L12_WaitingTime_After - Lecture 12 Waiting Time Problems...

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Slide 1 Lecture 12 Waiting Time Problems
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Slide 2 Le a rning  O b je c tive s Analyze stationary arrival queues Predict waiting times and quantify performance metrics Redesign service systems to reduce variability
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Slide 3 Waiting time in a Stationary Queue
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Slide 4 Arrival On average, a flow unit arrives every  a  time units, or  1/a  flow units arrives per  unit of time .    The standard deviation of the inter-arrival time is  σ a . The coefficient of variation is  CV a a /a . Process There are   servers (resource) in the system It takes an average of  p  units of time to serve a flow unit, or  1/p  flow units can  be processed per unit of time. The standard deviation of the processing time is   p . The coefficient of variation is  CV p /p . Assumptions   Stationary Queuing System
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Slide 5 Performance Measures  Stationary Queuing System Stable   average demand rate = 1/a < m/p  = average capacity I q   = flow units (inventory) in the queue or queue length I p  = flow units (inventory) in service I  = flow units (inventory) in the system T q  = waiting time while in the queue T  = flow time Arrival (demand) a  (time/unit) 1/a   (unit/time) CV a   σ a /a Process (resource) m  servers/resources p   (time / unit)   1/p   (unit / time) CV p /p ……
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Slide 6 Example: RentAPhone The company currently has 80 phones. There are , on average, 25 customers per day  (24 hours) requesting a phone.  The corresponding coefficient of variation is 1.  Customers keep their phones on average 72 hours. The standard deviation of this time  is 100 hours.  a  = 24/25=0.96 hr/customer 1/a  = 25/24 customer/hr CV a  =1 m  = 80   p  = 72 hr/customer 1/p  = 1/72 customer/hr CV p  = 100/72=1.3889 % 75 . 93 72 80 24 25 / 1 = = = p m a u Utilization : …… hr customer a r / 04 . 1 1 = = Flow rate:   93.75%  X  80 = 75 phones on rent      5 phones in company on average
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Slide 7 Example: RentAPhone cont customers T a T r I q q q 3 . 10 89 . 9 24 25 1 = × = × = × = customers mu a p p r I p 75 / = = = × = Waiting time : hrs CV CV u u m p T p a m q 89 . 9 2 1 2 2 1 ) 1 ( 2 = + - - + Flow time : hrs p T T q 89 . 81 72 89 . 9 = + = + = Inventory in queue : Total inventory : Inventory in service : q p I I customers T a T r I + = = × = × = × = 3 . 85 89 . 81 24 / 25 1 a  = 24/25=0.96 hr/customer 1/a  = 25/24 customer/hr CV a  =1 m  = 80   p  = 72 hr/customer 1/p  = 1/72 customer/hr CV p  = 100/72=1.3889 ……
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Slide 8 Example: RentAPhone cont 7419.87/mo $ 30 24 3 . 10 / 3 . 10 3 . 10 1 = × × = = × = hr hI q  Waiting inventory incurs a carrying cost of $1 per unit per hour
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L12_WaitingTime_After - Lecture 12 Waiting Time Problems...

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