The shortest queue problem

40 Pages·2002·0.31 MB·English

by  Ivo Adan

(cid:3)(cid:4)
The shortest queue problem
Ivo Adan
March 19, 2002
(cid:1)(cid:2)
JJJJ JJ NN II IIII 11//4400
(cid:3)(cid:4)
queue 1
m
l
queue 2
join the
m
shortest queue
Where:
• Poisson arrivals with rate λ
• Exponential service times with mean 1/µ
(cid:1)(cid:2)
JJJJ JJ NN II IIII 22//4400
(cid:3)(cid:4)
queue 1
m
l
queue 2
randomly m
assign
Variants:
• Randomly assign customers, with probability 1/2 to either queue
• Randomly assign customers and ‘social servers’: when the queue is
empty, the server will serve a customer in the other queue
(cid:1)(cid:2)
JJJJ JJ NN II IIII 33//4400
(cid:3)(cid:4)
Join the shortest queue
Markov proces with states (i,j) where
i = length of queue 1
j = length of queue 2
Transition rate diagram: (symmetric in the diagonal)
• • • • • • •
j
µ λ
• • • • •λ •
λ µ µ
• • • • •
µ
• • • •
• • •
•λ •
•
i
(cid:1)(cid:2)
JJJJ JJ NN II IIII 44//4400
(cid:3)(cid:4)
Exploit the symmetry by using the state description (m,n) where
m = length of the shortest queue
n = difference between the length of the longest and shortest queue
Transition rate diagram:
n
m
m l m l
l 2m l
m
(cid:1)(cid:2)
JJJJ JJ NN II IIII 55//4400
(cid:3)(cid:4)
Randomly assign with probability 1/2
Markov proces with states (m,n) where
m = length of queue 1
n = length of queue 2
Transition rate diagram:
n
λ/2 λ/2
λ/2 µ
λ/2
µ µ
λ/2 λ/2
λ/2 µ λ/2
m
(cid:1)(cid:2)
JJJJ JJ NN II IIII 66//4400
(cid:3)(cid:4)
Randomly assign and social servers
Markov proces with states (m,n) where
m = length of queue 1
n = length of queue 2
Transition rate diagram:
n
λ/2 λ/2
λ/2 µ
λ/2
2µ µ
λ/2 λ/2
λ/2 2µ λ/2
m
(cid:1)(cid:2)
JJJJ JJ NN II IIII 77//4400
(cid:3)(cid:4)
Equilibrium distribution
Let p(m,n) be the equilibrium probability of state (m,n).
They are the unique normalized solution of the balance equations.
Balance equations for the shortest queue problem (flow out = flow in):
p(m,n)(λ + 2µ) = p(m − 1,n + 1)λ + p(m,n + 1)µ + p(m + 1,n − 1)µ
m > 0,n > 1;
p(0,n)(λ + µ) = p(0,n + 1) + p(1,n − 1)
n > 1;
p(0,1)(λ + µ) = p(0,2) + p(1,0) + p(0,0)λ
p(m,1)(λ + 2µ) = p(m − 1,2)λ + p(m,2)µ + p(m + 1,0)2µ + p(m,0)λ
m > 0;
p(m,0)(λ + 2µ) = p(m − 1,1)λ + p(m,1)µ
m > 0;
p(0,0)λ = p(1,0)µ
(cid:1)(cid:2)
JJJJ JJ NN II IIII 88//4400
(cid:3)(cid:4)
Methods to solve the balance equations:
• Generating function methods
∞ ∞
XX
P(x,y) = p(m,n)xmym
n=0 m=0
Transform the balance equations into a functional equation for P(x,y)
and solve this equation.
• Separation of variables
Try to find a solution in the form of a sum of product-form solutions.
(cid:1)(cid:2)
JJJJ JJ NN II IIII 99//4400
(cid:3)(cid:4)
Sums of product forms as a solution
Is this always possible?
If not, under which conditions?
And how are we going to find these product forms?
Three directions:
1. Exactly one product form as solution
2. Finite sum of product forms as solution
3. Infinite sum of product forms as solution
(cid:1)(cid:2)
JJJJ JJ NN II IIII 1100//4400