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

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

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A queuing system

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

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Elements of a queuing system - Arrival process - Service process - Departure process - Queue discipline - Balking

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

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

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

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

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Modeling of the Arrivals and Service Processes Verify the previous result by substituting x = -λt. Then, dt = -(1/λ)dx. Therefore, 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/λ 2 for exponential dist Mean can be shown to be 1/λ and Variance can be shown to be 1/λ 2 for exponential dist

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Property 1 of the Exponential Distribution Memoryless Memoryless But But Hence, Hence,

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

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Property 2 of the Exponential Distribution Assume that N 0 = 0 and the arrival process satisfies the following three conditions. Assume that N 0 = 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, is a small quantity in comparison to especially as tends to zero. In other words,

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Property 2 of the Exponential Distribution Assume that N 0 = 0 and the arrival process satisfies the following three conditions. Assume that N 0 = 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.

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

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

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Property 2 of the Exponential Distribution Mean of the Poisson distribution given by E{N t } = λt is equal to the variance Mean of the Poisson distribution given by E{N t } = λt is equal to the variance Exponential is the only distribution to have such a property 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 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 Notice that the notation used for the arrival process parameter is λ, whereas for the service process it is μ

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Kendall-Lee notation for Queuing models A / B / C / D / E / F, where 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)

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

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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 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 N t = 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 λ n, μ n, respectively, for n = 0, 1, 2, … Consider a system at some state at time t, in which we have N t = 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 λ n, μ n, respectively, for n = 0, 1, 2, …

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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 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 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 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 Analysis of most birth-death processes computationally feasible only when the system has reached steady state

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

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Birth-death queuing model P j is steady-state probability of being in state j P j is steady-state probability of being in state j For state 0, μ 1 P 1 = λ 0 P 0. Therefore, P 1 = (λ 0 /μ 1 )P 0 For state 0, μ 1 P 1 = λ 0 P 0. Therefore, P 1 = (λ 0 /μ 1 )P 0 For state 1, μ 2 P 2 + λ 0 P 0 = λ 1 P 1 + μ 1 P 1 Therefore, For state 1, μ 2 P 2 + λ 0 P 0 = λ 1 P 1 + μ 1 P 1 Therefore, - P 2 = (λ 1 /μ 2 )P 1 + (1/μ 2 )(μ 1 P 1 - λ 0 P 0 ) = (λ 1 /μ 2 )(λ 0 /μ 1 )P 0 + (λ 0 /μ 2 )(P 0 - P 0 ) = (λ 1 /μ 2 )(λ 0 /μ 1 )P 0 For state 2, μ 3 P 3 + λ 1 P 1 = λ 2 P 2 + μ 2 P 2 Therefore, For state 2, μ 3 P 3 + λ 1 P 1 = λ 2 P 2 + μ 2 P 2 Therefore, - P 3 = (λ 2 /μ 3 )P 2 + (1/μ 3 )(μ 2 P 2 - λ 1 P 1 ) = (λ 2 /μ 3 )(λ 1 /μ 2 )(λ 0 /μ 1 )P 0 … … … … … … Let c n = (λ n-1 /μ n )(λ n-2 /μ n-1 )…(λ 0 /μ 1 ) for n = 1, 2, … Let c n = (λ n-1 /μ n )(λ n-2 /μ n-1 )…(λ 0 /μ 1 ) for n = 1, 2, … Then, P n = c n P 0, for n=1, 2,... Then, P n = c n P 0, for n=1, 2,...

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Birth-death queuing model L=average number of customers in system L q =average number of customers in queue L s =average number of customers in service W=average time a customers spends in system W q =average time a customer spends in queue W s =average time a customer spends in service λ=mean arrival rate μ=mean service rate

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Birth-death queuing model Using Little’s Law, which states that L = λW, we can calculate W, W q and W s Using Little’s Law, which states that L = λW, we can calculate W, W q and W s

