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 -- 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:
k!
t)( t)in time arrivalsP(k
k te
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:
k!
t)( t)in time services potentialP(k
k te
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
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 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 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
• 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.
02
1
0
0
!1
!
1
!
1
1
pkk
L
k
k
kn
p
k
k
n
kn
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 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 = 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.
• Prob.0 customers in the system -- p0 = .044944
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
FINITE QUEUES
• 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
• Thus the effective arrival rate is e = 1 - pF
• Use e to calculate L, Lq, W, and Wq
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