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Flow Time 4 -Dynamics Based on the book: Managing Business Process Flows. Flow Time Example .
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Flow Time 4-Dynamics Based on the book: Managing Business Process Flows
Flow Time Example The “Steal a Deal” gift shop specializes in heavily discounted merchandise for the holiday season. The store prices are so appealing that lines of people queue up in the early AM hours in front of the store doors in order to secure a place in the line. On the morning after Thanksgiving, one of the most busy days of the year, the store opened its doors at 8:00AM. From 7:45 to 8:00, 120 customers arrive and are already waiting in line. From 8:00AM to 10:00AM, new customers arrive at the rate of 5 per minute. After 10:00AM, the rate reduces to 2 per minute. The store admits customers at the rate of 4 per minute. Ardavan Asef-Vaziri, Sep. 2013
Schematic Representation of the Flow Dynamics Rp<Ra=5/min Rp= 4/min Ra-Rp=1/min 120 Rp-Ra=2/min 7:15 8:00 10:00 12:00 9:00 Flow Time Dynamics Ardavan Asef-Vaziri, Sep. 2013 3
Dynamics of Inventory and Waiting Time a) How many customers are in the waiting line at 10 AM? 120 at 8:00 Buffer increases at rate of 1/min. At 10:00 thee are 120+1(120) = 240 b) Jacob plans to arrive to the store at 9:00AM. How long should he expect to wait? 120 at 8:00 Buffer increases at rate of 1/min. At 9:00 thee are 120+1(60) = 180 Flow Time Dynamics Ardavan Asef-Vaziri, Sep. 2013 4
Dynamics of Inventory and Waiting Time Buffer is served at rate of 4/min 180/4 = 45 min. It takes Jacob 45 min to get in 9:00 + 45 min = 9:45 c) Rachel does not want to wait more than 15 minutes. When should she show up (in terms of the number of customers in the waiting line)? Buffer is served at rate of 4/min In 15 min they can serve 15(4) = 60. When there are 60 customers in line. Flow Time Dynamics Ardavan Asef-Vaziri, Sep. 2013 5
Dynamics of Inventory and Waiting Time d) Rachel does not want to wait more than 15 minutes. When should she show up (in terms of time)? At 10:00 there are 240 customers in the waiting line. Rachel should arrive when there are no more than 60 customers in line because those in front of her are served at rate of 4/ min. 2 new customers per minutes arrive after 10:00 AM. Customers are reduced at rate of (4-2) = 2/min (240-60)/2 = 90 min 10:00 + 90 min = 11:30 Flow Time Dynamics Ardavan Asef-Vaziri, Sep. 2013 6
Average Inventory e) Compute the average inventory from 7:45AM to 12:00PM. Inventory starts from 0, goes up to 120 in 15 min. Therefore during 0.25 hour the average inventory is (0+120)/2 = 60. Inventory then starts from 120, goes up to 240 in 2 hours. Therefore, during 2 hour the average inventory is (120+240)/2 = 180. Then inventory goes down from 240 to 0 in 240/2 = 120 min = 2 hour; at 12:00. Therefore during 2 hours the average inventory is (240+0)/2 = 120. The average inventory 0.25(60) + 2(180) + 2(120) = 615 Flow Time Dynamics Ardavan Asef-Vaziri, Sep. 2013 7
Average Inventory We should divide this by the sum of the relative weights; 0.25 + 2 + 2 = 4.25 The average inventory (customers in the line) = 615/4.25 = 144.7 Average inventory is the same, regardless the unit of time used (for hours, minutes and in this case, 4.25 hours). Flow Time Dynamics Ardavan Asef-Vaziri, Sep. 2013 8
Average Throughput f) Compute the average throughput from 7:45AM to 12:00PM. From 7:45 to 8:00; 0.25 hours 120 customers arrive From 8:00 to 10:00; 2 hours (2×60)(5) = 600 customers arrive From 10:00 to 12:00; 2 hours (2×60)(2) = 240 customers arrive A total of 120+5(60×2)+2(120) = 960 arrive over the time interval of 7:45 to 12:00PM, that is 4 hours and 15 min. or in 4.25 hours. R is 960 customers per 4.25 hours. R = 960/4.25 = 225.882 per hour. R = 225.882/60 = 3.76 per min. Flow Time Dynamics Ardavan Asef-Vaziri, Sep. 2013 9
Average Flow Time g) Compute the average flow time from 7:45AM to 12:00PM. RT = I, R = 3.76 per min, I = 144.7 T = 144.7/3.76 = 38.44 min. Flow Time Dynamics Ardavan Asef-Vaziri, Sep. 2013 10