Queuing Networks

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Queuing Networks

Wha Sook Jeon

Mobile Computing & Communications Lab.

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Time Reversibility (1)

• Time reversibility

− Statistical characteristic of forward process is the same as that of backward process

− The arrival process of the forward process is the arrival process of the backward process, which is the departure process of the

forward process ⇒ The arrival process of time reversible process has the same statistical characteristic as its own departure process

1

F

𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑃𝑃𝐴𝐴𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃

B

𝐷𝐷𝑃𝑃𝐷𝐷𝐴𝐴𝐴𝐴𝐷𝐷𝐷𝐷𝐴𝐴𝑃𝑃 𝑃𝑃𝐴𝐴𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃

arrival process_F = arrival process_𝐵𝐵 = departure process_𝐹𝐹 = departure process_𝐵𝐵

𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑃𝑃𝐴𝐴𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃

𝐷𝐷𝑃𝑃𝐷𝐷𝐴𝐴𝐴𝐴𝐷𝐷𝐷𝐷𝐴𝐴𝑃𝑃 𝑃𝑃𝐴𝐴𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃

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Time Reversibility (2)

DTMC

•

Forward process

− Transition probability from state i to state j: 𝐷𝐷_{𝑖𝑖𝑖𝑖} 𝐷𝐷_{𝑖𝑖𝑖𝑖} = Pr {𝑋𝑋_𝑛𝑛+1 = 𝑗𝑗|𝑋𝑋_𝑛𝑛 = 𝐴𝐴}

•

Backward process

− Transition probability from state i to state j: 𝑞𝑞_{𝑖𝑖𝑖𝑖}

_𝑞𝑞_{𝑖𝑖𝑖𝑖} _{= Pr {𝑋𝑋}_𝑛𝑛 ₌ _{𝑗𝑗|𝑋𝑋}_𝑛𝑛+1 ₌ _𝐴𝐴}

− 𝑞𝑞_{𝑖𝑖𝑖𝑖} = ^{Pr {𝑋𝑋}_{Pr {𝑋𝑋}^𝑛𝑛^{=𝑖𝑖, 𝑋𝑋}^𝑛𝑛+1^=𝑖𝑖}

𝑛𝑛+1=𝑖𝑖} = ^{Pr {𝑋𝑋}^𝑛𝑛+1_{Pr {𝑋𝑋}^{=𝑖𝑖|𝑋𝑋}^𝑛𝑛^{=𝑖𝑖}Pr {𝑋𝑋}^𝑛𝑛^=𝑖𝑖}

𝑛𝑛+1=𝑖𝑖} = ^𝜋𝜋^𝑗𝑗_𝜋𝜋^𝑝𝑝^{𝑗𝑗𝑖𝑖}

𝑖𝑖

•

Necessary and sufficient condition for time reversibility:

𝑞𝑞_{𝑖𝑖𝑖𝑖} = 𝐷𝐷_{𝑖𝑖𝑖𝑖}

− 𝑞𝑞_{𝑖𝑖𝑖𝑖} = ^𝜋𝜋^𝑗𝑗_𝜋𝜋^𝑝𝑝^{𝑗𝑗𝑖𝑖}

𝑖𝑖 = 𝐷𝐷_{𝑖𝑖𝑖𝑖} ⇒ 𝜋𝜋_𝑖𝑖𝐷𝐷_{𝑖𝑖𝑖𝑖} = 𝜋𝜋_𝑖𝑖 𝐷𝐷_{𝑖𝑖𝑖𝑖}

• Time reversible DTMC, 𝜋𝜋_𝑖𝑖𝐷𝐷_{𝑖𝑖𝑖𝑖} = 𝜋𝜋_𝑖𝑖𝐷𝐷_{𝑖𝑖𝑖𝑖}

2

i j

𝐷𝐷_{𝑖𝑖𝑖𝑖}

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Time Reversibility (3)

Time reversible CTMC : 𝜋𝜋

_𝑖𝑖

𝐴𝐴

_{𝑖𝑖𝑖𝑖}

= 𝜋𝜋

_𝑖𝑖

𝐴𝐴

_{𝑖𝑖𝑖𝑖}

• Birth & death process is time reversible

− Since M/M/c queuing system is a special case of birth & death process, M/M/c is time reversible

− Arrival process of M/M/c queuing system is the same as its

departure process. Thus, departure process of M/M/c is a Poisson process

3

M/M/c

Poisson process

with rate λ Poisson process

with rate λ

i j

𝐴𝐴_{𝑖𝑖𝑖𝑖}

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Open Queuing Networks (1)

• Open Queuing networks with product form solution

− Poisson arrivals from outside source

− All servers have exponentially distributed service time

− A job from device i joins device j with (routing) probability 𝑞𝑞_{𝑖𝑖𝑖𝑖}

− Each device is modeled as M/M/c, being independent of each other. ⁴ 1

Poisson arrival with rate λ

2

4 3

5

S

^𝑞𝑞^𝑠𝑠𝑠

𝑞𝑞_𝑠𝑠1

𝑞𝑞_𝑠𝑠𝑠

M/M/c

𝑞𝑞₁₅

𝑞𝑞_𝑠5

𝑞𝑞_𝑠4

𝑞𝑞₄₅

M/M/c

Poisson

Population source

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Open Queuing Networks (2)

