《土木与环境工程》（英文版） Spatially Distributed Queues II

Spatially Distributed QueuesⅢ M/G/1 2 Servers n servers approximations

Spatially Distributed Queues II M/G/1 2 Servers N servers Approximations

M/G1 Directions Of travel 0,Y Ambulance

M/G/1 Ambulance (0,0) (0,Y0) Directions Of Travel XA YA (X0,0)

Two-Server“ Hypercube Queueing Model a Distinguishable servers a Different workloads(due to geography) a Can appear with or without queueing AWith--usually FCFS AWithout--usually means a backup contract service is in place

Two-Server “Hypercube” Queueing Model aDistinguishable servers aDifferent workloads (due to geography) aCan appear with or without queueing `With -- usually FCFS `Without -- usually means a backup contract service is in place

Poisson arrivals from any sub-region a 入(B-A)入2 B-A (4)=A1 B="Service region =x1+A2

Poisson Arrivals from any sub-region A A B-A λ ( A)= λ1 λ (B-A)= λ 2 B = “Service Region” λ = λ1+ λ 2

Balance of flow equations 1.0 0.0

Balance of Flow Equations 1,0 λ 2 µ 0,0 0,1 λ 1,1 λ µ µ λ1 µ

Balance of flow equations 1+12)=Po4+P1 (+=P1+P Etc +p 00Tr10101+11

Balance of Flow Equations P00 (λ1 + λ2) = P01 µ + P10µ P01 (λ + µ) = P11 µ + P00 λ1 Etc. P00 +P10+P01+P11 = 1

Workload and imbalances p1=W1=P1+P p2=W2= 10 F11 o Workload Imbalance AW=W,-W2l

Workload and Imbalances a ρ1 = W1 = P01 + P11 a ρ2 = W2 = P10 + P11 aWorkload Imbalance = ∆W = |W1 - W2|

Average System-Wide Travel time 7(A)G(A)+272BA) }(B-A)+272(A) Q queueing

Average System-Wide Travel Time T( A)=f11 T1 ( A) + f22 T2 (B-A ) +f12 T1 (B-A) + f21 T2 ( A ) Queueing

Average System-Wide Travel time 7(A)子(A)+2B-A) +2(B-A)+2(A Geometry

Average System-Wide Travel Time T( A)=f11 T1 ( A) + f22 T2 (B-A ) +f12 T1 (B-A) + f21 T2 ( A ) Geometry

How do we obtain the f is? o Consider a long time interval T 02=(# requests that assign unit 1 to area 2) (total requests answered) 06 Total requests answered=(1-P1nT oB Average #f requests that are server 1 to area 2is n2TP10. Why? 06 Therefore f2=(n2TP10/[1-Piil7 2(1-P1)}P0

How do we obtain the fnj’s? aConsider a long time interval T a f12=(# requests that assign unit 1 to area 2)/ (total # requests answered)\ aTotal # requests answered = (1-P11) λT aAverage # requests that are “server 1 to area 2” is λ2TP10. Why? aTherefore f12 =( λ2TP10 / [1-P11]λT) = { λ2/(1-P11) λ)} P10