1. MGTSC 352 Lecture 23: Congestion Management Introduction: Asgard Bank example Simulating a queue Types of congested systems, queueing template Ride’n’Collide example MEC example Manufacturing example

2. Analyzing a Congested System (pg. 174)  System Description Measures of Quality of Service Measures important to Servers Model of the System Inputs Outputs

3. Asgard Bank: Times Between Arrivals (pg. 173) pg. 168

4. Asgard Bank: Service Times

5. Including Randomness: Simulation Service times: Normal distribution, mean = 57/3600 hrs, stdev = 10/3600 hrs. MAX(NORMINV(RAND(),57/3600,10/3600),0) Inter-arrival times: Exponential distribution, mean = 1/ 60 hrs. – (1/60)*LN(RAND()) To Excel …

6. Simulated Lunch Hour 1: 71 arrivals

7. Simulated Lunch Hour 2: 50 arrivals

8. Simulated Lunch Hour 3: Unused capacity

9. Causes of Congestion Higher than average number of arrivals Lower than average service capacity Lost capacity due to timing Lesson: For a service where customers arrive randomly, it is not a good idea to operate the system close to its average capacity

10. Template.xls Does calculations for –M/M/s –M/M/s/s+C –M/M/s/  /M –M/G/1 Want to know more? Go to http://www.bus.ualberta.ca/aingolfsson/qtp/ Asgard Bank Data –Model: M/G/1 –Arrival rate: 1 per minute –Average service time: 57/60 min. –St. dev of service time: 10/60 min.

11. Asgard Conclusions The ATM is busy 95% of the time. Average queue length = 9.3 people Average no. in the system = 10.25 (waiting, or using the ATM) Average wait = 9.3 minutes What if the service rate changes to … –1.05 / min.? –1.06 / min.?

12. Ride’n’Collide (pg. 178) Repair personnel cost: $10 per hour Average repair duration: 30 minutes Lost income: $50 per hour per car Number of cars: 20 cars A car will function for 10 hours on average from the time it has been fixed until the next time it needs to be repaired. How many repair-people should be hired?

13. Ride’n’Collide Customers = Servers = Average number in system = Lost revenue per hour = Arrival rate = Service rate = Model to use:

14. Waiting Line Analysis Template: Which Model to Use? Who are the customers? Who are the servers? Where is the queue? … not always obvious

15. If you are told how many customers there are … then you should consider using the “finite population” template waiting room = queue potential customers parallel servers Number is small enough to worry about

16. If you are told the maximum number of customers that can wait (the size of the waiting room) … then you should consider using the “finite Q” template waiting room = queue potential customers parallel servers Capacity is small enough to worry about

17. If you are told the standard deviation of the service times, and there is 1 server … then you should consider using the “MG1” template waiting room = queue potential customers one server, non- exponential service time distribution

18. If you are told nothing about the size of the pool of potential customers, or the maximum number that will wait, or the standard deviation of the service times, … then you should probably use the “MMs” template

19. MEC (p. 181) One operator, two lines to take orders –Average call duration: 4 minutes exp –Average call rate: 10 calls per hour exp –Average profit from call: $24.76 Third call gets busy signal How many lines/agents? –Line cost: $4.00/ hr –Agent cost: $12.00/hr –Avg. time on hold < 1 min.

20. Modeling Approaches Simulation Waiting line analysis template We’ll use both for this example To Excel …

21. Manufacturing Example (p. 184) Machine (1.2 or 1.8/minute) 1/minute Poisson arrivals Exponential service times

22. Manufacturing Example Arrival rate for jobs: 1 per minute Machine 1: –Processing rate: 1.20/minute –Cost: $1.20/minute Machine 2: –Processing rate: 1.80/minute –Cost: $2.00/minute Cost of idle jobs: $2.50/minute Which machine should be chosen? To Excel …

23. Manufacturing Example Cost of machine 1 = $1.20 / min. + ($2.50 / min. / job)  (5.00 jobs) = $13.70 / min. Cost of machine 2 = $2.00 / min. + ($2.50 / min. / job)  (1.25 jobs) = $5.13 / min.  Switching to machine 2 saves money – reduction in lost revenue outweighs higher operating cost.

24. Cost of waiting (Mach. 1) Method 1: –Unit cost × L = ($2.50 / min. job)  (5.00 jobs) = $13.70 / min Method 2: –Unit cost × × W  = ($2.50 / min. job)  (5.00 min)  (1 job/min) = $13.70 / min Little’s LawL = × W