First Come First Served Discipline 1 3 2 - Isaac Newton …€¦ · · 2014-05-31First Come First Served Discipline Gideon Weiss University of Haifa Newton Institute, SCS program,

Queues with Multi-Type Jobs and Multi-Type Servers

under

First Come First Served Discipline

Gideon Weiss

University of Haifa

Newton Institute, SCS program, 2010

joint work and work by:

Ivo Adan, Rene Caldentey, Cor Hurkens,

Ed Kaplan, Rishi Talreja, Jeremy Visschers, Ward Whitt

Gideon Weiss, University of Haifa, Infinite Matching, ©2008 1

Multi-type queues:

Customer types i=a,b,c, . . . S(i) servers of i

Server types j=1,2,. . .,K C(j) customers of j

Bipartite system graph

• Loss systems: Servers m1, m2, . . ., mK (Adan Hurkens W, ‘10)

Intenet traffic, switches, wireless,

• Stable FCFS queues: (Visschers ‘00, Visschers Adan W, ‘10)

Manufacturing

• Overloaded system with Reneging: : Fa, Fb, Fc, . . . (Talreja Whitt ‘07)

Call centers with skill based routing

• Infinite matching (Kaplan ‘88, Caldentey Kaplan ‘02, Caldentey Kaplan W ‘09),

Public housing, adoptions, kidney transplants

C ab c

1 3 2


Multi-type queues:

• Loss systems: Servers m1, m2, . . ., mK

• Stable FCFS queues:

• Overloaded system with Reneging: : Fa, Fb, Fc, . . .

• Infinite matching

m1 m3 m2

a c

m1 m2

ccb/c b/cb/c a/ca/b

m3

aa/b/ca/b/ca/b/c

b c a a b

3 1 2 2


Overloaded system with reneging (Talreja Whitt, ManSci’08)

Overloaded systems, arrival rates ! "i > ! µj

general arrival streams, general service times,

reneging patience distributions Fi

Uniform acceleration regime:

servers are busy almost all the time

queues force almost all customers to wait until almost reneging

on a fluid scale GLOBAL FCFS is achieved

- everyone waits approximately W.

Question: How many type i are served by type j rij

equivalently: what is the rate at which of i by j service #ij

!1 !2 !3

µ1µ2 µ3

!iFi (W )" = µ j"


Service rates in overloaded system

!iFi (W )" = µ j"

Question: How many type i are served by type j rij

equivalently: what is the rate at which of i by j service #ij

Solve for W from:

Then solve

!iji"C( j)# = µ j , j = 1,…, J

!ijj"S(i)# = $iFi (W ), i = 1,…, I

These may have no positive solution, unique solution of many solutions

Solved for trees, complete graphs, hybrids

1 2 3

1 2 3

1 2 3

1 2 3

4

4

Non-

Unique


Infinite matching (Kaplan ‘84, Caldentey Kaplan ‘02,Caldentey, Kaplan, W ‘09)

“. . . Households applying for public housing are allowed to specify those housing

projects in which they are willing to live; when a public housing unit becomes newly

available, of those households willing to live in the associated housing project, the

one that has been waiting the longest is o!ered the unit. . . .”

This is not a customer server situation ---

Housing units leave with the households

Other examples:

Adoptions,

Kidney Transplants,

Auctions ?

In assigning housing to entitled families several housing projects are available

(Kaplan, ‘84.’88).

b c a a b

3 1 2 2


FCFS Infinite matching

Customer types i=1,2,. . .,I S(i) servers of i

Server types j=1,2,. . .,J C(j) customers of j

System graph G

Customer types and server types are i.i.d from $, %

Matching is FCFS

1 2 3

1 2

C

S

!1 !2 !3

"1 "2

c1

c2

c3

cn!3

cn!2

cn!1

cn

s1

s2

s3

sn!3

sn!2

sn!1

sn

Fraction of i->j matches in first n :

Question: when does

what is

ri, jn

ri, jn a.s.! "! ri, j

ri, j


Existence of rates:

Balance equations: Matching rates must satisfy:

ri, jj!S(i)

" =#i , ri, ji!C( j)

" = $ j ,

Necessary condition: Nonnegative solution exists iff

for every subset C and every subset S

!(C) " #(S(C)), #(S) "!(C(S)), *

!

!

C S

a b

i

j

!"i

# j

Proof

max-flow / min-cut

But: solution may be non-unique


Markov chain

Consider the matching for s1 :

he will take the first cm in c1 ,c2,. . . that matches him,

leave a geometric number of unmatched c from C(s1)

Xn

i = 1,…, I

Define: the Markov chain of ordered unmatched customers

state space is words in alphabet of

Consider then sn+1 he will first look at those c left by s1 . . . sn

If he finds a match he will take the first, and leave one less unmatched behind.

Else he will look for a match among those not considered previously,

until he finds a match, adding a geometric number of c from C(sn+1) .

