Queueing Theory (Part 3) - University of...

Queueing Theory-1

Queueing Theory (Part 3)

M/M/s Queueing Systems with Variations (M/M/s, M/M/s//K, M/M/s///N)

Queueing Theory-2

M/M/s Queueing System •  We define

λ = mean arrival rate µ = mean service rate s = number of servers (s > 1) ρ = λ / sµ = utilization ratio

•  We require λ < sµ , that is ρ < 1 in order to have a steady state

Rate Diagram

0 1 2

λ λ λ

µ 2µ 3µ

s-1 s

λ λ λ

(s-1)µ sµ sµ

s+1

λ

sµ

… 3 4

λ λ

4µ 5µ

…

Queueing Theory-3

M/M/s Queueing System Steady-State Probabilities

!!!!

"

!!!!

#

$

++=%%&

'(()

*µ+

=%%&

'(()

*µ+

=µµµ+++

=

,

,

,,

... 2,s 1,s!

s ..., 2, 1,!

......

11

021

nss

nn

C

sn

n

n

nn

nnn

)(

1

/11

!)/(

1

0!)/(

0

µ!"µ!

"

=

µ! +=

# ss

s

nn

snP

and Pn = CnP0

where

Use Birth Death Processes Rate In = Rate Out Coefficients are easy to remember if you think of rate diagram Example: s = 3

0 1 2

λ λ λ

µ 2µ 3µ

3 4

λ λ

3µ 3µ

…

!

C0 =1

C1 ="µ

C2 ="µ"2µ

="µ

#

$ % &

' (

212!

C3 ="µ

#

$ % &

' (

313!

!

C4 ="µ

#

$ % &

' (

41313!

C5 ="µ

#

$ % &

' (

513#

$ % &

' ( 2 13!

Cn ="µ

#

$ % &

' (

n1s#

$ % &

' ( n)s 1s!

Queueing Theory-4

M/M/s Queueing System L, Lq, W, Wq

21

10

20

)()!1(

)1(!)/(

!"µµ"

!=

#"

#µ!=

"

+

ssP

sPL

s

s

s

q

!"

#$%

&''(

)**+

,

--

-

-+=>

----

µ./µ.

0µ.µ

µ

/11

)1(!)/(1)(

)/1(0

se

sPetP

stst

How to find L? W? Wq? Use Lq to find Wq (Lq = λWq):

Wq = Lq/λ Use Wq to find W:

W = Wq + 1/µ Use L=λW to find L in terms of Lq:

L = λW = λ(Wq + 1/µ) = λ(Lq/λ + 1/µ) = Lq + λ/µ

tss

nnq ePtP )1(

1

01)( !µ" ##

#

=

$%

&'(

)#=> *

If s – 1 – λ /µ = 0 then this term is (µt)

Queueing Theory-5

M/M/s Example: A Better ER •  As before, we have

–  Average arrival rate = 1 patient every ½ hour λ = 2 patients per hour

–  Average service time = 20 minutes to treat each patient µ = 3 patients per hour

•  Now we have 2 doctors s = 2

•  Utilization ρ = λ/2µ = 2/6 = 1/3 (Before s=1, ρ=2/3)

Queueing Theory-6

M/M/s Example: ER Questions

In steady state, what is the… 1.  probability that both doctors are idle?

probability that exactly one doctor is idle?

2.  probability that there are n patients?

3.  expected number of patients in the ER?

Queueing Theory-7


In steady state, what is the… 1.  probability that both doctors are idle?

probability that exactly one doctor is idle?

2.  probability that there are n patients?

3.  expected number of patients in the ER?

!

P0 =1

"µ

#

$ % &

' (

0

0!+

"µ

#

$ % &

' (

1

1!+

"µ

#

$ % &

' (

2

2!1

1) *

=1

1+23

+49 + 2

12 3

=12

!

P1 =" µ( )1

1!P0 =

2312

=13

!

Pn =

"µ

#

$ % &

' (

n

n!P0 =

23#

$ % &

' ( n 1n!

12

if 0 ) n < 2

"µ

#

$ % &

' (

n

s!sn*sP0 =

23#

$ % &

' ( n 1

2!12#

$ % &

' ( n*2 1

2=

13#

$ % &

' ( n

if n + 2

,

-

.

.

.

/

.

