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Facilities Design

Facilities Design. S.S. Heragu Industrial Engineering Department University of Louisville. Appendix: Introduction to Queuing, Queuing Network, and Simulation Modeling. A queuing system. Queuing Models are descriptive models. What is the expected number of parts waiting in a queue?

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Facilities Design

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  1. Facilities Design S.S. Heragu Industrial Engineering Department University of Louisville

  2. Appendix:Introduction to Queuing, Queuing Network, and Simulation Modeling

  3. A queuing system

  4. Queuing Models are descriptive models • What is the expected number of parts waiting in a queue? • What is the expected time a part spends waiting in a queue? • What is the probability that a machine will be idle? • What is the probability of a queue being filled to capacity?

  5. Elements of a queuing system • Arrival process • Service process • Departure process • Queue discipline • Balking

  6. Modeling of the Arrivals and Service Processes • Ti - ithinterarrival time (or service time for the ith customer) • Assume Ti to be an independent, continuous random variable F. • Let the probability density function (pdf), expected value and variance of F be f(t), E(F) and V(F), respectively. Then,

  7. Modeling of the Arrivals and Service Processes • Suppose that F follows an exponential distribution with parameter λ. • The pdff(t) for an exponential distribution is a strictly decreasing function of t (t 0). So, • It is not difficult to infer from this property that F is likely to take on a very small value, perhaps near zero.

  8. Modeling of the Arrivals and Service Processes • Let F represent service time • Time to serve a customer is typically small, but an occasional customer requires extensive service • This service time distribution can be approximated as an exponential distribution • Many, but not all systems exhibit such a characteristic

  9. Modeling of the Arrivals and Service Processes • PDF of an exponential distribution • Cumulative probabilities for an exponential distribution

  10. Modeling of the Arrivals and Service Processes • Verify the previous result by substituting x = -λt. Then, dt = -(1/λ)dx. Therefore, • Mean can be shown to be 1/λ and Variance can be shown to be 1/λ2 for exponential dist

  11. Property 1 of the Exponential Distribution • Memoryless • But • Hence,

  12. Property 2 of the Exponential Distribution • If interarrival time is exponentially distributed, then the arrival process is Poisson • Consider an arrival process {Nt = n}, where Nt = n is the number of arrivals occur during time interval t (t 0). • Assume that N0 = 0 and the arrival process satisfies the following three conditions

  13. Property 2 of the Exponential Distribution • Assume that N0 = 0 and the arrival process satisfies the following three conditions. • (1) The probability of an arrival occurring between times t and depends only upon the length and does not depend upon either the number of arrivals occurring until time t or the specific value of t. The probability is equal to is a small quantity in comparison to especially as tends to zero. In other words,

  14. Property 2 of the Exponential Distribution • Assume that N0 = 0 and the arrival process satisfies the following three conditions. • (2) The probability of > 1 arrival occurring during a very small time interval = • (3) The number of arrivals occurring in nonoverlapping intervals are independent. E.g., the number of arrivals between the first twenty time units has no bearing on the number of arrivals between the next twenty, thirty or forty time units.

  15. Property 2 of the Exponential Distribution • Assume P{Nt = n} is the probability of Nt = n arrivals occurring during time interval t • Probability of –ve arrivals is zero. Then, • LHS above is nothing but differentiation of P{Nt = n} with respect to t. These infinite linear, first order differential equations can be solved and P{Nt = n} can be shown to be

  16. Property 2 of the Exponential Distribution • Given the arrival process described by P(Nt = n), assume F is the random variable for the time between successive arrivals – the interarrival time • To show that F follows an exponential distribution, note that P{tT} = P{NT = 0}; i.e., the probability of the next arrival exceeding time any specific value T is equivalent to having no arrival in time T. • Taking derivatives of both sides, we get: f(t) = λe-λT, which is an exponential distribution