• System state: (𝑛𝑛₁,𝑛𝑛_𝑠,𝑛𝑛_𝑠,𝑛𝑛₄,𝑛𝑛₅)

− 𝑛𝑛_𝑖𝑖: number of jobs in device i

• Jackson’s decomposition theorem

𝑃𝑃(𝑛𝑛₁,𝑛𝑛_𝑠,𝑛𝑛_𝑠,𝑛𝑛₄,𝑛𝑛₅) = 𝑃𝑃₁ 𝑛𝑛₁ 𝑃𝑃_𝑠 𝑛𝑛_𝑠 𝑃𝑃_𝑠 𝑛𝑛_𝑠 𝑃𝑃₄ 𝑛𝑛₄ 𝑃𝑃₅ 𝑛𝑛₅

• 𝑃𝑃(𝑛𝑛₁,𝑛𝑛_𝑠,𝑛𝑛_𝑠,𝑛𝑛₄,𝑛𝑛₅): System state probability

• 𝑃𝑃_𝑖𝑖 𝑛𝑛_𝑖𝑖 : Probability of 𝑛𝑛_𝑖𝑖 jobs in device i

• When all devices are M/M/1

𝑃𝑃_𝑖𝑖 𝑛𝑛_𝑖𝑖 = 𝜌𝜌_𝑖𝑖^𝑛𝑛^𝑖𝑖 1− 𝜌𝜌_𝑖𝑖

𝑃𝑃(𝑛𝑛₁,𝑛𝑛_𝑠,𝑛𝑛_𝑠,𝑛𝑛₄,𝑛𝑛₅) = ∏⁵_i=1𝜌𝜌_𝑖𝑖^𝑛𝑛^𝑖𝑖(1 − 𝜌𝜌_𝑖𝑖)

− 𝜌𝜌_𝑖𝑖 = ^λ^𝑖𝑖

𝜇𝜇_𝑖𝑖

− λ₁ =λ𝑞𝑞_𝑠𝑠1, λ_𝑠 =λ𝑞𝑞_𝑠𝑠𝑠 , λ_𝑠 = λ𝑞𝑞_𝑠𝑠𝑠,

− λ₄ =λ_𝑠 +λ_𝑠𝑞𝑞_𝑠4 , λ₅ =λ₁+λ_𝑠𝑞𝑞_𝑠5+λ₄ 5 1

λ

2

4

3

5

𝑞𝑞_𝑠𝑠𝑠 𝑞𝑞_𝑠𝑠1

𝑞𝑞_𝑠𝑠𝑠

M/M/1

𝑞𝑞₁₅

𝑞𝑞_𝑠5

𝑞𝑞_𝑠4 𝑞𝑞_𝑠4

𝑞𝑞₄₅ M/M/1

M/M/1

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Open Queuing Networks (3)

• Performance measure

< Device i >

− Utilization of device i: 𝜌𝜌_𝑖𝑖 = _𝜇𝜇^λ^𝑖𝑖

𝑖𝑖

− Mean number of jobs in device i: 𝑁𝑁�_𝑖𝑖 = _1−𝜌𝜌^𝜌𝜌^𝑖𝑖

𝑖𝑖

< System >

− Mean number of jobs: 𝑁𝑁� = ∑^𝑀𝑀_𝑖𝑖=1𝑁𝑁�_𝑖𝑖

• 𝑀𝑀: the number of devices in the network

− Mean sojourn time of a job in the network: 𝑇𝑇� = ^𝑁𝑁�_λ

6

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Exercise1: Open Queuing Networks

A machine shop has four machines, A, B, C, and D. The numbers of servers in the machines A, B, C and D are one, one, two, and three, respectively. Service time distributions of the servers in the machines A, B, C and D are exponential at their respective rates µ_A, µ_B, µ_C, and µ_D. The shop gets four types of jobs, numbered 1 through 4, where each type requires service on machines in a

particular sequence; type 1: ABDA, type 2: CABC, type 3: ACB, type 4: BCAD. The arrival process of type i jobs is Poisson at rate λ_i.

1) Under what conditions is this system stable?

2) What is the joint stationary distribution of the number of jobs at each machine?

λ₁ λ₂ λ₃ λ₄

1) λ_A< µ_A , λ_B< µ_B , λ_C< 2µ_C , λ_D< 3µ_D

2) Note that the machines A, B, C, and D are an M/M/1. M/M/1, M/M/2, and M/M/3 respectively.

