Matching Supply with Demand: An Introduction to Operations Management, 4e (Cachon)
Chapter 09 Variability and Its Impact on Process Performance: Waiting Time Problems
[The following information applies to questions 1-2.]
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UP Fitness will install 3 new trapezoidal gel-resistance toners, or “trappies,” which are the latest exercise gadget specifically designed to contour upper back muscles. Clients who want to use these machines arrive at the rate of 85 per hour. The coefficient of variation of the interarrival times is 3. If all three machines are busy, UP Fitness clients use other pieces of equipment, waiting for one of the “trappies” to become available. The trappy experience is intense—120 seconds of explosive exercise intensity. The standard deviation of the usage times (measured in seconds) is quite small—only 30.
1) On average, how many UP Fitness clients are using a trappy?
60*60=3600 seconds in one hour
3600/120*3=90 clients per hour
Utilization 85/90 = 0.944
0.944 *3 = 2.83
The capacity of UP Fitness is 3600 sec (1 hour)/120 sec * 3 trappies = 90 clients per hour
The utilization of UP Fitness is 85 arrivals per hour / capacity 90 clients per hour = 0.944
That is, on average 3 machines * 0.944 = 2.83 machines are in use. In other words, 2.83 clients are using a trappy on average.
2) On average, how many UP Fitness clients are waiting to use a trappy?
Activity time = 120 sec = 2 min
M = number of servers = 3
Utilization = 0.944
CVp = Standard deviation of activity times/ Average activity time = 30/120 = 0.25
Time in queue = 48.98 min
85/60*48.98 = 69.39
The number of clients waiting to use a trappy, is inventory in queue.
The processing tine is 120 sec, or 2 min. utilization is 0.944 from Q1
From Little’s Law,
T * R = inventory
T : time in queue, R: arrival rate at 85 per hour = 85/60 per minute
48.98*(85/60) = 69.39 clients
3) approves study abroad documents for Penn. Students must wait in line with their forms outside Max’s office. One student at a time is allowed in his office, and Max takes precisely 25 minutes to evaluate each student’s set of documents. On average, 2.2 students per hour go to his office, and they spend on average 160 minutes trying to get their forms approved (time waiting in queue plus time in Max’s office for him to evaluate their documents). On average, how many students are waiting outside of Max’s office? (Note: Ignore start-of-the-day and end-of-the-day effects.)
Processing time p = 25
Arrival a = 60 min/2.2 per hour = 27.27
Time in queue Tq = total time in system T – processing time Tp = 160 -25 = 135
Use little law,
Inventory in queue Iq = R * T = (1/a) * Tq = (1/27.27)*135 = 4.95
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