!(C) < "(S(C)), "(S) <!(C(S)), **

Theorem: If the Markov chains are ergodic then ri, jn

n!" a.s.# !### ri, j

Conjecture: The Markov chain X is ergodic iff for all non-trivial S,C:


Almost complete system graph

Customer types i=0,1, . . ., I

Server types j=0,1, . . .,I

Type 0 match everything,

Type i matches all except i

!X(ekx) = "k

#k(1$ "k

%&'

()*

x

1+#i"i

1$#i $ "ii=1

I

+%

&'(

)*

1 2 3

1 2

C

S

!1 !2 !3

"1 "2

3

"3

Z!I

The unmatched of the X process

will always be customers of a single type

X process moves on the integer axes in I dimensions

Ergodic iff

ek the k unit vector:

ri, j =!i" j (1#!i )(1# " j )#! j"i$% &'(1#!i # "i )(1#! j # " j )

1+!i"i

1#!i # "ii=1

I

()

*+,

-.

1!"i ! #i


`NN’ model states and transitions

!3 <"1, "3 < !1, "1 + !1 <1Necessary condition for ergodicity:

111

111

11

11

1

10

3

3

33

33

333

333

3333

3333

33333

33333

3

2

33

32

333

332

3333

3332

33333

33332

33

2x

333

32x

3333

332x33333

3332x

333

2xx

3333

32xx

33333

332xx

2

3

32

33

332

333

3332

3333

33332

33333

2x

33

32x

333

332x

3333

3332x

33333

333333

3332xx

3333333

33332xx

33333333

333332xx

333333

33332x

3333333

333332x

333333

333332

33332x

333333

333332x

3333333

333332

333333

!3"3

!1"1 !1"1

!1"1

!3"3

!1"1

(1#!1)(1# "1 )

(1#! 3)(1 # "1 )

!3"2

!2"3

!3 (1 #"1)

(1#! 3)"1 !2"1

!1 (1 #"1)

!3 (1 #"1)

!2"1!2"3

!1"2!3"2

!2 (1 # "1 )

!"2

1# !"2

333333

333333

!2"1 1# !"21

!"2 ="2

"2 + "3

1 2

23

C

S

!1

3

1

!2 !3

"3 "2 "1

State space and transitions:


A Lyapunov function of Fayolle, Malyshev and Menshkov.

To show positive recurrence we need to show that

E(!f (x, y)) " #h < 0

0 5 10 15 20

-10

-5

0

5

10

We were able to constructthe right f (choice of u,v,w)


1st attempt to solve: Caldentey-Kaplan ‘02 Algorithm

Let indicator that customer type i matches server type j,

is the incidence matrix of the bipartite system graph

Let

AijA I ! J

Li, j =!i Ai, j

!(C( j)), Mi, j =

" jAi, j

"(S(i))

For the complete graph matching rates are Fi, jall

=!i" j

Start with Fi, j0= Gi, j

0= Fi, j

allAi, j

It does not satisfy the balance equations,

fi0=!i " Fi, j

0

j

# row shortfalls are:

g j0= ! j " Fi, j

0

i

#column shortfalls are:

f!1

= 0,

g!1

= 0

fk+1

= gkLT

gk+1

= fkM

Fi, jk+1

= Fi, jk+ (!1)

k!1fik!1

Mi, j + (!1)kg jkLi, j

Converges to a solution of balance equations: Fi, jk

k!"# !## Fi, j


2nd attempt to solve: Quasi independent cells

Kaplan considered a hydraulic analogue: pipes connect customers and server types,

with as pressures. Using a limiting case non-Newtonian `power law’ fluid,

reduced to the model of `quasi independence’ of contingency tables.

!,"

In two dimensional contingency tables one often considers the hypothesis of independence of the 2 factors

If and are the row and column marginal frequencies, cell (i,j) should have frequency !i ! j !i" j

Often tables have missing cells, because of missing observations, forbidden or impossible combinations

Aijxiy j =!i ,j

" Aijxiy j = # j ,i

"

One postulates the existence of abstract full table marginals, and obtains quasi-independent model:

Then: solves the balance equations hi, j = xiy j


Attempts failed - examplesExample 1: the almost complete 3 type graph - Caldentey Kaplan works, Quasi-independent fails

1 2 3

1 2

C

S

!1 !2 !3

"1 "2

3

"3

Example 2: one more customer type - Quasi-independent seems to work, Caldentey Kaplan fails

1 2 3

1 2

C

S

!0 !2 !3

"1 "2

3

"3

!1

0


Back to Queueing - Exact asymptotics for the `N’ model

Adan, Foley, and Macdonnald, ‘08 analyze the queueing model for the `N’ system graph:

Arrivals and processing are memoryless,

Service is FCFS.