P(n_A , n_B , n_C , n_D ) = 𝜌𝜌_𝐴𝐴^𝑛𝑛^𝐴𝐴(1− 𝜌𝜌_𝐴𝐴) 𝜌𝜌_𝐵𝐵^𝑛𝑛^𝐵𝐵(1− 𝜌𝜌_𝐵𝐵) 2𝜌𝜌_𝐶𝐶^𝑛𝑛^𝐶𝐶𝑃𝑃_𝑐𝑐(0) ⁹

𝑠𝜌𝜌_𝐷𝐷^𝑛𝑛^𝐷𝐷𝑃𝑃_𝐷𝐷(0) n_A = ^𝜆𝜆^𝐴𝐴

𝜇𝜇_𝐴𝐴−𝜆𝜆_𝐴𝐴, n_B = ^𝜆𝜆^𝐵𝐵

𝜇𝜇_𝐵𝐵−𝜆𝜆_𝐵𝐵, n_C = ^4𝜇𝜇^𝐶𝐶^𝜆𝜆^𝐶𝐶

4𝜇𝜇_𝐶𝐶²−𝜆𝜆_𝐶𝐶²,

n_D: calculate by yourselves, using performance measure equation of an M/M/3 system 7

A B C D

λ_A=2λ₁ +λ₂ + λ₃+λ₄ λ_B=λ₁ +λ₂ + λ₃+λ₄ λ_C=2λ₂ + λ₃+λ₄ λ_D=λ₁ +λ₄

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Exercise2: Open Queuing Networks

• 자장면 전문체인점에 자장면을 먹으러 오는 고객들이 도착율 a의 포아송 프 로세스로 도착한다. 고객은 도착하자마자 주문을 한 후 자장면이 만들어질 때까지 기다린 후 자장면이 만들어지면 먹고 계산대에서 계산을 한 뒤 떠난 다. 따라서 임의의 고객은 기다리든지 먹든지 계산을 하든지 셋 중 하나의 상 태에 있다. 자장면을 만드는 요리사의 수와 는 계산원은 각각 1명이다. 요리 사가 자장면을 만드는 시간과 먹는 시간, 그리고 계산하는데 걸리는 시간은 각각 평균이 s1, s2, s3인 지수분포를 따른다. 체인점은 충분히 넓다고 가정 하자. 고객이 체인점에 머무는 시간을 구하라.

− 𝑛𝑛₁ = _{1−𝑎𝑎×𝑠𝑠1}^{𝑎𝑎×𝑠𝑠1} , 𝑇𝑇₁= _{1−𝑎𝑎×𝑠𝑠1}^𝑠𝑠1 , 𝑇𝑇_𝑠 = 𝑃𝑃𝑠, 𝑛𝑛_𝑠 = _{1−𝑎𝑎×𝑠𝑠𝑠}^{𝑎𝑎×𝑠𝑠𝑠} , 𝑇𝑇_𝑠= _{1−𝑎𝑎×𝑠𝑠𝑠}^𝑠𝑠𝑠

− 𝑇𝑇 = 𝑇𝑇₁ + 𝑇𝑇_𝑠+ 𝑇𝑇_𝑠

8

…

a a a a

M/M/1

M/M/∞

M/M/1 s1

s2

s2 s3

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Closed Queuing Networks (1)

 Queuing network with no jobs from the outside

 The total number of jobs within the system is fixed.

⁻ N: the total number of jobs in the network

− M: the number of devices in the network

•

System state: (

𝑛𝑛₁,𝑛𝑛_𝑠, … ,𝑛𝑛_𝑀𝑀

)

− 𝑛𝑛_𝑖𝑖: number of jobs in device i

− 𝑁𝑁 = ∑^𝑀𝑀_𝑖𝑖=0𝑛𝑛_𝑖𝑖

9

2 3 1

𝑞𝑞₁

𝑞𝑞_𝑠

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Closed Queuing Networks (2)

• Assumptions for product form solution

− The system is in steady state

− All servers have exponentially distributed service time

− Jobs are stochastically independent of each other

− A job from device i joins device j with the (routing) probability 𝑞𝑞_{𝑖𝑖𝑖𝑖}

 Gordon and Newell’s decomposition theorem

𝑃𝑃 ( 𝑛𝑛

₁

, 𝑛𝑛

_𝑠

, … , 𝑛𝑛

_𝑀𝑀

) =

_𝐺𝐺¹

𝐹𝐹

₁

𝑛𝑛

₁

𝐹𝐹

_𝑠

𝑛𝑛

_𝑠

… 𝐹𝐹

_𝑀𝑀

(𝑛𝑛

_𝑀𝑀

)

• ∑n_{∈𝑺𝑺(𝑀𝑀,𝑁𝑁)} 𝑃𝑃

(

𝑛𝑛₁,𝑛𝑛_𝑠, … , 𝑛𝑛_𝑀𝑀

)

= 1

 n = (𝑛𝑛₁,𝑛𝑛_𝑠, … ,𝑛𝑛_𝑀𝑀)

