You can easily change this to hours by multiplying r

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was in terms of minutes. You can easily change this to hours by multiplying R by 60 minutes to get hours (2 x 60 = 120 per hour. 5/120 = .0417 hours) or to days, or to months to find the rate that best fits your needs, but REMEMBER: INVENTORY (I) REMAINS CONSTANT. Note, that the Little’s Law of T=I/R is nothing more than a unit conversion, converting numbers into time. It turned 5 units of inventory into 2.5 minutes of inventory. Suppose we have 100 units of item A, and 1000 units of item B. What item do we have more of? By the count dimension, item B has a higher inventory. Now suppose we use 4 units of item A per day (Ra= 4/ day ), and 200 units of item B per day (Rb=200/ day ). By the time dimension we have 25 days of inventory for item A (100/4 = T), and 5 days of inventory for item B (1000/200 = T). The inventory of item A, from the time perspective, is larger than that of item B.
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Now, suppose there are two lines. Suppose R is still 2 customers per minute and still on average there are 5 customers in the first line to pay for their order and get their non-exotic coffee. In addition, suppose that 4 out of the five customers place their order and then go on to wait in the exotic order (latte, cappuccino, etc.) line, i.e. 40% of the customers place exotic orders. What is the flow of time for a person who orders latte, cappuccino, etc.? Answer: What is the throughput time (T) of customers in the 1 st line? Each customer spends 2.5 minutes in the first line (determined by you from the first problem). What is the throughput rate (R) of customers in the exotic order line? Throughput rate of the second line is 40% of 2 customers per minute, or .8 customers per minute [R = 0.4(2) = 0.8]. What is the inventory (I) of the exotic order line? Inventory of the second line is 4. You can now determine on average how long an exotic order customer waits using the Little’s Law. Units Value Inventory Time 40% 60% 40% Exotic order customers All customers Ordinary order customers
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R x T = I .8 x T = 4 T(second line)= 5 T(simple order) = 2.5 minutes. T(exotic order) = 2.5 + 5 = 7.5 minutes. What is the flow time of all customers? This is the average flow time of all customers that enter your coffee shop, both simple order and exotic order. Answer: Procedure 1 - 60% simple order: T = 2.5 40% exotic order: T=2.5+5= 7.5 An average customer is 60% a simple order person (2.5 min.) and 40% an exotic order person (7.5 min). T = 0.6(2.5) + 0.4(7.5) = 4.5 minutes Procedure 2 - Everyone goes through the first process and spends 5 minutes. 60% spend no additional time and leave. 40% spend 5 additional minutes. 0.6(0) ordinary order customers + 0.4(5) exotic order customers = 2 minutes 2.5 (every customer) + 2 (exotic orders) = 4.5 minutes. Procedure 3 – (Simplest solution) Throughput rate of the system is 2 per minute. There are 9 people in the system (5 at the register and 4 in the second line). R x T= I 2 x T = 9 T = 4.5 minutes Throughput rate of the coffee shop (system) is 2 per minute or 120 per hour or 720 per day (assuming 6 hours per day). But inventory in the system is ALWAYS 9.
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Throughput time of the coffee shop (system) is 4.5 minutes per customer (inventory).
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