  17. Property 2 of the Exponential Distribution • Mean of the Poisson distribution given by E{Nt} = λt is equal to the variance • Exponential is the only distribution to have such a property • Much of the above discussion holds for the service process also. To verify, the reader has to replace the terms interarrival time and arrival with service time and service completion, respectively • Notice that the notation used for the arrival process parameter is λ, whereas for the service process it is μ

  18. Kendall-Lee notation for Queuing models • A / B / C / D / E / F, where • A denotes the nature of the arrival process • B denotes the service time distribution • C is the number of parallel servers (S1) • D is the queue discipline (FCFS, LCFS, SIRO, GD, etc.) • E is the max number of customers allowed in the system (C1) • F denotes the size of the calling population (finite or infinite)

  19. Queuing models that can be solved exactly • The calling population is infinite • Some finite population models, like the M/M/1/GD/K/K model - typically called the machine repairman model • The queue capacity is infinite • For some special models, analytic solution is possible for only FCFS and priority disciplines • The probability distribution of the interarrival and service times are exponential

  20. Birth-death queuing model • The birth-death process is a continuous time Markov chain or stochastic process in which the state of the system at any continuous time (rather than discrete points in time) is a nonnegative integer • Consider a system at some state at time t, in which we have Nt = n customers. Assume that the pdf of the remaining time until next arrival (birth) and pdf of the remaining time until next service completion (death or departure) are exponential with parameters λn, μn, respectively, for n = 0, 1, 2, …

  21. Birth-death queuing model • Only one event (either a birth or a death) can occur in any period of time Δt. Birth and death rates are independent of each other, but state dependent • Because of the exponential assumption, birth and death occur randomly • Figure above (rate diagram) shows an infinite birth-death process in which birth and death rates are state dependent • Analysis of most birth-death processes computationally feasible only when the system has reached steady state

  22. Birth-death queuing model • Mean rate of entering any state n = Mean rate of leaving that state • Balance equation or flow conservation equation • Denote En(t), Ln(t) as the number of times the process enters and leaves state n by time t • We have |En(t)-Ln(t)| <1 • Dividing the left- and right-hand sides by t and letting t-> infinity, we get . Hence,

  23. Birth-death queuing model • Pj is steady-state probability of being in state j • For state 0, μ1P1 = λ0P0. Therefore, P1 = (λ0/μ1)P0 • For state 1, μ2P2 + λ0P0 = λ1P1 + μ1P1 Therefore, • P2 = (λ1/μ2)P1 + (1/μ2)(μ1P1 - λ0P0) = (λ1/μ2)(λ0/μ1)P0 + (λ0/μ2)(P0 - P0) = (λ1/μ2)(λ0/μ1)P0 • For state 2, μ3P3 + λ1P1 = λ2P2 + μ2P2 Therefore, • P3 = (λ2/μ3)P2 + (1/μ3)(μ2P2 - λ1P1) = (λ2/μ3)(λ1/μ2)(λ0/μ1)P0 • … … … • Let cn = (λn-1/μn)(λn-2/μn-1)…(λ0/μ1) for n = 1, 2, … • Then, Pn = cnP0, for n=1, 2, ...

  24. Birth-death queuing model L = average number of customers in system Lq = average number of customers in queue Ls = average number of customers in service W = average time a customers spends in system Wq= average time a customer spends in queue Ws = average time a customer spends in service λ = mean arrival rate μ = mean service rate

