代做ETF5650: Business Optimisation Skills Assignment 2, 2024代做Python编程

Hourly admission  ()	Observed frequency ()	Hourly admission ()	Observed frequency ()
0	0	8	14
1	0	9	13
2	1	10	11
3	4	11	8
4	6	12	5
5	10	13	3
6	13	14	2
7	14

Table 1

Show that the over-all average number of admissions during the peak period on Saturday is 7.8654 per hour. HINT: Suppose number of admissions in an hour is  with frequency . Then the mean of the frequency distribution is computed as  

1.3 Draw a plot of the frequency distribution given in Table 1.

1.4 Does the sample data resemble a Poisson distribution?

HINT: First, plot the observed frequency distribution (data given in Table 1) and overlay it with the theoretically obtained expected frequency distribution. Then, see if the two plots are similar or not. If they are similar, then the fit is good.

NOTE: How to obtain expected frequency?

Let  = number of arrivals within one hour. In this context, probability distribution of  is Poisson.

Poisson probability distribution of  is given by   where  is the mean of the Poisson distribution which is assumed 7.8654 (average number of admissions per hour and is given in part 1.2).

Then the expected frequency of  is  computed as . This gives how many of 104 observations we may expect to have hourly admission at .

1.5 At Colombo Community Hospital, having more than one receiving doctor is not feasible. The standard protocol dictates that upon admission, each patient undergoes examination by the receiving doctor, who then assigns the patient to the suitable facility within the emergency treatment section.

             

ETF5650 students want to analyse this situation as a single-server queuing process (M/M/1/ ). They know that a service rate is required. Hence, they recorded actual service times on four consecutive Saturdays between 6pm and 8pm. A total of 71 observations was included in the sample.

Assit, a member of the ETF5650 group tasked with analysing the operation at Colombo Community Hospital, raised the question:

"We observed 71 service completions over 8 hours. So, shall we use these 71 service times and compute the service rate as 71/8 = 8.8750 per hour?"

However, the other students in the group did not support this approach. What are your thoughts on this?

1.6 The ETF5650 group calculated the average service time as 4.7578 minutes. They plotted a histogram of the observed service times and noticed that the pattern satisfies the condition required for application of the M/M/1/ queuing system.

(a) What conclusions can be drawn from traffic intensity or server utilisation factor of this queuing system?

(b) What is the expected number of patients in the system?

(c) What is the expected time a patient spends waiting for the service? 

1.7 Do you agree with ETF5650 students that the given situation may be assumed as an M/M/1/ queuing system? 

QUESTION 2                                                                                    [3+(2+6+2) = 13 marks]

Consider the business process chart shown in Figure 2. The parenthesis gives, for each activity, the expected waiting time and the expected processing time in minutes in that order. Assume that the loop may be repeated only once if needed.

(a)  Calculate the average cycle time of this business process.

(b) There are 10 units of resource R1, 12 units of resource R2 and 12 units of resource R3. Table 2 shows the type of resource that each activity requires.