.

.

!

L = "W = " Lq " +1 µ( ) = Lq + " µ =1 12 + 2 3 = 3 4

Queueing Theory-8


In steady state, what is the… 4.  expected number of patients waiting for a doctor?

5. expected time in the ER?

6.  expected waiting time?

7.  probability that there are at least two patients waiting in queue?

probability that a patient waits more than 30 minutes?

Queueing Theory-9


In steady state, what is the… 4.  expected number of patients waiting for a doctor?

5. expected time in the ER? W = L/λ = (3/4)/2 = 3/8 hour ≈ 22.5 minutes

6.  expected waiting time? Wq = Lq/λ = (1/12)/2 = 1/24 hour ≈ 2.5 minutes

7.  probability that there are at least two patients waiting in queue?

P(≥ 4 patients in system) = 1 – P0 – P1 – P2 – P3 = 1 – ½ - 1/3 – 1/9 – 1/27 ≈ 0.0185

8.  probability that a patient waits more than 30 minutes?

!

Lq =P0 " µ( )s#s! 1$ #( )s

=1 2( ) 2 /3( )2 1 3( )2! 2 /3( )2

=112

!

P "q > t( ) = 1# P0 # P1( )e#2µ 1#$( ) t = 1# 12#13

%

& '

(

) * e#2(3) 2 3( )t =

16e#4 t

P "q > 30min( ) = P "q >12hour

%

& '

(

) * + 0.022

Queueing Theory-10

Performance Measurements

s = 1 s = 2

ρ 2/3 1/3

L 2 3/4

Lq 4/3 1/12

W 1 hr 3/8 hr

Wq 2/3 hr 1/24 hr

P(at least two patients waiting in queue)

0.296 0.0185

P(a patient waits more than 30 minutes)

0.404 0.022

Queueing Theory-11

Travel Agency Example •  Suppose customers arrive at a travel agency according to a

Poisson input process and service times have an exponential distribution

•  We are given –  λ= 0.10/minute, that is, 1 customer every 10 minutes –  µ =0.08/minute, that is, 8 customers every 100 minutes

•  If there was only one server, what would happen? λ/µ > 1 Customers would balk at long lines – never reach steady state - lose customers - go out of business?

•  How many servers would you recommend? Calculate P0, Lq and Wq for s=2, s=3, and s=4

Queueing Theory-12

P(ω>t) =

P(ωq>t) =

Queueing Theory-13

P(ω>t) =

P(ωq>t) =

Queueing Theory-14

P(ω>t) =

P(ωq>t) =

Queueing Theory-15

Single Queue vs. Multiple Queues •  Would you ever want to keep separate queues for separate

servers?

Single queue

Multiple queues

vs.

Queueing Theory-16

Bank Example •  Suppose we have two tellers at a bank •  Compare the single server and multiple server models •  Assume λ = 2, µ = 3,

L Lq W Wq P0 ρ

0.75 0.083 0.375 0.042 0.5 λ/2µ =1/3

1.0 0.334 0.5 0.167 0.4449

λ`/µ =(λ/µ)/3

=1/3

Queueing Theory-17

Bank Example Continued

•  Suppose we now have 3 tellers •  Again, compare the two models

M/M/3 Three M/M/1 queues λ=2, µ=3 λ’ = λ/3 = 2/3, µ=3 ρ=λ/(sµ) = 2/9 M/M/1: ρ=λ’/3 = 2/9 ρ is the same L= 0.676 L=0.286 3L = 0.858 Lq = 0.009 Lq=0.063 3Lq = 0.189 W = 0.338 W = 0.429 Wq = 0.005 Wq = 0.095 P0 = 0.5122 P0 = 0.7778 (P0)3 = 0.47

Queueing Theory-18

M/M/s//K Queueing Model (Finite Queue Variation of M/M/s)

•  Now suppose the system has a maximum capacity, K •  We will still consider s servers •  Assuming s ≤ K, the maximum queue capacity is K – s •  Some applications for this model:

Trunk lines for phone – call center Warehouse with limited storage Parking garage

•  Draw the rate diagram for this problem:

0 1

λ λ

µ 2µ

… s-1 s s+1

λ λ λ

sµ sµ sµ

K

0 λ

(s-1)µ

λ

sµ

…

Queueing Theory-19


Balance equations: Rate In = Rate Out State 0: µP1 = λP0

State 1: λP0 + 2µP2 = (λ+µ)P1

State 2: λP1 + 3µP3 = (λ+2µ)P2

State 3: λP2 + 3µP4 = (λ+3µ)P3

State K-1: λPK-2 + 3µPK = (λ+3µ)PK-1

State K: λPK-1 = 3µPK

0 1 2 3=s 4

λ λ λ

2µ 3µ 3µ

λ

µ

K

0 λ

3µ

… λ

3µ

…

!