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M/M/1/GD/ inf/inf queuing model Let λ and μ be the arrival and service rate 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 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 ρ must be less than 1 λ j = λ; μ j = μ, for i=1, 2,... and μ 0 = 0 λ j = λ; μ j = μ, for i=1, 2,... and μ 0 = 0 P 1 = (λ/μ)P 0, P 2 = (λ/μ) 2 P 0,..., P n = (λ/μ) n P 0. Substituting ρ = (λ/μ) and assuming 0 < ρ < 1, we get P 1 = ρP 0, P 2 = ρ 2 P 0,..., P n = ρ n P 0, … P 1 = (λ/μ)P 0, P 2 = (λ/μ) 2 P 0,..., P n = (λ/μ) n P 0. Substituting ρ = (λ/μ) and assuming 0 < ρ < 1, we get P 1 = ρP 0, P 2 = ρ 2 P 0,..., P n = ρ n P 0, … P 0 +P 1 +…+P n +… = 1. So, P 0 (1 + ρ + ρ 2 + … + ρ n + …) = 1 P 0 +P 1 +…+P n +… = 1. So, P 0 (1 + ρ + ρ 2 + … + ρ n + …) = 1 To evaluate (1 + ρ + ρ 2 + … + ρ n + …), let S = (1 + ρ + ρ 2 + … + ρ n + …). Then ρS = (ρ + ρ 2 + … + ρ n + …) To evaluate (1 + ρ + ρ 2 + … + ρ n + …), let S = (1 + ρ + ρ 2 + … + ρ n + …). Then ρS = (ρ + ρ 2 + … + ρ n + …) Clearly, ρS - S = 1. Hence, S = 1/(1-ρ). Thus, P 0 (1+ρ+ρ 2 + …+ρ n +…) = P 0 /(1-ρ) = 1. Therefore, P 0 = 1-ρ Clearly, ρS - S = 1. Hence, S = 1/(1-ρ). Thus, P 0 (1+ρ+ρ 2 + …+ρ n +…) = P 0 /(1-ρ) = 1. Therefore, P 0 = 1-ρ Similarly, P 1 = ρ(1-ρ), P 2 = ρ 2 (1-ρ), …, P n = ρ n (1-ρ), … Similarly, P 1 = ρ(1-ρ), P 2 = ρ 2 (1-ρ), …, P n = ρ n (1-ρ), …

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

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M/M/1/GD/ inf/inf queuing model Using Little's formula, we can then determine, W, W q and W s as Using Little's formula, we can then determine, W, W q and W s as

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

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Example 1 Solution M/M/1 queuing system M/M/1 queuing system λ = 10 per hour and μ = 12 per hour λ = 10 per hour and μ = 12 per hour Utilization factor is ρ = λ/μ = 0.833 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. P 0 = 1-ρ = 0.167 or the press is idle 16.7% of the time The probability that the press is idle is the probability that there are no jobs being served or waiting. P 0 = 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 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 L q = ρ 2 /(1-ρ) = 4.167 parts The average queue length is L q = ρ 2 /(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 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 W q = ρ/(μ(1-ρ)) = 0.4167 hours = 25 minutes The amount of time spent in the queue is W q = ρ/(μ(1-ρ)) = 0.4167 hours = 25 minutes

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Property 3 of the Exponential Distribution The minimum of independent exponential random variables is also exponential The minimum of independent exponential random variables is also exponential Let F 1, F 2, …, F n be independent exponential random variables with parameters μ 1, μ 2, … μ n Let F 1, F 2, …, F n be independent exponential random variables with parameters μ 1, μ 2, … μ n Let F min = min {F 1, F 2, …, F n }. Then for any t > 0, Let F min = min {F 1, F 2, …, F n }. Then for any t > 0, P(F min > t) = P{F 1 >t}P{F 2 >t}…P{F n >t} = P(F min > t) = P{F 1 >t}P{F 2 >t}…P{F n >t} =

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Property 3 of the Exponential Distribution Suppose n types of customers, with the i th type of customer having an exponential interarrival time distribution with parameter λ i, arrive at a queuing system Suppose n types of customers, with the i th type of customer having an exponential interarrival time distribution with parameter λ i, arrive at a queuing system Assume that an arrival has just taken place 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 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 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

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

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M/M/S/GD/ inf/inf queuing model The second summation in the denominator is of the form 1+x+x 2 +x 3 +… where x = λ/(Sμ). This summation is equal to 1/(1-x) or 1/(1-(λ/Sμ)). Hence, The second summation in the denominator is of the form 1+x+x 2 +x 3 +… where x = λ/(Sμ). This summation is equal to 1/(1-x) or 1/(1-(λ/Sμ)). Hence, Substituting ρS=(λ/μ), we get Substituting ρS=(λ/μ), we get Thus, Thus,

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M/M/S/GD/ inf/inf queuing model We now find L q We now find L q The last summation above denoted as S’ in the M/M/1 model was previously shown to be equal to ρ/(1-ρ) 2 The last summation above denoted as S’ in the M/M/1 model was previously shown to be equal to ρ/(1-ρ) 2 W q = L q /λ, W=W q +1/μ, L=λW = λ(W q +1/μ)= L q +λ/μ W q = L q /λ, W=W q +1/μ, L=λW = λ(W q +1/μ)= L q +λ/μ Results for the basic M/M/1 and M/M/S models are summarized in the Table A.1 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, P n =c n P 0. Formulae for c n are summarized below 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, P n =c n P 0. Formulae for c n are summarized below Single server model: c n =ρ n, for n=1,2,…, P 0 = 1-ρ Single server model: c n =ρ n, for n=1,2,…, P 0 = 1-ρ Multiple server model: Multiple server model:

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Shortcut formula - M/M/S/GD/ inf/inf model Note that the P 0 calculation which itself is required to calculate L q and W q is somewhat tedious Note that the P 0 calculation which itself is required to calculate L q and W q is somewhat tedious Sakasegawa (1977) came up with an approximate formula to calculate W q from which L q can be obtained using Little’s law. The W q formula is shown next Sakasegawa (1977) came up with an approximate formula to calculate W q from which L q can be obtained using Little’s law. The W q formula is shown next

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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 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 Reverse also true

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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 queue length. The average number of parts in the queuing system (WIP). The average number of parts in the queuing system (WIP). The expected amount of time spent in the queue. The expected amount of time spent in the queue. The throughput of the system. The throughput of the system.