  25. Birth-death queuing model • Using Little’s Law, which states that L = λW, we can calculate W, Wq and Ws

  26. M/M/1/GD/inf/infqueuing model • Let λ and μ be the arrival and service rate • If λ>μ, the queue will grow infinitely; ρ = λ/μ is called the traffic intensity or serve utilization • ρ must be less than 1 • λj = λ; μj = μ, for i=1, 2, ... and μ0 = 0 • P1 = (λ/μ)P0, P2 = (λ/μ)2P0, ..., Pn = (λ/μ)nP0. Substituting ρ = (λ/μ) and assuming 0 <ρ < 1, we get P1 = ρP0, P2 = ρ2P0, ..., Pn = ρnP0, … • P0+P 1+…+Pn+… = 1. So, P0(1 + ρ + ρ 2 + … + ρn + …) = 1 • To evaluate (1 + ρ + ρ 2 + … + ρn + …), let S = (1 + ρ + ρ 2 + … + ρn + …). Then ρS = (ρ + ρ 2 + … + ρn + …) • Clearly, ρS - S = 1. Hence, S = 1/(1-ρ). Thus, P0(1+ρ+ρ 2+ …+ρn +…) = P0/(1-ρ) = 1. Therefore, P0 = 1-ρ • Similarly, P1 = ρ(1-ρ), P2 = ρ2(1-ρ), …, Pn = ρn(1-ρ), …

  27. M/M/1/GD/inf/infqueuing model • The summation can be determined by substituting • Then • RHS of the above = ρS and S was shown to be = 1/(1-ρ) • Hence, we get Therefore, • Lq can also be determined in this way:

  28. M/M/1/GD/inf/infqueuing model • Using Little's formula, we can then determine, W, Wq and Ws as

  29. Example 1 Find the following values: • 1) The percentage of the time that the press is idle. • 2) The average number of parts in the queuing system. • 3) The average queue length. • 4) The throughput time of the system. • 5) The amount of time spent in the queue.

  30. Example 1 Solution • M/M/1 queuing system • λ = 10 per hour and μ = 12 per hour • Utilization factor is ρ = λ/μ = 0.833 • The probability that the press is idle is the probability that there are no jobs being served or waiting. P0 = 1-ρ = 0.167 or the press is idle 16.7% of the time • The average number of parts in the queuing system is L = ρ/(1-ρ) = 5 parts. This is also the average Work-In-Process (WIP) inventory • The average queue length is Lq = ρ2/(1-ρ) = 4.167 parts • The throughput time (or average time spent by a part in the system) is W = 1/(μ(1-ρ)) = 0.5 hours • The amount of time spent in the queue is Wq = ρ/(μ(1-ρ)) = 0.4167 hours = 25 minutes

  31. Property 3 of the Exponential Distribution • The minimum of independent exponential random variables is also exponential • Let F1, F2, …, Fn be independent exponential random variables with parameters μ1, μ2, …μn • Let Fmin = min {F1, F2, …, Fn}. Then for any t> 0, • P(Fmin > t) = P{F1>t}P{F2>t}…P{Fn>t} =

  32. Property 3 of the Exponential Distribution • Suppose n types of customers, with the ith type of customer having an exponential interarrival time distribution with parameter λi, arrive at a queuing system • Assume that an arrival has just taken place • Then, from the no-memory property, it follows that the time remaining until the next arrival is also exponential • Using property 3, it is easy to see that the interarrival time for the entire queuing system (which is the minimum amongst all interarrival times) has an exponential distribution with parameter

  33. M/M/S/GD/inf/infqueuing model • Let λ and μ be the arrival and service rate • If n<S, all customers are in service • If n > S, all S servers are busy and n-S customers are waiting • From property 3, service rate for the entire queuing system is • The utilization factor for an M/M/S system is ρ = λ/Sμ • Clearly, μn = μ{min{n,S}} • Hence, μn = nμ for n = 1, 2, …, S and μn = Sμ for n = S+1, S+2, …, Also, λn = λ for n = 0, 1, 2, …, Hence, Pn = P0cn, for n = 1, 2, …S. We get

  34. M/M/S/GD/inf/infqueuing model • The second summation in the denominator is of the form 1+x+x2+x3+… where x = λ/(Sμ). This summation is equal to 1/(1-x) or 1/(1-(λ/Sμ)). Hence, • Substituting ρS=(λ/μ), we get • Thus,

