The arrival rate for a waiting-line system obeys Posson distribution with a mean of 0.5 unit per period. It is required that the probability of one or more units un the system not exceed 0.20. If the cost of waiting is $5.00 per unit period, and the service facility cost $2.50 per unit served, what is the total system cost? If the service rate is Doubled at a total cost of $3.50 per unit served what is the total system cost?
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It is estimated that cars arrive according to a Poisson distribution at the rate of 3 every 6 minutes and that there is enough space to accommodate a line of 12 cars. Other arriving cars can wait outside this space, if necessary. It takes 2.5 minutes on the average to fill an order, but the service time actually varies according to an exponential distribution. Determine the following: The probability that the facility is not idle; The expected number of customers waiting but currently not being served; The expected waiting time until a customer can place his order at the window;Consider a two-channel waiting line with Poisson arrivals and exponential service times. The meanarrival rate is 28 units per hour, and the mean service rate is 20 units per hour for each channel.A. What is the probability that no units are in the system?(Please show full calculation)
- BBA Bank (a fictitious one) has a drive-through teller window and observed that 15 customers arrive for service per hour, on an average, and the average service time per customer is 3 minutes. Assume inter-arrival time and service time follow a negative Exponential distribution. The bank hires you as a consultant. You guess that this is an M|M|1 system and you are required to determine the following: Probability (teller is busy) Probability (teller is idle) Probability of 3 customers in the system. Average number of customers waiting for service, that is, number of autos in the line excluding the one at the teller window. Average number of customers in the system, that is, number of autos in the line including the one at the teller window. Average time a customer spends in the system, that is, waiting time plus service time. Average time a customer spends in the waiting line before reaching the teller window.Below are the data of the gasoline station. Number of gas pumps = 1 Policy of the gasoline station is "If you want to wait, pay PhP45 per liter; if you do't want to wait, pay PhP55/liter." Service mean time (exponential distribution) = 5 minutes Arrival rate (Poisson distribution) = 10 per hour. Customers usually wait to buy gasoline. a) What is the probability (in decimal) that customer will wait? Answer in 4 decimal places.b) Determine the probability that the customer will not wait. Answer in 4 decimal places.c) How many customers are falling in line? Answer in integer value.d) Determine the expected price of gasoline per liter.Neve Commercial Bank is the only bank in the town of York, Pennsylvania. On a typical Friday, an average of 10 customers per hour arrive at the bank to transact business. There is currently one teller at the bank, and the average time required to transact business is 4 minutes. It is assumed that service times may be described by the negative exponential distribution. If asingle teller is used, find:a) The average number in the system.b) The probability that the bank is empty.c)CEO Benjamin Neve is considering adding a second teller (who would work at the same rate as the first) to reduce the waiting time for customers. A single line would be used, and the customer at the front of the line would go to the first available bank teller. He assumes that this will cut the waiting time in half. If a second teller is added, find the new answers to parts (a) to (e).