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