Heavy traffic approximation

In queueing theory, a discipline within the mathematical theory of probability, a heavy traffic approximation (sometimes heavy traffic limit theorem ^[1] or diffusion approximation) is the matching of a queueing model with a diffusion process under some limiting conditions on the model's parameters. The first such result was published by John Kingman who showed that when the utilisation parameter of an M/M/1 queue is near 1 a scaled version of the queue length process can be accurately approximated by a reflected Brownian motion. ^[2]

Heavy traffic condition

Heavy traffic approximations are typically stated for the process X(t) describing the number of customers in the system at time t. They are arrived at by considering the model under the limiting values of some model parameters and therefore for the result to be finite the model must be rescaled by a factor n, denoted ^{[3] : 490}

${\displaystyle {\hat {X}}_{n}(t)={\frac {X(nt)-\mathbb {E} (X(nt))}{\sqrt {n}}}}$

and the limit of this process is considered as n → ∞.

There are three classes of regime under which such approximations are generally considered.

1. The number of servers is fixed and the traffic intensity (utilization) is increased to 1 (from below). The queue length approximation is a reflected Brownian motion. ^{[4] [5] [6]}
2. Traffic intensity is fixed and the number of servers and arrival rate are increased to infinity. Here the queue length limit converges to the normal distribution. ^{[7] [8] [9]}
3. A quantity β is fixed where

${\displaystyle \beta =(1-\rho ){\sqrt {s}}}$

with ρ representing the traffic intensity and s the number of servers. Traffic intensity and the number of servers are increased to infinity and the limiting process is a hybrid of the above results. This case, first published by Halfin and Whitt is often known as the Halfin–Whitt regime ^{[1] [10] [11]} or quality-and-efficiency-driven (QED) regime. ^[12]

Results for a G/G/1 queue

Theorem 1. ^[13] Consider a sequence of G/G/1 queues indexed by ${\displaystyle j}$.
For queue ${\displaystyle j}$ let ${\displaystyle T_{j}}$ denote the random inter-arrival time, ${\displaystyle S_{j}}$ denote the random service time; let ${\displaystyle \rho _{j}={\frac {\lambda _{j}}{\mu _{j}}}}$ denote the traffic intensity with ${\displaystyle {\frac {1}{\lambda _{j}}}=E(T_{j})}$ and ${\displaystyle {\frac {1}{\mu _{j}}}=E(S_{j})}$; let ${\displaystyle W_{q,j}}$ denote the waiting time in queue for a customer in steady state; Let ${\displaystyle \alpha _{j}=-E[S_{j}-T_{j}]}$ and ${\displaystyle \beta _{j}^{2}=\operatorname {var} [S_{j}-T_{j}];}$

Suppose that ${\displaystyle T_{j}{\xrightarrow {d}}T}$, ${\displaystyle S_{j}{\xrightarrow {d}}S}$, and ${\displaystyle \rho _{j}\rightarrow 1}$. then

${\displaystyle {\frac {2\alpha _{j}}{\beta _{j}^{2}}}W_{q,j}{\xrightarrow {d}}\exp(1)}$

provided that:

(a) ${\displaystyle \operatorname {Var} [S-T]>0}$

(b) for some ${\displaystyle \delta >0}$, ${\displaystyle E[S_{j}^{2+\delta }]}$ and ${\displaystyle E[T_{j}^{2+\delta }]}$ are both less than some constant ${\displaystyle C}$ for all ${\displaystyle j}$.

Heuristic argument

• Waiting time in queue

Let ${\displaystyle U^{(n)}=S^{(n)}-T^{(n)}}$ be the difference between the nth service time and the nth inter-arrival time; Let ${\displaystyle W_{q}^{(n)}}$ be the waiting time in queue of the nth customer;

Then by definition:

${\displaystyle W_{q}^{(n)}=\max(W_{q}^{(n-1)}+U^{(n-1)},0)}$

After recursive calculation, we have:

${\displaystyle W_{q}^{(n)}=\max(U^{(1)}+\cdots +U^{(n-1)},U^{(2)}+\cdots +U^{(n-1)},\ldots ,U^{(n-1)},0)}$