C0 =1

C1 ="µ

C2 ="2

2µ2

C3 ="3

3!µ3 (s = 3)

C4 =1

3!#3$

% &

'

( ) "µ

$

% & '

( )

4

!

Cn =1

3!#3 n*s( )

$

% &

'

( ) "µ

$

% & '

( )

n

for s + n + K

CK +1 = 0

P0 =1

Cnn=0

K

!

Pn =CnP0

Queueing Theory-20


Solving the balance equations, we get the following steady state probabilities:

Kn

nP

ss

nP

n

P nsn

n

n

n

n

>

+=µ

!

=µ

!

="

0

K ..., 1,s s, for!

s ..., 2, 1, for!

0

0

Verify that these equations match those given in the text for the single server case (M/M/1//K)

!

P0 =1

1+ (" /µ )n

n!n=1

s

# + (" /µ )s

s!"sµ( )

n$s

n= s+1

K

#

Queueing Theory-21


!!"

#$$%

&'++=

µ(=))')'')')'

)µ(=

**'

=

'

=

''

1

0

1

0

20

1

/ where)],1()(1[)1(!)/(

s

nnq

s

nn

sKsKs

q

PsLnPL

ssKsPL

WL != qq WL != )1(0

Kn

nn PP !"="=" #$

=

To find W and Wq:

Although L ≠ λW and Lq ≠ λWq because λn is not equal for all n,

and where

Also, because there is a finite number of states, the steady state equations do hold, even if ρ>1

Queueing Theory-22

M/M/s///N Queueing Model (Finite Calling Population Variation of M/M/s)

•  Now suppose the calling population is finite, N •  We will still consider s servers •  Assuming s ≤ N, the maximum number in the queue capacity is

N – s, so K ≥ N does not affect anything If N is the entire population, then the maximum number in system is N. Assume N ≤ K and s ≤ N

•  Application for this model: Machine replacement

•  Draw the rate diagram for this problem:

0 1

Nλ (N-1)λ

µ 2µ

… s s+1

(N-s)λ (N-(s+1))λ

sµ sµ

N

0 (N-(s-1))λ

sµ

λ

sµ

…

Queueing Theory-23

M/M/s///N Queueing Model (Finite Calling Population Variation of M/M/s)

Queueing Theory-23

Balance equations: Rate In = Rate Out State 0: µP1 = λP0 è P1 = (Nλ/µ)P0 State 1: NλP0 + 2µP2 = ((N-1)λ+µ)P1 è P2 = (1/2)(Nλ/µ) ((N-1)λ/µ)P0

0 1 2 3 4

(N-1)λ (N-2)λ (N-3)λ

2µ 3µ 3µ

Nλ

µ

N

0 λ

3µ

… (N-4)λ

3µ

…

!

C0 =1

C1 = N "µ

#

$ % &

' (

C2 =N N )1( )2

"µ

#

$ % &

' (

2

C3 =N N )1( ) N ) 2( )

3!"µ

#

$ % &

' (

3

Queueing Theory-24

M/M/s///N Results

! !"

= =" #

$%

&'(µ)

"+#

$%

&'(µ)

"

=1

0

0

!)!(!

!)!(!

1s

n

N

sn

n

sn

n

ssnNN

nnNN

P

!!!

"

!!!

#

$

>

%%&&'

())*

+

µ

,

-

=&&'

())*

+

µ

,

-

=-

Nn for0

Nn for!)!(

!

,...,1,0 for!)!(

!

0

0

sPssnN

N

snPnnN

N

Pn

sn

n

n

!=

"=N

snnq PsnL )( !!

"

#$$%

&'++= ((

'

=

'

=

1

0

1

01

s

nnq

s

nn PsLnPL

Queueing Theory (Part 3) - University of...

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