Example-1Customers arrive at a bakery at an average rate of 18 per hour on weekday mornings. The arrival distribution can be described by a Poisson distribution with a mean of 18. Each clerk can serve a customer in an average of four minutes; this time can be described by an exponential distribution with a mean of 4.0 minutes. A) What are the arrival and service rates? B) Compute the average number of customers being served at any time (assuming that system utilization is less than 1). C) Suppose it has been determined that the average number of customers waiting in line is 3.6. Compute the average number of customers in the system (i.e. waiting in line or being served), the average time customers wait in line, and the average time in the system. D) Determine the system utilization for M=2, 3 and 4 servers. Example-1: Solution Given : λ = 18 customers/hr,1 / µ = 4 minutes A) Arrival Rate λ = 18 customers/hr, Service Rate µ = 60 / 4 = 15 customers/hr B) r = r= λ µ 18 15 = average number of customers being served = 1.2 customers C) Lq = 3.6 customers, L s , Wq , Ws = ? L s = L q + r = 3.6 + 1.2 = 4.8 customers 3.6 = 0.20 hrs 18 1 1 Ws = Wq + = 0.20 + = 0.267 hrs 15 µ Wq = Lq D) ρ = λ Mµ λ = 18 = 0.60 2(15) 18 = 0.40 M = 3 ⇒, ρ = 3(15) 18 = 0.30 M = 4 ⇒, ρ = 4(15) M = 2 ⇒, ρ = Example-2 An Airline is planning to open a satellite ticket desk in a new shopping plaza, staffed by one ticket agent. It is estimated that requests for tickets and information will average 15 per hour, and request rates will have a Poisson distribution. Service time is assumed to be exponentially distributed. Previous experience with similar satellite operations suggests that mean service time should average about three minutes per request. Determine each of the following: A) B) C) D) E) System utilization. Percent of time the server (agent) will be idle. The expected number of customers waiting to be served. The average time customers will spend in the system. the probability of zero customers in the system and the probability of four customers in the system. Example-2: Solution Given : λ = 15 requests/hr, (1 / µ ) = 3 minutes A) Arrival Rate λ = 15 requests/hr, Service Rate µ = 60 / 3 = 20 requests/hr λ 15 ρ= = = 0.75 Mµ 1(20) B) Idle time percentage = 1 - ρ = 1 - 0.75 = 0.25 C) Lq = D) Ws = 15 2 λ2 = = 2.25 requests µ ( µ − λ ) 20( 20 − 15) Lq λ + 1 µ = 2.25 1 + = 0.20 hrs 15 20 15 λ E) P0 = 1 − = 1 − = 0.25 20 µ ⎛λ⎞ ⎛ 15 ⎞ and P4 = P0 ⎜ ⎟ = 0.25⎜ ⎟ = 0.079 ⎜µ⎟ ⎝ 20 ⎠ ⎝ ⎠ 4 4 2) . B) Probability of zero customers in the system.6 requests/hr.786 7(1.6 Lq λ = 1. (1 / µ ) = 50 minutes. C) Average waiting time for an arrival not immediately served. customers request cabs at a rate that follows the Poisson distribution with a mean of 6. D) Probability that an arrival will have to wait for service.Example-3 Alpha Taxi and Hauling Company has seven cabs stationed at the airport.456 E)ρ = λ Mµ = 6 .2) − 6. Service time is exponential with a mean of 50 minutes per customer.2 requests/hr.4. Assume that there is one customer per cab and that taxis return to the airport.2536 hrs = 0.5 µ 1 .674 requests λ 6. The company has determined that during the late-evening hours on weeknights. M = 7 taxis(servers) A) Service Rate µ = 60 / 50 = 1. E) System Utilization.4.556) = 0. Find each of the following: A) Average