Flow Line Model: Practice Problems - University of Washington



Flow Line Model: A Brief IntroductionOrganization of most work can be thought of as dividing a given amount of work between a number of resources to produce an output. Flow line model offers a way to think about organization of work when work is repetitive, that is, when operations are set to produce or serve roughly similar work/customer/demand one after the other. As we will discuss later, this basic model applies to a great variety of situations but, for now, let us focus on something simple we are all familiar with, a Subway sandwich line.InputsLet us start with imagining Subway as a sandwich factory where sandwiches are continuously produced for stocking rather than for customers. Let us say there are three types of resources behind the counter: starter, stuffer and packer/cashier. Total work of producing a meatballs sub (a unit of output or a flow-unit) is divided into activities or tasks and these activities are assigned to different resources: starter employees prepare the bread, stuffer employees put in meatballs and other stuff you want, and packer/cashier employees packages the sandwich. The assignment of certain activities to one type of resource is referred to as a station. The three stations here are: starter, stuffer, packer. The sum of times for activities carried at each station for each flow unit is service time at that station. Let us assume that there is one employee at each station and that the service time for starting is 1 min per sandwich, service time for stuffing is 1.5 min per sandwich and service time for packaging is 0.5 min per sandwich. We use the term service rate to express what one employee at a station can do in a given time (say, an hour). One employee at starting station has a service rate of 60 sandwiches per hour (60 minutes per hour divided by 1 min per sandwich). Check that service rates at other two stations are 40 per hour (60/1.5) and 120 per hour (60/0.5) respectively.There can be more than one employee at each station. We say that multiple employees on a station work in parallel; that is, two employees on stuffing will be working on two different sandwich in parallel. Alternatively, we can think of multiple employees at a station as equally dividing the service time between themselves. Either way, two employees at stuffing station, for example, will double the capacity of the stuffing station from 40 per hour to 80 per hour. If there are m employees on a station, we say:Resource-type or station capacity = m / service time.OutputsWe focus on performance measures from the perspective of the flow-unit, a sandwich. First, how much time did a sandwich spend in line? This is called Lead time. Clearly, this is equal to the sum of service times plus any waiting time a sandwich spends in the system. For now, we are going to leave thinking of waiting time till the next chapter. In reality, it is possible that line is run in a way to create some waiting time. For example, starter station can work through the night to produce an inventory buffer for packer who starts in the morning resulting in waiting time for sandwiches for packing. We ignore these possibilities for now and say minimum lead time is sum of service times = 1+1.5+0.5=3 mins in this case.The next performance measure is the capacity of the whole line; we call this max throughput of the line. Note that this is different from the capacity of each station. We would like to answer the question, how many finished sandwiches can be produced by the whole line in a given time (say, an hour)? We make the observation that the whole line can not do things at a higher rate than any one of its stations. That is, the capacity of the line can only be as much as the least capacity of any station in the line. Assuming only one employee at stuffing, stuffing station has the least capacity of all three stations, the line max throughput is equal to the stuffing station capacity of 40 sandwiches per hour. Line can also be run at a throughput less than its maximum throughput. The station or resource type with least capacity, stuffing in this case, is called bottleneck.The next measure is Cycle time. Assuming that the line has been running for some time, cycle time is the time between two consecutive output units at the end of the line. If cycle time is x min, then it tells us that one unit comes out of the line every x min. It is closely related to line throughput. If the line is running at its max throughput of 40 per hour than the one sandwich comes out every (60 minutes per hour / 40 sandwich per hour) = 1.5 minutes. Thus the cycle time is 1.5 minutes corresponding to max throughput. We will define “actual” throughput below and, unless question specifically asks otherwise, we will compute cycle time corresponding to actual throughput.Cycle time = 1 / actual Line throughput (Th)Work in process measures the average amount of unfinished work in the flow line. Intuitively, one feels that it should be connected to the time a units spends in the system (lead time) and how many units are produces (line Th). The relationship is known as Little’s law: WIP = Th * Lead TimeSince Line Th is 40 per hour and Lead time is 3 minutes (=1/20 hr); in this caseWIP = 40 per hr * (1/20) hr = 2 unitsUtilization of each resource type is simply equal to the ratio of what it actually produces and the maximum it can produce. What it actually produces is Line Th. Resource type utilization = Actual Line Th / Station capacity.If the line is run at its max Th with only one employee at stuffing, utilization at starting is 40/60 = 66.67%, utilization at stuffing is 60/60=100% and utilization at packing is 40/120=33.33%.Workload ArrivalsWe had assumed that the sandwich line was running as