1 / 13

M/G/1 queue

Shanghai Jiao Tong University. M/G/1 queue. M/G/1 queue. The M/G/1 queue. G eneral service time distribution - i.i.d (identical independent distribution) Service time is independent from arrival = 1/ μ , average service time : Second moment of service time, <∞ .

tayte
Télécharger la présentation

M/G/1 queue

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Shanghai Jiao Tong University M/G/1 queue

  2. M/G/1 queue The M/G/1 queue • General service time distribution • - i.i.d (identical independent distribution) • Service time is independent from arrival • = 1/μ, average service time • : Second moment of service time, <∞ Poisson Arrival rate λ: Markovian Poisson arrivals M/G/1 General independent service times

  3. Pollaczek-Khinchin (P-K) formula Applying Little’s Theorem W: Average waiting time in queue , Number of customers in queue Average system time: queueing delay + service time Again Little’s Theorem, we get the number of customers in system

  4. M/G/1 examples • Example 1: M/M/1 • Example 2: M/D/1 Deterministic service time: 1/μ • Waiting time of M/D/1 is half of M/M/1, in fact the results of M/G/1 is a lower bound for all M/G/1 with same λ and μ

  5. Proof of P-K formula • Suppose customer i arrives to the system and finds • Ni customers waiting in queue • Up to one customer receiving service • And also define: • Ri: the residual service time seen by i • Wi: the waiting time in queue of customer i Xi Customer i arrives Xi-1 Xi-2 Xi-3 Xi-4 Ni =3

  6. Proof of P-K formula (Cont.) Customer i starts to receive service Customer i arrives Wi Xi Xi-1 Xi-2 Xi-3 Xi-4 Ri Ni =3 By Little’s Theorem, ,

  7. R – The average residual service time R(t) residual service time • Average residual service time is the sum of area of the triangles divided by time t • X3 • X1 • X2 • X4 t • X3 • X4 • X1 • X2 Let M(t) = the number of customers served by time t As is the average departure rage, and is equal to arrival rate λ Hence, The P-K formula is then proved.

  8. R in M/M/1 • For M/M/1, we already know that • So,R for M/M/1 is: • Why it is not equal to 1/μ, given the PASTA property? R(t) residual service time • X3 • X1 • X2 • X4 t • X3 • X4 • X1 • X2

  9. M/G/1 with vacations • Once the system is empty, the server takes a vacation • If system is still empty after the vacation, the server takes another vacation • Useful in analyzing some polling and reservation systems • Vacation times are i.i.d and independent of service times and arrival times • The only impact on analysis is that a customer may enter to find the server on vacation, and must wait until end of that vacation Customer i arrives Wi Xi Xi-1 Xi-2 Xi-3 Vi Ni =3

  10. R(t) with vacations R(t) residual service time • A customer will always experience some residual time, either because server is on vacation, or is serving a customer • X3 • X1 • X2 • X4 t • X3 • X4 • X1 • X2 • V1 • V2 • V3 Let M(t) = the number of customers served by time t L(t) = the number of vacations taken by time t

  11. R(t) with vacations (cont.) • From the server’s point of view, And we have , ? • Hence, And

  12. Example: Slotted M/D/1 system • Fixed slot duration 1/ μ • Service can begin only at start of a slot • If a customer misses the slot start, it must wait until next start (slot in vacation) 1/μ E[X] = E[V] = 1/μ E[X2] = E[V2] = 1/μ2 • This part of time is spent waiting for the slot (syncronizing)

  13. Example. Text problem 3.29

More Related