Stochastic processing networks: steady-state 23 diffusion approximations Jim Dai School of Operations Research and Information Engineering Cornell University and Institute for Data and Decision Analytics (iDDA) Chinese University of Hong Kong, Shenzhen May 14, 2018 Jim Dai 1 / 37
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Stochastic processing networks: steady-state 23 diffusion ...Stochastic processing networks: steady-state 23 diffusion approximations Jim Dai School of Operations Research and Information
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Assuming (preemptive-resume) LBFS policy, one has the Markovianrepresentation: X = {X (t)}t≥0 is Markov process, where state at time t is
X (t) = (L(t),Re(t),Rs(t)).
Li (t) = number of customers in buffer i at time t (including the one inservice)Re(t) = residual time until next exogenous arrivalRs,i (t) = residual time until next service completion in buffer iThe reentrant in the figure has a 7-dimensional representation:(
L1(t),Re(t),Rs,1(t)),(L2(t),Rs,2(t)
),(L3(t),Rs,3(t)
)
Jim Dai 11 / 37
A sample result
When λ = 1− r < 1, Markov process X is positive Harris recurrent (Dai1995): as t →∞,
Lr (t) =⇒ Lr (∞) = (Lr1(∞), Lr
2(∞), Lr3(∞)).
Theorem 1
rLr (∞) =⇒(
L∗1(∞), L∗2(∞), 0)
as r → 0.
rLr3(∞)⇒ 0, state space collapse (SSC).(
rLr1(∞), rLr
2(∞))
=⇒(
L∗1(∞), L∗2(∞)).
Jim Dai 12 / 37
State space collapse
The pre-limit is a K -dimensional problem, and the limit is a d-dimensionalproblem, where K is the number of buffers and d is the number of stations.
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Jim Dai 13 / 37
Strong SSC
In proving a version of Theorem 1 for general multiclass queueing networksunder priority policies, we need the following strong SSC:
rEν [Lr3(∞)]→ 0.
Theorem 2Assume interarrival and service times have phase-type distributions. Assume thatthe critically loaded fluid model has SSC. Then
where µi = 1/mi is service rate for class i jobs.H = {3}. When L3(t) > 0, T3(t) = 1, which implies
L3(t) = µ2T2(t)− µ3 ≤ µ2 − µ3 < 0.
L3(t) = 0 for t ≥ L3(0)/(µ3 − µ2).Jim Dai 15 / 37
LBFS priority policy?
- -
-
-
-
-
m1 m2
m3 m4
m5
Station 1 Station 2α1
Under the LBFS-FBFS priority policy and λ(m2 + m5) > 1,
Time0 4000 8000 12000
Job
coun
t
0
1000
2000
3000
4000Station 1Station 2
Jim Dai 16 / 37
Miyazawa’s mgf approach
For a family of nonnegative random vector L(r)(∞) ∈ Rd+,
L(r)(∞)⇒ L(∞) for some random vector L(∞)
if and only if the mgf converges
φ(r)(θ) = E[e〈θ,L(r)(∞)〉]→ E[e〈θ,L(∞)〉] for all θ ≤ 0.
The family {L(r)(∞)} is tight iff for any sequence rn → 0,
limn→∞
φ(rn)(θ)→ φ(θ) ∀θ ≤ 0 implies that
φ(0−) = φ(0−, . . . , 0−) = 1. (1)
Equation (1) says φ(·) is left continuous at 0.
Jim Dai 17 / 37
GI/GI/1 queue
S1B1 departuresarrivals
{Te(i), i ≥ 1} iid interarrival times; λ = 1/E[Te(i)];{Ts(i), i ≥ 1} iid service times; µ = 1/E[Ts(i)].Heavy traffic condition
λ = µ− r with r ↓ 0.
X = {X (t), t ≥ 0} is a Markov process, where
X (t) = (L(t),Re(t),Rs(t)),
whereL(t) is the number of customers in system,Re(t) is remaining interarrival times,Rs(t) is the remaining service times.