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Example 2 Solution M/M/2 model: Rate diagram shown above 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 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 P 0 = 1/[(ρS) S /S!(1-ρ) + Σ((ρS) n /n!)] = 7/17 = 0.412, or 41.2% of the time The new idle time of the press is P 0 = 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 L q = [ρ(ρS) S P 0 ]/[S!(1-ρ) 2 ] = 7/40 = 0.175 parts The expected queue length in the two server system is L q = [ρ(ρS) S P 0 ]/[S!(1-ρ) 2 ] = 7/40 = 0.175 parts The expected WIP is L = L q + λ/μ = 121/120 = 1.0083 parts The expected WIP is L = L q + λ/μ = 121/120 = 1.0083 parts The expected amount of time spent in the queue is W q = [(ρS) S P 0 ]/[S!Sμ(1-ρ) 2 ] = 7/400 = 0.0175 hours = 1.05 minutes The expected amount of time spent in the queue is W q = [(ρS) S P 0 ]/[S!Sμ(1-ρ) 2 ] = 7/400 = 0.0175 hours = 1.05 minutes The throughput time of the system is W = W q + 1/μ = 0.10083 hours = 6.05 minutes The throughput time of the system is W = W q + 1/μ = 0.10083 hours = 6.05 minutes Veriy results using Sakasegawa (1977) formula Veriy results using Sakasegawa (1977) formula

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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. 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 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? 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 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

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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. 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.

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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. 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 (L q ) are: W =1/(μ(1-ρ)) = 1/(μ-λ); L q =λ 2 /μ(μ-λ). For an M/M/1 system with parameters λ and μ, the waiting time in a queue (W) and length of the queue (L q ) are: W =1/(μ(1-ρ)) = 1/(μ-λ); L q =λ 2 /μ(μ-λ). For an M/M/2 system, P 0 =1/[(2ρ) 2 /(2!(1-ρ)) + Σ 1 n=0 ((2ρ) n /n!)] =1/[(2ρ 2 )/(2!(1-ρ) + (2ρ) 0 /0! + (2ρ) 1 /1!] =[2(1-ρ)]/[2+4ρ-2ρ-4ρ 2 +4ρ 2 )] = (1-ρ)/(1+ρ); L q =[(2ρ) 2 P 0 ρ]/[2!(1- ρ) 2 ] = [(2ρ) 2 (1-ρ)ρ]/[2!(1+ρ)(1-ρ) 2 ] = 2ρ 3 /(1-ρ 2 ) = λ 3 /[μ(4μ 2 -λ 2 )]; W q =L q /λ = λ 2 /[μ(4μ 2 - λ 2 )]; W=W q + 1/μ = λ 2 /[μ(4μ 2 -λ 2 )]+ 1/μ = 4μ/(4μ 2 -λ 2 ) For an M/M/2 system, P 0 =1/[(2ρ) 2 /(2!(1-ρ)) + Σ 1 n=0 ((2ρ) n /n!)] =1/[(2ρ 2 )/(2!(1-ρ) + (2ρ) 0 /0! + (2ρ) 1 /1!] =[2(1-ρ)]/[2+4ρ-2ρ-4ρ 2 +4ρ 2 )] = (1-ρ)/(1+ρ); L q =[(2ρ) 2 P 0 ρ]/[2!(1- ρ) 2 ] = [(2ρ) 2 (1-ρ)ρ]/[2!(1+ρ)(1-ρ) 2 ] = 2ρ 3 /(1-ρ 2 ) = λ 3 /[μ(4μ 2 -λ 2 )]; W q =L q /λ = λ 2 /[μ(4μ 2 - λ 2 )]; W=W q + 1/μ = λ 2 /[μ(4μ 2 -λ 2 )]+ 1/μ = 4μ/(4μ 2 -λ 2 ) For system (a), therefore, W (a) = 1/(μ-λ/2) = 2/(2μ-λ). L q (a) = λ 2 /4μ(μ-λ/2) = λ 2 /2μ(2μ- λ). For system (b), W (b) = 1/(2μ-λ). L q (b) = λ 2 /2μ(2μ-λ). For system (c), W (c) = 4μ/(4μ 2 -λ 2 ). L q (c) = λ 3 /[μ(4μ 2 -λ 2 )]. To prevent an infinite queue, λ must be less than 2μ (i.e. ρ 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. For system (a), therefore, W (a) = 1/(μ-λ/2) = 2/(2μ-λ). L q (a) = λ 2 /4μ(μ-λ/2) = λ 2 /2μ(2μ- λ). For system (b), W (b) = 1/(2μ-λ). L q (b) = λ 2 /2μ(2μ-λ). For system (c), W (c) = 4μ/(4μ 2 -λ 2 ). L q (c) = λ 3 /[μ(4μ 2 -λ 2 )]. To prevent an infinite queue, λ must be less than 2μ (i.e. ρ 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 L q (c) = [λ 2 /(2μ(2μ-λ))][2λ/(2μ+λ)] < [λ 2 ]/[2μ(2μ-λ)] = L q (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). The WIP in the queue of system (c) is given by L q (c) = [λ 2 /(2μ(2μ-λ))][2λ/(2μ+λ)] < [λ 2 ]/[2μ(2μ-λ)] = L q (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).