If system is empty, and type a arrives, machine 1 will take him w.p

m2

jobs

machines

!a

m1

!b

µ2µ1

a b

"1#"

!

States:

(0,0) -empty

(1,0) -one job, machine 2 working on it

(0,y) - machine 1 working, machine 2 idle y-1 type b jobs waiting

(x,y) - both machines working, y-1 type b waiting head of line,

x-1 additional unspecified waiting

Results: steady state exact asymptotics, for large (x,y)

with explicit values of the constants for various

combinations of parameter values and various directions

!(x, y) ~ c(x / y)"a + "bµ1 + µ2

#$%

&'(

x"bµ1

#$%

&'(

y1 2

12

C

S

! 1"!

1" # #


Procduct form solution for the `N’ queueing system

m2

jobs

machines

!a

m1

!b

µ2µ1

a b

"1#"

When this has a product form solution:

µ1

n

m

(n,m2 ,m,m1)!a + !b

(m,m1)

µ1

!a

!b

µ2p0

µ2p1

µ2p2

µ2p3

µ2"4

µ1

(n,m1,0,m2 )

!a + !b

µ2p0

µ2p1

µ2p2

µ2p3

(0,m1)

(0,m2 )

(0)

µ1

(n,m2 ,0,m1)

!a + !b

µ2p0

µ2p1

µ2p2

µ2p3

µ2"4

(0,m1,0,m2 ) (0,m2 ,0,m1)

µ2

µ1

!a# + !b!a (1$#)

!a + !b µ2µ1µ1

!b!a

µ2

! =!* ="a

2"a + "b


A reversible multi-type loss system (Adan, Hurkens, W ‘10)

Servers j=1, . . .,J working at rates µj exponential

Customers of types i=1,. . . I arriving at rates "i Poisson

Graph G : C(j) customer types served by server j

This is a loss system:

State: set of busy machines {M1, M2,. . ., Mk}

Completions: {M1, M2,. . ., Mk}--> {M1, M2,. . ., Mk-1} rate µMk

Arrivals: {M1, M2,. . ., Mk-1} --> {M1, M2,. . ., Mk} rate "Mk{M1, M2,. . ., Mk-1}

Assume MC reversible. Detailed balance:

!Mk{M1,…,Mk"1} = !iP(i

i#C(Mk )

$ chooses server Mk | {M1,…,Mk"1} busy)

!{M1,…,Mk}µMk

= !{M1,…,Mk"1}#Mk{M1,…,Mk"1}

Solution:

!{M1,…,Mk}µMk

= !{"}

!M1{"}!M2

{M1}!!Mk{M1,…,Mk#1}

µM1 µM2 ! µMk


Assignment condition

The solution:

!{M1,…,Mk}µMk

= !{"}

!M1{"}!M2

{M1}!!Mk{M1,…,Mk#1}

µM1 µM2 ! µMk

Makes sense only if for all permutations:

!M1

{"}!M2{M1}!!Mk

{M1,…,Mk#1} = !M1{"}!

M2{M1}!!

Mk{M1,…,Mk#1}

!M (S) = !ii"C(S)# 1+

! j (S$M )

!M (S$ j)j%S#

&

'()

*+

This uniquely determines

Question: Can you find

!M (S) = P(ii"C(M )

# , M | S)

Answer: Yes, always, maxflow of does it

This maxflow exists by monotonicity

!M (R) " !M (S), M #R $ S

a b

i

M

SC(S)

!i

!M (S)

"

P(i,M | S)

!ii"C(S)

#


Product form for multi-type FCFS (Visschers ‘02,Visschers, Adan, W ‘10)

Servers j=1, . . .,J working at rates µj exponential

Customers of types i=1,. . . I arriving at rates "i Poisson

Graph G : C(j) customer types served by server j

FCFS service, random assignment to idle servers.

Thm: if assignment is as in loss system, then MC obeys partial balance, and has

product form stationary distribution

! s( ) = ! 0( )"M1

{#}"M2{M1}!"Mk

{M1,…,Mk$1}

µM1 (µM1 +µM2 ) !(µM1 +!+ µMk)%1n1%2

n2!%

k

nk

%l ="U (M1,…,Ml )

µM1 +!+ µMl

M1M2Mk

n1

! ! !

nk

U (M1)U (M1,M2) U (M1,…,Mk )

n2


Conclusion and future work

(1) Loss system may be insensitive

(2) In the product form model, when r-->1, we approach the !"matching model

(3) r-->1 in product form solution may solve the !"matching model

(4) In the overloaded with reneging, when r-->1, we approach the !"matching

model

(5) Solution of !"matching model may solve the overload reneging model

(6) The !"matching model is independent of processing and arrival distributions

Documents

First Come First Served Discipline 1 3 2 - Isaac Newton …€¦ · · 2014-05-31First Come First Served Discipline Gideon Weiss University of Haifa Newton Institute, SCS program,