 𝑺𝑺 𝑀𝑀,𝑁𝑁 = {(𝑛𝑛₁,𝑛𝑛_𝑠, … ,𝑛𝑛_𝑀𝑀)|𝑛𝑛₁ +𝑛𝑛₁ +⋯+ 𝑛𝑛_𝑀𝑀 = 𝑁𝑁}

• Normalization factor 𝐺𝐺 = ∑n_{∈𝑺𝑺(𝑀𝑀,𝑁𝑁)}∏^𝑀𝑀_𝑖𝑖=1𝐹𝐹_𝑖𝑖 𝑛𝑛_𝑖𝑖

10

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Closed Queuing Networks (3)

11

• 𝑃𝑃 ( 𝑛𝑛

₁

, 𝑛𝑛

_𝑠

, … , 𝑛𝑛

_𝑀𝑀

) =

_𝐺𝐺¹

𝐹𝐹

₁

𝑛𝑛

₁

𝐹𝐹

_𝑠

𝑛𝑛

_𝑠

… 𝐹𝐹

_𝑀𝑀

(𝑛𝑛

_𝑀𝑀

)

• 𝐹𝐹_𝑖𝑖(𝑛𝑛_𝑖𝑖) = � 1 , 𝑛𝑛_𝑖𝑖 = 0 𝑉𝑉_𝑖𝑖 ⨯ 𝑃𝑃_𝐴𝐴(𝑛𝑛_𝑖𝑖) ⨯ 𝐹𝐹_𝑖𝑖 (𝑛𝑛_𝑖𝑖 − 1) , 𝑛𝑛_𝑖𝑖 ≥ 1

− 𝑉𝑉_𝑖𝑖 : Visit ratio of device i (relative input rate)

• 𝑋𝑋_𝑖𝑖 :Throughput (Input rate) of device i

− 𝑃𝑃_𝐴𝐴(𝑛𝑛_𝑖𝑖) ∶ the service time of device i when there are 𝑛𝑛_𝑖𝑖 jobs in device i

Insight: Remind that, for open queueing network,

𝑃𝑃(𝑛𝑛₁,𝑛𝑛_𝑠, … ,𝑛𝑛_𝑀𝑀) = 𝑃𝑃₁ 𝑛𝑛₁ 𝑃𝑃_𝑠 𝑛𝑛_𝑠 …𝑃𝑃_𝑀𝑀 𝑛𝑛_𝑀𝑀 𝑃𝑃_𝑖𝑖(𝑛𝑛_𝑖𝑖) = �𝑃𝑃_𝑖𝑖(0) , 𝑛𝑛_𝑖𝑖 = 0

𝜆𝜆_𝑖𝑖 ⨯ 𝑃𝑃_𝑖𝑖⨯ 𝑃𝑃_𝑖𝑖 𝑛𝑛_𝑖𝑖 − 1 , 𝑛𝑛_𝑖𝑖 ≥ 1 in M/M/1

(13)

12

• Derivation of 𝑉𝑉_𝑖𝑖,

− For any appropriate link, 𝑉𝑉₀ =1.

− Then, calculate other 𝑉𝑉_𝑖𝑖 values : 𝑉𝑉_𝑖𝑖 = ∑^𝑀𝑀_𝑖𝑖=0𝑉𝑉_𝑖𝑖 𝑞𝑞_{𝑖𝑖𝑖𝑖}

2 3 1

𝑽𝑽_𝟎𝟎 =1

𝟏𝟏�𝟒𝟒

𝟐𝟐�𝟒𝟒

𝟏𝟏�𝟒𝟒

< Example >

𝑉𝑉₀ = 1 , 𝑉𝑉₀ = 𝑉𝑉_𝑠 + 𝑉𝑉_𝑠 ,

𝑉𝑉₁ = ¹₄𝑉𝑉₁ +𝑉𝑉₀ , 𝑉𝑉_𝑠 = ^𝑠₄𝑉𝑉₁ , 𝑉𝑉_𝑠 = ¹₄𝑉𝑉₁

⇒ 𝑉𝑉₁= ⁴_𝑠 , 𝑉𝑉_𝑠 = ^𝑠_𝑠 , 𝑉𝑉_𝑠 = ¹_𝑠

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 Derivation for the product form solution

< Notation >

− n ≔ ( 𝑛𝑛

₁

, 𝑛𝑛

_𝑠

, … , 𝑛𝑛

_𝑖𝑖

, … , 𝑛𝑛

_𝑖𝑖

, … , 𝑛𝑛

_𝑀𝑀

)

− n

_ij

≔ ( 𝑛𝑛

₁

, 𝑛𝑛

_𝑠

, … , 𝑛𝑛

_𝑖𝑖

− 1, … , 𝑛𝑛

_𝑖𝑖

+ 1, … , 𝑛𝑛

_𝑀𝑀

)

− 𝐴𝐴(𝒏𝒏 → 𝒌𝒌) : transition rate from state n to state k

1. Steady State Assumption (in-rate = out-rate)

• ∑ 𝑃𝑃 𝒍𝒍 𝐴𝐴(_𝒍𝒍 𝒍𝒍 → n) = 𝑃𝑃(𝒏𝒏)∑ 𝐴𝐴(_𝒌𝒌 𝒏𝒏 → 𝒌𝒌)