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Finite Queue Size Models Single server model: Single server model: Multiple server model: Multiple server model: ModelM/M/1/GD/C/infM/M/S/GD/C/inf PoPo (1-ρ)/(1-ρ C+1 )1/[Σ S n=1 (ρS) n /n! + ((ρS) S /S!)(Σ C n=S+1 ρ n-S ] Lρ/(1-ρ) - (C+1)ρ C+1 /(1-ρ C+1 )L q +λ(1-P C )/μ LqLq L+P 0 -1(ρS) S P 0 ρ[1-ρ C-S -(C-S)ρ C-S (1-ρ)]/[S!(1-ρ) 2 ] WL/(λ(1-P C )) WqWq L q /(λ(1-P C ))

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Finite Source Models Single server model: Single server model: Multiple server model: Multiple server model: ModelM/M/1/GD/N/NM/M/S/GD/N/N P0P0 1/[Σ N n=0 (λ n N!/(μ n (N-n)!)]1/[Σ S-1 n=0 (λ n N!)/(μ n n!(N-n)!)+Σ N n=S (λ n N!)/(μ n S!S n-S (N-n)!)] LN-μ(1-P 0 )/λL q +Σ S-1 n=0 nP n + S(1-Σ S-1 n=0 P n ) LqLq N-(λ+μ)(1-P 0 )/λΣ N n=S (n-S)P n WL/(λ(N-L)) WqWq L q /(λ(N-L))

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Nonexponential Models M/G/1: M/G/1: M/E k /1: M/E k /1: G/G/m: G/G/m:

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Queuing Networks Open Open Closed Closed Semi-Open Semi-Open

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Open Networks Jackson network has the following properties. Jackson network has the following properties. The arrival process from the external node to node i is Poisson with mean rate of γ i. The arrival process from the external node to node i is Poisson with mean rate of γ i. The service time at node i follows an exponential distribution with parameter μ i. The service time at node i follows an exponential distribution with parameter μ i. p i0 is the probability that a part will exit the network after completion of processing at node i and p ij is the probability that a part will visit node j after completion of processing at node i. p i0 and p ij are assumed to be known and independent of the state of the system. p i0 is the probability that a part will exit the network after completion of processing at node i and p ij is the probability that a part will visit node j after completion of processing at node i. p i0 and p ij are assumed to be known and independent of the state of the system. Two-machine System Two-machine System

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Two-node, Open Jackson Network The flow balance (rate in equals rate out) equations for any {n 1, n 2 } pair greater than 0 (λ+μ 1 +μ 2 )P(n 1,n 2 ) = λP(n 1 -1,n 2 ) + μ 2 P(n 1,n 2 +1) + μ 1 P(n 1 +1,n 2 -1) For the remaining states (0,0), (n 1,0), (0,n 2 ), the following flow equations are also easy to derive. λP(0,0) = μ 2 P(0,1) (λ+μ 1 )P(n 1,0) = μ 2 P(n 1,1) + λP(n 1 -1,0) (λ+μ 2 )P(0,n 2 ) = μ 1 P(1,n 2 -1) + μ 2 P(0,n 2 +1) Also, Σ Σ P(n 1,n 2 ) = 1

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Two-node, Open Jackson Network P(n 1,n 2 ) = ρ 1 n1 (1-ρ 1 )ρ 2 n2 (1-ρ 2 ) = P n 1 P n 2 P(0,0) = (1-ρ 1 )(1-ρ 2 ) = P 0 P 0

