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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
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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
1. Customers arrive according to some arrival pattern The Queue The System 2. Customers may have to wait in a queue 3. Customers are served according to some service distribution and depart The Queuing Process 4 customers in the queue 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 FORSTEADY 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 STATEPERFORMANCE 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/1PERFORMANCE 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 SHOESPERFORMANCE 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
EXAMPLELITTLETOWN 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
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 Select MMkF Worksheet e = (1-.06667)(10) = 9.33333
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