(NOTE:NEED HELP FOR FIVE, ANS FOR FOUR BELOW. THANK YOU) A petroleum company is considering expansion of its one unloading facility at its Australian refinery. Due to random variations in weather, loading delays, and other factors, ships arriving at the refinery to unload crude oil arrive according to a Poisson distribution with a rate of 15 ships per week. Service time is exponential with an average service rate of 22 ships per week. a) What are and ? b) What is the utilisation of the facility? c) What is the average number of ships waiting to gain access to the single unloading facility? d) What is the average time a ship must wait before beginning to deliver its cargo to the refinery? e) What is the average total time (waiting plus delivery in the system) that a ship spends at the refinery? (15 marks) 5. Based on the scenario from task 4 above, the company that could rent a second unloading berth for $38,000 per week (ships waiting in a queue can be served by either berth upon availability). The service time for this berth is also exponential with the same service rate of 22 ships per week. Assume the opportunity cost for a ship to wait is $45,000 per week. a) What type of queue is this? (1 marks) b) What is the new utilisation rate for the facility? (2 marks) c) If the second berth is rented, how many ships on average will be waiting in the queue? (3 marks) d) How much time does a ship spend waiting on average if there are two berths? (3 marks) e) Is the benefit in reduced waiting time for the ships worth the rental cost of the second berth? (6 marks)

(have the answers for task 4 if needed, just need help with 5)

1. = 15 ships/week, =service rate = 22 ships / week

2. Mean utilisation rate of facility: 15 ships per week / 22 ships per week = 0.681 (68.1% busy)

3. Avg numbers waiting Lq = ^2 / ( - ) = 225 / (22 x 4) = 225/88 = 2.55 ships

4. Avg waiting time = Lp/ = 2.55 / 15= 0.17(weeks)

5. Avg total time spent in the system = 1/ - = 1 / 22-15=0.14 week