is 10.5, and t is 0.5, and write the code to display the average acceleration. If you get a logical or runtime error, please refer https://liangpy.pearsoncmg.com/faq.html.During an eight-hour shift, 750 non-defective parts are desired as a result of a manufacturing operation. The default operation time is 15 minutes. As the operators of machine are inexperienced, the actual time they take to perform the operation is 20 minutes, and, on average, a fifth of the parts that start to be manufactured are lost. Assuming that each one of the machines used in this operation will not be available for one hour in each shift, determine the number of machines needed.
- Consider a service system in which each entering customer must be served first by server 1, then by server 2, and finally by server 3. The amount of time it takes to be served by server i is an exponential random variable with rate μi, i = 1, 2, 3. Suppose that you enter the system when it contains a single customer who is being served by server 3. Find the probability that server 3 will still be busy when you move over to server 2. Find the probability that server 3 will still be busy when you move over to server 3. Find the expected amount of time you spend in the system. (whenever you encounter a busy server, you must wait for the service in progress to end before you can enter the service) Suppose that you enter the system when it contains a single customer who is being served by server 2. Find the expected amount of time that you spend in the system.A service facility consists of one server who can serve an average of two customers per hour (service times are exponential). An average of three customers per hour arrive at the facility (interarrival times are assumed to be exponential). The system capacity is three customers: two waiting and one being served. What is the probability that the server is busy at a typical point in time? If needed, round your answer to a whole percentage. The excel formula breaks the answer down to 0.8769. However 0.87 or 1% is showing as wrong answer.Wizard, a popular brand of electric garage door opener, includes two 40-watt bulbs that go onwhen the garage door is opened. A bulb will generally last about one year in normal operation.Three neighbors, James, Smith, and Walker, each has a Wizard opener in their respectivegarages. Each time a bulb burns out, James replaces both bulbs. Smith, on the other hand, replaces only the bulb that has burned out, and Walker replaces both bulbs only after both haveburned out. Assume that light bulbs fail according to an exponential law.b. What percentage of the time will Walker have only one bulb burning?
- Wizard, a popular brand of electric garage door opener, includes two 40-watt bulbs that go onwhen the garage door is opened. A bulb will generally last about one year in normal operation.Three neighbors, James, Smith, and Walker, each has a Wizard opener in their respectivegarages. Each time a bulb burns out, James replaces both bulbs. Smith, on the other hand, replaces only the bulb that has burned out, and Walker replaces both bulbs only after both haveburned out. Assume that light bulbs fail according to an exponential law.c. Is there any advantage of James’s strategy over Smith’s?Wizard, a popular brand of electric garage door opener, includes two 40-watt bulbs that go onwhen the garage door is opened. A bulb will generally last about one year in normal operation.Three neighbors, James, Smith, and Walker, each has a Wizard opener in their respectivegarages. Each time a bulb burns out, James replaces both bulbs. Smith, on the other hand, replaces only the bulb that has burned out, and Walker replaces both bulbs only after both haveburned out. Assume that light bulbs fail according to an exponential law.a. Over a 10-year period, how many bulbs, on average, will each neighbor require?A car repair can be performed either on time or late, and either satisfactoryily or unsatisfactorily. The probability of a repair being on time and satisfactory is 0.26. The probability of a repair being on time is 0.74. The probability of a repair being satisfactory is 0.41. What is the probabilityof a repair being late and unsatisfactory?