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General Open Jackson Network For each node i, i=1, 2, …, m, let: For each node i, i=1, 2, …, m, let: γ i mean rate of the Poisson arrival process from the external world to node i γ i mean rate of the Poisson arrival process from the external world to node i 1/μ i mean (exponentially distributed) service time at node i 1/μ i mean (exponentially distributed) service time at node i S i number of identical servers at node i S i number of identical servers at node i p i0 probability that a part will exit the network after completion of processing at node i p i0 probability that a part will exit the network after completion of processing at node i p ij probability that a part will visit node j after completion of processing at node i. p ij 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: 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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Example 4 Consider a system consisting of a pre-heat furnace, a forge press, and an inspection station. An ingot arrives at the furnace to be heated and then sent to the press. After pressing, the ingot is transferred to inspection. Ingots are delivered to the furnace at a rate of 12 per hour. The furnace heats the ingots at a rate of 15 per hour. The press and inspection each have capacities of 18 and 21 ingots per hour. All interarrivals are assumed to be exponential. The three machine network is shown below. Consider a system consisting of a pre-heat furnace, a forge press, and an inspection station. An ingot arrives at the furnace to be heated and then sent to the press. After pressing, the ingot is transferred to inspection. Ingots are delivered to the furnace at a rate of 12 per hour. The furnace heats the ingots at a rate of 15 per hour. The press and inspection each have capacities of 18 and 21 ingots per hour. All interarrivals are assumed to be exponential. The three machine network is shown below. Determine the expected number of jobs in the furnace (F), press (P), and inspection (I) as well as the WIP and processing time of the system. Determine the expected number of jobs in the furnace (F), press (P), and inspection (I) as well as the WIP and processing time of the system. Suppose the capacity of the queue in front of the press is 5 ingots, is the current design sufficient? Suppose the capacity of the queue in front of the press is 5 ingots, is the current design sufficient?

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Solution to Example 4 The arrival rate for both P and I is λ = 12 ingots per hour The arrival rate for both P and I is λ = 12 ingots per hour λ F = 12 per hourμ F = 15 per hourρ F = 0.80 λ F = 12 per hourμ F = 15 per hourρ F = 0.80 λ P = 12 per hourμ P = 18 per hourρ P = 0.67 λ P = 12 per hourμ P = 18 per hourρ P = 0.67 λ I = 12 per hourμ I = 21 per hourρ I = 0.57 λ I = 12 per hourμ I = 21 per hourρ I = 0.57 Using the equations in Table A.1, we find the operating characteristics as follows. Using the equations in Table A.1, we find the operating characteristics as follows. W F = 1/(μ F (1-ρ F )) = 1/(15(1 - 0.8)) = 0.33 hours = 20 minutes W F = 1/(μ F (1-ρ F )) = 1/(15(1 - 0.8)) = 0.33 hours = 20 minutes L F = ρ F /(1-ρ F ) = 0.8/(1 - 0.8) = 4 ingots L F = ρ F /(1-ρ F ) = 0.8/(1 - 0.8) = 4 ingots W P = 1/(μ P (1-ρ P )) = 1/(18(1 - 0.67)) = 0.167 hours = 10 minutes W P = 1/(μ P (1-ρ P )) = 1/(18(1 - 0.67)) = 0.167 hours = 10 minutes L P = ρ P /(1-ρ P ) = 0.67/(1 - 0.67) = 2 ingots L P = ρ P /(1-ρ P ) = 0.67/(1 - 0.67) = 2 ingots W I = 1/(μ I (1-ρ I )) = 1/(21(1 - 0.57)) = 0.11 hours = 6.67 minutes W I = 1/(μ I (1-ρ I )) = 1/(21(1 - 0.57)) = 0.11 hours = 6.67 minutes L I = ρ I /(1-ρ I ) = 0.57/(1 - 0.57) = 1.33 ingots L I = ρ I /(1-ρ I ) = 0.57/(1 - 0.57) = 1.33 ingots The system WIP and processing time is the sum of the individual values. The system WIP and processing time is the sum of the individual values. L = L F + L P + L I = 4 + 2 + 1.33 = 7.33 ingots L = L F + L P + L I = 4 + 2 + 1.33 = 7.33 ingots W = W F + W P + W I = 20 + 10 + 6.67 = 36.67 minutes W = W F + W P + W I = 20 + 10 + 6.67 = 36.67 minutes

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Solution to Example 4 To find out if a design with a maximum of 5 ingots is sufficient, we must find the probability that more than 5 ingots are in the queue. Recall that P n, the probability of n jobs being at a machine in steady state is given by the equation, P n =(1-ρ)ρ n, n 1. Therefore, To find out if a design with a maximum of 5 ingots is sufficient, we must find the probability that more than 5 ingots are in the queue. Recall that P n, the probability of n jobs being at a machine in steady state is given by the equation, P n =(1-ρ)ρ n, n 1. Therefore, P n>5 = 1-P n 5 = 1-P n<5 = 1-(1-0.67)(1.00+0.67+0.67 2 +0.67 3 +0.67 4 +0.675 5 ) = 0.09046. This design will result in the queue in front of the press being at full capacity 9.05% of the time. When the queue is full there will be a blocking problem in the queuing system and the steady-state results in part 1 will not hold because they were derived under the assumption that each machine has an infinite queue in front of it. This design will result in the queue in front of the press being at full capacity 9.05% of the time. When the queue is full there will be a blocking problem in the queuing system and the steady-state results in part 1 will not hold because they were derived under the assumption that each machine has an infinite queue in front of it.