a factory producing at its full capacity. Thinking about a regular Subway, sandwiches are only produced in response to a customer demand. We must now think of incorporating arrival of work/customers. The main observation here is that in this case line will produce at a rate to match the customer arrival rate. For example, if customers arrive only at the rate of 20 per hour then the line will have an actual throughput of 20 per hour even though it is capable of a maximum throughput of 40 per hour (see above). Note that utilization calculation should use this actual line Th in the numerator. Also, if the question is not specifically asking for cycle time corresponding to max Th, please compute cycle time for actual th; that is cycle time = 1/actual th (unless the question specifies otherwise).Other ExtensionsEven though sometimes reality is more complicated than the model described above, the model can be used to provide insights at some level of aggregation. For example, one common complication is variety. There are many different types of flow-units and, at a station, different types of flow-units need different service times. In such case, we model an average or typical flow-unit and modify the service times at each station to an average service time. See practice problems for more detail. Another example is when rather than a single unit flowing through the line, multiple units flow in a batch. Again, we can redefine the unit of flow to be a “average batch” and then compute service times for one unit of this “average batch.”Other complexities come from situations where it is not clear how to define stations. For one, there may not be any physical stations. Another problem comes from when resources can move between different resource types or stations. We will discuss a few such situations in class.SummaryA unit flows through stations of different resource-types, each of which can have multiple resources. Each resource-type station performs part of the work on a unit.Service time is the time taken to complete work on a unit by one resource on a station.Service rate of one resource on a station = 1 / service timeStation capacity = number of resources m / service time = m* service rateMin Lead time = sum of service times on all stations (+waiting time, see next chapter)Max throughput of line = capacity of the station that has least capacity of all (bottleneck)Actual throughput can be less than max throughput. For example, if line is run to match the rate of workload or customer arrivals, actual throughput = workload arrival rate.Cycle time of line = 1 / actual throughputWork-in-process WIP = actual throughput * leadtime [ known as Little’s law]Station or resource-type’s Utilization = a017780000ctual throughput / station capacityFlow Line Model: Practice Problems1. A fast-food restaurant processes an average of 5000 kilograms (kg) of hamburgers per week. Typical inventory of raw meat in cold storage is 2500 kg. On an average, how long does a kg of meat spend in the cold storage?2. A branch office of an insurance company processes 10,000 claims per year (50 weeks per year). Average processing time is 3 weeks. How many claims, on an average, are being processed at any given point? On an average what is the pace of the line, that is, the time between two consecutive outputs?3. A manufacturer sells $300 million worth of cellular equipment per year. The average amount in account receivable is $45 million. On an average, how much time elapses from the time a customer is billed to the time payment is received?4. Let us imagine a simple production line, as shown, for making a table. An order flows in to workstation A where a worker takes 3 hours to create table legs. The order and table legs then move on to workstation B where a worker takes 5 hours to create a table top. The order, legs, and top then moves to Workstation C where a worker takes 2 hours to assemble the table. Each workstation has only one worker devoted to the assigned activity. One order arrives every 7.5 hours. What are the performance measures: Min leadtime, Max Th and corresponding cycle time, Actual Th, WIP and resource utilization at each workstation? 5. If there are multiple types of units flowing through the process, analysis is slightly more involved. Here is an example: An employment verification agency receives resumes from consulting firms and law firms with the request to validate information provided by their job candidates. The process flow diagram below describes the process of handling the resumes. The three customer types share the first step and the last step in the process but differ with respect to intermediate steps. The capacity of the process crucially depends on the product mix. The demand is a total of 180 applications a day of which there are: 30 for consulting positions, 110 for staff and 40 for internship positions. Assume that the working day is 10 hours. 34290015430500Other data:Resource typeActivity Time(min./appli.)