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

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 -- L q Average customer time in the system -- W Average customer waiting time in the queue -- W q Probability there are n customers in the system -- p n 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 -- L q Average customer time in the system -- W Average customer waiting time in the queue -- W q Probability there are n customers in the system -- p n Average number of busy servers (utilization rate) - 

Little’s Laws and Other Relationships Little’s Laws relate L to W and L q to W q by: L = W L q = W q Also, (# in Sys) = (# in queue) + (# being served) Thus E(# in Sys) = E(# in queue) + E(# being served) L = L q +  Thus knowing one of L, W, L q and W q 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

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 -- L q = L - /  Average customer time in the system -- W = L/ Average customer waiting time in the queue -- W q = L q / Probability 0 customers in the system -- p 0 = 1- /  Probability n customers in the system -- p n =( /  ) n p 0 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 miuntes – = (60min./hr)1/12 cust/min. = 60/12 = 5/hr. Service times are exponentialand average 8 min. –  (service rate) = (60min/hr)(1/8cust./min.) = 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 -- L q = L - /  = 2 - (2/3) = 4/3 Avg. customer time in the system -- W = L/ = 2/5 hrs. Avg cust.time in the queue - W q = L q / = (4/3)/5 = 4/15 hrs. Prob.0 customers in the system -- p 0 = 1- /  1-(2/3) = 1/3 Prob. n customers in the system -- p n =( /  ) n p 0 =(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 We could program formulas into EXCEL cells WINQSB gives us results

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 Formulas much more complex e.g.

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 WINQSB, with = 100,  = 40, k = 3 Average system utilization rate = /k  = 100/120=.83 Avg # of busy servers =  = /  =(100/40) = 2.5 Average # in the system -- L = Average # in the queue -- L q = Avg. customer time in the system -- W =.0601 hrs. Avg cust.time in the queue - W q =.0351hrs. Prob.0 customers in the system -- p 0 =

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 p n 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

FINITE QUEUES Frequently there are systems that have limits to the maximum number of customers in the system F Thus with probability p F the system is FULL and an arriving customer cannot join the queue-- i.e. we lose p F portion of the potential customers Thus the effective arrival rate is e = 1 - p F Use e to calculate L, L q, W, and W q

M/M/1 QUEUES WITH FINITE CALLING POPULATIONS 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

ECONOMIC ANALYSES Each problem is different 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