number of customers in line.674 / 6.2536 /(0. Example-3: Solution Given : λ = 6.2 B) From Table 19 .6 per hour. P0 = 0.6 = 0.003 C) Wa = D) Wq = PW = Wq Wa 1 1 = = 0.6 = = 5 .556 hrs Mµ − λ 7(1. Lq = 1. From Table 19 .6 = 0. and the service rate (including return time to the rail station) is expected to be 1.8(20 / 60) = 1.6 requests For M = 4 and r = 3. . µ = 1.2 ⇒ Lq = 0.2 µ ⇔ Lq = λ Wq = 4. How many cabs will be needed to achieve an average time in line of 20 minutes or less? Example-4: Solution Given : λ = 4.5 requests/hr. M = ? taxis(servers) for Wq = 20 min Solution : r= λ = 4 .4.386 For M = 5 and r = 3. Lq ≤ 1.8 requests/hr. So M = 5 cabs are needed for Wq ≤ 20 min .Example-4 Alpha Taxi and Hauling Company also plans to have cabs at the new rail station. The expected arrival rate is 4.5 per hour.2 ⇒ Lq = 2.8 / 1 .5 = 3 .6 requests Wq = 20 min From Table 19 .8 customers per hour. for Wq ≤ 20 min.513. Total Cost =Customer Waiting Cost+ Capacity Cost Example-5: Solution Given : λ = 15 requests/hr.48 371.36 402. truck and driver cost is $120 per hour.099 3.36 5=(2)+(4) Total Cost 943.36 902. Crew Cost = 100$/hr.354 3.88 363.528 3. Truck Waiting Cost = 120 $/hr M = ? crews(servers) for minimum Cost Solution : λ r = = 15 / 5 = 3 µ 1 Crew Size 4 5 6 7 2=(1)x(100$) Crew Cost 400 500 600 700 Solution 3=Lq+r Ls 4. The high unloading rate is due to cargo being containerized. The new rates are: crew and dock cost is $100 per hour.Example-5 Trucks arrive at a warehouse at a rate of 15 per hour during business hours. µ = 5 requests/hr.88 1063.36 . Recent changes in wage rates have caused the warehouse manager to reexamine the question of of how many crews to use.028 4=(3)x(120$) Truck Cost 543. Crews can unload the trucks at a rate of five per hour.48 971. 9 5=(2)+(4) Total Cost 279 51. Example-6: Solution Given : λ = 18 requests/hr. The manager of the shop must decide on the number of attendants needed to staff the crib. Based on previous experience. the manager estimates requests for parts will average 18 per hour with a service capacity of 20 requests per hour per attendant. Mechanic Cost = 30 $/hr M = ? attendant(servers) for minimum Cost Solution : 18 r= = 0. µ = 20 requests/hr.90 20 1 # of Attendants 1 2 3 Solution 2=(1)x(9$) Attendant Cost 9 18 27 3=Lq+r Ls 9 1. Attendants will receive $9 per hour in salary and fringe benefits. Attendant Cost = 9 $/hr.87 27.93 4=(3)x(30$) Mechanic Cost 270 33.Example-6 One of the features of a new machine shop will be a well-stocked crib.129 0. Mechanics’ time will be worth $30 per hour. How many attendants should be on duty if the manager is willing to assume that arrival and service rates will be Poisson-distributed? (Assume the number of mechanics is very large.9 .87 54. which includes salary and fringe benefits plus two lost work time caused by waiting for parts. so an infinite-source model is appropriate). Case 3. since we simply convert TIME into RATES.833 jobs / hr µ = 1 /(24 x 2) = 0. B) Would the calculated rates be different if there were inter-arrival times rather than service times? Example-7: Solution Case 1. .2 hours 2 days Determine the service rate for each operation.Example-7 The following is a list of service times for three different operations: Operations A B C A) Service Time 8 minutes 1.0208 jobs / hr No. Case 2. which