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Queuing Models Economic Analyses. ECONOMIC ANALYSES Each problem is different Examples –To determine the minimum number of servers to meet some service.

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Presentation on theme: "Queuing Models Economic Analyses. ECONOMIC ANALYSES Each problem is different Examples –To determine the minimum number of servers to meet some service."— Presentation transcript:

1 Queuing Models Economic Analyses

2 ECONOMIC ANALYSES Each problem is different Examples –To determine the minimum number of servers to meet some service criterion (e.g. an average of < 4 minutes in the queue) -- trial and error with M/M/k systems –To compare 2 or more situations -- Consider the total (hourly) cost for each system and choose the minimum

3 Example 1 Determining Optimal Number of Servers Customers arrive according to a Poisson process to an electronics store at random at an average rate of 100 per hour. Service times are exponential and average 5 min. How many servers should be hired so that the average time of a customer waits for service is less than 30 seconds? –30 seconds =.5 minutes =.00833 hours

4 GOAL: Average time in the queue, W Q <.00833hrs. ……….. How many servers? Arrival rate = 100/hr. Average service time 1/  = 5 min. = 5/60 hr.  = 60/5 = 12/hr.

5 .003999 First time W Q <.008333 12 12 servers needed Input values for and  Go to the MMk Worksheet

6 Example 2 Determining Which Server to Hire Customers arrive according to a Poisson process to a store at night at an average rate of 8 per hour. The company places a value of $4 per hour per customer in the store. Service times are exponential and the average service time that depends on the server. ServerSalary Average Service Time – Ann$ 6/hr.6 min. – Bill$ 10/hr.5 min. – Charlie$ 14/hr.4 min. Which server should be hired?

7 Ann 1/  = 6 min.  A = 60/6 = 10/hr. Hourly Cost = $6 + $4L Ann = 8/hr L Ann = ? ANN

8 Ann 1/  = 6 min.  A = 60/6 = 10/hr. Hourly Cost = $6 + 4L Ann L Ann = 8/hr L Ann = 4 Hourly Cost = $6 + $4(4) = $22

9 Bill 1/  = 5 min.  B = 60/5 = 12/hr. Hourly Cost = $10 + $4L Bill = 8/hr L Bill = ? BILL

10 Bill 1/  = 5 min.  B = 60/5 = 12/hr. Hourly Cost = $10 + 4L Bill L Bill = 8/hr L Bill = 2 Hourly Cost = $10 + $4(2) = $18

11 Charlie 1/  = 4 min.  C = 60/5 = 15/hr. Hourly Cost = $14 + $4L Charlie = 8/hr L Charlie = ? CHARLIE

12 Charlie 1/  = 4 min.  C = 60/4 = 15/hr. Hourly Cost = $14 + 4L Charlie L Charlie 1.14 = 8/hr L Charlie = 1.14 Hourly Cost = $14 + $4(1.14) = $18.56

13 Optimal Ann--- Total Hourly Cost = $22 Bill--- Total Hourly Cost = $18 Charlie--- Total Hourly Cost = $18.56 Bill Hire Bill

14 Example 3 What Kind of Line to Have A fast food restaurant will be opening a drive-up window food service operation whose service distribution is exponential. Customers arrive according to a Poisson process at an average rate of 24/hr. Three systems are being considered. Customer waiting time is valued at $25/hr. Each clerk makes $6.50/hr. Each drive-thru lane costs $20/hr. to operate Which of the following systems should be used?

15 System 1 -- 1 clerk, 1 lane Store = 24/hr. 1/  = 2 min.  = 60/2 = 30/hr. Total Hourly Cost Salary + Lanes + Wait Cost $6.50 + $20 + $25L Q

16 System 1 -- 1 clerk, 1 lane Store = 24/hr. 1/  = 2 min.  = 60/2 = 30/hr. Total Hourly Cost Salary + Lanes + Wait Cost $6.50 + $20 + $25L Q L Q = 3.2 Total Hourly Cost Salary + Lanes + Wait Cost $6.50 + $20 + $25(3.2) = $106.50

17 System 2 -- 2 clerks, 1 lane = 24/hr. 1 Service System 1/  = 1.25 min.  = 60/1.25 = 48/hr. Total Hourly Cost Salary + Lanes + Wait Cost 2($6.50) + $20 + $25L Q Store