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Example 5 Consider the open network queuing system in Example 4. Suppose that the rejection rate at inspection is 10%. Rejected parts are sent back to the furnace and are reworked through the entire system. This new network is shown in Figure A.11. Determine the expected number of jobs in the furnace (F), press (P), and inspection (I) as well as the WIP and processing time of the system. Consider the open network queuing system in Example 4. Suppose that the rejection rate at inspection is 10%. Rejected parts are sent back to the furnace and are reworked through the entire system. This new network is shown in Figure A.11. Determine the expected number of jobs in the furnace (F), press (P), and inspection (I) as well as the WIP and processing time of the system.

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Solution to Example 5 Step 1: Calculate the arrival rate at each center. λ F =γ F +λ R, where λ F = effective arrival rate to the furnace λ F =γ F +λ R, where λ F = effective arrival rate to the furnace γ F = external arrival rate γ F = external arrival rate λ R = internal arrival rate (rejected parts) = P(R)λ F λ R = internal arrival rate (rejected parts) = P(R)λ F We also have, We also have, P(R) = the probability of rejection; P(R) = the probability of rejection; P(A) = the probability of acceptance = 1-P(R) P(A) = the probability of acceptance = 1-P(R) Therefore, Therefore, λ F = γ F /(1-P(R))= γ F /P(A) λ F = γ F /(1-P(R))= γ F /P(A) The arrival and service rates and utilization at each center are: The arrival and service rates and utilization at each center are: λ F =13.33 per hourμ F =15 per hourρ F =0.889 λ F =13.33 per hourμ F =15 per hourρ F =0.889 λ P =13.33 per hourμ P =18 per hourρ P =0.741 λ P =13.33 per hourμ P =18 per hourρ P =0.741 λ I =13.33 per hourμ I =21 per hourρ I =0.635 λ I =13.33 per hourμ I =21 per hourρ I =0.635 Utilization factors are now calculated using the effective arrival rates Utilization factors are now calculated using the effective arrival rates

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Solution to Example 5 Step 2: Analyze each machine in the system independently. W F =1/(μ F (1-ρ F )) = 1/(15(1 - 0.889)) = 0.6 hours = 36 minutes W F =1/(μ F (1-ρ F )) = 1/(15(1 - 0.889)) = 0.6 hours = 36 minutes L F =ρ F /(1-ρ F ) = 0.889/(1 - 0.889) = 8 ingots L F =ρ F /(1-ρ F ) = 0.889/(1 - 0.889) = 8 ingots W P =1/(μ P (1-ρ P )) = 1/(18(1 - 0.741)) = 0.214 hours = 12.9 minutes W P =1/(μ P (1-ρ P )) = 1/(18(1 - 0.741)) = 0.214 hours = 12.9 minutes L P =ρ P /(1-ρ P ) = 0.741/(1 - 0.741) = 2.86 ingots L P =ρ P /(1-ρ P ) = 0.741/(1 - 0.741) = 2.86 ingots W I =1/(μ I (1-ρ I )) = 1/(21(1 - 0.635)) = 0.124 hours = 7.5 minutes W I =1/(μ I (1-ρ I )) = 1/(21(1 - 0.635)) = 0.124 hours = 7.5 minutes L I =ρ I /(1-ρ I ) = 0.635/(1 - 0.635) = 1.74 ingots L I =ρ I /(1-ρ I ) = 0.635/(1 - 0.635) = 1.74 ingots Step 3: Combine the results from each center to analyze the performance of the entire system. v F =λ F /γ F = 13.33/12 = 1.111v P =v I = 1.111 v F =λ F /γ F = 13.33/12 = 1.111v P =v I = 1.111 L=L F +L P +L I = 8 + 2.86 + 1.74 = 12.6 ingots L=L F +L P +L I = 8 + 2.86 + 1.74 = 12.6 ingots W=v F W F +v P W P +v I W I = 1.111(36 + 12.9 + 7.5) = 62.7 minutes W=v F W F +v P W P +v I W I = 1.111(36 + 12.9 + 7.5) = 62.7 minutes The "rework" loop increases the WIP and cycle time by 71.9%, and 70.9%, respectively The "rework" loop increases the WIP and cycle time by 71.9%, and 70.9%, respectively

