Queueing Theory Chapter 17 1 Basic Queueing Process

Queueing Theory Chapter 17 1

Basic Queueing Process Arrivals • Arrival time distribution • Calling population (infinite or finite) Queue Service • Capacity (infinite or finite) • Queueing discipline • Number of servers (one or more) • Service time distribution “Queueing System” 2

Examples and Applications • • Call centers (“help” desks, ordering goods) Manufacturing Banks Telecommunication networks Internet service Intelligence gathering Restaurants Other examples…. 3

Labeling Convention (Kendall-Lee) / Interarrival time distribution M D Ek G / Service time distribution Markovian (exponential interarrival times, Poisson number of arrivals) Deterministic Erlang with shape parameter k General / Number of servers / Queueing discipline FCFS first come, first served LCFS last come, first served SIRO service in random order GD general discipline / System capacity Calling population size Notes: 4

Labeling Convention (Kendall-Lee) Examples: M/M/1 M/M/5 M/G/1 M/M/3/LCFS Ek/G/2//10 M/M/1///100 5

Terminology and Notation • State of the system Number of customers in the queueing system (includes customers in service) • Queue length Number of customers waiting for service = State of the system - number of customers being served • N(t) = State of the system at time t, t ≥ 0 • Pn(t) = Probability that exactly n customers are in the queueing system at time t 6

Terminology and Notation • n = Mean arrival rate (expected # arrivals per unit time) of new customers when n customers are in the system • s = Number of servers (parallel service channels) • n = Mean service rate for overall system (expected # customers completing service per unit time) when n customers are in the system Note: n represents the combined rate at which all busy servers (those serving customers) achieve service completion. 7

Terminology and Notation When arrival and service rates are constant for all n, = mean arrival rate (expected # arrivals per unit time) = mean service rate for a busy server 1/ = expected interarrival time 1/ = expected service time = /s = utilization factor for the service facility = expected fraction of time the system’s service capacity (s ) is being utilized by arriving customers ( ) 8

Terminology and Notation Steady State When the system is in steady state, then Pn = probability that exactly n customers are in the queueing system L = expected number of customers in queueing system = … Lq = = expected queue length (excludes customers being served) … 9

Terminology and Notation Steady State When the system is in steady state, then = waiting time in system (includes service time) for each individual customer W = E[ ] q = Wq waiting time in queue (excludes service time) for each individual customer = E[ q] 10

Little’s Formula Demonstrates the relationships between L, W, Lq, and Wq • Assume n= and n= (arrival and service rates constant for all n) Intuitive Explanation: • In a steady-state queue, 11

Little’s Formula (continued) • This relationship also holds true for (expected arrival rate) when n are not equal. Recall, Pn is the steady state probability of having n customers in the system 12

Heading toward M/M/s • The most widely studied queueing models are of the form M/M/s (s=1, 2, …) • What kind of arrival and service distributions does this model assume? • Reviewing the exponential distribution…. • If T ~ exponential(α), then • A picture of the distribution: 13

Exponential Distribution Reviewed If T ~ exponential( ), then E[T] = ______ Var(T) = ______ 14

Property 1 Strictly Decreasing The pdf of exponential, f. T(t), is a strictly decreasing function • A picture of the pdf: f. T(t) t 15

Property 2 Memoryless The exponential distribution has lack of memory i. e. P(T > t+s | T > s) = P(T > t) for all s, t ≥ 0. Example: P(T > 15 min | T > 5 min) = P(T > 10 min) The probability distribution has no memory of what has already occurred. 16

Property 2 Memoryless • Prove the memoryless property • Is this assumption reasonable? – For interarrival times – For service times 17

Property 3 Minimum of Exponentials The minimum of several independent exponential random variables has an exponential distribution If T 1, T 2, …, Tn are independent r. v. s, Ti ~ expon( i) and U = min(T 1, T 2, …, Tn), U ~expon( ) Example: If there are n servers, each with exponential service times with mean , then U = time until next service completion ~ expon(____) 18

Property 4 Poisson and Exponential If the time between events, Xn ~ expon( ), then the number of events occurring by time t, N(t) ~ Poisson( t) Note: E[X(t)] = αt, thus the expected number of events per unit time is α 19

Property 5 Proportionality For all positive values of t, and for small t, P(T ≤ t+ t | T > t) ≈ t i. e. the probability of an event in interval t is proportional to the length of that interval 20

Property 6 Aggregation and Disaggregation The process is unaffected by aggregation and disaggregation Aggregation Disaggregation N 1 ~ Poisson( 1) N 2 ~ Poisson( 2) … Nk ~ Poisson( k) N 1 ~ Poisson( p 1) N ~ Poisson( ) = 1+ 2+…+ k N ~ Poisson( ) p 1 p 2 pk N 2 ~ Poisson( p 2) … Nk ~ Poisson( pk) Note: p 1+p 2+…+pk=1 21