18 System 2 -- 2 clerks, 1 lane = 24/hr. 1 Service System 1/  = 1.25 min.  = 60/1.25 = 48/hr. Total Hourly Cost Salary + Lanes + Wait Cost 2($6.50) + $20 + $25L Q Store L Q =.5 Total Hourly Cost Salary + Lanes + Wait Cost 2($6.50) + $20 + $25(.5) = $45.50

19 System 3 -- 2 clerks, 2 lanes Store = 24/hr. Total Hourly Cost Salary + Lanes + Wait Cost 2($6.50) + $40 + $25L Q Store 1/  = 2 min.  = 60/2 = 30/hr.

20 System 3 -- 2 clerks, 2 lanes Store = 24/hr. Total Hourly Cost Salary + Lanes + Wait Cost 2($6.50) + $40 + $25L Q Store 1/  = 2 min.  = 60/2 = 30/hr. L Q =.152 Total Hourly Cost Salary + Lanes + Wait Cost 2($6.50) + $40 + $25(.152) = $56.80

21 Optimal System 1 --- Total Hourly Cost = $106.50 System 2 --- Total Hourly Cost = $ 45.50 System 3 --- Total Hourly Cost = $ 58.80 Best option -- System 2

22 Example 4 Which Store to Lease Customers are expected to arrive by a Poisson process to a store location at an average rate of 30/hr. The store will be open 10 hours per day. The average sale grosses $25. Clerks are paid $20/hr. including all benefits. The cost of having a customer in the store is estimated to be $8 per customer per hour. Clerk Service Rate = 10 customers/hr. (Exponential) Should they lease a Large Store ($1000/day, 6 clerks, no line limit) or a Small Store ($200/day, 2 clerks – maximum of 3 in store)?

23 Large Store = 30/hr. … 6 Servers Unlimited Queue Length All customers get served! Lease Cost = $1000/day = $1000/10 = $100/hr.

24 Small Store = 30/hr. 2 Servers Maximum Queue Length = 1 Lease Cost = $200/day = $200/10 = $20/hr. Will join system if 0,1,2 in the system Will not join the queue if there are 3 customers in the system

25 Hourly Profit Analysis Large Small Arrival Rate = 30 e = 30(1-p 3 ) Hourly Revenue $25(Arrival Rate) (25)(30)=$750 $25 e Hourly Costs Lease Lease $100 $20 Server Server $20(#Servers) $120 $40 Waiting Waiting $8(Avg. in Store) $8L l $8L s Net Hourly Profit?? Net Hourly Profit ? ?

26 Large Store -- M/M/6 3.099143 LlLl

27 Small Store -- M/M/2/3 LsLs p3p3 e = (1-.44262)(30) = 16.7213

28 Hourly Profit Analysis Large Small Arrival Rate = 30 e = 16.7213 Hourly Revenue $25(Arrival Rate) $750 $25 e =$418 Hourly Costs Lease Lease $100 $20 Server Server $20(#Servers) $120 $40 Waiting Waiting $8(Avg. in Store) $25 $17 Net Hourly Profit$505 $341 Net Hourly Profit $505 $341 Lease the Large Store

29 Example 5 Which Machine is Preferable Jobs arrive according to a Poisson process to an assembly plant at an average of 5/hr. Service times do not follow an exponential distribution. Two machines are being considered –(1) Mean service time of 6 min. (  = 60/6 = 10/hr.) standard deviation of 3 min. (  = 3/60 =.05 hr.) –(2) Mean service time of 6.25 min.(  = 60/6.25 = 9.6/hr.); std. dev. of.6 min. (  =.6/60 =.01 hr.) Which of the two M/G/1designs is preferable?

30 Machine 1

31 Machine 2

32 Machine Comparisons Machine1 Machine 2.5000 Prob (No Wait) -- P 0.5000.4792 6 min. Average Service Time 6 min. 6.25 min. 8065 Average # in System.8125.8065.2857 Average # in Queue.3125.2857.1613 hr. Average Time in System.1625 hr..1613 hr. 9.68 min. 9.75 min. 9.68 min..0571 hr. Average Time in Queue.0625 hr..0571 hr. 3.43 min. 3.75 min. 3.43 min. Machine 2 looks preferable

33 Review List Components of System Develop a model Use templates to get parameter estimates Select “optimal” design

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