Queuing Models M/M/k Systems

CLASSIFICATION OF QUEUING SYSTEMS Recall that queues are classified by (Arrival Dist.)/(Service Dist.)/(# servers) Designations for Arrival/Service distributions include: –M = Markovian (Poisson process) –D = Deterministic (Constant) –G = General M/M/1We begin with the basic model, the M/M/1 system.

M/M/1 An 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/  hours 1 = one server Simplest system -- like EOQ for inventory -- a good starting point

M/M/1 PERFORMANCE MEASURES M/M1 systemFor the M/M1 system the performance measures are given by these simple formulas: L /(  - ) L = Average # of customers in the system = /(  - ) L Q L - /  L Q = Average # of customers in the queue = L - /  W L/ W = Average customer time in the system = L/ W Q L q / W Q = Average customer time in the queue = L q / p 0 1- /  p 0 = Probability 0 customers in the system = 1- /  p n ( /  ) n p 0 p n = Probability n customers in the system = ( /  ) n p 0 ρ /  ρ = utilization rate or Average number customers being served = / 

EXAMPLE -- Mary’s Shoes Customers arrive according to a Poisson Process about once every 12 minutes Service times are exponential and average 8 min. One server This is an M/M/1 system with: – 5/hr. – = (60min./hr)/(12 min./customer) = 60/12 = 5/hr. –  7.5/hr. –  (service rate) = (60min/hr)/(8min./customer) = 7.5/hr. Will steady state be reached? YES – = 5 <  = 7.5/hr. YES

MARY’S SHOES PERFORMANCE MEASURES Avg # of busy servers (utilization rate) or  2/3 Avg # customers being served =  = /  =(5/7.5) = 2/3 L2Average # in the system -- L = /(  - ) = 5/(7.5-5) = 2 L q 4/3Average # in the queue -- L q = L - /  = 2 - (2/3) = 4/3 W 2/5 hrs.Avg. customer time in the system -- W = L/ = 2/5 hrs. W q 4/15 hrs.Avg cust.time in the queue - W q = L q / = (4/3)/5 = 4/15 hrs. p 0 1/3Prob. 0 customers in the system -- p 0 = 1- /  =1-(2/3) = 1/3 p nProb. 3 customers in the system -- p n =( /  ) 3 p 0 =(2/3) 3 (1/3) =8/81

COMPUTER SOLUTION The formulas for an M/M/1 are very simple, but those for other models can be quite complex We can use a queuing template to calculate the steady state quantities for any number of servers, k M/M/1 modelM/M/k worksheetFor the M/M/1 model use the M/M/k worksheet in Queue Template –Since k = 1, the results are in the row corresponding to 1 server

Steady State Results P n ’s Input and  Go to the MMk Worksheet p3p3

M/M/k SYSTEMS M/M/k An M/M/k 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/  hours regardless of the server k = k IDENTICAL servers steady stateλ < kμTo reach steady state: λ < kμ

M/M/k PERFORMANCE MEASURES

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 –  = 40/hr. –Since λ < 3μ, i.e. 100 < 120, –STEADY STATE will be reached

Solution Using the formulas, with = 100,  = 40, k = 3, it can be shown that: p 0 = Prob.0 customers in the system -- p 0 = L = Average # in the system -- L = L q = Average # in the queue -- L q = W =.0601 hrs.Avg. customer time in the system -- W =.0601 hrs. W q =.0351hrs.Avg cust.time in the queue - W q =.0351hrs.  /k .83Average system utilization rate  = /k  = 100/120 =.83 ρ 2.5Avg # of busy servers = kρ = /  =(3X0.83) = 2.5

Input and  Performance Measures for 3 servers P n ’s Go to the MMk Worksheet

M/M/k/F Systems M/M/k/F An M/M/k/F 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/  hours regardless of the server k = k IDENTICAL servers F = maximum number of customers that can be in the system at any time Because the queue cannot build up indefinitely, steady state will be achieved regardless of the values of λ and μ! Formulas for steady state quantities are complex – use template.

Basic Concept of M/M/k/F Systems The number of customers that can be in the system is 0, 1, 2, …,F –If 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. effective arrival rate, λ e λ e = λ(1-p F ).Thus the effective arrival rate, λ e, the average number of arrivals per hour that actually join the system is: λ e = λ(1-p F ).

EXAMPLE RYAN’S ROOFING The 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. M/M/1/3This is an M/M/1/3 system with: – = 10/hr. –μ 20/hr. –μ = 60/3 = 20/hr.

USING THE M/M/k/F TEMPLATE F > kThe 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 > k To determine the effective arrival rate, we find p F on the right side of the output. Then in a cell (or by hand) we can calculate: Effective Arrival Rate λ e = λ(1-p F )

Go to the MMkF Worksheet Input, , k and F Steady State Results P n ’s p F = p 3 Effective Arrival Rate λ e = λ(1-p F ) =C4*(1-P12) Excel = 10( ) =

Review M/M/k systems are ones with: –a Poisson arrival distribution –an exponential service distribution –k identical servers Steady state formulas for M/M/k model Finite queuing models –Always reach steady state –Effective arrival rate, λ e = λ(1-p F ) Use of Templates –M/M/k –M/M/k/F