Queuing -- Mmk Systems (1)

Queuing ModelsM/M/k Systems

CLASSIFICATION OF QUEUING SYSTEMSRecall that queues are classified by

(Arrival Dist.)/(Service Dist.)/(# servers)Designations for Arrival/Service distributions include:M = Markovian (Poisson process)D = Deterministic (Constant)G = GeneralWe begin with the basic model, the M/M/1 system.

M/M/1An M/M/1 system is one with: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/ hours1 = one server

Simplest system -- like EOQ for inventory -- a good starting point

M/M/1PERFORMANCE MEASURESFor the M/M1 system the performance measures are given by these simple formulas:L = Average # of customers in the system = /(- )LQ = Average # of customers in the queue = L - / W = Average customer time in the system = L/ WQ = Average customer time in the queue = Lq/ p0 = Probability 0 customers in the system = 1-/pn = Probability n customers in the system = (/)n p0 = utilization rate or Average number customers being served = /

EXAMPLE -- Marys ShoesCustomers arrive according to a Poisson Process about once every 12 minutesService times are exponential and average 8 min.One serverThis is an M/M/1 system with: = (60min./hr)/(12 min./customer) = 60/12 = 5/hr. (service rate) = (60min/hr)/(8min./customer) = 7.5/hr.Will steady state be reached? = 5 < = 7.5/hr. YES

MARYS SHOESPERFORMANCE MEASURESAvg # of busy servers (utilization rate) orAvg # customers being served = = / =(5/7.5) = 2/3Average # in the system -- L = /(- ) = 5/(7.5-5) = 2Average # 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) = 1/3Prob. 3 customers in the system -- pn=(/)3 p0 =(2/3)3(1/3) =8/81

COMPUTER SOLUTIONThe formulas for an M/M/1 are very simple, but those for other models can be quite complexWe can use a queuing template to calculate the steady state quantities for any number of servers, kFor the M/M/1 model use the M/M/k worksheet in Queue TemplateSince k = 1, the results are in the row corresponding to 1 server

M/M/k SYSTEMSAn M/M/k system is one withM = 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 serverk = k IDENTICAL servers

To reach steady state: < k

M/M/k PERFORMANCE MEASURES

EXAMPLELITTLETOWN POST OFFICEBetween 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 = 3This is an M/M/3 system = 100/hr = 40/hr.Since < 3, i.e. 100 < 120, STEADY STATE will be reached

SolutionUsing the formulas, with = 100, = 40, k = 3, it can be shown that:Prob.0 customers in the system -- p0 = .044944Average # in the system -- L = 6.0112Average # in the queue -- Lq = 3.5112 Avg. customer time in the system -- W = .0601 hrs.Avg cust.time in the queue - Wq = .0351hrs.Average system utilization rate = /k = 100/120 = .83Avg # of busy servers = k = / =(3X0.83) = 2.5

M/M/k/F SystemsAn M/M/k/F system is one withM = 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 serverk = k IDENTICAL serversF = maximum number of customers that can be in the system at any timeBecause the queue cannot build up indefinitely, steady state will be achieved regardless of the values of and !Formulas for steady state quantities are complex use template.

Basic Concept of M/M/k/F SystemsThe number of customers that can be in the system is 0, 1, 2, ,FIf an arriving customer finds < F customers in the system when he arrives, he will join the system.If an arriving customer finds F customers in the system when he arrives, he cannot join the system, he will leave, and his service is lost.Thus the effective arrival rate, e, the average number of arrivals per hour that actually join the system is: e = (1-pF).

EXAMPLERYANS ROOFINGThe average number of customers that call the company per hour is 10.There is 1 operator who averages 3 minutes per call.Both calls and operator time conform to Poisson processes.There are 3 phone lines so 2 calls could be on hold. A caller that calls when all 3 lines are busy, gets the busy signal and does not join the system.This is an M/M/1/3 system with: = 10/hr. = 60/3 = 20/hr.

USING THE M/M/k/F TEMPLATEThe template is designed to be used for the case where a queue is possible that is the maximum number of customers in the system is greater than the number of servers, i.e. F > kTo determine the effective arrival rate, we find pF on the right side of the output. Then in a cell (or by hand) we can calculate: Effective Arrival Rate

e = (1-pF)

Effective Arrival Rate e= (1-pF)=C4*(1-P12) Excel= 10(1-.06667) = 9.3333

ReviewM/M/k systems are ones with:a Poisson arrival distributionan exponential service distributionk identical servers Steady state formulas for M/M/k modelFinite queuing modelsAlways reach steady stateEffective arrival rate, e = (1-pF)Use of TemplatesM/M/k M/M/k/F

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