Jim Dai 18 / 37
Piecewise deterministic Markov process (PDMP)
The process X = (L,Re ,Rs) is a piecewise deterministic Markov process(PDMP); Davis (1981)A sample path of a PDMP is composed of two parts, deterministic andcontinuous sections and (random) jumps due to expiration of remainingtimes.
0t
Te(1) Te(2) Te(3) Te(4)
Ts(4)
Ts(3)
Ts(2)
Te(5)
Ts(1)
L(t)
Re(t)Rs(t)
Figure: A sample path of remaining times for GI/G/1 queue
Jim Dai 19 / 37
Change of variables for PDMP
Consider function f (x) = f (z , u, v) : Z+ × R+ × R+ → R.Define the jump size ∆f (X (s)) = f (X (s))− f (X (s−)),
f (X (t))− f (X (0)) =∫ t
0
dds f (X (s)) ds +
∑si∈(0,t]
∆f (X (si ))
= −∫ t
0
[ ∂∂u f (X (s))− 1{Z (s) > 0} ∂
∂v f (X (s))]ds (continuous)
+[∫ t
0∆f (X (s))dNA(s) +
∫ t
0∆f (X (s))dND(s)
](jumps)
NA – arrival process.ND – departure process.
Jim Dai 20 / 37
Full BAR in GI/GI/1 setting
Assume X has stationary distribution ν.
Basic Adjoint Relationship (BAR)
0 =tEν[− ∂
∂u f (X (0))− 1{Z (0) > 0} ∂∂v f (X (0)
]+ (continuous)
+Eν
[∫ t
0∆f (X (s))dNA(s) +
∫ t
0∆f (X (s))dND(s)
](jumps)
Jump terms are intractable; getting rid of arrival-jump term requires
ETe
[f (L + 1,Te ,Rs)− f (L, 0,Rs)
]= 0 (2)
for any given L and Rs .Similarly,
ETs
[f (L− 1,Re ,Ts)− f (L,Re , 0)
]= 0 (3)
for any given L ≥ 1 and Re .Jim Dai 21 / 37
Exponential functions (I)
Fix a θ ≤ 0, take
f (θ; z , u, v) = eθz+a(θ)u+bθv .
Then
E[f (z + 1,Te , v)− f (z , 0, v)] = 0,Eeθ(z+1)+a(θ)Te+b(θ)v = eθz+b(θ)v ,
E[eθ+a(θ)Te
]= 1.
For each θ ≤ 0, find a = a(θ) such that
E[ea(θ)Te
]= e−θ. (4)
Jim Dai 22 / 37
Exponential functions (II)
Similarly, for each θ ≤ 0, find b(θ) such that
E[eb(θ)Ts ] = eθ. (5)
Then
E[f (z − 1, u,Ts)− f (z , u, 0)] = 0 for z ≥ 1.
Jim Dai 23 / 37
Exponential functions (III): Summary
Define a(θ) and b(θ) via (4) and (5). Set
f θ; z , u, v) = eθz+a(θ)u+b(θ)v .
Then,
∂
∂u f (θ; z , u, v) = a(θ)f (θ; z , u, v),
∂
∂v f (θ; z , u, v) = b(θ)f (θ; z , u, v).
Thus, BAR becomes
Eν[− a(θ)f (θ; X (0))− b(θ)1{L(0)>0}f (θ; X (0))
]= 0,
or equivalently
[a(θ) + b(θ)]Eν [f (θ; X (0))]− b(θ)Eν[1{L(0)=0}f (θ; X (0))
]= 0. (6)
Jim Dai 24 / 37
Prelimit (restricted) BAR
From (6), for θ ≤ 0,
[a(θ) + b(θ)]Eν [f (θ; X (0))]− b(θ)Pν(L(0) = 0)Eν[f (θ; X (0))|L(0) = 0
]= 0.
[a(θ) + b(θ)]Eν [f (θ; X (0))]− b(θ)(1− λ/µ)Eν[f (θ; X (0))|L(0) = 0
]= 0.
Scaling: changing θ to rθ for any θ ≤ 0 to get pre-limit BAR,