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## Probability Distribution of Time Between Arrivals

An emergency room of a hospital in Sydney has one receptionist, one doctor, and one nurse on duty. The emergency room opens 24 hours per day. Patients arrive at the emergency room according to the probability distribution as given in Table Q1.1.The attention needed by a patient who comes to the emergency room is defined by the following
probability distribution as shown in Table Q1.2.

Table Q1.1 Distribution of time between Table Q1.2 Probability distribution of arrivals (min) attention needed Time between arrivals (min)
Probability Patient needs to see Probability
5 0.10 Doctor alone 0.40
10 0.15 Nurse alone 0.35
15 0.22 Both 0.25
20 0.25 1.00
25 0.18
30 0.10
1.00
2
If a patient needs to see both the doctor and the nurse, he or she cannot see one before the other that is, the patient must wait to see both together.
The length of the patient’s visit (in minute) is defined by the following probability distributions:Doctor Probability Nurse Probability Both Probability
10 0.20 5 0.10 15 0.08
15 0.30 10 0.25 20 0.15
20 0.25 15 0.50 25 0.21
25 0.15 20 0.15 30 0.28
30 0.10 1.00 35 0.18
1.00 40 0.10
1.00
1) Simulate the arrival of 50 patients to the emergency room (assume there is no patient when the simulation starts).
2) Compute the probability that a patient must wait and the average waiting time.
3) Suppose the process continues with enough number of patients, what is the optimum minimum number of simulation trials (i.e. number of patients)?
4) Based on the simulation results, does it appear that this system provides adequate patient care?
Problem 2
The IEEE Association in Power Electronics (IAPE) is going to hold its annual conference at the Hilton Hotel in Hong Kong in 2019. Based on historical data, the IAPE believes the number of rooms it will need for its members attending the conference is normally distributed,with a mean of 700 and a standard deviation of 120. The IAPE can reserve rooms now (one year prior to the conference) for \$120; however, for any rooms not reserved now, the cost will be at
the hotel’s regular room rate of \$200. The IAPE guarantees the room rate of \$120 to its members. If its members reserve fewer than the number of rooms it reserves, IAPE must pay the hotel for the difference with a discount, at the \$80 room rate.

1) Using Crystal Ball, determine whether the IAPE should reserve 500, 600, 700, 800 or 900 rooms in advance to realize the lowest total cost.
2) Can you determine a more exact value for the number of rooms to reserve to minimize cost? National Australia Bank at city central is trying to determine whether it should install one or two drive-through teller windows. The following probability distributions for arrival intervals and

2 0.25
3 0.45
4 0.20
5 0.10
Service Time (minute) Probability
2 0.10
3 0.40
4 0.20
5 0.20
6 0.10
1.00
Assume that in the two-server system, an arriving car will join the shorter queue. When the queues are of equal length, there is a 50 – 50 chance the driver will enter the queue for either window.
1) Simulate both the one- and two-teller systems. Computer the average queue length, waiting time, and percentage utilization for each system.
2) Discuss your results in 1), and answer the minimal simulation trials and suggest the degree to which they could be used to make a decision about which system to be employed.

Problem 4
A new car registration and license issuing process for NSW Roads and Maritime Services (RMS) is being tested in Wollongong city. During the peak demand hours of 10:00 am to 1:00 pm, there is one customer arriving in average every 4 minutes (on average, according to an exponential distribution).Typically, 30% of the customers want to register their cars (only), 45% want to renew their licenses (only), and 25% want to do both (register their cars and renew their licences).

