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Queuing Models

Queuing Models. Economic Analyses. 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 --

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Queuing Models

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  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 1Determining 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. Average service time1/ = 5 min. = 5/60 hr. = 60/5 =12/hr. ……….. How many servers? Arrival rate = 100/hr. GOAL: Average time in the queue, WQ < .00833hrs.

  5. Input values for  and  First time WQ < .008333 12 servers needed .003999 Go to the MMkWorksheet

  6. Example 2Determining 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. Server Salary Average Service Time • Ann $ 6/hr. 6 min. • Bill $ 10/hr. 5 min. • Charlie $ 14/hr. 4 min. • Which server should be hired?

  7. Ann1/ = 6 min. A = 60/6 = 10/hr. LAnn = ?  = 8/hr ANN Hourly Cost = $6 + 4LAnn

  8. LAnn = 4 Ann1/ = 6 min. A = 60/6 = 10/hr. LAnn Hourly Cost = $6 + 4LAnn  = 8/hr Hourly Cost = $6 + $4(4) = $22

  9. Bill1/ = 5 min. A = 60/5 = 12/hr. LBill = ?  = 8/hr BILL Hourly Cost = $10 + 4LBill

  10. LBill = 2 Bill1/ = 5 min. B = 60/5 = 12/hr. LBill Hourly Cost = $10 + 4LBill  = 8/hr LBill Hourly Cost = $10 + $4(2) = $18

  11. Charlie1/ = 4 min. A = 60/5 = 15/hr. LCharlie = ?  = 8/hr CHARLIE Hourly Cost = $14 + 4LCharlie

  12. LCharlie = 1.14 Charlie1/ = 4 min. C = 60/4 = 15/hr. LCharlie Hourly Cost = $14 + 4LCharlie  = 8/hr LCharlie 1.14 Hourly Cost = $14 + $4(1.14) = $18.56

  13. Optimal HireBill • Ann --- Total Hourly Cost = $22 • Bill --- Total Hourly Cost = $18 • Charlie --- Total Hourly Cost = $18.56

  14. Example 3What 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 1/ = 2 min.  = 60/2 = 30/hr. Store  = 24/hr. Total Hourly Cost Salary + Lanes + Wait Cost $6.50 + $20 + $25LQ

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

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

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

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

  20. System 3 -- 2 clerks, 2 lanes 1/ = 2 min.  = 60/2 = 30/hr. LQ = .152 Store Store  = 24/hr. Total Hourly Cost Salary + Lanes + Wait Cost 2($6.50) + $40 + $25LQ 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 4Which 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 6Servers Unlimited QueueLength …  = 30/hr. Lease Cost = $1000/day= $1000/10 = $100/hr. All customersget served!

  24. Small Store 2Servers Maximum QueueLength = 1 Will join system if0,1,2 in the system Will not join thequeue if there are 3 customers in the system  = 30/hr. Lease Cost = $200/day= $200/10 = $20/hr.

  25. Hourly Profit Analysis Large Small Arrival Rate  = 30 e = 30(1-p3) Hourly Revenue $25(Arrival Rate) (25)(30)=$750 $25e Hourly Costs Lease$100 $20 Server$20(#Servers) $120 $40 Waiting$8(Avg. in Store) $8L $8L Net Hourly Profit??

  26. Large Store -- M/M/6 L 3.099143

  27. Small Store -- M/M/2/3 p3 L 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$100 $20 Server$20(#Servers) $120 $40 Waiting$8(Avg. in Store) $25 $17 Net Hourly Profit$505 $341 Lease the Large Store

  29. Example 5Which 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 Prob (No Wait) -- P0 .5000.4792 Average Service Time6 min.6.25 min. Average # in System.8125.8065 Average # in Queue .3125.2857 Average Time in System.1625 hr..1613 hr. 9.75 min.9.68 min. Average Time in Queue.0625 hr..0571 hr. 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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