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 -- Consider the total (hourly) cost for each system and choose the minimum

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

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

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

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. 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?

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

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

BILL Bill 1/ = 5 min. A = 60/5 = 12/hr. LBill = ? Hourly Cost = $10 + 4LBill  = 8/hr

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

CHARLIE Charlie 1/ = 4 min. A = 60/5 = 15/hr. LCharlie = ? Hourly Cost = $14 + 4LCharlie  = 8/hr

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

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

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?

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

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

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

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

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

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

Optimal Best option -- System 2 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

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)?

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

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

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) $8Ll $8Ls Net Hourly Profit ? ?

Large Store -- M/M/6 3.099143 Ll

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

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

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?

Machine 1

Machine 2

Machine 2 looks preferable 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. Machine 2 looks preferable

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