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