2. Exponential Server Assumption (in Steady State)

• ∑ 𝑃𝑃n_ij n_ij 𝐴𝐴( n_ij → n) = 𝑃𝑃(𝒏𝒏) ∑ 𝐴𝐴(n_ij 𝒏𝒏 → n_ij)

13

(15)

3. Independent Routing of each job

• 𝐴𝐴(𝒏𝒏 → n_ij) = 𝑞𝑞_{𝑖𝑖𝑖𝑖} _𝑠𝑠^{𝛿𝛿(𝑛𝑛}^𝑖𝑖⁾

𝑖𝑖(𝑛𝑛_𝑖𝑖) where 𝛿𝛿(𝑛𝑛_𝑖𝑖)=�1 𝑛𝑛_𝑖𝑖 > 0 0 𝑛𝑛_𝑖𝑖 = 0

 Exponential Server and Independent Routing of each job

𝑃𝑃(𝒏𝒏)∑ 𝐴𝐴(n_ij 𝒏𝒏 → n_ij) = ∑ 𝑃𝑃n_ij n_ij 𝐴𝐴(n_ij → n)

− 𝑃𝑃(𝒏𝒏) ∑ ∑ 𝑞𝑞

_{𝑖𝑖𝑖𝑖} _𝑠𝑠^{𝛿𝛿(𝑛𝑛}^𝑖𝑖⁾

𝑖𝑖(𝑛𝑛_𝑖𝑖)

j

i

= ∑ ∑

^{𝛿𝛿(𝑛𝑛}_𝑠𝑠^𝑖𝑖^{)𝛿𝛿(𝑛𝑛}^𝑗𝑗⁺¹⁾

𝑗𝑗(𝑛𝑛_𝑗𝑗+1)

j

i

𝑞𝑞

_{𝑖𝑖𝑖𝑖}

𝑃𝑃 n

_ij

𝑃𝑃(𝒏𝒏) ∑

_𝑠𝑠^{𝛿𝛿(𝑛𝑛}^𝑖𝑖⁾

i

= ∑ ∑

_𝑠𝑠 ^{𝛿𝛿(𝑛𝑛}^𝑖𝑖⁾

j

i

𝑞𝑞

_{𝑖𝑖𝑖𝑖}

𝑃𝑃 n

_ij

14

𝑛𝑛_𝑖𝑖−1≥0, 𝑛𝑛_𝑖𝑖+ 1 > 0

𝑛𝑛_𝑖𝑖 > 0

(16)

 Define

 𝐹𝐹_𝑖𝑖(𝑛𝑛_𝑖𝑖) = � 1 , 𝑛𝑛_𝑖𝑖 = 0 𝑦𝑦_𝑖𝑖 ⨯ 𝑃𝑃_𝐴𝐴(𝑛𝑛_𝑖𝑖) ⨯ 𝐹𝐹_𝑖𝑖 (𝑛𝑛_𝑖𝑖 − 1) , 𝑛𝑛_𝑖𝑖≥ 1 where 𝑦𝑦_𝑖𝑖’s are unknown parameters

 Assume: P( 𝑛𝑛

₁

, 𝑛𝑛

_𝑠

, … , 𝑛𝑛

_𝑀𝑀

) = 𝐶𝐶 𝐹𝐹

₁

𝑛𝑛

₁

𝐹𝐹

_𝑠

𝑛𝑛

_𝑠

… 𝐹𝐹

_𝑀𝑀

(𝑛𝑛

_𝑀𝑀

) From 𝑃𝑃 (𝒏𝒏) ∑

i

= ∑ ∑

_𝑠𝑠 ^{𝛿𝛿(𝑛𝑛}^𝑖𝑖⁾

j

i

𝑞𝑞

_{𝑖𝑖𝑖𝑖}

𝑃𝑃 n

_ij

,

𝐶𝐶 𝐹𝐹

₁

𝑛𝑛

₁

𝐹𝐹

_𝑠

𝑛𝑛

_𝑠

… 𝐹𝐹

_𝑀𝑀

(𝑛𝑛

_𝑀𝑀

) ∑

𝛿𝛿(𝑛𝑛_𝑖𝑖) 𝑠𝑠_𝑖𝑖(𝑛𝑛_𝑖𝑖)

i

= ∑ ∑

𝛿𝛿(𝑛𝑛_𝑖𝑖) 𝑠𝑠_𝑗𝑗(𝑛𝑛_𝑗𝑗+1)

j

i

𝑞𝑞

_{𝑖𝑖𝑖𝑖}

𝐶𝐶 𝐹𝐹

₁

𝑛𝑛

₁

𝐹𝐹

_𝑠

𝑛𝑛

_𝑠

…

_𝑦𝑦^{𝛿𝛿 𝑛𝑛}^𝑖𝑖

𝑖𝑖 𝑠𝑠_𝑖𝑖 𝑛𝑛_𝑖𝑖

𝐹𝐹

_𝑖𝑖

(𝑛𝑛

_𝑖𝑖

)

