1 / 27

Cost Optimization in Service Industries: Queuing Analysis

Understand economic models in service industries for optimal cost decisions, including servers, hiring choices, drive-thru systems, store leasing, and machine selection. Examples and calculations provided for practical application.

zeus-alston
Download Presentation

Cost Optimization in Service Industries: Queuing Analysis

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Module C8 Queuing Economic/Cost Models

  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 to an electronics store at random at an average rate of 100 per hour. • Service times are random but 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

  6. Example 2Determining Which Server to Hire • Customers arrive at random 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 random. 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. LAnn = 4 Ann1/ = 6 min. A = 60/6 = 10/hr. LAnn Hourly Cost = $6 + 4LAnn  = 8/hr LAnn 4 Hourly Cost = $6 + $4(4) = $22

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

  9. 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

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

  11. Example 3What Kind of Line to Have • .A fast food operation is considering opening a drive-up window food service operation. • Customers arrive at random at an average rate of 24 per hour. Three systems are being considered. • Customer waiting time is valued at $25 per hour. • Each clerk makes $6.50 per hour. • Each drive-thru lane costs $20 per hour to operate. • Which system should be used?

  12. 1/ = 2 min.  = 60/2 = 30/hr. LQ = 3.2 Store  = 24/hr. Option 1 -- 1 clerk, 1 lane 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

  13. 1 Service System 1/ = 1.25 min.  = 60/1.25 = 48/hr. LQ = .5 Store  = 24/hr. Option 2 -- 2 clerks, 1 lane 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

  14. 1/ = 2 min.  = 60/2 = 30/hr. LQ = .152 Store Store  = 24/hr. Option 3 -- 2 clerks, 2 lanes 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

  15. BestOption 2 Optimal • Option 1 --- Total Hourly Cost = $106.50 • Option 2 --- Total Hourly Cost = $ 45.50 • Option 3 --- Total Hourly Cost = $ 58.80

  16. Example 4Which Store to Lease • Customers are expected to arrive to a store location at random at an average rate of 30 per hour. • The store will be open 10 hours per day. • Service times can be considered random. • 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. • Should they lease a Large Store or Small Store?

  17. Large Store 6Servers Unlimited QueueLength …  = 30/hr. Lease Cost = $1000/day= $1000/10 = $100/hr. All customersget served!

  18. Small Store 2Servers Maximum QueueLength = 1 Will join system if0 or 1 in the queue Will not join thequeue if there is 1 customer in the queue  = 30/hr. Lease Cost = $200/day= $200/10 = $20/hr.

  19. Hourly Profit Analysis Large Small Arrival Rate  = 30 e = 30(1-p3) Revenue $25(Arrival Rate) (25)(30)=$750 $25e Costs Lease $100 $20 Server $20(#Servers) (20)(6) =$120 (20)(2) =$40 Waiting $8(Avg. in Store)=$8L $8L Net Profit ? ?

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

  21. p3 L Small Store -- M/M/2/3 e = (1-.44262)(30) = 16.7213

  22. Hourly Profit Analysis 30 16.7213 Large Small Arrival Rate  = 30 e = 30(1-p3) Revenue $25(Arrival Rate) (25)(30)=$750 $25e Costs Lease $100 $20 Server $20(#Servers) (20)(6) =$120 (20)(2) =$40 Waiting $8(Avg. in Store)=$8L $8L Net Profit ? ? $418 $750 $100 $ 20 $ 40 $120 $8(3.099143) $ 25 $ 17 $8(2.11475) Lease the large store $750-$100-$120-$25 $418-$20-$40-$17 $505 $341

  23. Example 5Which Machine is Preferable • Jobs arrive randomly to an assembly plant at an average rate of 5 per hour. • 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.) standard deviation of .6 min. ( = .6/60 = .01 hr.) • Which of the two designs seems preferable?

  24. Machine 1

  25. Machine 2

  26. Machine Comparisons Machine1 Machine 2 Prob (No Wait) -- P0.5000 .4792 Average Service Time 6 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.

  27. Module C8 Review • List Components of System • Develop a model • Use templates to get parameter estimates • Select “optimal” design

More Related