Question
Homework 5 Problem 1 (25 points) The time between incoming phone calls to customer service is Exponential with a mean of 14 minutes. Each call
Homework 5
Problem 1 (25 points)
The time between incoming phone calls to customer service is Exponential with a mean of 14 minutes. Each call typically takes min=5, max=20, and most likely = 10 minutes, (Triangularly distributed). Simulate the system based on the following scenarios (run each scenario for 6 hours, with 50 replications).
- There are two operators, and the system has only a single queue. Report the histogram and mean for the average waiting time[1].
- There are two operators, and each operator has its own queue. Phone calls will be directed to the random queue (policy = "random"). Report the histogram and mean for the average waiting time.
- Compare the average waiting time of parts a and b. Is there any meaningful difference? Explain.
Problem 2 (25 points)
This problem is the revised version of problem 1. Suppose the time between incoming phone calls to customer service is Exponential with a mean of 5 minutes. And, suppose there are 3 operators and each operator has its own queue. Phone calls will be directed to queues randomly. Operators 1 and 2 respond to calls typically between min=5 to max=20 minutes (Uniformly distributed). The third operator is chattier, and her service takes more time, which is Normally distributed with a mean of 15 and a standard deviation of 3 minutes. Run your model for 6 hours, with 50 replications.
- Report a utilization plot for each operator. Discuss what you learned from the plot.
- Report a waiting time for each operator. Discuss your results.
- Find a 95% confidence interval for the average waiting time for each operator.
Problem 3 (25 points)
Consider a manufacturing system comprising two different machines and two operators.
- Each operator is assigned to run a single machine.
- Parts arrive with a Truncated-Normally distributed interarrival time with a min of 0.5 minutes, mean of 4 minutes, and stdev of 2 minutes. The arriving parts are one of two types.
- Sixty percent of the arriving parts are Type 1 and are processed on Machine 1. These parts require the assigned operator for a 1-minute setup operation.
- The remaining 40% of the parts are Type 2 and are processed on Machine 2. These parts require the assigned operator for a 1.5-minute setup operation.
- The service times (excluding the setup time) are Normally distributed with a mean of 4.5 minutes and a standard deviation of 1 minute for Type 1 parts and a mean of 7.5 minutes and a standard deviation of 1.5 minutes for Type 2 parts.
Run your model for 2,000 minutes, with 50 replications.
- Plot the utilization, usage, flow_time, and waiting_time. Discuss your findings.
- Report the average total time spent in the system (flow time) for each type of part.
Problem 4 (25 points)
Consider an Urgent Care (UC) comprising three doctors and one nurse to serve patients during the day.
- On a typical day, the interarrival time is Exponentially distributed with a mean of 6 minutes.
- 25% of patients are high-priority, and the remaining are low-priority.
- Upon arrival at UC, the patients are triaged by a nurse into one of the two types of patients. The service time for triage is distributed by Triangular distribution with min = 3, max = 10, and the most likely value = 5 minutes.
- Then, the patients wait in the waiting room and get called to visit doctors on a firstcome-first-served basis. If more than 10 people are waiting for service, an arriving patient will exit before being triaged.
- Finally, low-priority patients may depart if they have to wait longer than205 minutes (Uniformly distributed) after triage.
- The doctor service time distributions are given as follows.
Priority | Service Time Distribution (in Minutes) |
High | Normal(mean = 40, sd = 5) |
Low | Gamma(shape = 15, rate = 1) |
Assuming that the UC opens at 8 hours, simulate the process for 50 replications. The UC would like to estimate the following:
(a)the average flow time of each type of patient. (b)the probability that low-priority patients balk.
[1] You might have NAs in your waiting time, because some people start the service but do not finish it before the end of simulation clock. Therefore, in R, you can use mean(waitTime, na.rm=T).
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