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## BUAD 311 Week 3 Notes

by: Emily Laurienti

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Emily Laurienti
USC

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These notes cover week 3 lectures on Little's Law and Waiting Queues
COURSE
Operations Management
PROF.
TYPE
Class Notes
PAGES
4
WORDS
CONCEPTS
operations, Management
KARMA
25 ?

## Popular in Operations Management

This 4 page Class Notes was uploaded by Emily Laurienti on Wednesday September 14, 2016. The Class Notes belongs to BUAD 311 at University of Southern California taught by Prof. Hamid Nazer-Zadeh in Fall 2016. Since its upload, it has received 9 views. For similar materials see Operations Management in Business Administration at University of Southern California.

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Date Created: 09/14/16
9.6.16 Little’s Law ???????????? ???????????????? ???????????????? = ???????????????? ????????????????  Flow Rate—rate at which units enter the system  Flow Time—average time a unit stays in the system  Bank Teller Example o WIP = 6 customers o Flow Rate = 12 customers per hour o Flow Time = 6/12 = .5 o Each customer spends about .5 hours in the bank  This is an average, not every customer spends that long in the bank  Restaurant Example o Average # of customers in a restaurant = 50  This number is WIP o Average # of customers arriving per hour = 30  This number is flow rate o What is the Flow Time?  Plug into Little’s Law  50/30 = 1.66 hours or 100 minutes that the average customer spends in the restaurant o How can we increase revenue?  Revenue is proportional to flow rate  ???????????????? ???????????????? = ???????????? ???????????????? ????????????????  Increase flow rate (capacity) by:  Increasing WIP—adding more chairs and tables  Decreasing Flow Time—providing faster service  Admission Flow Example o 1000 applications received per month (flow rate) o Stage 1—preliminary classification  50% advance to secondary, 50% rejected o 200 applications in the wait line for preliminary review (WIP 1) o 50 applications in the wait-line for the secondary review (WIP 2) o What is the average processing time?  You can break it down into group A and B  Group B denied in preliminary o FT = 200/1000 = .2 month = 6 days  Group A moves on to secondary review o 500 applicants moved on = flow rate o FT = 50/500 = .1 month = 3 days  Total average time = .5 * 6 + .5 * 9 = 7.5 days  You can also simplify this by taking the total WIP which is 250 and plugging into Little’s Law:  250/1000= .25 month = 7.5 days  Clinic Example o A patient can be treated in exactly 15 minutes o 2 patients arrive at 15 minutes past the hour, one patient at 45 minutes past the hour  Flow Rate = 3 / hour o What is the flow time?  ???????? = ????????+????????+???????? = ???????? ???????????????????????????? ???? o What is WIP? ????+????+????+????  ???????????? = ???? = ???? ???????? ???????????????????????????? o Why do patients wait?  Arrivals are not evenly spread out. If they were evenly spread out, we could increase capacity to four customers per hour  Practice Problem o Call center employs 1000 agents—WIP o Every month 125 leave and 125 are hired—flow rate o How long on average does an agent work for this call center? ????????????????  ???????????????? ???????????????? = ???????????? = ???? ???????????????????????? ???????? ???????????????????????????? o Cost of hiring and training a new agent = \$2500 o Want to increase average work time to 20 months o How much do the hiring and training costs decrease?  What is the new flow rate if flow time is 20 months?  ???????? = ???????????????? ????  ???? = ???????? ???????????????????????????????????? ???????????? ????????????????????  What is the current cost of training?  125 * 2500 = 312,500  What would be the new cost of training?  50 * 2500 = 125,000  Total Decrease = 312,500 – 125,000 = 187,500 o What is the increase in salary the manager can afford to offer?  Total savings is the additional amount the manager could afford  ???????????????????????? = ????.???????? ???????????????????????????????????????? ???????????? ???????????????? ????????????????∗????∗???????? 9.8.16 Waiting Line Management  Characteristics of a queuing system o Customer arrival time (interarrival time) o Service time  Both are unknown in advance o Number of lines/ servers o Service discipline—FIFO vs. priority classes o Reneging/Balking—leaving a long line  Performance measures of queuing systems o Average time in line and in the system o Average queue length--# of customers in line o Average # of customers in the system—WIP o Average utilization rate of the server  No Uncertainty Example o Interarrival time = 10 min o Service time = 8 min o The average queue is 0 minutes because interarrival is longer than time in the system  Uncertain Demand Example o Interarrival is either 5 or 15 minutes each w/ probability .5  Mean interarrival time = .5 * 15 + .5 * 5 = 10 minutes o Service time is 8 minutes o Let’s say they arrive 5 minutes, 5 minutes, 15 minutes  Total time = 33 minutes  Average wait = 4 minutes o Variability causes idleness in addition to waiting time  Measure Variability o Coefficient of variation o ???????? = ???????????????????????????????? ???????????????????????????????????? ????????????????  Managerial Insight o Higher variability in interarrivals increases the length of the queue  Notations o a = interarrival time o 1/a = average arrival rate—flow rate o p = average service time o 1/p = average service rate o m = number of servers ????  ???????????????????????????????????????????? = ( ????∗????) ????  ???????????? ???????? ???????????????????????????? = ∗ ???? = ???? ∗ ???? ???? ????  ???????????? ???????? ???????????????????? = ???????? ∗ ???? ????  ???????????? = (???????? + ????) ????  Queue time with one server: ???????? ????????????+???????????? o ???????? = ????−????∗ ????  Gap Example o 10 customers per hour for the checkout line o Flow rate = 1/a = 10 customers/hour o a = 6 minutes o standard deviation of interarrival time = 5 o average service time for checkout: p = 5 min; 1/p = 12 customers/hour o standard deviation of service times = 1 min o one cashier; m = 1 o What is the waiting time in line? ???????? ???????? ????????  ????−???? ∗ (???? + ???? )/????)  = 9.18 average wait time o What is the flow time?  FT = Tq + p = 9.18 + 5 = 14.18 o What is WIP?  Customers in service = m*u = .83  Customers in queue = Tq/a = 9.18/6 = 1.53  Total WIP = .83 + 1.53 = 2.36 people

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