Back to Queueing • Remember that N(t), t ≥ 0, describes the state of the system: The number of customers in the queueing system at time t • We wish to analyze the distribution of N(t) in steady state 22

Birth-and-Death Processes • • • If the queueing system is M/M/…/…, N(t) is a birth-and-death process A birth-and-death process either increases by 1 (birth), or decreases by 1 (death) General assumptions of birth-and-death processes: 1. Given N(t) = n, the probability distribution of the time remaining until the next birth is exponential with parameter λn 2. Given N(t) = n, the probability distribution of the time remaining until the next death is exponential with parameter μn 3. Only one birth or death can occur at a time 23

Rate Diagrams 24

Steady-State Balance Equations 25

M/M/1 Queueing System • • • Simplest queueing system based on birth-and-death We define = mean arrival rate = mean service rate = / = utilization ratio We require < , that is < 1 in order to have a steady state – Why? Rate Diagram 1 0 2 3 … 4 26

M/M/1 Queueing System Steady-State Probabilities Calculate Pn, n = 0, 1, 2, … 27

M/M/1 Queueing System L, Lq, W, Wq Calculate L, Lq, W, Wq 28

M/M/1 Example: ER • • • Emergency cases arrive independently at random Assume arrivals follow a Poisson input process (exponential interarrival times) and that the time spent with the ER doctor is exponentially distributed Average arrival rate = 1 patient every ½ hour = • Average service time = 20 minutes to treat each patient = • Utilization = 29

M/M/1 Example: ER Questions What is the… 1. probability that the doctor is idle? 2. probability that there are n patients? 3. expected number of patients in the ER? 4. expected number of patients waiting for the doctor? 5. expected time in the ER? 6. expected waiting time? 7. probability that there at least two patients waiting? 8. probability that a patient waits more than 30 minutes? 30

Car Wash Example • • • Consider the following 3 car washes Suppose cars arrive according to a Poisson input process and service follows an exponential distribution Fill in the following table Car Wash A Car Wash B Car Wash C 0. 1 0. 5 car/min 0. 11 car/min 0. 1 car/min L Lq W Wq P 0 What conclusions can you draw from your results? 31

M/M/s Queueing System • • We define = mean arrival rate = mean service rate s = number of servers (s > 1) = / s = utilization ratio We require < s , that is < 1 in order to have a steady state Rate Diagram 0 1 2 3 4 … 32

M/M/s Queueing System Steady-State Probabilities and P n = C n. P 0 where 33

M/M/s Queueing System L, Lq, W, Wq How to find L? W? Wq? 34

M/M/s Example: A Better ER • As before, we have – – • • Average arrival rate = 1 patient every ½ hour = 2 patients per hour Average service time = 20 minutes to treat each patient = 3 patients per hour Now we have 2 doctors s= Utilization = 35

M/M/s Example: ER Questions What is the… 1. probability that both doctors are idle? a) probability that exactly one doctor is idle? 2. probability that there are n patients? 3. expected number of patients in the ER? 4. expected number of patients waiting for a doctor? 5. expected time in the ER? 6. expected waiting time? 7. probability that there at least two patients waiting? 8. probability that a patient waits more than 30 minutes? 36

Performance Measurements s=1 s=2 ρ 2/3 1/3 L 2 3/4 Lq 4/3 1/12 W 1 hr 3/8 hr Wq 2/3 hr 1/24 hr P(at least two patients waiting in queue) 0. 296 0. 0185 P(a patient waits more than 30 minutes) 0. 404 0. 022 37

Travel Agency Example • • Suppose customers arrive at a travel agency according to a Poisson input process and service times have an exponential distribution We are given – – • • =. 10/minute = 1 customer every 10 minutes =. 08/minute = 8 customers every 100 minutes If there were only one server, what would happen? How many servers would you recommend? 38

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Single Queue vs. Multiple Queues • Would you ever want to keep separate queues for separate servers? Single queue vs. Multiple queues 42

Bank Example • • • Suppose we have two tellers at a bank Compare the single server and multiple server models Assume = 2, = 3 L Lq W Wq P 0 43

Bank Example Continued • • Suppose we now have 3 tellers Again, compare the two models 44

M/M/s//K Queueing Model (Finite Queue Variation of M/M/s) • • Now suppose the system has a maximum capacity, K We will still consider s servers Assuming s ≤ K, the maximum queue capacity is K – s List some applications for this model: • Draw the rate diagram for this problem: 45