# of workersFile31Contact clients/supers202Contact employers153Benchmark grades82Confirmation letters21Which resource type is the bottleneck? Flow Line Model: Practice Problems Solutions1. A fast-food restaurant processes an average of 5000 kilograms (kg) of hamburgers per week. Typical inventory of raw meat in cold storage is 2500 kg. On an average, how long does a kg of meat spend in the cold storage?Flow boundaries: cold storage inside restaurantFlow unit: kg of meatWIP = 2500 Kg, Th = 5000 Kg/weekUsing Little’s LawAverage Lead time = 2500 kg / 5000 kg per week =0.5 week2. A branch office of an insurance company processes 10,000 claims per year (50 weeks per year). Average processing time is 3 weeks. How many claims, on an average, are being processed at any given point? On an average what is the pace of the line, that is, the time between two consecutive outputs?Flow boundaries: branch officeFlow unit: claimTh = 10000 claims per year = 10000/50 per week = 200 per weekLead time = 3 weeksAverage inventory WIP = 10000/50 per week * 3 weeks = 600 claimsTime between two consecutive outputs = cycle time = (1/Th) = (1/200) week.3. A manufacturer sells $300 million worth of cellular equipment per year. The average amount in account receivable is $45 million. On an average, how much time elapses from the time a customer is billed to the time payment is received?Process boundaries: accounts-receivable deptFlow unit: a dollarThroughput: 300 million / yearAverage WIP = 45 millionAverage lead time = 45 / 300 = 0.15 years = 1.8 months4. Let us imagine a simple production line, as shown, for making a table. An order flows in to workstation A where a worker takes 3 hours to create table legs. The order and table legs then move on to workstation B where a worker takes 5 hours to create a table top. The order, legs, and top then moves to Workstation C where a worker takes 2 hours to assemble the table. Each workstation has only one worker devoted to the assigned activity. One order arrives every 7.5 hours. [IMPORTANT: units in the following figure should be Time Units: hour. In each of the three boxes, “mins” should be “hour”.]What are the performance measures: Min leadtime, Max Th and cycle time corresponding to Max Th, Actual Th, WIP and resource utilization at each workstation?Min leadtime=3+5+2=10 hoursStation capacities: A=1/3 =0.33 per hr; B: 1/5 =0.2 per hr; C: 1/2 =0.5 per hr;Resource at station B is the bottleneck.Max Th = min of station capacities = 0.2 per hr;Cycle time corresponding to max th= 1/ max Th = 1/0.2 hr = 5 hoursActual Th = arrival rate = 1/7.5 = 0.133 per hrCycle time corresponding to actual th= 1/ actual Th = 7.5 hoursWIP = actual Th * Leadtime = 0.133 per hr *10 hr = 1.33 ordersUtilization of resource at station A: 0.133/0.33=40.3%, B:0.133/0.2=66.5%, C: 0.133/0.5=26.6%5. If there are multiple types of units flowing through the process, analysis is slightly more involved. Here is an example: An employment verification agency receives resumes from consulting firms and law firms with the request to validate information provided by their job candidates. The process flow diagram below describes the process of handling the resumes. The three customer types share the first step and the last step in the process but differ with respect to intermediate steps. The capacity of the process crucially depends on the product mix. The demand is a total of 180 applications a day of which there are: 30 for consulting positions, 110 for staff and 40 for internship positions. Assume that the working day is 10 hours. Other data:Resource-typeActivity Time(min./appli.)# of workersFile31Contact clients/supers202Contact employers153Benchmark grades82Confirmation letters21Which resource-type is the bottleneck?To handle such a case, the main idea is to think in terms of a “typical” unit of flow where this “typical” unit is an application for consulting with probability 30/180, for staff with a probability of 110/180 and for internship with probability of 40/180. Given a 10 hour day, there are, on average per hour, 3 consulting, 11 staff and 4 intern applications adding up to “typical” 18 applications per hour. Now, we will determine the service times of this “typical” flow-unit for each resource-type. This is done by simply taking a weighted average of times of different customer types. In this setting, each activity is assigned to one resource-type so we can refer to both by the same name. For example resource-type filing clerk performs only one activity “file.” We can refer to both by “file.” In other settings, it is possible that one resource-type performs multiple activities. In such a case, the service time for this resource-type will be the sum of average times for all activities. File: since all types go through this step, the “typical” unit takes [(3/18)*3+(11/18)*3+(4/18)*3] =3 minutes and the capacity per hour is 60 minutes/3 min. = 20 per hour, giving a utilization of 18/20=90%.Contact clients/supers: Since only consulting applications go through this step, we can consider the time for staff and intern applications to be zero. The “typical” unit time is [(3/18)*20+(11/18)*0+(4/18)*0]=60/18. Because there are two resources, the capacity per hour is 2*[60 minutes/(60/18)] min. = 36 per hour, giving a utilization of 18/36=50%.Contact employers: The “typical” unit time is [(3/18)*5+(11/18)*5+(4/18)*0]=70/18. The capacity per hour is 60 minutes/(70/18) min. = 108/7 per hour, giving a utilization of 18/(108/7)=116.66%. Well, more than 100% utilization means that they got to run some overtime. Grade/School analysis: The “typical” unit time is [(3/18)*0+(11/18)*0+(4/18)*8]=32/18=1.78 min. The capacity per hour is 2*[ 60 minutes/(32/18)] min. = 1080/16 per hour = 67.5 per hour, giving a utilization of 18/(1080/16)=26.66%.Confirmation letter: The “typical” unit time is [(3/18)*2+(11/18)*2+(4/18)*2]=2 min. The capacity per hour is 60 minutes/2 min. = 30 per hour, giving a utilization of 18/30=60%.Clearly, “Contact employers” is the bottleneck. Note that this approach would also work in the more general case where each different type of unit would take different time at each step.Suppose, just as an example, the resource devoted to confirmation letter leaves. We ask the resource-type working on activity grades to also work on activity confirmation letters. The service time for this resource-type is now the sum of two activity times (32/18) min + 2 min = 3.78 min and the revised capacity for this resource-type is 2*[60/3.78]=31.75 per hour. ................
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