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Queueing Theory. The study of queues – why they form, how they can be evaluated, and how they can be optimized. Building blocks – arrival process and.

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Presentation on theme: "Queueing Theory. The study of queues – why they form, how they can be evaluated, and how they can be optimized. Building blocks – arrival process and."— Presentation transcript:

1 Queueing Theory

2 The study of queues – why they form, how they can be evaluated, and how they can be optimized. Building blocks – arrival process and a service process.

3 Arrival process – individually/in groups, independent/correlated, single source/multiple sources, infinite/finite population, limited/unlimited capacity. Service process – single/multiple servers, single/multiple stages, individually/in groups, independent/correlated, service discipline (FCFS/priority). Some characteristics of arrival and service processes

4 GX/GY/k/NGX/GY/k/N A Common notation G : distribution of inter- arrival times X : distribution of arrival batch (group) size G : distribution of service times Y : distribution of service batch size k : number of servers N : maximum number of customers allowed

5 Common examples M / M /1 M / G /1 M / M / k M / M /1/ N M X / M / 1 GI / M /1 M / M / k/k

6 Fundamental quantities L : expected number of customers in the system, L = E ( n ). L Q : expected number of customers waiting in queue. W : expected time a customer spends in the system. W Q : expected time a customer spends waiting in queue E [ S ]: expected time customer spends in service. : customer arrival rate, = lim t  ∞ N ( t )/ t, where N ( t ) is the number of arrivals up to time t.

7 Fundamental relationships L = L Q + N s W = W Q + E(S) L = W L Q = W Q N s = E(S) The relationship L = W is often referred to as Little’s law.

8 t 1 t 2 t 3 t 4 t 5 t 6 t 7 T Number in system 3 2 1 A heuristic proof

9 L = [1( t 2 - t 1 )+2( t 3 - t 2 )+1( t 4 - t 3 )+2( t 5 - t 4 )+3( t 6 - t 5 )+2( t 7 - t 6 )+1(T- t 7 )]/ T = (area under curve)/ T = (T+ t 7 + t 6 - t 5 - t 4 + t 3 - t 2 - t 1 )/ T W = [( t 3 - t 1 )+( t 6 - t 2 )+( t 7 - t 4 )+(T- t 5 )]/4 = (T+ t 7 + t 6 - t 5 - t 4 + t 3 - t 2 - t 1 )/4 = (area under curve)/ N(T)

10 L = (area under curve)/ T, W = (area under curve)/ N(T)  LT = WN ( T )  L = WN ( T )/T Since as T  ∞, N ( T )/ T , L = W as T  ∞. A similar heuristic proof can be used to show L Q = W Q and N s = E(S).

11 For a single server queue:

12 Case 1 Customers arrive at regular & constant intervals Service times are constant Arrival rate < service rate ( <  ) W Q = 0 W = W Q + E(S) = E(S) L = W = E(S) L Q = W Q = 0  E(S) TH (output/throughput rate) = Why do queues form?

13 Case 2 Customers arrive at regular & constant intervals Service times are constant Arrival rate > service rate ( >  ) W Q = ∞ W = W Q + E(S) = ∞ L = W = ∞ L Q = W Q = ∞  > 1 (but utilization is actually 1) TH (output/throughput rate) =  Why do queues form?

14 Case 3 Customers arrive at regular & constant intervals Service times are not constant Arrival rate < service rate ( <  ) Example: Inter-arrival time = 8 min  Average service time = 6 min

15  = 6/8 = 0.75 TH = 1/8 parts/min = 7.5 parts/hour W Q = ? L q = ?

16 Conclusion 1: If customers arrive at a faster rate than the service rate, the system becomes instable and infinitely large queues will form. Conclusion 2: In the presence of variability, customers will generally wait for processing and a queue in front of the processing unit will build up.

17 In managing queueing systems, we must alway strive to reduce variability while allowing for enough capacity

18 Measuring Variability

19 The M/M/1 queue

20 Example Example: Customers arrive according to a Poisson process with rate of 1 per every 12 minutes and that the service time is exponential at a rate of one service per 8 minutes. What is L and W ? What happens if arrival rate increases by 20%? If there is a waiting cost of $2 per minute a customer spends in the system, what is the total cost per minute incurred in both cases?

21 Example A 20% increase in arrival rate leads to a 100% increase in number of customers!

22 The M/M/1/N queue

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24 Example: A service facility with a finite queue size of N has service rate  and an arrival rate. Each customer that is served generates $ A. Service rate can be increased. However, there is a cost $c  per unit time for operating a facility with rate . What is the optimal choice of  ?

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26 The G/G/1 queue If (1) < , (2) the distributions of customer service time and inter-arrival times are stationary, and (3) customers are served on a first come, first served (FCFS) basis, then average waiting time in the queue can be approximated as follows: waiting time in a M/M/1 queue

27 The G/G/m queue When there are m parallel servers, then average waiting time can be approximated as follows:

28 Examples m = 1 C A = C S = 1  = 1 Case 1:  = 0.50  W = 2, L = 1 Case 2:  = 0.66  W = 3, L = 1.98 Case 3:  = 0.75  W = 4, L = 3 Case 4:  = 0.80  W = 5, L = 4 Case 5:  = 0.90  W = 10, L = 9 Case 6:  = 0.95  W = 20, L = 19 Case 7:  = 0.99  W = 100, L = 99

29 Examples m = 1 C A = 1  = 1  = 0.8 Case 1: c S = 0  W = 3, L = 2.4 Case 2: c S = 0.5  W = 4, L = 3.2 Case 1: c S = 1  W = 5, L = 4 Case 1: c S = 1.5  W = 6, L = 4.8 Case 1: c S = 2  W = 7, L = 5.6

30 Network of Queues Two servers in series Customers arrive to server 1 according to a Poisson process with rate Service times are exponentially distributed at servers 1 and 2 with rates  1 and  2, respectively There is always enough waiting room between the two servers

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32 Server 1 alone is simply an M/M/1 queue. Then If server 2 alone is also an M/M/1 queue (this is actually true), then If the number of customers at servers 1 and 2 are independent (this is also true), then

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36 The results generalize to k servers in series. Each server behaves like an M/M/1 queue and number of customers at each server are independent of the number of customers at other servers.


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