In: Accounting
A clinic has one receptionist, one doctor and one nurse. The clinic opens at time zero and patients begin to arrive some time later. Patients arrive at the clinic according to the following probability distribution:
Time between arrivals (minutes) |
Probability |
5 |
0.09 |
10 |
0.12 |
15 |
0.27 |
20 |
0.26 |
25 |
0.15 |
30 |
0.11 |
A patient may need to see only the doctor, only the nurse, or both the doctor and the nurse together. If a patient needs to see both the doctor and nurse, the patient must wait until both the doctor and nurse are free. The attention needed by a patient who comes to the clinic is defined by the following probability distribution:
Patient Needs to See |
Probability |
Doctor only |
0.62 |
Nurse only |
0.10 |
Both Doctor and Nurse |
0.28 |
The “service time” for a patient is the length of the time that the patient spends in consultation with the doctor/nurse/both, depending on the patient’s needs. The service time EXCLUDES the time that the patient spends waiting to see the doctor/nurse/both if they are busy. Based on whether the patient sees the doctor, the nurse, or both, the service time of each patient (in minutes) is defined by the following probability distributions.
Doctor |
Probability |
Nurse |
Probability |
Both |
Probability |
||
10 |
0.12 |
5 |
0.21 |
15 |
0.16 |
||
15 |
0.21 |
10 |
0.56 |
20 |
0.25 |
||
20 |
0.32 |
15 |
0.15 |
25 |
0.36 |
||
25 |
0.18 |
20 |
0.08 |
30 |
0.17 |
||
30 |
0.17 |
35 |
0.05 |
||||
40 |
0.01 |
Design a SIMULATION BY HAND of this clinic to simulate the arrival and departure of 8 patients using the following random numbers ranging 00 to 99 generated for each random variable:
64 34 15 50 94 81 27 79
48 58 97 88 75 94 69 56
Part a
For each of the 8 patients, simulate the three random variables from above:
Part b
Assume that the clinic opens at 9:00am and the first arrival occurs after the “time between arrivals” of the first patient (e.g. if the time between arrivals of patient 1 is 5 mins, then patient 1 arrives at 9:05am). Calculate each patient’s arrival time and departure time by completing the following table:
Patient |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
Arrival Time |
||||||||
Departure Time |
Part c
From the results in (b), determine the following performance measures based on the 8 patients:
Part a)
i) & ii)
iii)
Part b)
Patient | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
Attention by | Doctor Only | Both Doctor & Nurse | Both Doctor & Nurse | Doctor Only | Nurse Only | Doctor Only | Both Doctor & Nurse | Both Doctor & Nurse |
RN for Service Time | 48 | 58 | 97 | 88 | 75 | 94 | 69 | 56 |
Arrival Time | 9.20 Am | 9.35 AM | 9.45AM | 10.05 AM | 10.35AM | 11.00AM | 11.15AM | 11.40AM |
Service Time | 20 | 25 | 35 | 30 | 10 | 30 | 25 | 25 |
Service Start Time | 9.20 Am | 9.40 Am | 10.05AM | 10.40AM | 11.10AM | 11.20 AM | 11.50 AM | 12.15 PM |
Departure Time | 9.40 Am | 10.05AM | 10.40AM | 11.10AM | 11.20 AM | 11.50 AM | 12.15 PM | 12.40 Pm |
Patient Waiting time | 0 | 5 | 20 | 35 | 35 | 20 | 35 | 35 |
Partc)
i) Average time that patient must wait = sum of patient waiting time divided by No of patients
= 185/8 = 23.125 minutes
ii) Probabilty that the patient needs to wait for less than 30 minutes before service starts
= 4/8= 0.5
DC