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Queueing theory Birth-death Analysis

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Presentation on theme: "Queueing theory Birth-death Analysis"β€” Presentation transcript:

1 Queueing theory Birth-death Analysis
M/M/1 queues (intro to M/M/c)

2 Birth Death Processes Consider The M/M/1 queue
State 0 – the queue and server is empty State 1 – the server is in use and the queue is empty State 2 – the server is in use and 1 is in the queue State 3 – the server is in use and 2 in the queue n 1 2 3 πœ† πœ‡

3 Thoughts on Birth death processes
The rate of moving from state 2 into state 3 should be the same as the rate from moving from state 3 into state 2. Over the course of a day if the queuing system moved from state 2 to /day then the rate that the system goes from state 3 to state 2 must also be 5000/day(or maybe 4999/day). The probability of being in one state i is 𝑃 𝑖 . The probability of being in state i and going to state i+1 is equal to the probability of doing from state i+1 and going to state i. 𝑃 𝑖+1 (π‘‘π‘’π‘Žπ‘‘β„Ž 𝑖+1 )= 𝑃 𝑖 (π‘π‘–π‘Ÿπ‘‘β„Ž 𝑖 )

4 Birth-Death for M/M/1 queues continued
𝑃 0 =1βˆ’πœŒ=1βˆ’ πœ† πœ‡ πœ†π‘ƒ 0 = πœ‡π‘ƒ 1  𝑃 1 = πœ† πœ‡ 𝑃 0 𝑃 𝑛+1 = πœ† πœ‡ 𝑃 𝑛 π‘“π‘œπ‘Ÿ π‘Žπ‘™π‘™ 𝑛 𝑃 𝑛 = πœ† πœ‡ 𝑛 𝑃 0 𝑖=0 ∞ 𝑃 𝑛 =1  𝑖=1 ∞ πœ† πœ‡ 𝑛 𝑃 0 =1

5 Mathematical Summation
Recall from Calculus II 𝑖=0 ∞ π‘₯ 𝑖 = 1 1βˆ’π‘₯ π‘“π‘œπ‘Ÿ 0<π‘₯<1 When πœ†< πœ‡ π‘‘β„Žπ‘’π‘› 0< πœ† πœ‡ =𝜌<1 1= 𝑖=0 ∞ πœ† πœ‡ 𝑖 𝑃 0 = 1 1βˆ’πœŒ 𝑃 0

6 Expected Values of a roll of dice
𝐸= 𝑖=1 6 𝑖 𝑃 𝑖 = (5) 1 6 +(6) 1 6 = 21 6 =3.5

7 Expected # customers in M/M/1 queue
𝐸= 𝑖=0 ∞ 𝑖 𝑃 𝑖 = 𝑖=0 ∞ 𝑖 πœ† πœ‡ 𝑖 𝑃 0 = 𝑃 0 𝑖=0 ∞ 𝑖 πœ† πœ‡ 𝑖 Recall from Calculus II 𝑖=0 ∞ 𝑖π‘₯ π‘–βˆ’1 = 1 1βˆ’π‘₯ 2 π‘“π‘œπ‘Ÿ 0<π‘₯<1 𝐸= 𝑃 0 𝑖=0 ∞ 𝑖 𝜌 𝜌 π‘–βˆ’1 =(1βˆ’πœŒ) 𝜌 𝑖=0 ∞ 𝑖 𝜌 π‘–βˆ’1 𝐸= (1βˆ’πœŒ) 𝜌 1βˆ’πœŒ 2 = 𝜌 1βˆ’πœŒ = πœ† πœ‡ 1βˆ’ πœ† πœ‡

8 Example M/M/1 queue πœ†=2/ sec π‘Žπ‘›π‘‘ πœ‡=3/𝑠𝑒𝑐 What is the total time in the queueing system? Time in system = time waiting in queue + service time. 𝑇 𝑠 = π‘Š π‘ž +1/πœ‡ 𝑇 𝑠 = 𝐸 # (1/πœ‡)+1/πœ‡ 𝑇 𝑠 = πœ† πœ‡ 1βˆ’ πœ† πœ‡ (1/πœ‡)+1/πœ‡ 𝑇 𝑠 = 2 3 1βˆ’ (1) 1 3 = =1 second

9 M/M/3 queue birth death analysis
State 0 – system is empty State 1 – 1 server in use, 2 servers idle, queue empty State 2 – 2 servers in use, 1 server idle, queue empty State 3 – 3 servers in use, queue empty State 4 – 3 servers in use, 1 customer in the queue State 5 – 3 servers in use, 2 customer in the queue X

10 Birth death equations for M/M/3
πœ†π‘ƒ 0 = πœ‡π‘ƒ 1 πœ†π‘ƒ 1 = 2πœ‡π‘ƒ 2 πœ†π‘ƒ 2 = 3πœ‡π‘ƒ 3 πœ†π‘ƒ 3 = 3πœ‡π‘ƒ 4 πœ†π‘ƒ 𝑛 = 3πœ‡π‘ƒ 𝑛+1 π‘Žπ‘™π‘™ 𝑛>3 n 1 2 3 πœ† πœ‡ 2πœ‡ 3πœ‡


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