would be same for inter-arrival times.2 = 0.5 jobs / hr µ = 1 / 1. µ = 60 / 8 = 7. Requests for ambulances during non-holiday weekends average 0. µ = 1 requests/hr.152 µ 1 Lq C) Wq = D)Wa = PW = Wq Wa λ = 0. Example-8: Solution Given : λ = 0. M = 2 ambulances(servers) A) ρ = B) r = λ Mµ = 0.8 1 1 = = 0.40 or 40% 2(1) λ 0.152 = 0.19 / 0. The average number of customer waiting. System utilization.8 = 0.8 = = 0.190 hr 0.228 . Find: A. C. B.833 = 0.8 per hour and tend to be Poisson distributed.8.833 Mµ − λ 2(1) − 0.Example-8 A small town with one hospital has two ambulances to supply ambulance service.4 ⇒ L q = 0. D. The probability that both ambulances will be busy when a call comes in. Travel and assistance time averages one hour per call and follows an exponential distribution. from table 19 . The average time customers wait for an ambulance.8 = 0.8 requests/hr. M = 2 inspectors(servers) A) Lq = 2.m.711 1 1 = = 0. each of whom can inspect 25 trucks an hour: A. Trucks arrive at the station at the rate of 40 an hour between 7 p.444 trucks µ λ µ C) Wa = PW = Wq Wa L 1 1 = = 0.0711 / 0. including those being inspected? B.1111 hrs 40 25 λ = 2.Example-9 Trucks are required to pass through a weighing station so that they can be checked for weight violations.884 1 + = 0.m.0711 hrs Mµ − λ 2(25) − 40 λ = 0. so Ls = Lq + B) Ws = Lq + 1 = 2. Currently two inspectors are on duty during those hours.1 and Wq = q = 2.844 from Table19 − 4. and 9 p. µ = 25 trucks/hr. about how many minutes could the driver expect to be at the station? C. How many minutes. If a truck was just arriving at the station.1 hr Mµ − λ 2(25) − 40 D) Wa = .844 + 1. How many trucks would you expect to see at the weighing station. on average.844 / 40 = 0. would a truck that is not immediately inspected have to wait? Example-9: Solution Given : λ = 40 trucks/hr.1 = 0.6 = 4. What is the probability that both inspectors would be busy at the same time? D. 15 Improvement Cost 100 200 5=(2)+(4) Total Cost 1900. A.80 1 Dock Capacity 1 2 2=(1)x(1100$) Dock Cost 1100 2200 Solution(A) 3=Lq+r Ls 4 0. How many docks should be requested if trucks arrive at the rate of four per day. The equipment would cost $100 per day for each dock.334 1.71 trucks/day 1 Dock Capacity 1 2 2=(1)x(1100$) Dock Cost 1100 2200 Solution(B) 3=Lq+r Ls 2.69 5=(2)+(4) Total Cost 2300 2485.15 . µ = 5 trucks/day.71 trucks per day. Should the manager invest in the new equipment? Example-10: Solution Given : λ = 4 trucks/day. and both rates are Poisson? B. each dock can handle five trucks per day.2 2904.Example-10 The manager of a regional warehouse must decide on the number of loading docks to request for a new facility in order to minimize the sum of dock costs and driver-truck costs.6805 4=(3)x(300$) Truck Cost 700.100 per day. An employee has proposed adding new equipment that would speed up the loading rate to 5. M = ? docks(servers) for minimum Cost Truck Waiting Cost = 300$/day.100$/day Solution : r = 0.69 µ = 5.2 504.9523 4=(3)x(300$) Truck Cost 1200 285. Dock Cost = 1. The manager has learned that each driver-truck combination represents a cost of $300 per day and that each dock plus loading crew represents a cost of $1. from Table19 − 4 ⇒ Lq = 0.228 1 1 = = 0.40 Mµ 2(15) (1 − ρ ) = 1 − 0.01266 hrs λ Mµ − λ 2(15) − 12 = 0.152 + 0. D. What is the