1. M/G/1
Cheng-Fu Chou

2. Residual Life
Inter-arrival time of bus is exponential w/ rate l while hippie arrives at an arbitrary instant in time
Question: How long must the hippie wait, on the average , till the bus comes along?
Answer 1: Because the average inter-arrival time is 1/l, therefore 1/2l
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3. Residual Life (cont.)
Answer 2: because of memoryless, it has to wait 1/l
General Result
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4. Derivation
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5. P. 5

6. M/G/1 M/G/1 a(t) = le –lt b(t) = general
Describe the state [N(t), X0(t)]
N(t): the no. of customers present at time t
X0(t): service time already received by the customer in service at time t
Rather than using this approach, we use “the method of the imbedded Markov Chain”
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7. Imbedded Markov Chain [N(t), X0(t)]
Select the “departure” points, we therefore eliminate X0(t)
Now N(t) is the no. of customer left behind by a departure customer. (HW)
For Poisson arrival pk(t) = rk(t)
If in any system (even in non-Markovian) where N(t) makes discontinuous changes in size (plus or minus) one, then
rk = dk = prob[departure leaves k customers behind]
Therefore, for M/G/1
rk = dk = pk
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10. Mean Queue Length
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17. This is the famous Pollaczek – Khinchin Mean Value Formula

18. Examples
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19. Mean Residual Service Time
Wi: waiting time in queue of the i-th customer
Ri: residual service time seen by the i-th customer
Xi: service time of the i-th customer
Ni: # of customers found waiting in the queue by the i-th customer upon arrival
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20. P. 20

21. Residual Service Time rt x1
time t x1 x2
xM(t)
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23. Distribution of Number in the System
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24. P. 24

25. Ex
Q(z) for M/M/1
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26. Sol.
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27. Waiting Time Distribution
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28. Ex Response time distribution for M/M/1
Waiting time distribution for M/M/1
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29. Response Time Distribution

30. Waiting Time Distribution
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