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Queuing Theory For Dummies
Jean-Yves Le Boudec 1
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All You Need to Know About Queuing Theory
16/04/2017 Queuing is essential to understand the behaviour of complex computer and communication systems In depth analysis of queuing systems is hard Fortunately, the most important results are easy We will study only simple concepts 2
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1. Deterministic Queuing
16/04/2017 Easy but powerful Applies to deterministic and transient analysis Example: playback buffer sizing 3
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Use of Cumulative Functions
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Solution of Playback Delay Pb
16/04/2017 A(t) A’(t) D(t) time bits d(0) d(0) - D d(0) + D (D1): r(t - d(0) + D) (D2): r (t - d(0) - D) d(t) A. 5
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2. Operational Laws Intuition:
Say every customer pays one Fr per minute present Payoff per customer = R Rate at which we receive money = N In average λ customers per minute, N = λ R 6
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Little Again 16/04/2017 Consider a simulation where you measure R and N. You use two counters responseTimeCtr and queueLengthCtr. At end of simulation, estimate R = responseTimeCtr / NbCust N = queueLengthCtr / T where NbCust = number of customers served and T=simulation duration Both responseTimeCtr=0 and queueLengthCtr=0 initially Q: When an arrival or departure event occurs, how are both counters updated ? A: queueLengthCtr += (tnew - told) . q(told) where q(told) is the number of customers in queue just before the event. responseTimeCtr += (tnew - told) . q(told) thus responseTimeCtr == queueLengthCtr and thus N = R x NbCust/T ; now NbCust/T is our estimator of 7
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Other Operational Laws
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The Interactive User Model
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Network Laws 10
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Bottleneck Analysis Apply the following two bounds Example (1) (2) 17
16/04/2017 Apply the following two bounds Example (1) (2) 17 11
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Throughput Bounds 16/04/2017 12
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Bottlenecks 16/04/2017 A 13
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DASSA Intuition: within one busy period: to every departure we can associate one arrival with same number of customers left behind 14
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3. Single Server Queue 15
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i.e. which are event averages (vs time averages ?)
16/04/2017 i.e. which are event averages (vs time averages ?) 16
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Non Linearity of Response Time
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Impact of Variability 16/04/2017 21
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Optimal Sharing Compare the two in terms of Response time Capacity 22
16/04/2017 Optimal Sharing Compare the two in terms of Response time Capacity 22
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The Processor Sharing Queue
Models: processors, network links Insensitivity: whatever the service requirements: Egalitarian 23
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PS versus FIFO PS FIFO 24
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4. A Case Study Impact of capacity increase ? Optimal Capacity ? 25
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Methodology 26
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4.1. Deterministic Analysis
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Deterministic Analysis
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4.2 Single Queue Analysis Assume no feedback loop: 29
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4.3 Operational Analysis A refined model, with circulating users
Apply Bottleneck Analysis ( = Operational Analysis ) Z/(N-1) -Z 1/c waiting time 30
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5. Networks of Queues Stability
Queuing networks are frequently used models The stability issue may, in general, be a hard one Necessary condition for stability (Natural Condition) server utilization < at every queue 33
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Instability Examples Poisson arrivals ; jobs go through stations 1,2,1,2,1 then leave A job arrives as type 1, then becomes 2, then 3 etc Exponential, independent service times with mean mi Priority scheduling Station 1 : 5 > 3 >1 Station 2: 2 > 4 Q: What is the natural stability condition ? A: λ (m1 + m3 + m5 ) < 1 λ (m2 + m4) < 1 34
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If λ (m1 + … + m5 ) < 1 network is stable; why?
λ = 1 m1 = m3 = m4 = 0.1 m2 = m5 = 0.6 Utilization factors Station 1: 0.8 Station 2: 0.7 Network is unstable ! If λ (m1 + … + m5 ) < 1 network is stable; why? 35
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Bramson’s Example 1: A Simple FIFO Network
Poisson arrivals; jobs go through stations A, B,B…,B, A then leave Exponential, independent service times Steps 2 and last: mean is L Other steps: mean is S Q: What is the natural stability condition ? A: λ ( L + S ) < 1 λ ( (J-1)S + L ) < 1 Bramson showed: may be unstable whereas natural stability condition holds
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Bramson’s Example 2 A FIFO Network with Arbitrarily Small Utilization Factor
m queues 2 types of customers λ = 0.5 each type routing as shown, … = 7 visits FIFO Exponential service times, with mean as shown S S L S S L S S L S S L Utilization factor at every station ≤ 4 λ S Network is unstable for S ≤ 0.01 L ≤ S8 m = floor(-2 (log L )/L) 37
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Take Home Message The natural stability condition is necessary but may not be sufficient There is a class of networks where this never happens. Product Form Queuing Networks 38
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Product Form Networks Customers have a class attribute
Customers visit stations according to Markov Routing External arrivals, if any, are Poisson 2 Stations Class = step, J+3 classes Can you reduce the number of classes ? 39
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Chains Customers can switch class, but remain in the same chain ν 40
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Chains may be open or closed
Open chain = with Poisson arrivals. Customers must eventually leave Closed chain: no arrival, no departure; number of customers is constant Closed network has only closed chains Open network has only open chains Mixed network may have both 41
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3 Stations 4 classes 1 open chain 1 closed chain ν 42
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Bramson’s Example 2 A FIFO Network with Arbitrarily Small Utilization Factor
2 Stations Many classes 2 open chains Network is open 43
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Visit Rates 44
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Visit rates θ11 = θ13 = θ15 = θ22 = θ24 = λ θsc = 0 otherwise
2 Stations 5 classes 1 chain Network is open Visit rates θ11 = θ13 = θ15 = θ22 = θ24 = λ θsc = 0 otherwise 45
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ν 46
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Constraints on Stations
Stations must belong to a restricted catalog of stations See Section 8.4 for full description We will give commonly used examples Example 1: Global Processor Sharing One server Rate of server is shared equally among all customers present Service requirements for customers of class c are drawn iid from a distribution which depends on the class (and the station) Example 2: Delay Infinite number of servers No queuing, service time = service requirement = residence time 47
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Example 3 : FIFO with B servers
FIFO queueing Service requirements for customers of class c are drawn iid from an exponential distribution, independent of the class (but may depend on the station) Example of Category 2 (MSCCC station): MSCCC with B servers FIFO queueing with constraints At most one customer of each class is allowed in service Examples 1 and 2 are insensitive (service time can be anything) Examples 3 and 4 are not (service time must be exponential, same for all) 48
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Say which network satisfies the hypotheses for product form
B (FIFO, Exp) C (Prio, Exp) 49
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The Product Form Theorem
If a network satisfies the « Product Form » conditions given earlier The stationary distrib of numbers of customers can be written explicitly It is a product of terms, where each term depends only on the station Efficient algorithms exist to compute performance metrics for even very large networks For PS and Delay stations, service time distribution does not matter other than through its mean (insensitivity) The natural stability condition holds 50
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Conclusions Queuing is essential in communication and information systems M/M/1, M/GI/1, M/G/1/PS and variants have closed forms Bottleneck analysis and worst case analysis are usually very simple and often give good insights Queuing networks may be very complex to analyze except if product form – be able to recognize them ! 53
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