… 𝑦𝑦_𝑖𝑖 𝑃𝑃_𝑗𝑗(𝑛𝑛_𝑖𝑖 + 1) 𝐹𝐹_𝑖𝑖 𝑛𝑛_𝑖𝑖 …𝐹𝐹_𝑀𝑀(𝑛𝑛_𝑀𝑀)

15 𝐹𝐹_𝑖𝑖(𝑛𝑛_𝑖𝑖−1)

𝐹𝐹_𝑖𝑖(𝑛𝑛_𝑖𝑖+1)

(17)

 ∑

𝛿𝛿(𝑛𝑛_𝑖𝑖) 𝑠𝑠_𝑖𝑖(𝑛𝑛_𝑖𝑖)

i

= ∑ ∑

j

i

𝑞𝑞

_{𝑖𝑖𝑖𝑖} ^𝑦𝑦_𝑦𝑦^𝑗𝑗

𝑖𝑖

∑

i

1 − ∑

^𝑦𝑦_𝑦𝑦^𝑗𝑗

j 𝑖𝑖

𝑞𝑞

_{𝑖𝑖𝑖𝑖}

= 0 Thus, 1 = ∑

^𝑦𝑦_𝑦𝑦^𝑗𝑗

j 𝑖𝑖

𝑞𝑞

_{𝑖𝑖𝑖𝑖}

𝑦𝑦

_𝑖𝑖

= ∑

^𝑀𝑀_𝑖𝑖=1

𝑦𝑦

_𝑖𝑖

𝑞𝑞

_{𝑖𝑖𝑖𝑖}

 Note that the 𝑦𝑦

_𝑖𝑖

’s can be anything as long as they satisfy the above equation.

• Applying to the throughput, 𝑋𝑋_𝑖𝑖= ∑^𝑀𝑀_𝑖𝑖=1𝑋𝑋_𝑖𝑖𝑞𝑞_{𝑖𝑖𝑖𝑖}

• Applying to the visit ratio 𝑉𝑉_𝑖𝑖 := 𝑋𝑋_𝑖𝑖/ 𝑋𝑋₀, 𝑉𝑉_𝑖𝑖= ∑^𝑀𝑀_𝑖𝑖=1𝑉𝑉_𝑖𝑖𝑞𝑞_{𝑖𝑖𝑖𝑖}

16

(18)

 In summary

𝑃𝑃 ( 𝑛𝑛

₁

, 𝑛𝑛

_𝑠

, … , 𝑛𝑛

_𝑀𝑀

) =

_𝐺𝐺¹

𝐹𝐹

₁

𝑛𝑛

₁

𝐹𝐹

_𝑠

𝑛𝑛

_𝑠

… 𝐹𝐹

_𝑀𝑀

(𝑛𝑛

_𝑀𝑀

)

− 𝐺𝐺 = ∑ n

_{∈𝑺𝑺(𝑀𝑀,𝑁𝑁)}

∏

^𝑀𝑀_𝑖𝑖=1

𝐹𝐹

_𝑖𝑖

𝑛𝑛

_𝑖𝑖

− 𝐹𝐹_𝑖𝑖(𝑛𝑛_𝑖𝑖) = � 1 , 𝑛𝑛_𝑖𝑖 = 0 𝑉𝑉_𝑖𝑖 ⨯ 𝑃𝑃_𝐴𝐴(𝑛𝑛_𝑖𝑖) ⨯ 𝐹𝐹_𝑖𝑖 (𝑛𝑛_𝑖𝑖 − 1) , 𝑛𝑛_𝑖𝑖≥ 1

− 𝑉𝑉_𝑖𝑖 = ∑^𝑀𝑀_𝑖𝑖=0𝑉𝑉_𝑖𝑖 𝑞𝑞_{𝑖𝑖𝑖𝑖} (For any appropriate link, 𝑉𝑉₀ =1)

Note that the number of feasible states can be too many to calculate G

17

(19)

18

 Buzen’s Recursive Algorithm for simply calculating 𝐺𝐺

− Let 𝑔𝑔_𝑚𝑚(𝑛𝑛) ≔ ∑n^{∈𝑺𝑺(𝑚𝑚,𝑛𝑛)}∏^𝑚𝑚_𝑖𝑖=1𝐹𝐹_𝑖𝑖 𝑛𝑛_𝑖𝑖

where n = (𝑛𝑛₁,𝑛𝑛_𝑠, … ,𝑛𝑛_𝑚𝑚) , S 𝑚𝑚,𝑛𝑛 = (𝑛𝑛₁,𝑛𝑛_𝑠, … ,𝑛𝑛_𝑚𝑚)|𝑛𝑛₁ + 𝑛𝑛₁ + ⋯+ 𝑛𝑛_𝑚𝑚 = 𝑛𝑛

