Transcript Slide 1

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
• We 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
• For 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
ρ = 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
• Service times are exponential and average 8
min.
• One server
• This 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
MARY’S SHOES
PERFORMANCE 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. 3 customers in the system -- pn=(/)3 p0 =(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
• For 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
Input  and 
Steady State Results
Pn’s
p3
Go to the MMk
Worksheet
M/M/k SYSTEMS
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
• To reach steady state: λ < kμ
M/M/k PERFORMANCE
MEASURES
• Formulasmuch morecomplexe.g.
1
p0 
n
k
k 1
1
1     k 
     


k!     k   
n  0 n!   
k
   



L
p0   
2
k  1!k   

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:
•
•
•
•
•
•
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 = ρ/k = /k =
100/120 = .83
Input  and 
Performance Measures
for 3 servers
Pn’s
Go to the MMk
Worksheet
M/M/k/F Systems
•
•
•
•
•
•
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 acchieved 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.
• Thus the effective arrival rate, λe, the
average number of arrivals per hour that
actually join the system is: λe = λ(1-pF).
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.
• This is an M/M/1/3 system with:
–  = 10/hr.
– μ = 60/3 = 20/hr.
USING THE M/M/k/F TEMPLATE
• The 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 pF on the right side of the output.
Then in a cell (or by hand) we can
calculate:
Effective Arrival Rate
λe = λ(1-pF)
Input , , k and F
Effective Arrival Rate λe= λ(1-pF)
=C4*(1-P12) Excel
= 10(1-.06667) = 9.3333
Steady State Results
Pn’s
pF = p3
Go to the MMkF
Worksheet
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/1 model
• Finite queuing models
– Always reach steady state
– Effective arrival rate, λe = λ(1-pF)
• Use of Templates
– M/M/k
– M/M/k/F