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311_Session12-08

# 311_Session12-08 - Session12:WaitingLines PreviousClass...

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Operations Management Session 12:  Waiting Lines

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Session 12 Operations Management 2 Previous Class Understanding the phenomenon of waiting  Impact of variability/uncertainty & utilization rate
Session 12 Operations Management 3 Objectives Classification of the waiting-line systems Formula/macro for waiting-line calculation Practices waiting time using formulas

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Session 12 Operations Management 4 Terminology and Classification  of Waiting Lines Terminology : The characteristics of a queuing  system is captured by five parameters: Arrival pattern Service pattern Number of server Restriction on queue capacity The queue discipline
Session 12 Operations Management 5 Terminology and Classification  of Waiting Lines M/G/12/23 Exponential interarrival times General service times 12 severs Queue capacity is 23

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Session 12 Operations Management 6 M/M/1 Performance Evaluation Notation: Arrival rate: λ   (10 per hour) Average interarrival time: 1/ λ (1/10 = 6 min 29 Service rate: μ  (15 per hour) Average service time: 1/ μ  (4 min)
Session 12 Operations Management 7 M/M/1 Performance Evaluation   ) ( ) ( line in ting number wai Average 2 λ μ μ λ - = l n ) ( ) ( system in ting number wai Average λ μ λ - = s n ) ( ) ( line in waiting time Average λ μ μ λ - = l t ) ( 1 ) ( system in waiting time Average λ μ - = s t

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Session 12 Operations Management 8 M/M/1 Performance Evaluation rate Service rate Arrival = = μ λ ρ μ λ - = = 1 system in the customers 0 exactly are y there Probabilit 0 p ) 1 ( ) ( system in the customers exactly are y there Probabilit μ λ μ λ - = = n n n p
Session 12 Operations Management 9 M/M/1 Performance Evaluation Example:  The arrival rate to a GAP store is            6 per hour ( λ ).  (Poisson) The service time is 5 min per  customer (1/ μ )  (Exponential) In average how many customers are in the system  (queue)? In average how long do they stay in the system (queue)?

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Session 12 Operations Management 10 M/M/1 Performance Evaluation customers 5 . 0 72 36 ) 6 12 ( 12 6 ) ( 2 2 = = - = - = λ μ μ λ l n customers 1 6 12 6 ) ( = - = - = λ μ λ s n min 5 12 1 72 6 ) 6 12 ( 12 6 ) ( = = = - = - = λ μ μ λ l t min 10 6 1 6 12 1 ) ( 1 = = - = - = λ μ s t
Session 12 Operations Management 11 M/M/1 Performance Evaluation   What is the probability of having exactly zero customers in the  system? 5 . 0 12 6 = = = μ λ ρ 5 . 0 5 . 0 1 1 0 = - = - = ρ p

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Session 12 Operations Management 12 M/M/1 Performance Evaluation What is the probability of having in the GAP store more than  1 customer? Probability of having more than one customer = 1 –  (probability of having zero customers + probability of having  one customer) = 1 – (P 0  + P 1 ) = 1 – (0.5 + P 1 ) Probability of having more than one customer = 1-0.75 = 0.25  25 . 0 5 . 0 * 5 . 0 ) 1 ( ) ( 1 1 = = - = μ λ μ λ p
Session 12 Operations Management 13 M/M/1 Performance Evaluation  What if the arrival rate is 11 per hour?

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311_Session12-08 - Session12:WaitingLines PreviousClass...

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