probability that a mechanic would have to wait for service? If a mechanic has to wait. how long would the average wait be? What percentage of time are the clerks idle? E.0555 hr Mµ − λ 2(15) − 12 12 λ = = 0.8 = 0. including those being served? B. M = 2 clerks(servers) A) λ 12 = = 0. Suppose there are two clerks at the counter.152 µ 15 λ so Ls = Lq + = 0. and this can be modeled by a Poisson distribution that has a mean of 15. The time between requests can be modeled by a negative exponential distribution that has a mean of five minutes. C.80.60 = %60 idle D) ρ = B) Wa = PW = Wq Wa C) Wa = . On average. what number of clerks would be optimal in terms of minimizing total cost? Example-11: Solution Given : λ = 12 requests/hr. A clerk can handle requests at a rate of 15 per hour. A.152 / 12 = 0.Example-11 The parts department of a large automobile dealership has a counter used exclusively for mechanics’ requests for parts.01266 / 0.0555 = 0. how many mechanics would be at the counter. If clerks represent a cost of $20 per hour and mechanics a cost of $30 per hour.0555 and Wq = q = 0.953 requests µ L 1 1 = = 0.4 = 0. µ = 15 requests/hr. What is the probability that an arriving student (just before entering the Administrative Services Office) will find at least one other student waiting in line? . How much time.952 0. Judy Gumshoes.56 84.Example-11: Solution (contd. What percentage of time is Judy idle? B. The service counter is staffed by one clerk. who works eight hours per day. A. Assume Poisson arrivals and exponential service times. and their requests take an average of 12 minutes to be processed. on average.57 5=(2)+(4) Total Cost 140 68. How long is the (waiting) line on average? D. does a student spend waiting in line? C.819 4=(3)x(30$) Mechanics Cost 120 28.) 1 # of Clerks 1 2 3 2=(1)x(20$) Clerk Cost 20 40 60 3=Lq+r Ls 4 0.56 24.57 Solution Example-12 Students arrive at the Administrative Services Office at an average of one every 15 minutes. 2 = 20% idle Mµ 1(5) 4 λ = = 0. µ = 5 requests/hr. They are currently considering the following two options: A.64 ⎜µ⎟ ⎝ ⎠ 2 Example-12 Extension The managers of the Administrative Services Office estimate that the time a student spends waiting in line costs them (due to goodwill loss and so on) $10 per hour. is Judy right to prefer the hired help? Assume Poisson arrival and Exponential service times. for example).8(4) = 3. . M = 1 A) ρ = 4 λ = = 0. they know that they need to improve Judy’s processing time (previous example). B. with which Judy expects to be able to complete a student request 40 percent faster (from 2 minutes per request to 1 minute and 12 second. Hire another temporary clerk. who will work at the same rate as Judy? If the computer costs $99.8 hr µ ( µ − λ ) 5(5 − 4) B) Wq = C) L q = Wq λ = 0. To reduce the time a student spends waiting. Install a computer system. (1 − ρ ) = 0.82 = 0.Example-12: Solution Given : λ = 4 students/hr.50 to operate per day.8.2 students D) Arriving Student will find at least one other student waiting ≡ there are at least 2 students in the system = P≥ 2 ⎛λ⎞ P≥ 2 = 1 − P< 2 = 1 − (1 − ⎜ ⎟ ) = 0. while the company clerk gets paid $13 per hour. 6)). M = 1 OPTION 2 . with exponential service times.Extra Clerk = 243.8 µ Example-13 The manager of a grocery store in the retirement community of