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Example 6 The figure below shows a four-machine open network queuing system consisting of a furnace (F), a press (P), a rolling mill (M), and an inspection area (I). Each machine has a single server and the rejection rate is 5%. The part routing matrix is shown below. Given that the external arrival rate is λ=12, and μ F =15, μ P =12, μ M =18, and μ I =21, determine the WIP and processing time of the system The figure below shows a four-machine open network queuing system consisting of a furnace (F), a press (P), a rolling mill (M), and an inspection area (I). Each machine has a single server and the rejection rate is 5%. The part routing matrix is shown below. Given that the external arrival rate is λ=12, and μ F =15, μ P =12, μ M =18, and μ I =21, determine the WIP and processing time of the system From/ToFPMIExit F-0.30.7-- P---1.0- M-0.4-0.6- I0.05----0.95

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Solution to Example 6 Step 1: Calculate the effective arrival rate. λ F =λ+0.05λ I ; λ P =0.3λ F +0.4λ M ; λ M =0.7λ F ; λ I =1.0λ P +0.6λ M =λ F λ F =λ+0.05λ I ; λ P =0.3λ F +0.4λ M ; λ M =0.7λ F ; λ I =1.0λ P +0.6λ M =λ F From Example 5, we know that λ F = λ/P(A) = 12/0.95 = 12.63. Solving these equations yields the effective arrival rates presented next. The service rates and utilizations at each machine center are also shown. From Example 5, we know that λ F = λ/P(A) = 12/0.95 = 12.63. Solving these equations yields the effective arrival rates presented next. The service rates and utilizations at each machine center are also shown. λ F =12.63μ F =15ρ F =0.842 λ F =12.63μ F =15ρ F =0.842 λ P =7.33μ P =12ρ P =0.611 λ P =7.33μ P =12ρ P =0.611 λ M =8.84μ M =18ρ M =0.491 λ M =8.84μ M =18ρ M =0.491 λ I =12.63μ I =21ρ I =0.602 λ I =12.63μ I =21ρ I =0.602 Step 2: Analyze each machine in the system independently. All the machines have only one server, therefore, we will use Table A.1 for the operating characteristics of an M/M/1 queue. W F =1/(μ F (1-ρ F )) = 1/(15(1 - 0.842)) = 0.422 hours = 25.33 minutes W F =1/(μ F (1-ρ F )) = 1/(15(1 - 0.842)) = 0.422 hours = 25.33 minutes L F =ρ F /(1-ρ F ) = 0.842/(1 - 0.842) = 5.33 ingots L F =ρ F /(1-ρ F ) = 0.842/(1 - 0.842) = 5.33 ingots W P =1/(μ P (1-ρ P )) = 1/(15(1 - 0.842)) = 0.422 hours = 25.33 minutes W P =1/(μ P (1-ρ P )) = 1/(15(1 - 0.842)) = 0.422 hours = 25.33 minutes L P =ρ P /(1-ρ P ) = 0.611/(1 - 0.611) = 1.57 ingots L P =ρ P /(1-ρ P ) = 0.611/(1 - 0.611) = 1.57 ingots W M =1/(μ M (1-ρ M )) = 1/(18(1 - 0.491)) = 0.109 hours = 6.6 minutes W M =1/(μ M (1-ρ M )) = 1/(18(1 - 0.491)) = 0.109 hours = 6.6 minutes L M =ρ M /(1-ρ M ) = 0.491(1 - 0.491) = 0.97 ingots L M =ρ M /(1-ρ M ) = 0.491(1 - 0.491) = 0.97 ingots W I =1/(μ I (1-ρ I )) = 1/(21(1 - 0.602)) = 0.119 hours = 7.2 minutes W I =1/(μ I (1-ρ I )) = 1/(21(1 - 0.602)) = 0.119 hours = 7.2 minutes L I =ρ I /(1-ρ I ) = 0.602(1 - 0.602) = 1.51 ingots L I =ρ I /(1-ρ I ) = 0.602(1 - 0.602) = 1.51 ingots

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Solve two key equations simultaneously Parametric Decomposition (P-D) Method Parametric Decomposition (P-D) Method - Two parameters – mean and scv - Can be applied for any general, but known distribution Decompose network into stochastically independent queues Decompose network into stochastically independent queues - Capture network interaction effects - Solve two sets of equations simultaneously Solve each node separately Solve each node separately - Apply G/G/m model Combine results to get network-specific performance measures Combine results to get network-specific performance measures

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Key Network Operations Departures Departures Merging of Arrivals Merging of Arrivals Splitting of Departures Splitting of Departures Batching Batching d departure a split i a superposition a i

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Key Network Operations Mean arrival rate into each node: Mean arrival rate into each node: SCV of inter-arrival times into each node: SCV of inter-arrival times into each node: where where

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Closed Queuing Network

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Mean Value Analysis (MVA) Algorithm Step 1: The first step is to estimate an initial value of L ij Step 2: Determine the throughput time W ij using the equation Step 3: Determine the production rate X j for each machine j Step 4: Determine the queue length L ij using the equation L ij =X j (v ij W ij ) Step 5: Compare the L ij value calculated in step 4 with the previous value. If the new value is within a desired range of the previous value, stop. If not, go to step 2.