• Random walk

Let ${\displaystyle P^{(k)}=\sum _{i=1}^{k}U^{(n-i)}}$, with ${\displaystyle U^{(i)}}$ are i.i.d; Define ${\displaystyle \alpha =-E[U^{(i)}]}$ and ${\displaystyle \beta ^{2}=\operatorname {var} [U^{(i)}]}$;

Then we have

${\displaystyle E[P^{(k)}]=-k\alpha }$

${\displaystyle \operatorname {var} [P^{(k)}]=k\beta ^{2}}$

${\displaystyle W_{q}^{(n)}=\max _{n-1\geq k\geq 0}P^{(k)};}$

we get ${\displaystyle W_{q}^{(\infty )}=\sup _{k\geq 0}P^{(k)}}$ by taking limit over ${\displaystyle n}$.

Thus the waiting time in queue of the nth customer ${\displaystyle W_{q}^{(n)}}$ is the supremum of a random walk with a negative drift.

• Brownian motion approximation

Random walk can be approximated by a Brownian motion when the jump sizes approach 0 and the times between the jump approach 0.

We have ${\displaystyle P^{(0)}=0}$ and ${\displaystyle P^{(k)}}$ has independent and stationary increments. When the traffic intensity ${\displaystyle \rho }$ approaches 1 and ${\displaystyle k}$ tends to ${\displaystyle \infty }$, we have ${\displaystyle P^{(t)}\ \sim \ \mathbb {N} (-\alpha t,\beta ^{2}t)}$ after replaced ${\displaystyle k}$ with continuous value ${\displaystyle t}$ according to functional central limit theorem. ^{[14] : 110} Thus the waiting time in queue of the ${\displaystyle n}$th customer can be approximated by the supremum of a Brownian motion with a negative drift.

• Supremum of Brownian motion

Theorem 2. ^{[15] : 130} Let ${\displaystyle X}$ be a Brownian motion with drift ${\displaystyle \mu }$ and standard deviation ${\displaystyle \sigma }$ starting at the origin, and let ${\displaystyle M_{t}=\sup _{0\leq s\leq t}X(s)}$

if ${\displaystyle \mu \leq 0}$

${\displaystyle \lim _{t\rightarrow \infty }P(M_{t}>x)=\exp(2\mu x/\sigma ^{2}),x\geq 0;}$

otherwise

${\displaystyle \lim _{t\rightarrow \infty }P(M_{t}\geq x)=1,x\geq 0.}$

Conclusion

${\displaystyle W_{q}^{(\infty )}\thicksim \exp \left({\frac {2\alpha }{\beta ^{2}}}\right)}$ under heavy traffic condition

Thus, the heavy traffic limit theorem (Theorem 1) is heuristically argued. Formal proofs usually follow a different approach which involve characteristic functions. ^{[4] [16]}

Example

Consider an M/G/1 queue with arrival rate ${\displaystyle \lambda }$, the mean of the service time ${\displaystyle E[S]={\frac {1}{\mu }}}$, and the variance of the service time ${\displaystyle \operatorname {var} [S]=\sigma _{B}^{2}}$. What is average waiting time in queue in the steady state?

The exact average waiting time in queue in steady state is given by:

${\displaystyle W_{q}={\frac {\rho ^{2}+\lambda ^{2}\sigma _{B}^{2}}{2\lambda (1-\rho )}}}$

The corresponding heavy traffic approximation:

${\displaystyle W_{q}^{(H)}={\frac {\lambda ({\frac {1}{\lambda ^{2}}}+\sigma _{B}^{2})}{2(1-\rho )}}.}$

The relative error of the heavy traffic approximation:

${\displaystyle {\frac {W_{q}^{(H)}-W_{q}}{W_{q}}}={\frac {1-\rho ^{2}}{\rho ^{2}+\lambda ^{2}\sigma _{B}^{2}}}}$

Thus when ${\displaystyle \rho \rightarrow 1}$, we have :

${\displaystyle {\frac {W_{q}^{(H)}-W_{q}}{W_{q}}}\rightarrow 0.}$

External links

• The G/G/1 queue by Sergey Foss

References

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