− 𝐺𝐺 = 𝑔𝑔_𝑀𝑀(𝑁𝑁)

− 𝑔𝑔₁ 𝑛𝑛 = 𝐹𝐹₁ 𝑛𝑛

− 𝑔𝑔_𝑚𝑚 0 = ∏^𝑚𝑚_𝑖𝑖=1𝐹𝐹_𝑖𝑖 0 = 1

− 𝑔𝑔_𝑚𝑚 𝑛𝑛 = ∑𝑛𝑛 𝐹𝐹_𝑚𝑚 𝑘𝑘 ∑ _𝑛𝑛₁_…𝑛𝑛_𝑚𝑚−1 ∈𝑺𝑺(𝑚𝑚−1,𝑛𝑛−𝑘𝑘)∏^𝑚𝑚−1_𝑖𝑖=1 𝐹𝐹_𝑖𝑖 𝑛𝑛_𝑖𝑖

𝑘𝑘=0 , _𝑛𝑛 _{> 0,}_𝑚𝑚 _{> 1}

= ∑^𝑛𝑛_𝑘𝑘=0𝐹𝐹_𝑚𝑚 𝑘𝑘 𝑔𝑔_𝑚𝑚−1 𝑛𝑛 − 𝑘𝑘

𝑔𝑔_𝑚𝑚 𝑛𝑛 can be calculated in a recursive fashion

(20)

19

1 1 … 1 _⨯F_m(n) 1 … 1

F₁(1) 𝑔𝑔_𝑠(1) … 𝑔𝑔_𝑚𝑚−1(1)

F₁(n-1) 𝑔𝑔_𝑠 (n-1) … 𝑔𝑔_𝑚𝑚−1(n-1)

F₁(n) 𝑔𝑔_𝑠(n) … 𝑔𝑔_𝑚𝑚−1(n) 𝑔𝑔_𝑚𝑚(𝑛𝑛)

𝑔𝑔_𝑀𝑀(N-1)

F₁(N) 𝑔𝑔_𝑠(N) 𝑔𝑔_𝑀𝑀(𝑁𝑁)

0 1

. . .

n-1

N n

1 2 m-1 m M

. .

. . . .

. .

. . . .

⨯ F_m(0) = +

+ +

. . .

= 𝑮𝑮

• Calculation of 𝑔𝑔

_𝑚𝑚

(𝑛𝑛) :

⨯ F_m(1)

⨯F_m(n-1)

𝑔𝑔_𝑚𝑚 𝑛𝑛 = 𝑔𝑔_𝑚𝑚−1 0 𝐹𝐹_𝑚𝑚 𝑛𝑛 + 𝑔𝑔_𝑚𝑚−1 1 𝐹𝐹_𝑚𝑚 𝑛𝑛 − 1 + 𝑔𝑔_𝑚𝑚−1 2 𝐹𝐹_𝑚𝑚 𝑛𝑛 − 2 + ··· +𝑔𝑔_𝑚𝑚−1 𝑛𝑛 − 1 𝐹𝐹_𝑚𝑚 1 + 𝑔𝑔_𝑚𝑚−1 𝑛𝑛 𝐹𝐹_𝑚𝑚 0

(21)

20

− When the service rate of each device is constant (a single server)

• 𝑃𝑃_𝐴𝐴(𝑛𝑛_𝑖𝑖) = 𝑃𝑃_𝑖𝑖 , ∀𝑛𝑛_𝑖𝑖 ≥ 1 ⇒ 𝐹𝐹_𝑚𝑚 𝑘𝑘 = 𝑉𝑉_𝑚𝑚𝑃𝑃_𝑚𝑚𝐹𝐹_𝑚𝑚(𝑘𝑘 − 1)

− 𝑔𝑔_𝑚𝑚 𝑛𝑛 = 𝐹𝐹_𝑚𝑚 0 𝑔𝑔_𝑚𝑚−1 𝑛𝑛 + ∑^𝑛𝑛_𝑘𝑘=1𝐹𝐹_𝑚𝑚 𝑘𝑘 𝑔𝑔_𝑚𝑚−1 𝑛𝑛 − 𝑘𝑘 = 𝑔𝑔_𝑚𝑚−1 𝑛𝑛 + 𝑉𝑉_𝑚𝑚𝑃𝑃_𝑚𝑚∑^𝑛𝑛_𝑘𝑘=1𝐹𝐹_𝑚𝑚 𝑘𝑘 − 1 𝑔𝑔_𝑚𝑚−1 𝑛𝑛 − 𝑘𝑘 = 𝑔𝑔_𝑚𝑚−1 𝑛𝑛 + 𝑉𝑉_𝑚𝑚𝑃𝑃_𝑚𝑚 𝑔𝑔_𝑚𝑚 𝑛𝑛 − 1

1 1 … 1 1 … 1

F₁(1) 𝑔𝑔_𝑠(1) … 𝑔𝑔_𝑚𝑚−1(1) 𝑔𝑔_𝑚𝑚(1)

F₁(n-1) 𝑔𝑔_𝑠 (n-1) … 𝑔𝑔_𝑚𝑚−1(n-1) 𝑔𝑔_𝑚𝑚(n-1)

F₁(n) 𝑔𝑔_𝑠(n) … 𝑔𝑔_𝑚𝑚−1(n) 𝑔𝑔_𝑚𝑚 𝑛𝑛

F₁(N) 𝑔𝑔_𝑠(N)

0 1

. . .

n-1

N n

1 2 m-1 m M

. .