Sunnyville is interested in providing good service to the senior citizens who shop at his store.With Computer : λ = 4 students/hr. the store has a separate checkout counter for senior citizens. µ = 5 requests/hr. Time spent in the system. D.With an additional Clerk(8hr/day × $13/hr = $104/day) : λ = 4 students/hr.55$ Net Savings = Waiting Time saving . Number of customers in line. according to a Poisson distribution. and are served at an average rate of 35 customers per hour. .84$/day OPTION 2 is better! λ = 0.333(8. Waiting time in line.333 requests/hr (60/(12x0.8 .0 = 139. 30 senior citizens per hour arrive at the counter. On average.038 Savings in Waiting Cost = 32 students x 10$ x (0. Utilization of the checkout clerk. Number of customers in the system.99. Presently. Probability of zero customers in the system. E.152 / 4 = 0. M = 2 Option 1-) Wq = 4 λ = = 0.84 .8 .84$ Net Savings = Waiting Time saving .1108 hr µ ( µ − λ ) 8.Example-12 Extension: Solution Given : 2 Options (currently 32 students arrive per day) OPTION 1 .0. Find the following operating characteristics: A.55 .Computer Cost = 220. B.038) = 243.1108) = 220.05$/day Option 2-) Lq = 0. C. µ = 8.0.152 for M = 2 and r = Wq = Lq / λ = 0.104. F.5 = 121.333 − 4) Savings in Waiting Cost = 32 students x 10$ x (0. 857 = 5 .857 Mµ 1(35) C ) Ls = λ µ −λ = 30 = 6 citizens 35 − 30 D ) Lq = Ls − λ = 6 − 0 . what is the probability of having more than four customers in the system? C.143 ⎜µ⎟ ⎝ 35 ⎠ ⎝ ⎠ 30 λ B) ρ = = = 0.857 = 0. For that service rate.143 = 0. µ = 35 citizens/hr ⎛λ⎞ ⎛ 30 ⎞ A) P0 = 1 − ⎜ ⎟ = 1 − ⎜ ⎟ = 1 − 0. What service rate would be required to have customers average only eight minutes in the system? B.1714 hr 30 Lq λ Example-14 The manager of the Sunnyville grocery in the above example wants answers to the following questions: A.Example-13: Solution Given : λ = 30 citizens/hr.2 hr 30 5. What service rate would be required to have only a 10 percent chance of exceeding four customers in the system? . 143 citizens µ E ) Ws = F ) Wq = Ls λ = = 6 = 0. and each crew costs $30 per hour.55 citizens/hr 5 5 µ Example-15 The management of the American Parcel Service terminal in Verona. What is the total hourly cost of operating the system? .3277 ⎜µ⎟ ⎝ 37.5 ⎠ ⎝ ⎠ C ) P> 4 = 0.5 citizen/hr µ − λ µ − 30 5 5 ⎛λ⎞ ⎛ 30 ⎞ B) P> 4 = 1 − P≤ 4 = 1 − P<5 = 1 − (1 − ⎜ ⎟ ) = ⎜ ⎟ = 0. Average utilization of the server.10 ⇒ µ = ? ⎛ 30 ⎞ ⎛λ⎞ P> 4 = 1 − P<5 = 1 − (1 − ⎜ ⎟ ) = ⎜ ⎟ = 0. Average number of customers in line & in the system. On average a crew can unload a semi-trailer rig in one hour.Example-14: Solution Given : λ = 30 citizens/hr. Average waiting time in line & the system. The estimated cost of an idle truck is $50 per hour. Each bay requires a crew of two employees. Probability of zero customers in the system. E. waiting to be loaded.10 = ⇒ µ = 47. C. B. is concerned about the amount of time the company’s trucks are idle. with exponential service times. A. D. Wisconsin. µ = ? A) Ws = 8 min = 8/60 hr = 1 1 = ⇒ µ = 37. Trucks arrive at an average rate of three per hour. The terminal operates with four unloading bays.10 ⎜µ⎟ ⎜µ⎟ ⎝ ⎠ ⎝ ⎠ 30 5 0. according to a Poisson distribution. 5093 + = 1.. Lq = 1.0 /hr Total Cost = $226.1. ρ ( K + 1) ρ K +1 Ls = − 1..0 /hr = $466.528 + = 4.528 trucks 1 µ E ) Total Cost = Waiting Cost + Capacity Cost Waiting Cost = $50 × Ls = 50 × 4. µ = 1 truck/hr (from 1/µ = 1 hr/truck).