(NOTE:NEED HELP FOR SIX, ANS FOR FOUR and FIVE BELOW. THANK YOU) PLEASE DON'T ANSWER WITH THE SAME ANSWERS FROM OTHER QUESTION BECAUSE THIS HAS DIFFERENT VALUES COMPARED TO THE OTHER QUESTIONS 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)

6. Based on the scenarios from tasks 4 and 5 above, the company could instead use only one berth, but incorporate process improvements through BPR to speed up the unloading process. After these process improvements, the new service time would still be exponential, but with a service rate of 26 ships per week. The incremental cost for this BPR project would be $25,600 per week. a) What are and ? (1 marks) b) What is the utilisation () for the one berth? (2 marks) c) Compute the average number of ships waiting. (3 marks) d) Compute the expected waiting time and the average total time (waiting plus delivery in the system) for the ships. (3 marks) e) Is the benefit in reduced waiting time for the ships worth the cost for the BPR project? (6 marks)

(have the answers for task 4 and 5 if needed, just need help with 6. If you think my answers for 4 and 5 are questionable. please tell me. Thank you)

4)

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

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

c. Avg numbers waiting Lq = ^2 / ( - )

= 225 / (22 x 4)

= 225/88

= 2.55 ships

d. Avg waiting time = Lp/ = 2.55 / 15

= 0.17(weeks)

e. Avg total time spent in the system = 1/ - = 1 / 22-15

=0.14 week

5)

a. This is an M/M/s type queue in which one queue serves a number of servers (2 in this case). It can be called a M/M/2 single line 2 server queue.

b. The new utilisation rate for the facility is (rho) = / (c x ) where c= numbers of servers.

The means the new utilisation is rho = 15 / (2 x 22)

= 0.34 (34% busy)

c. Utilisation of one server rho= /

= 0.682

Probability of 0 customers in system P(0) = 1/ [ 1+ rho+ rho^2 / ( 2-) ]

= 1/[ 1+0.682+0.46487 x 22 /(44-15)] = 0.4915

Number of ships in line Lq = P(0)x[ x x rho^2 / ( 2-)^2]

= 0.4918 x 22 x 15 x 0.46487 /(29)^2

= 0.0897

d. Avg wait time = Lp / = 0.0987 / 15

= 0.00598 week

Wait time for all ships per week = x Wq = 0.00598 x 15

= 0.0897 weeks

Wait time for all ships per week = x Wq = 0.00598*15

= 0.0897 weeks

e. Wait cost per week = 0.0897 x 45000= 4036.92

SInce this cost is much less than the rent of a second berth, the investment is not worth the money.

(I KNOW THIS IS A SIMILAR QUESTION TOWARDS OTHER QUESTION BYT MY QUESTION HAS DIFFERENT VALUES THEN THE OTHER QUESTION)