Q3 With Solution
RSM 270 Introduction to Operations Management
This is the sample questions for lecture 3 for your preparation.
Problem 1 Johnny has just slipped on a banana peel and he thinks he may have broken a hip. His friend drives him to the Emergency Room and they see 24 patients in the waiting room. According to Little’s law, what ratio can be used to find the waiting time of patients who are waiting in the ER for a consult? Further, if the doctor is seeing 1 patient every 15 minutes, how long do they have to wait until Johnny is seen by a physician?
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by Little’s Law, we have I = R × T. Therefore, the ratio of the number of patient waiting over the rate at which the patients are seen by the physician define the average waiting time for patients. Since the doctor sees one patientevery
15 minutes, R = ! patients per minute. 15 Therefore the waiting time is T = # %&!” $
=!/!” =24×15minutesor6hours.
Problem 2 For a coffee shop that opens in the morning at 7am, we consider a five-minute time interval from 8am-8:05am.For this time interval, we want to consider possible relation between input rate, capacity rate, throughput rate.
The input rate could be larger but also smaller than the throughput rate. The throughput rate could be smaller or also equal to the capacity rate. The capacity rate could be smaller or also larger than the input rate.
Problem 3 The market at the UTSC farm opens every Saturday morning. They offer fresh organic products that mostly come from the farm itself. The counter opens at 9 am. However, customers start lining up at 8:30 AM. Customers show up at a rate of 30/hr until 9:30 AM and then at a rate of 15/hr until 10:30 AM. The counter can serve at a rate of 20/hr, and the counter works till all customers are serves.
A. Draw a graph of the numbers of customers waiting in the line> Start the graph at 8:30 AM and show the numbers of waiting customers until the line is empty.
B. Using the inventory build-up diagram, calculate the average number of customers in line. The average number of customers in the system is 12.95. Probably the easiest way to calculate this number is by calculating the area under the graph and then by dividing by the entire time. The area under the graph is a summation of two triangles and two trapezoids: 15×0.5/2+(15+20)×0.5/2+(20+15)×1/2+15×0.75/2 = 35.625. [Actually even easier is (if you don’t know how to calculate the area of a trapezoid), to determine the area under the graph as the summation of four rectangular triangles and one rectangle.] Now, divide 35.625 by 2.75 hrs and we find that on average there are 12.95 customers in the system.
C. How long – on average – does a person stay in the line? (Hint: Little’s Law. 2nd Hint: Make sure you calculate the throughput rate correctly.) Throughput rate = Actual output rate = 45 customers per 2.75 hrs = 16.36 customers per hr. Using the fact that throughput rate = min {capacity rate, input rate}, you can also calculate as follows: For the first half an hour, your throughput rate is actually 0, because your capacity rate is 0. For the remaining time your throughput rate is 20 cust/hr (draw down inventory from the queue). So, you have (0.5 × 0 + 2.25 × 20)/2.75 = 16.36 cust/hr. Little’s Law: Average Flow time = Average Inventory/Throughput rate = 12.95/16.36 = 0.79 hr = 47.5 minutes.
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