Queuing Models Basic Concepts Module C7 Queuing Models Basic Concepts

QUEUING MODELS Queuing theory is the analysis of waiting lines It can be used to: Determine the # checkout stands to have open at a store Determine the type of line to have at a bank Determine the seating procedures at a restaurant Determine the scheduling of patients at a clinic Determine landing procedures at an airport Determine the flow through a production process Determine the # toll booths to have open on a bridge

COMPONENTS OF QUEUING MODELS Arrival Process Waiting in Line Service/Departure Process Queue -- The waiting line itself System -- All customers in the queuing area Those in the queue Those being served

7 customers in the system The Queuing Process 1. Customers arrive according to some arrival pattern The Queue 4 customers in the queue 3. Customers are served according to some service distribution and depart 2. Customers may have to wait in a queue The System 7 customers in the system

ARRIVAL PROCESS Deterministic or Probabilistic (how?) Determined by # customers in system/balking? Single or batch arrivals Priority or homogeneous customers

THE WAITING LINE One long line or several smaller lines Jockeying allowed? Finite or infinite line length Customers leave line before service? Single or tandem queues

THE SERVICE PROCESS Single or multiple servers Deterministic of probabilistic (how?) All servers serve at same rate? Speed of service depends on line length? FIFO/LIFO or some other service priority

OBJECTIVE To design systems that optimize some criteria Maximizing total profit Minimizing average wait time for customers Meeting a desired service level

TYPICAL SERVICE MEASURES Average Number of customers in the system -- L Average Number of customers in the queue -- Lq Average customer time in the system -- W Average customer waiting time in the queue -- Wq Probability there are n customers in the system -- pn Average number of busy servers (utilization rate) -- 

POISSON ARRIVAL PROCESS REQUIRED CONDITIONS Orderliness at most one customer will arrive in any small time interval of t Stationarity for time intervals of equal length, the probability of n arrivals in the interval is constant Independence the time to the next arrival is independent of when the last arrival occurred

NUMBER OF ARRIVALS IN TIME t Assume  = the average number of arrivals per hour (THE ARRIVAL RATE) For a Poisson process, the probability of k arrivals in t hours has the following Poisson distribution:

Time Between Arrivals The average time between arrivals is 1/ For a Poisson process, the time between arrivals in hours has the following exponential distribution: f(x) = e-t This means: P(next arrival occurs > t hours from now) = e-t P(next arrival occurs within the next t hours) = 1- e-t

POISSON SERVICE PROCESS REQUIRED CONDITIONS Orderliness at most one customer will depart in any small time interval of t Stationarity for time intervals of equal length, the probability of completing n potential services in the interval is constant Independence the time to the completion of a service is independent of when it started IS THIS A GOOD ASSUMPTION?

NUMBER OF POTENTIAL SERVICES IN TIME t Unlike the arrival process, there must be customers in the system to have services Assume  = the average number of potential services per hour (SERVICE RATE) For a Poisson process, the probability of k potential services in t hours has the following Poisson distribution:

THE SERVICE TIME The average service time is 1/ For a Poisson process, the service time has the following exponential distribution: f(x) = e-t This means: P(the service will take t additional hours) = e-t P(the remaining service will take longer than t hours) = 1- e-  t

TRANSIENT vs. STEADY STATE Steady state is the condition that exists after the system has been operational for a while and wild fluctuations have been “smoothed out” Until steady state occurs the system is in a transient state -- transiting to steady state It is the long run steady state behavior that we will measure

CONDITIONS FOR STEADY STATE For any queuing system to be stable the overall arrival rate must be less than the overall potential service rate, i.e. For one server:  <  For k servers with the same service rate:  < k For k servers with different service rates:  < 1 + 2 + 3 + …+ k

STEADY STATE PERFORMANCE MEASURES We’ve mentioned these before: Average Number of customers in the system -- L Average Number of customers in the queue -- Lq Average customer time in the system -- W Average customer waiting time in the queue -- Wq Probability there are n customers in the system -- pn Average number of busy servers (utilization rate) - 

Little’s Laws and Other Relationships Little’s Laws relate L to W and Lq to Wq by: L = W Lq = Wq Also, (# in Sys) = (# in queue) + (# being served) Thus E(# in Sys) = E(# in queue) + E(# being served) L = Lq +  Thus knowing one of L, W, Lq and Wq allows us to find the other values.

CLASSIFICATION OF QUEUING SYSTEMS Queuing systems are typically classified using a three symbol designation: (Arrival Dist.)/(Service Dist.)/(# servers) Designations for Arrival/Service distributions include: M = Markovian (Poisson process) D = Deterministic (Constant) G = General Sometimes the designation is extended to 4 or 5 symbols to indicate Max queue length and # in population

M/M/1 M = Customers arrive according to a Poisson process at an average rate of  / hr. M = Service times have an exponential distribution with an average service time = 1/ hours 1 = one server Simplest system -- like EOQ for inventory -- a good starting point

M/M/1 PERFORMANCE MEASURES Average Number of customers in the system -- L = /(- ) Average Number of customers in the queue -- Lq = L - / Average customer time in the system -- W = L/  Average customer waiting time in the queue -- Wq = Lq/  Probability 0 customers in the system -- p0 = 1-/ Probability n customers in the system -- pn =(/)n p0 Average number of busy servers (utilization rate) or Average number customers being served =  = /