ρ K +1 ) r = 1.528 = $226.4 /hr + $240.ρ ) ρ n Pn = (1.P0 n = 0..4 with M = 4.ρ (1.528 trucks µ Lq Wq = λ = 1.Example-14: Solution Given : λ = 3 trucks/hr.PK) (Why?) . K customers (1.4 /hr Capacity Cost = (2 × 30) × 4 = $240.528 = 0.4 /hr 1 = 0. P0 = 0.5093 hr 3 D) Ls = Lq + Ws = Wq + 1 3 λ = 1.ρ K +1 ) Other measures from the basic formulas by replacing λ with λeff= λ(1.75 = 75% utilization of each server 4(1) λ = 3 from Table 19 . K Extended Model-1 formulas are applicable even if λ > µ.038 µ B) ρ = C) λ Mµ = 3 = 0.4 with M = 4.5093 hr 1 µ Extending Model 1 Max. M = 4 λ A) = 3 from Table 19 . but when there are more than 10 waiting. B.λ /µ 1 − (λ/µ ) K +1 Ls λ (1 − PK ) 1 Lq = Ls − . Customers are willing to wait. Ws = . 1 − (λ / µ ) K +1 Pn = ( P0 )( ) n for n ≤ K. A.Ch19: Extended Model 1 (Finite Queue Length) P0 = Ls = 1− λ / µ . λ µ λ/µ ( K + 1)(λ/µ ) K +1 − . 1 . they are turned away. The arrival rate is 20 per hour. Wq = Ws − . C. What percent of customers are turned away? What is the rate at which customers are turned away? What is the average number of customers in line & in the system? What is the average waiting time in line and in the system? . exponentially distributed. µ λ (1 − PK ) µ Little’s Law Updated for Effective Arrival Rate (λeffective) λeffective = λ (1 − PK ) = Ls Lq = Ws Wq Example-15 A popular attraction at the Montreal Old Port is a street artist who will paint a caricature in about 5 minutes. D. P = P = [(1 . and µ = 12/hr. ρ = λ/µ = 20/12 = 1. Determine the average waiting time. the average queue length.6 min. The maximum number of vehicles in the system is four. s q s Example-16 Slick’s Quick Lube is a one-bay service facility next to a busy highway. λeff = (1. Ls = ρ/(1 . P λ = 0. . a. The mean time required to perform to lube operation is 2 minutes.4*20 = 8/hr. Both the inter-arrival times and the service times are exponentially distributed.53/12 = 0.4 of arrivals are turned away.71hr.53 Since 0.79.ρK+1 ) = 9. * 60 ≈ 42. The facility has space for only one vehicle in service and three vehicles lined up to wait for service. W = W .Solution λ = 20/hr.1/µ = 0. There is no space for cars to line up on the busy adjacent highway.79 – 1/12 = 0. The mean time between arrivals for customers seeking lube service is 3 minutes.67.P )λ = 12/hr. K = 11 (10 waiting in the line plus one being painted).ρ)* ρK ]/[1. so if the waiting line is full (3 cars). prospective customers must drive on. K W = Ls / λeff = 9.ρK+1 ] ≈ 0. K c. K 11 b.4 (40% of time customers have to be turned away).ρ) – (K+1) ρK+1 /(1.Example-15. and the probability that a customer will have to drive on. 758 = 1.076 = 7.076) Wq = Ws − 1 = 0. and K = 4 (= 3 waiting + 1 service) 1− λ / µ 1 − 20 / 30 = = 0.2 / 3 1 − (2 / 3) K +1 λeffective 1 Ls = λ (1 − P4 ) Ls = 1.03387 hr µ 30 Lq = λeffective × Wq = λ (1 − P4 ) × Wq = 20(1 − 0.242 = 0.03387 = 0.242 cars = − − K +1 1 .λ/µ 1 − ( λ /µ ) 1 .0672 hr 20(1 − 0.384 1 − (λ / µ ) K +1 1 − (20 / 30) 4+1 λ 20 P4 = ( P0 )( ) 4 = 0.076) × 0. 1/µ = 2 min /car So λ = 20 cars/hr and µ = 30 cars/hr.6% drives on µ 30 P0 = Ls = Ws = λ/µ ( K + 1)(λ/µ ) K +1 2 / 3 (4 + 1)(2 / 3) 4+1 = 2 − 0.Example-16: Solution Given :1 / λ = 3 min/car.0672 - .384 × ( ) 4 = 0.626 cars = 0.