. . . .

. .

. . . .

+ =

. . .

⨯ 𝑉𝑉_𝑚𝑚𝑃𝑃_𝑚𝑚

∑^𝑛𝑛−1_𝑎𝑎=0𝐹𝐹_𝑚𝑚 𝐴𝐴 𝑔𝑔_𝑚𝑚−1 𝑛𝑛 −1− 𝐴𝐴

(22)

21

• Performance measure

− Throughput of device M^:𝑋𝑋_𝑀𝑀

• 𝑋𝑋_𝑀𝑀 = ∑ 𝑃𝑃_𝑀𝑀(𝑘𝑘)_𝑠𝑠 ¹

𝑀𝑀(𝑘𝑘) 𝑁𝑁𝑘𝑘=1

 𝑃𝑃_𝑀𝑀(𝑘𝑘) : Probability that there are k jobs in the device M

 𝑃𝑃_𝑀𝑀(𝑘𝑘) = _∑_(𝑛𝑛₁_,𝑛𝑛₂_{,…,𝑛𝑛}_𝑀𝑀−1)∈𝑺𝑺(𝑀𝑀−1,𝑁𝑁−𝑘𝑘)_{𝑃𝑃(𝑛𝑛}₁_{, … ,}_𝑛𝑛_𝑀𝑀−1_,_𝑘𝑘)

= _∑ ¹

𝐺𝐺𝐹𝐹₁ 𝑛𝑛₁ …𝐹𝐹_𝑀𝑀−1 𝑛𝑛_𝑀𝑀−1 𝐹𝐹_𝑀𝑀 𝑘𝑘

(𝑛𝑛₁,𝑛𝑛₂,…,𝑛𝑛_𝑀𝑀−1)∈𝑺𝑺(𝑀𝑀−1,𝑁𝑁−𝑘𝑘)

= ¹

𝐺𝐺𝐹𝐹_𝑀𝑀 𝑘𝑘 𝑔𝑔_𝑀𝑀−1 𝑁𝑁 − 𝑘𝑘

⇒ 𝑋𝑋_𝑀𝑀 = ∑ _𝐺𝐺¹ 𝐹𝐹_𝑀𝑀 𝑘𝑘 𝑔𝑔_𝑀𝑀−1 𝑁𝑁 − 𝑘𝑘 _𝑠𝑠 ¹

= ∑ _𝐺𝐺¹ 𝑉𝑉_𝑀𝑀𝑃𝑃_𝑀𝑀(𝑘𝑘)𝐹𝐹_𝑀𝑀(𝑘𝑘 − 1) 𝑔𝑔_𝑀𝑀−1 𝑁𝑁 − 𝑘𝑘 _𝑠𝑠 ¹

= _𝐺𝐺¹ 𝑉𝑉_𝑀𝑀 ∑^𝑁𝑁_𝑘𝑘=1𝐹𝐹_𝑀𝑀 𝑘𝑘 − 1 𝑔𝑔_𝑀𝑀−1 𝑁𝑁 − 𝑘𝑘 = ¹

𝐺𝐺 𝑉𝑉_𝑀𝑀 𝑔𝑔_𝑀𝑀 𝑁𝑁 − 1

∑^𝑁𝑁−1_𝑚𝑚=0𝐹𝐹_𝑀𝑀 𝑚𝑚 𝑔𝑔_𝑀𝑀−1 𝑁𝑁 − 1− 𝑚𝑚

(23)

Closed Queuing Networks (14)

22

Since ^𝑋𝑋^𝑖𝑖

𝑋𝑋_𝑗𝑗 = ^𝑉𝑉^𝑖𝑖

𝑉𝑉_𝑗𝑗 for any devices i , j

− System throughput : 𝑋𝑋₀ 𝑋𝑋₀= ^𝑋𝑋^𝑀𝑀

𝑉𝑉_𝑀𝑀 = ^𝑔𝑔^𝑀𝑀 ^𝑁𝑁−1

𝐺𝐺

− Throughput of any device i 𝑋𝑋_𝑖𝑖 = 𝑉𝑉_𝑖𝑖 𝑋𝑋₀

− System response time: T

N jobs ^𝑋𝑋⁰

𝑋𝑋₀ 𝑇𝑇

By Little’s Law, T = _𝑋𝑋^𝑁𝑁

0

2 3 1

𝑽𝑽_𝟎𝟎 =1

𝟏𝟏�𝟒𝟒

𝟐𝟐�𝟒𝟒

𝟏𝟏�𝟒𝟒