EXAMPLE -- Mary’s Shoes Customers arrive according to a Poisson Process about once every 12 minutes  = (60min./hr)/(12 min./customer) = 60/12 = 5/hr. Service times are exponential and average 8 min.  (service rate) = (60min/hr)/(8min./customer) = 7.5/hr. One server This is an M/M/1 system Will steady state be reached?  = 5 <  = 7.5/hr. YES

MARY’S SHOES PERFORMANCE MEASURES Avg # of busy servers (utilization rate) or Avg # customers being served =  = / =(5/7.5) = 2/3 Average # in the system -- L = /(- ) = 5/(7.5-5) = 2 Average # in the queue -- Lq = L - / = 2 - (2/3) = 4/3 Avg. customer time in the system -- W = L/  = 2/5 hrs. Avg cust.time in the queue - Wq = Lq/  = (4/3)/5 = 4/15 hrs. Prob.0 customers in the system -- p0 = 1-/ 1-(2/3) = 1/3 Prob. n customers in the system -- pn=(/)n p0 =(2/3) n(1/3)

COMPUTER SOLUTION The formulas for an M/M/1 are very simple, but those for other models can be quite complex Use M/M/k worksheet in Queue Template Results are in the row corresponding to 1 server

Input  and  Steady State Results Pn’s

M/M/k SYSTEMS M = Customers arrive according to a Poisson process at an average rate of  / hr. M = Service times have an exponential distribution with an average service time = 1/ hours regardless of the server k = k IDENTICAL servers

M/M/k PERFORMANCE MEASURES

EXAMPLE LITTLETOWN POST OFFICE Between 9AM and 1PM on Saturdays: Average of 100 cust. per hour arrive according to a Poisson process --  = 100/hr. Service times exponential; average service time = 1.5 min. --  = 60/1.5 = 40/hr. 3 clerks; k = 3 This is an M/M/3 system  = 100/hr < 3( = 40/hr.) i.e. 100 < 120 STEADY STATE will be reached

Solution Using the formulas, with  = 100,  = 40, k = 3 Prob.0 customers in the system -- p0 = .044944 Average # in the system -- L = 6.0112 Average # in the queue -- Lq = 3.5112 Avg. customer time in the system -- W = .0601 hrs. Avg cust.time in the queue - Wq = .0351hrs. Avg # of busy servers =  = / =(100/40) = 2.5 Average system utilization rate = /k = 100/120=.83

Input  and  Performance Measures for 3 servers Pn’s

M/G/1 Systems M = Customers arrive according to a Poisson process at an average rate of  / hr. G = Service times have a general distribution with an average service time = 1/ hours and standard deviation of  hours (1/ and  in same units) 1 = one server Cannot get formulas for pn but can get performance measures

Example -- Ted’s TV Repair Customers arrive according to a Poisson process once every 2.5 hours --  = 1/2.5 = .4/hr. Repair times average 2.25 hours with a standard deviation of 45 minutes  = 1/2.25 = .4444/hr.  = 45/60 = .75 hrs. Ted is the only repairman: k= 1 THIS IS AN M/G/1 SYSTEM

There are no formulas for the pn’s. Performance Measures P0 = 1-/ = 1-(.4/.4444) = .0991 L = (()2 + (/ )2)/(2(1-/ )) + / = ((.4)(.75)2 + (.4/.4444)2)/(2(.0991)) + (.4/.4444) = 5.405 Lq = L - / = 5.405 - .901 = 4.504 W = L/  = 5.405/.4 = 13.512 hrs. Wq = Lq/  = 4.504/.4 = 11.262 hrs. There are no formulas for the pn’s.

Input  (in customers/hr.)  (in customers/hr.)  (in hours) Performance Measures Select MG1 Worksheet

FINITE QUEUES (M/M/k/F) Frequently there are systems that have limits to the maximum number of customers in the system F Thus with probability pF the system is FULL and an arriving customer cannot join the queue-- i.e. we lose pF portion of the potential customers The effective arrival rate is then e = (1 - pF) Use e to calculate L, Lq, W, and Wq Steady state will always be achieved regardless of  and since the queue cannot build up indefinitely!!

Example -- Ryan’s Roofing Customer calls average 10/hr. 1 server -- average service time -- 3 minutes  = 60/3 = 20/hr. 3 lines so 2 can be on hold -- they will hold until they are served Arriving phone call when all 3 lines busy will not join the system

Input , , k, F Performance Measures pn’s W = L/ e e = (1-.06667)(10) = 9.33333 Select MMkF Worksheet

M/M/1 QUEUES WITH FINITE CALLING POPULATIONS (M/M/1//m) Maximum m school buses at repair facility, or m assigned customers to a salesman, etc. 1/ = average time between repeat visits for each of the m customers  = average number of arrivals of each customer per time period (day, week, mo. etc.) 1/ = average service time  = average service rate in same time units as 

Example -- Pacesetter Homes 4 projects m = 4 Average 1 work stoppage every 20 days/project “arrival” rate of work stoppages,  = 1/20 = .05/day Average 2 days to resolve work stoppage dispute “service” rate,  = 1/2 = .5/day

Input , , m Performance Measures pn’s Select MM1 m Worksheet

Module C7 Review Components of a queuing system Arrivals, Queue, Services Assumptions for Poisson (Markovian) distribution Requirements for Steady State Overall service rate > Overall arrival rate Steady State Performance Measures L, Lq, W, Wq, pn’s,  Queuing Systems M/M/1, M/M/k, M/G/1, M/M/k/F, M/M/1//m Use of the Queue Template