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Example 7 Consider a machine shop which manufactures two parts. Part 1 visits machines 1 and 3, and part 2 visits machines 1 and 2. The service rates for machines 1, 2, and 3 are 10 parts per hour, 15 parts per hour, and 12 parts per hour, respectively. The service rates are not affected by the part being serviced. JIT/Kanban considerations limit the WIP of parts 1 and 2 to 5 and 4 units respectively. Consider a machine shop which manufactures two parts. Part 1 visits machines 1 and 3, and part 2 visits machines 1 and 2. The service rates for machines 1, 2, and 3 are 10 parts per hour, 15 parts per hour, and 12 parts per hour, respectively. The service rates are not affected by the part being serviced. JIT/Kanban considerations limit the WIP of parts 1 and 2 to 5 and 4 units respectively. 1) Determine the amount of time each part spends waiting and being served at each machine. 1) Determine the amount of time each part spends waiting and being served at each machine. 2) Determine the number of each part waiting and being served at each machine. 2) Determine the number of each part waiting and being served at each machine. 3) Based on the solution to parts 1 and 2, what conclusions can be drawn about the system design? 3) Based on the solution to parts 1 and 2, what conclusions can be drawn about the system design?

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Solution to Example 7 Step 1: Determine initial values of L ij. The number of each part type in the system is N 1 =5, and N 2 =4. Each part type visits only 2 of the 3 machines. Therefore, L 11 =L 31 =N 1 /2=2.5 and L 12 =L 22 =N 2 /2=2. Because part type 1 does not visit machine 2, and part type 2 does not visit machine 3, L 21 =L 32 =0 Step 1: Determine initial values of L ij. The number of each part type in the system is N 1 =5, and N 2 =4. Each part type visits only 2 of the 3 machines. Therefore, L 11 =L 31 =N 1 /2=2.5 and L 12 =L 22 =N 2 /2=2. Because part type 1 does not visit machine 2, and part type 2 does not visit machine 3, L 21 =L 32 =0 Step 2: Determine the throughput time W ij Step 2: Determine the throughput time W ij - W 11 = 1/10 + [4/5][2.5/10] + 2/10 = 0.50 = 30 minutes - W 21 = 0 - W 31 = 1/12 + [4/5][2.5/12] = 0.25 = 15 minutes - W 12 = 1/10 + [3/4][2/10] + 2.5/10 = 0.50 = 30 minutes - W 22 = 1/12 + [3/4][2/15] = 0.25 = 0.167 = 10 minutes - W 32 = 0 Step 3: Determine the production rate X r. In this example, v ij =1 for all i and r. - Step 3: Determine the production rate X r. In this example, v ij =1 for all i and r. - - We get X 1 = 5/[30+15] = 1/9 = 0.111 parts per minute - X 2 = 4/[30+10] = 1/10 = 0.10 parts per minute Step 4: Determine the queue length L ij using L ij =X r (v ij W ij ) for all i and j. Step 4: Determine the queue length L ij using L ij =X r (v ij W ij ) for all i and j. - L 11 = (0.111)(30) = 3.33 parts; L 21 = 0; L 31 = (0.111)(15) = 1.67 parts - L 12 = (0.10)(30) = 3 parts; L 22 = (0.10)(10) = 1 part; L 32 = 0 - Notice that L 11 +L 31 =5, and L 12 +L 22 =4

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Solution to Example 7 Step 5: Compare the value of L ij calculated in step 4 with the previous value. The values of L ij warrant another iteration. Therefore, we return to step 2 and repeat until the new calculated values of L ij are within the recommended range (5%) of the previous values. The solution below is found after four additional iterations. Step 5: Compare the value of L ij calculated in step 4 with the previous value. The values of L ij warrant another iteration. Therefore, we return to step 2 and repeat until the new calculated values of L ij are within the recommended range (5%) of the previous values. The solution below is found after four additional iterations. - L 11 = 4.28 partsW 11 = 47.8 minutes - L 21 = 0W 21 = 0 - L 31 = 0.72 partsW 31 = 8.1 minutes - L 12 = 3.60 partsW 12 = 47.5 minutes - L 22 = 0.40 partsW 22 = 5.3 minutes - L 32 = 0W 32 = 0 The current design is creating a build up in front of machine 1. Approximately 88% [(4.28+3.60)/(5+4)] of the WIP is either waiting or being served at machine 1. The current design is creating a build up in front of machine 1. Approximately 88% [(4.28+3.60)/(5+4)] of the WIP is either waiting or being served at machine 1.

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Lecture 13 – Continuous-Time Markov Chains

Lecture 13 – Continuous-Time Markov Chains

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