## Attention Needed Probability Distribution

The process is as follows:
x Each customer gets into a line (with a maximum of 20 people allowed in that line).x Once the customer reaches the counter, he/she informs a clerk at the counter what he/ she needs.
x The time of this discussion with the clerk can be approximated by a normal distribution

with a mean time of 1.0 minute and a standard deviation of 0.5 minutes.x The clerk at the counter gives each customer a number and shows him/her where to sit (there is plenty of room for customers to sit and wait).x Those who want to do both (register their cars and renew their licences) get their licence renewed first.x There are specific officers dedicated to particular tasks, either licensing or registration:o Sharon is the officer dedicated to process licence renewals and o Peter is the officer dedicated to process car registrations.x Sharon processes a licence according to a normal distribution with a mean of 5.0.inutes and a standard deviation of 1.0 minute.x Peter processes a registration according to a uniform distribution with a minimum of 5.0 minutes and a maximum of 9.0 minutes.

1) Draw a flow diagram for the above process. (5 marks)
2) Develop a simulation model using SimQuick software for the peak time period (3-hour time period); perform 400 simulations to determine:
• The overall average waiting time in the process for customers;
• the number of customers served during the 3-hour time peak period;
• the number of customers who have already seen the clerk at the counter but waiting to be served at the end of the 3-hour peak period;
• the utilisation of these three employees, i.e.
o The clerk at the counter
o Sharon
o Peter
If the management is considering adding a new trained officer, Mary, who can renew licences as well as process a car registration. Mary’s approach is to keep an eye on both the registration and licencing queues and then serves the next person in the longer queue. The amount of time that Mary processes either a licence renewal or a car registration can be approximated by the following distribution:
Time taken by Mary to process either a licence or
registration for a car, y (min) 3 4 5 6 7
Probability, f(y) 0.15 0.30 0.25 0.20 0.10

1) Use SimQuick simulation model to decide whether or not the management team should employ Mary (use 500 simulations).
2) How best should the Mary’s time be utilised if the management team decides to employ
her?

In many instances the quality of the finished product is a function of the temperature and pressure at which the chemical reactions take place.It is to model the quality, Y, of a product as a function of temperature, x1 (qF), and the pressure x2 (psi) at which it is produced.Four inspectors independently assign a quality score (%) to each sample and the quality, Y, is evaluated by averaging the four (4) scores. The resultant data is given in the data table below.
Temperature, x1 (qF)
Pressure, x2 (psi)
50 60 70
200 52.8 58.7 55.4 92.4 90.9 90.9 74.5 73 71.2
250 63.4 61.6 63.4 93.8 92.1 97.4 70.9 68.8 71.3
300 46.6 49.1 46.4 69.8 72.5 73.2 38.7 42.5 41.4
1) Specify algebraically the complete second order model for the data.
2) Use JMP to determine the best model.
3) Produce two bivariate plots: Product quality versus temperature (x1) and product quality versus pressure (x2). Describe the shape of each graph and comment on what this indicates about the adequacy of the possible linear model
Y = b + b1 x1+ b2 x2 + H .
4) From the model obtained in 2), what temperature and/or pressure would you recommend?

Probability Distribution of time between arrivals:

 Time between arrivals(min) Probability Cumulative Probability 5 0.1 0.1 10 0.15 0.25 15 0.22 0.47 20 0.25 0.72 25 0.18 0.9 30 0.1 1

Probability distribution of attention needed:

 Patient needs to see Probability Cumulative Probability Doctor 0.4 0.4 Nurse 0.35 0.75 Both 0.25 1

The length of the patient’s visit (in minute) is defined by the following probability distributions:

 Doctor Probability Cumulative Probability 10 0.2 0.2 15 0.3 0.5 20 0.25 0.75 25 0.15 0.9 30 0.1 1
 Nurse Probability Cumulative Probability 5 0.1 0.1 10 0.25 0.35 15 0.5 0.85 20 0.15 1

Simulation:

 Simulation Results Element Element Statistics Overall types names Means Entrance(s) Door Objects entering process 45 Objects unable to enter 0 Service level 1 Work Station(s) Clerk Final status NA Final inventory (int. buff.) 0 Mean inventory (int. buff.) 0 Mean cycle time (int. buff.) 0 Work cycles started 45 Fraction time working 0 Fraction time blocked 0 Sharon Final status NA Final inventory (int. buff.) 0 Mean inventory (int. buff.) 0 Mean cycle time (int. buff.) 2 Work cycles started 10 Fraction time working 0 Fraction time blocked 0 Peter Final status NA Final inventory (int. buff.) 0 Mean inventory (int. buff.) 0 Mean cycle time (int. buff.) 0 Work cycles started 9 Fraction time working 0 Fraction time blocked 0 Buffer(s) Line Objects leaving 45 Final inventory 0 Minimum inventory 0 Maximum inventory 2 Mean inventory 0 Mean cycle time 0 License Objects leaving 10 Final inventory 10 Minimum inventory 0 Maximum inventory 11 Mean inventory 5 Mean cycle time 108 Registration Objects leaving 9 Final inventory 4 Minimum inventory 0 Maximum inventory 5 Mean inventory 2 Mean cycle time 53 Both Objects leaving 10 Final inventory 1 Minimum inventory 0 Maximum inventory 3 Mean inventory 1 Mean cycle time 15 Served Customer Objects leaving 0 Final inventory 19 Minimum inventory 0 Maximum inventory 19 Mean inventory 8 Mean cycle time 18 Decision Point(s) DP Objects leaving 44 Final inventory (int. buff.) 0 Mean inventory (int. buff.) 0 Mean cycle time (int. buff.) 0

Therefore, overall average waiting time in the process for customers = 18 minutes

The number of customers served during the 3-hour time peak period = 44

The number of customers who have already seen the clerk at the counter but waiting to be served at the end of the 3-hour peak period = 10+9+10 = 29

The utilisation of these three employees, i.e.

o The clerk at the counter = 45/45 = 100%

o Sharon = 10/45 = 22.22%

o Peter = 9/45 = 20%

Simulation result while one resource has been added:

 Simulation Results Element Element Statistics Overall types names Means Entrance(s) Door Objects entering process 45.08 Objects unable to enter 0.00 Service level 1.00 Work Station(s) Clerk Final status NA Final inventory (int. buff.) 0.00 Mean inventory (int. buff.) 0.00 Mean cycle time (int. buff.) 0.00 Work cycles started 45.02 Fraction time working 0.25 Fraction time blocked 0.00 Sharon Final status NA Final inventory (int. buff.) 0.29 Mean inventory (int. buff.) 0.25 Mean cycle time (int. buff.) 2.50 Work cycles started 10.14 Fraction time working 0.28 Fraction time blocked 0.25 Peter Final status NA Final inventory (int. buff.) 0.00 Mean inventory (int. buff.) 0.00 Mean cycle time (int. buff.) 0.00 Work cycles started 9.54 Fraction time working 0.34 Fraction time blocked 0.00 Buffer(s) Line Objects leaving 45.02 Final inventory 0.06 Minimum inventory 0.00 Maximum inventory 2.00 Mean inventory 0.05 Mean cycle time 0.21 License Objects leaving 10.14 Final inventory 10.19 Minimum inventory 0.00 Maximum inventory 11.09 Mean inventory 5.34 Mean cycle time 108.36 Registration Objects leaving 9.54 Final inventory 3.64 Minimum inventory 0.00 Maximum inventory 5.17 Mean inventory 2.18 Mean cycle time 45.34 Both Objects leaving 10.14 Final inventory 1.11 Minimum inventory 0.00 Maximum inventory 2.78 Mean inventory 0.79 Mean cycle time 14.87 Served Customer Objects leaving 0.00 Final inventory 19.00 Minimum inventory 0.00 Maximum inventory 19.00 Mean inventory 8.49 Mean cycle time Infinite Decision Point(s) DP Objects leaving 44.76 Final inventory (int. buff.) 0.00 Mean inventory (int. buff.) 0.00 Mean cycle time (int. buff.) 0.00 Resource(s) Mary Mean number in use 11.12

Second order algebraic Model:

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