  35. M/M/S/GD/inf/infqueuing model • We now find Lq • The last summation above denoted as S’ in the M/M/1 model was previously shown to be equal to ρ/(1-ρ)2 • Wq = Lq/λ, W=Wq+1/μ, L=λW = λ(Wq+1/μ)= Lq+λ/μ • Results for the basic M/M/1 and M/M/S models are summarized in the Table A.1 • It should be noted that ρ=λ/μ for the single server model, but ρ=λ/(μS) for the multiple server case. In both models, Pn=cnP0. Formulae for cn are summarized below • Single server model: cn=ρn, for n=1,2,…, P0 = 1-ρ • Multiple server model:

  36. Shortcut formula - M/M/S/GD/inf/infmodel • Note that the P0 calculation which itself is required to calculate Lq and Wqis somewhat tedious • Sakasegawa (1977) came up with an approximate formula to calculate Wqfrom which Lq can be obtained using Little’s law. The Wqformula is shown next

  37. Property 4 of the Exponential Distribution • If we have multiple arrivals in a queuing system and each interarrival time has an exponential distribution, the interarrival time for the system as a whole is also exponential with parameter equal to the sum of the parameters for each interarrival • Reverse also true

  38. Example 2 In an effort to reduce WIP inventory, the press in Example 1 is modified to have two servers. Find the following: • The average queue length. • The average number of parts in the queuing system (WIP). • The expected amount of time spent in the queue. • The throughput of the system.

  39. Example 2 Solution • M/M/2 model: Rate diagram shown above • λ = 10 per hour and μ = 12 per hour. The 2 servers result in a utilization factor of ρ = λ/Sμ = 0.417 • The new idle time of the press is P0 = 1/[(ρS)S/S!(1-ρ) + Σ((ρS)n/n!)] = 7/17 = 0.412, or 41.2% of the time • The expected queue length in the two server system is Lq = [ρ(ρS)SP0]/[S!(1-ρ)2] = 7/40 = 0.175 parts • The expected WIP is L = Lq + λ/μ = 121/120 = 1.0083 parts • The expected amount of time spent in the queue is Wq = [(ρS)SP0]/[S!Sμ(1-ρ)2] = 7/400 = 0.0175 hours = 1.05 minutes • The throughput time of the system is W = Wq + 1/μ = 0.10083 hours = 6.05 minutes • Veriy results using Sakasegawa (1977) formula

  40. Example 3 • A workstation in a company receives parts from two sources - a subcontractor and a machining center within the company. The arrival rate of each is Poisson with parameter λ/2. The company has purchased another identical workstation in order to increase throughput. Each has a service rate of μ. The company has the following three ways of designing the system. • Have all the parts from the subcontractor visit the first workstation, and all the parts from the internal machining center visit the recently purchased workstation for processing • Because the parts coming from the subcontractor are identical to those from the internal machining center, and the two workstations can be combined into one so as to have a service rate of 2μ, the company can have parts from both sources join a single queue and visit the combined workstation • Combine the two arrivals into a single queue but keep the workstations separate • Determine the effective arrival and service rates of the above systems • Determine which system would result in the most waiting time for the parts. Which system would result in the least waiting time? • If reducing WIP becomes a more important criterion that waiting time in the system, show that system (c) is preferred to system (b) above. Assume that WIP is measured as the mean number of parts in the queue

  41. Example 3 Solution • System (a) is an M/M/1 queue with parameters λ/2 and μ. System (b) is an M/M/1 queue with parameters λ and 2μ. System (c) is an M/M/2 queue with parameters λ and μ. The three systems are depicted below.