M/M/s//K Queueing Model (Finite Queue Variation of M/M/s) Rate Diagram 0 1 2 3 4 … Balance equations: Rate In = Rate Out 46

M/M/s//K Queueing Model (Finite Queue Variation of M/M/s) Solving the balance equations, we get the following steady state probabilities: Verify that these equations match those given in the text for the single server case (M/M/1//K) 47

M/M/s//K Queueing Model (Finite Queue Variation of M/M/s) To find W and Wq: Although L ≠ W and Lq ≠ Wq because n is not equal for all n, and where 48

M/M/s///N Queueing Model (Finite Calling Population Variation of M/M/s) • • Now suppose the calling population is finite We will still consider s servers Assuming s ≤ K, the maximum queue capacity is K – s List some applications for this model: • Draw the rate diagram for this problem: 49

M/M/s///N Queueing Model (Finite Calling Population Variation of M/M/s) Rate Diagram 0 1 2 3 4 … Balance equations: Rate In = Rate Out 50

M/M/s///N Results 51

Queueing Models with Nonexponential Distributions • M/G/1 Model – – – Poisson input process, general service time distribution with mean 1/ and variance 2 Assume = / < 1 Results 52

Queueing Models with Nonexponential Distributions • • M/Ek/1 Model – Erlang: Sum of exponentials – – Think it would be useful? Can readily apply the formulae on previous slide where Other models – – – M/D/1 Ek/M/1 etc 53

Application of Queueing Theory • • We can use the results for the queueing models when making decisions on design and/or operations Some decisions that we can address – – – Number of servers Efficiency of the servers Number of queues Amount of waiting space in the queue Queueing disciplines 54

Number of Servers Suppose we want to find the number of servers that minimizes the expected total cost, E[TC] – • Expected Total Cost = Expected Service Cost + Expected Waiting Cost (E[TC]= E[SC] + E[WC]) How do these costs change as the number of servers change? Expected cost • Number of servers 55

Repair Person Example • • • Sim. Inc has 10 machines that break down frequently and 8 operators The time between breakdowns ~ Exponential, mean 20 days The time to repair a machine ~ Exponential, mean 2 days Currently Sim. Inc employs 1 repair person and is considering hiring a second Costs: – – • • Each repair person costs $280/day Lost profit due to less than 8 operating machines: $400/day for each machine that is down Objective: Minimize total cost Should Sim. Inc hire the additional repair person? 56

Repair Person Example Problem Parameters • What type of problem is this? – – – – • M/M/1 M/M/s/K M/G/1 M/M/s finite calling population M/Ek/1 M/D/1 What are the values of and ? 57

Repair Person Example Rate Diagrams • Draw the rate diagram for the single-server and two-server case Single server 0 1 2 3 4 … 8 9 10 Two servers 0 1 2 3 4 … 8 9 10 • Expected service cost (per day) = E[SC] = • Expected waiting cost (per day) = E[WC] = 58

Repair Person Example Steady-State Probabilities • Write the balance equations for each case • How to find E[WC] for s=1? s=2? 59

Repair Person Example E[WC] Calculations N=n g(n) 0 s=1 s=2 Pn g(n) Pn 0 0. 271 0 0. 433 0 1 0 0. 217 0 0. 346 0 2 0 0. 173 0 0. 139 0 3 400 0. 139 56 0. 055 24 4 800 0. 097 78 0. 019 16 5 1200 0. 058 70 0. 006 8 6 1600 0. 029 46 0. 001 0 7 2000 0. 012 24 0. 0003 0 8 2400 0. 003 7 0. 00004 0 9 2800 0. 0007 0 0. 000004 0 10 3200 0. 00007 0 0. 0000002 0 E[WC] $281/day $48/day 60

Repair Person Example Results • • We get the following results s E[SC]: E[WC]: E[TC]: 1 $280/day $281/day $561/day 2 $560/day $48/day $608/day ≥ 3 ≥ $840/day What should Sim. Inc do? 61

Supercomputer Example • • • Emerald University has plans to lease a supercomputer They have two options Supercomputer Mean number of jobs per day Cost per day MBI 30 jobs/day $5, 000/day CRAB 25 jobs/day $3, 750/day Students and faculty jobs are submitted on average of 20 jobs/day, distributed Poisson i. e. Time between submissions ~ _____ • Which computer should Emerald University lease? 62

Supercomputer Example Waiting Cost Function • Assume the waiting cost is not linear: h( ) = 500 + 400 2 ( = waiting time in days) • What distribution do the waiting times follow? • What is the expected waiting cost, E[WC]? 63

Supercomputer Example Results • Next incorporate the leasing cost to determine the expected total cost, E[TC] • Which computer should the university lease? 64