  42. Example 3 Solution • The arrival rate is λ/2 + λ/2 = λ for (b) and (c). The effective arrival and service rates are λ and 2μ for all three systems. • For an M/M/1 system with parameters λ and μ, the waiting time in a queue (W) and length of the queue (Lq) are: W =1/(μ(1-ρ)) = 1/(μ-λ); Lq=λ2/μ(μ-λ). • For an M/M/2 system, P0 =1/[(2ρ)2/(2!(1-ρ)) + Σ1n=0 ((2ρ)n/n!)] =1/[(2ρ2)/(2!(1-ρ) + (2ρ)0/0! + (2ρ)1/1!]=[2(1-ρ)]/[2+4ρ-2ρ-4ρ2+4ρ2)] = (1-ρ)/(1+ρ); Lq=[(2ρ)2P0ρ]/[2!(1-ρ)2] = [(2ρ)2(1-ρ)ρ]/[2!(1+ρ)(1-ρ)2] = 2ρ3/(1-ρ2) = λ3/[μ(4μ2-λ2)]; Wq=Lq/λ = λ2/[μ(4μ2-λ2)]; W=Wq + 1/μ = λ2/[μ(4μ2-λ2)]+ 1/μ = 4μ/(4μ2-λ2) • For system (a), therefore, W(a) = 1/(μ-λ/2) = 2/(2μ-λ). Lq(a) = λ2/4μ(μ-λ/2) = λ2/2μ(2μ-λ). For system (b), W(b) = 1/(2μ-λ). Lq(b) = λ2/2μ(2μ-λ). For system (c), W(c) = 4μ/(4μ2-λ2). Lq(c) = λ3/[μ(4μ2-λ2)]. To prevent an infinite queue, λ must be less than 2μ (i.e. ρ<1). Therefore, W(c) = 4μ/(4μ2-λ2) = [4μ(2μ+λ)][1/(2μ-λ)] > 1/(2μ-λ) = W(b). Thus, W(a)>W(c) >W(b). Hence, system (a) would result in the most waiting time and system (b) the least. • The WIP in the queue of system (c) is given by Lq(c) = [λ2/(2μ(2μ-λ))][2λ/(2μ+λ)] < [λ2]/[2μ(2μ-λ)] = Lq(b). When WIP in the queue is the criterion system (c) is preferred to system (b), but when manufacturing lead time is considered, (b) is better than (c).

  43. Finite Queue Size Models • Single server model: • Multiple server model:

  44. Finite Source Models • Single server model: • Multiple server model:

  45. Nonexponential Models • M/G/1: • M/Ek/1: • G/G/m:

  46. Queuing Networks • Open • Closed • Semi-Open

  47. Open Networks • Jackson network has the following properties. • The arrival process from the external node to node i is Poisson with mean rate of γi. • The service time at node i follows an exponential distribution with parameter μi. • pi0 is the probability that a part will exit the network after completion of processing at node i and pij is the probability that a part will visit node j after completion of processing at node i. pi0 and pij are assumed to be known and independent of the state of the system. Two-machine System

  48. Two-node, Open Jackson Network The flow balance (rate in equals rate out) equations for any {n1, n2} pair greater than 0 (λ+μ1+μ2)P(n1,n2) = λP(n1-1,n2) + μ2P(n1,n2+1) + μ1P(n1+1,n2-1) For the remaining states (0,0), (n1,0), (0,n2), the following flow equations are also easy to derive. λP(0,0) = μ2P(0,1) (λ+μ1)P(n1,0) = μ2P(n1,1) + λP(n1-1,0) (λ+μ2)P(0,n2) = μ1P(1,n2-1) + μ2P(0,n2+1) Also, Σ Σ P(n1,n2) = 1

  49. Two-node, Open Jackson Network P(n1,n2) = ρ1n1(1-ρ1)ρ2n2(1-ρ2) = Pn1Pn2 P(0,0) = (1-ρ1)(1-ρ2) = P0P0

  50. General Open Jackson Network • For each node i, i=1, 2, …, m, let: • γi mean rate of the Poisson arrival process from the external world to node i • 1/μi mean (exponentially distributed) service time at node i • Si number of identical servers at node i • pi0 probability that a part will exit the network after completion of processing at node i • pij probability that a part will visit node j after completion of processing at node i. • Because the arrival at a node is made up of external and internal arrivals - the latter occurring as a result of service completions at other internal nodes - the arrival rate at node i, i=1, 2, …, m, is:

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